38148911-Solution-Manual-for-Applied-Statistics-and-Probability-for-Engineers-c02to11

38148911-Solution-Manual-for-Applied-Statistics-and-Probability-for-Engineers-c02to11

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Unformatted text preview: CHAPTER 2 Section 2-1 2-1. Let "a", "b" denote a part above, below the specification S = {aaa, aab, aba, abb, baa, bab, bba, bbb} 2-2. Let "e" denote a bit in error Let "o" denote a bit not in error ("o" denotes okay) ⎧eeee, eoee, oeee, ooee, ⎫ ⎪eeeo, eoeo, oeeo, ooeo,⎪ ⎪ ⎪ S =⎨ ⎬ ⎪eeoe, eooe, oeoe, oooe,⎪ ⎪eeoo, eooo, oeoo, oooo⎪ ⎩ ⎭ 2-3. Let "a" denote an acceptable power supply Let "f" ,"m","c" denote a supply with a functional, minor, or cosmetic error, respectively. S = {a, f ,m, c} 2-4. S = {0,1,2,...} = set of nonnegative integers 2-5. If only the number of tracks with errors is of interest, then S = {0,1,2, ...,24} 2-6. A vector with three components can describe the three digits of the ammeter. Each digit can be 0,1,2,...,9. Then S is a sample space of 1000 possible three digit integers, S = {000,001,...,999} 2-7. 2-8. S is the sample space of 100 possible two digit integers. Let an ordered pair of numbers, such as 43 denote the response on the first and second question. Then, S consists of the 25 ordered pairs {11,12,...,55} 2-9. S = {0,1,2,...,} in ppb. 2-10. S = {0,1,2,...,} in milliseconds 2-11. 2-12. s = small, m = medium, l = large; S = {s, m, l, ss, sm, sl, ….} 2-13 2-14. S = { .0,1.1,1.2, K14.0} 1 S = {0,1,2,...,} in milliseconds. automatic transmission with air red blue black white standard transmission with air without air red blue black white red blue black white 2-1 without air red blue black white 2-15. PRESS 1 2 CAVI TY 1 2 3 5 4 6 7 8 1 2 3 4 5 6 7 2-16. memory 4 8 12 disk storage 200 300 400 200 300 2-17. 2-18. c = connect, b = busy, S = {c, bc, bbc, bbbc, bbbbc, …} S = {s , fs , ffs , fffS, fffF S, f ff F F S, fffF F F A } 2-19 a.) b.) 2-2 400 200 300 400 8 c.) d.) e.) 2.20 a.) 2-3 b.) c.) d.) e.) 2-4 2-21. a) S = nonnegative integers from 0 to the largest integer that can be displayed by the scale. Let X represent weight. A is the event that X > 11 B is the event that X ≤ 15 C is the event that 8 ≤ X <12 S = {0, 1, 2, 3, …} b) S c) 11 < X ≤ 15 or {12, 13, 14, 15} d) X ≤ 11 or {0, 1, 2, …, 11} e) S f) A ∪ C would contain the values of X such that: X ≥ 8 Thus (A ∪ C)′ would contain the values of X such that: X < 8 or {0, 1, 2, …, 7} g) ∅ h) B′ would contain the values of X such that X > 15. Therefore, B′ ∩ C would be the empty set. They have no outcomes in common or ∅ i) B ∩ C is the event 8 ≤ X <12. Therefore, A ∪ (B ∩ C) is the event X ≥ 8 or {8, 9, 10, …} 2-22. a) A B C b) A B C c) 2-5 d.) A B C e.) 2-23. If the events are mutually exclusive, then A∩B is equal to zero. Therefore, the process would not produce product parts with X=50 cm and Y=10 cm. The process would not be successful Let "d" denoted a distorted bit and let "o" denote a bit that is not distorted. ⎧dddd, dodd, oddd, oodd, ⎫ ⎪ ⎪ ⎪dddo, dodo, oddo, oodo,⎪ a) S = ⎨ ⎬ ⎪ddod, dood, odod, oood, ⎪ ⎪ddoo, dooo, odoo, oooo ⎪ ⎩ ⎭ b) No, for example A 1 ∩ A 2 = {dddd, dddo, ddod, ddoo} c) ⎧dddd , dodd ,⎫ ⎪dddo, dodo ⎪ ⎪ ⎪ A1 = ⎨ ⎬ ⎪ddod , dood ⎪ ⎪ddoo, dooo ⎪ ⎩ ⎭ d) ⎧oddd , oodd ,⎫ ⎪oddo, oodo, ⎪ ⎪ ⎪ ′=⎨ A1 ⎬ ⎪odod , oood ,⎪ ⎪odoo, oooo ⎪ ⎩ ⎭ e) A1 ∩ A2 ∩ A3 ∩ A4 = {dddd } f) ( A1 ∩ A2 ) ∪ ( A3 ∩ A4 ) = {dddd , dodd , dddo, oddd , ddod , oodd , ddoo} 2-6 2-24. Let "d" denote a defective calculator and let "a" denote an acceptable calculator a a) b) c) d) e) S = {ddd , add , dda, ada, dad , aad , daa, aaa} A = {ddd , dda, dad , daa} B = {ddd , dda, add , ada} A ∩ B = {ddd , dda} B ∪ C = {ddd , dda, add , ada, dad , aad } 12 2 = 4096 A ∩ B = 70, A′ = 14, A ∪ B = 95 2-25. 2-26. a.) A′ ∩ B = 10, B ′ =10, A ∪ B = 92 b.) 2-27. Surface 1 G E Edge 1 G E G Surface 2 E Edge 2 E E G E E G E E E G G G G G G E E E G E G E G G 2-28. 2-29. A′ ∩ B = 55, B ′ =23, A ∪ B = 85 a) A′ = {x | x ≥ 72.5} b) B′ = {x | x ≤ 52.5} c) A ∩ B = {x | 52.5 < x < 72.5} d) A ∪ B = {x | x > 0} 2.30 a) {ab, ac, ad, bc, bd, cd, ba, ca, da, cb, db, dc} b) {ab, ac, ad, ae, af, ag, bc, bd, be, bf, bg, cd, ce, cf, cg, ef, eg, fg, ba, ca, da, ea, fa, ga, cb, db, eb, fb, gb, dc, ec, fc, gc, fe, ge, gf} 2.31 c) Let d = defective, g = good; S = {gg, gd, dg, dd} d) Let d = defective, g = good; S = {gd, dg, gg} Let g denote a good board, m a board with minor defects, and j a board with major defects. a.) S = {gg, gm, gj, mg, mm, mj, jg, jm, jj} b) S={gg,gm,gj,mg,mm,mj,jg,jm} 2-7 2-32.a.) The sample space contains all points in the positive X-Y plane. b) A 10 c) 20 B d) B 20 10 A 10 A e) B 20 2-8 2-33 a) b) c) d) 2-9 Section 2-2 2-34. All outcomes are equally likely a) P(A) = 2/5 b) P(B) = 3/5 c) P(A') = 3/5 d) P(A∪B) = 1 e) P(A∩B) = P(∅)= 0 2-35. a) P(A) = 0.4 b) P(B) = 0.8 c) P(A') = 0.6 d) P(A∪B) = 1 e) P(A∩B) = 0.2 2-36. a) S = {1, 2, 3, 4, 5, 6} b) 1/6 c) 2/6 d) 5/6 2-37. a) S = {1,2,3,4,5,6,7,8} b) 2/8 c) 6/8 2-38. x = 0.3, x = 6 20 2-39. a) 0.5 + 0.2 = 0.7 b) 0.3 + 0.5 = 0.8 2-40. a) 1/10 b) 5/10 2-41. a) 0.25 b) 0.75 2-42. Total possible: 1016, Only 108 valid, P(valid) = 108/1016 = 1/108 2-43. 3 digits between 0 and 9, so the probability of any three numbers is 1/(10*10*10); 3 letters A to Z, so the probability of any three numbers is 1/(26*26*26); The probability your license plate is chosen is then (1/103)*(1/263) = 5.7 x 10-8 2-44. a) 5*5*4 = 100 b) (5*5)/100 = 25/100=1/4 2-45. a) P(A) = 86/100 = 0.86 b) P(B) = 79/100 = 0.79 c) P(A') = 14/100 = 0.14 d) P(A∩B) = 70/100 = 0.70 e) P(A∪B) = (70+9+16)/100 = 0.95 f) P(A’∪B) = (70+9+5)/100 = 0.84 2-46. Let A = excellent surface finish; B = excellent length a) P(A) = 82/100 = 0.82 b) P(B) = 90/100 = 0.90 c) P(A') = 1 – 0.82 = 0.18 d) P(A∩B) = 80/100 = 0.80 e) P(A∪B) = 0.92 f) P(A’∪B) = 0.98 2-10 2-47. a) P(A) = 30/100 = 0.30 b) P(B) = 77/100 = 0.77 c) P(A') = 1 – 0.30 = 0.70 d) P(A∩B) = 22/100 = 0.22 e) P(A∪B) = 85/100 = 0.85 f) P(A’∪B) =92/100 = 0.92 2-48. a) Because E and E' are mutually exclusive events and E ∪ E ′ = S 1 = P(S) = P( E ∪ E ′ ) = P(E) + P(E'). Therefore, P(E') = 1 - P(E) b) Because S and ∅ are mutually exclusive events with S = S ∪ ∅ P(S) = P(S) + P(∅). Therefore, P(∅) = 0 c) Now, B = A ∪ ( A ′ ∩ B) and the events A and A ′ ∩ B are mutually exclusive. Therefore, P(B) = P(A) + P( A ′ ∩ B ). Because P( A ′ ∩ B ) ≥ 0 , P(B) ≥ P(A). Section 2-3 2-49. a) P(A') = 1- P(A) = 0.7 b) P ( A ∪ B ) = P(A) + P(B) - P( A ∩ B ) = 0.3+0.2 - 0.1 = 0.4 c) P( A ′ ∩ B ) + P( A ∩ B ) = P(B). Therefore, P( A ′ ∩ B ) = 0.2 - 0.1 = 0.1 d) P(A) = P( A ∩ B ) + P( A ∩ B ′ ). Therefore, P( A ∩ B ′ ) = 0.3 - 0.1 = 0.2 e) P(( A ∪ B )') = 1 - P( A ∪ B ) = 1 - 0.4 = 0.6 f) P( A ′ ∪ B ) = P(A') + P(B) - P( A ′ ∩ B ) = 0.7 + 0.2 - 0.1 = 0.8 A ∪ B ∪ C ) = P(A) + P(B) + P(C), because the events are mutually exclusive. Therefore, P( A ∪ B ∪ C ) = 0.2+0.3+0.4 = 0.9 b) P ( A ∩ B ∩ C ) = 0, because A ∩ B ∩ C = ∅ c) P( A ∩ B ) = 0 , because A ∩ B = ∅ d) P( ( A ∪ B ) ∩ C ) = 0, because ( A ∪ B ) ∩ C = ( A ∩ C ) ∪ ( B ∩ C ) = ∅ e) P( A′ ∪ B ′ ∪ C ′ ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1 2-50. a) P( 2-51. If A,B,C are mutually exclusive, then P( A ∪ B ∪ C ) = P(A) + P(B) + P(C) = 0.3 + 0.4 + 0.5 = 1.2, which greater than 1. Therefore, P(A), P(B),and P(C) cannot equal the given values. 2-52. a) 70/100 = 0.70 b) (79+86-70)/100 = 0.95 c) No, P( A ∩ B ) ≠ 0 2-53. a) 350/370 345 + 5 + 12 362 b) = 370 370 345 + 5 + 8 358 c) = 370 370 d) 345/370 2-54. a) 170/190 = 17/19 b) 7/190 2-55. a) P(unsatisfactory) = (5+10-2)/130 = 13/130 b) P(both criteria satisfactory) = 117/130 = 0.90, No 2-56. a) (207+350+357-201-204-345+200)/370 = 0.9838 b) 366/370 = 0.989 c) (200+145)/370 = 363/370 = 0.981 d) (201+149)/370 = 350/370 = 0.946 Section 2-4 2-11 2-57. a) P(A) = 86/100 b) P(B) = 79/100 P ( A ∩ B) 70 / 100 70 c) P( A B ) = = = P( B) 79 / 100 79 d) P( B A ) = P( A ∩ B) 70 / 100 70 = = P( A) 86 / 100 86 2-58.a) 0.82 b) 0.90 c) 8/9 = 0.889 d) 80/82 = 0.9756 e) 80/82 = 0.9756 f) 2/10 = 0.20 2-59. a) 345/357 b) 5/13 2-60. a) 12/100 b) 12/28 c) 34/122 2-61. Need data from Table 2-2 on page 34 a) P(A) = 0.05 + 0.10 = 0.15 P( A ∩ B ) 0.04 + 0.07 b) P(A|B) = = = 0.153 0.72 P( B) c) P(B) = 0.72 P( A ∩ B ) 0.04 + 0.07 = = 0.733 d) P(B|A) = 0.15 P( B) e) P(A ∩ B) = 0.04 +0.07 = 0.11 f) P(A ∪ B) = 0.15 + 0.72 – 0.11 = 0.76 2-62. a) 20/100 b) 19/99 c) (20/100)(19/99) = 0.038 d) If the chips are replaced, the probability would be (20/100) = 0.2 2-63. a) P(A) = 15/40 b) P( B A ) = 14/39 c) P( A ∩ B ) = P(A) P(B/A) = (15/40) (14/39) = 0.135 d) P( A ∪ B ) = P(A) + P(B) - P( A ∩ B ) = 15 + 14 − ⎛ 15 ⎞ ⎛ 14 ⎞ = 0.599 ⎜ ⎟⎜ ⎟ 40 39 ⎝ 40 ⎠ ⎝ 39 ⎠ 2-64. A = first is local, B = second is local, C = third is local a) P(A ∩ B ∩ C) = (15/40)(14/39)(13/38) = 0.046 b) P(A ∩ B ∩ C’) = (15/40)(14/39)(25/39) = 0.085 2-65. a) 4/499 = 0.0080 b) (5/500)(4/499) = 0.000080 c) (495/500)(494/499) = 0.98 2-66. a) 3/498 = 0.0060 b) 4/498 = 0.0080 c) ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 3 ⎞ = 4.82x10 −7 ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 500 ⎠ ⎝ 499 ⎠ ⎝ 498 ⎠ 2-67. a) P(gas leak) = (55 + 32)/107 = 0.813 b) P(electric failure|gas leak) = (55/107)/(87/102) = 0.632 c) P(gas leak| electric failure) = (55/107)/(72/107) = 0.764 2-12 2-68. No, if B ⊂ A , then P(A/B) = P( A ∩ B) P(B) = =1 P(B) P(B) A B 2-69. A B C Section 2-5 2-70. a) P( A ∩ B) = P( A B)P(B) = ( 0.4)( 0.5) = 0.20 b) P( A ′ ∩ B) = P( A ′ B)P(B) = (0.6)(0.5) = 0.30 2-71. P( A ) = P( A ∩ B) + P( A ∩ B ′) = P( A B)P(B) + P( A B ′)P(B ′) = (0.2)(0.8) + (0.3)(0.2) = 0.16 + 0.06 = 0.22 2-72. Let F denote the event that a connector fails. Let W denote the event that a connector is wet. P(F ) = P(F W )P( W ) + P(F W ′)P( W ′) = (0.05)(0.10) + (0.01)(0.90) = 0.014 2-73. Let F denote the event that a roll contains a flaw. Let C denote the event that a roll is cotton. P ( F) = P ( F C ) P ( C ) + P ( F C ′ ) P ( C ′ ) = ( 0. 02 )( 0. 70) + ( 0. 03)( 0. 30) = 0. 023 2-74. a) P(A) = 0.03 b) P(A') = 0.97 c) P(B|A) = 0.40 d) P(B|A') = 0.05 e) P( A ∩ B ) = P( B A )P(A) = (0.40)(0.03) = 0.012 f) P( A ∩ B ') = P( B' A )P(A) = (0.60)(0.03) = 0.018 g) P(B) = P( B A )P(A) + P( B A ')P(A') = (0.40)(0.03) + (0.05)(0.97) = 0.0605 2-13 2-75. Let R denote the event that a product exhibits surface roughness. Let N,A, and W denote the events that the blades are new, average, and worn, respectively. Then, P(R)= P(R|N)P(N) + P(R|A)P(A) + P(R|W)P(W) = (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15) = 0.028 2-76. Let B denote the event that a glass breaks. Let L denote the event that large packaging is used. P(B)= P(B|L)P(L) + P(B|L')P(L') = 0.01(0.60) + 0.02(0.40) = 0.014 2-77. Let U denote the event that the user has improperly followed installation instructions. Let C denote the event that the incoming call is a complaint. Let P denote the event that the incoming call is a request to purchase more products. Let R denote the event that the incoming call is a request for information. a) P(U|C)P(C) = (0.75)(0.03) = 0.0225 b) P(P|R)P(R) = (0.50)(0.25) = 0.125 2-78. a) (0.88)(0.27) = 0.2376 b) (0.12)(0.13+0.52) = 0.0.078 2-79. Let A denote a event that the first part selected has excessive shrinkage. Let B denote the event that the second part selected has excessive shrinkage. a) P(B)= P( B A )P(A) + P( B A ')P(A') = (4/24)(5/25) + (5/24)(20/25) = 0.20 b) Let C denote the event that the third chip selected has excessive shrinkage. P(C ) = P(C A ∩ B) P( A ∩ B) + P(C A ∩ B' ) P( A ∩ B' ) + P(C A'∩ B) P( A'∩ B) + P(C A'∩ B' ) P( A'∩ B' ) 3 ⎛ 4 ⎞⎛ 5 ⎞ 4 ⎛ 20 ⎞⎛ 5 ⎞ 4 ⎛ 5 ⎞⎛ 20 ⎞ 5 ⎛ 19 ⎞⎛ 20 ⎞ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ 23 ⎝ 24 ⎠⎝ 25 ⎠ 23 ⎝ 24 ⎠⎝ 25 ⎠ 23 ⎝ 24 ⎠⎝ 25 ⎠ 23 ⎝ 24 ⎠⎝ 25 ⎠ = 0.20 = 2-80. Let A and B denote the events that the first and second chips selected are defective, respectively. a) P(B) = P(B|A)P(A) + P(B|A')P(A') = (19/99)(20/100) + (20/99)(80/100) = 0.2 b) Let C denote the event that the third chip selected is defective. P( A ∩ B ∩ C ) = P(C A ∩ B) P( A ∩ B) = P(C A ∩ B) P( B A) P( A) 18 ⎛ 19 ⎞⎛ 20 ⎞ ⎟ ⎜ ⎟⎜ 98 ⎝ 99 ⎠⎝ 100 ⎠ = 0.00705 = Section 2-6 ≠ 2-81. Because P( A B ) 2-82. P(A') = 1 - P(A) = 0.7 and P( A ' B ) = 1 - P( A B ) = 0.7 P(A), the events are not independent. Therefore, A' and B are independent events. 2-83. P( A ∩ B ) = 70/100, P(A) = 86/100, P(B) = 77/100. Then, P( A ∩ B ) ≠ P(A)P(B), so A and B are not independent. 2-14 2-84. 2-85. P( A ∩ B ) = 80/100, P(A) = 82/100, P(B) = 90/100. Then, P( A ∩ B ) ≠ P(A)P(B), so A and B are not independent. a) P( A ∩ B )= 22/100, P(A) = 30/100, P(B) = 75/100, Then P( A ∩ B ) ≠ P(A)P(B), therefore, A and B are not independent. b) P(B|A) = P(A ∩ B)/P(A) = (22/100)/(30/100) = 0.733 2-86. If A and B are mutually exclusive, then P( A ∩ B ) = 0 and P(A)P(B) = 0.04. Therefore, A and B are not independent. 2-87. It is useful to work one of these exercises with care to illustrate the laws of probability. Let Hi denote the event that the ith sample contains high levels of contamination. ' ' ' ' ' ' ' ' ' ' a) P(H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) = P(H1)P(H2 )P(H3 )P(H4 )P(H5 ) by independence. Also, P(Hi' ) = 0.9 . Therefore, the answer is 0.9 5 = 0.59 ' ' ' b) A1 = (H1 ∩ H'2 ∩ H3 ∩ H4 ∩ H5 ) ' ' ' ' A 2 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) ' ' ' ' A 3 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) ' ' ' ' A 4 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) ' ' ' ' A 5 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) The requested probability is the probability of the union A 1 ∪ A 2 ∪ A 3 ∪ A 4 ∪ A 5 and these events are mutually exclusive. Also, by independence P( A i ) = 0.9 4 (0.1) = 0.0656 . Therefore, the answer is 5(0.0656) = 0.328. c) Let B denote the event that no sample contains high levels of contamination. The requested probability is P(B') = 1 - P(B). From part (a), P(B') = 1 - 0.59 = 0.41. 2-88. Let A i denote the event that the ith bit is a one. a) By independence P( A 1 ∩ A 2 ∩...∩ A 10 ) = P( A 1)P( A 2 )...P( A 10 ) = ( 1 )10 = 0.000976 2 ' ' ' c b) By independence, P ( A 1 ∩ A '2 ∩... ∩ A 10 ) = P ( A 1 ) P ( A '2 ) ... P ( A 10 ) = ( 1 ) 10 = 0. 000976 2 c) The probability of the following sequence is 1 ' ' ' ' ' P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ∩ A 6 ∩ A 7 ∩ A 8 ∩ A 9 ∩ A10 ) = ( )10 , by independence. The number of 2 () 10 10 ! sequences consisting of five "1"'s, and five "0"'s is 10 = = 252 . The answer is 252⎛ 1 ⎞ = 0.246 ⎜⎟ 5 5! 5! ⎝ 2⎠ 2-89. Let A denote the event that a sample is produced in cavity one of the mold. 1 a) By independence, P( A1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( )5 = 0.00003 8 b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually exclusive, P(B1 ∪ B 2 ∪...∪B 8 ) = P(B1) + P(B 2 )+...+P(B 8 ) 1 1 From part a., P(Bi ) = ( )5 . Therefore, the answer is 8( )5 = 0.00024 8 8 147 ' c) By independence, P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( ) ( ) . The number of sequences in 8 8 1 7 which four out of five samples are from cavity one is 5. Therefore, the answer is 5( ) 4 ( ) = 0.00107 . 8 8 2-15 2-90. Let A denote the upper devices function. Let B denote the lower devices function. P(A) = (0.9)(0.8)(0.7) = 0.504 P(B) = (0.95)(0.95)(0.95) = 0.8574 P(A∩B) = (0.504)(0.8574) = 0.4321 Therefore, the probability that the circuit operates = P(A∪B) = P(A) +P(B) − P(A∩B) = 0.9293 2-91. [1-(0.1)(0.05)][1-(0.1)(0.05)][1-(0.2)(0.1)] = 0.9702 2-92. Let Ai denote the event that the ith readback is successful. By independence, ' ' P ( A 1 ∩ A '2 ∩ A '3 ) = P ( A1 ) P ( A '2 ) P ( A '3 ) = ( 0. 02 ) 3 = 0. 000008. 2-93. a) P( B A ) = 4/499 and P( B) = P( B A) P( A) + P( B A' ) P( A' ) = (4 / 499)(5 / 500) + (5 / 499)(495 / 500) = 5 / 500 Therefore, A and B are not independent. b) A and B are independent. Section 2-7 2-94. Because, P( A B ) P(B) = P( A ∩ B ) = P( B A ) P(A), P( B A) = 2-95. P( A B) P( B) P( A) = 0.7(0.2) = 0.28 0.5 Let F denote a fraudulent user and let T denote a user that originates calls from two or more metropolitan areas in a day. Then, P(T F ) P( F ) 0.30(0.0001) P( F T ) = = = 0.003 P(T F ) P( F ) + P(T F ' ) P( F ' ) 0.30(0.0001) + 0.01(.9999) 2-96. main-storage backup 0.75 0.25 life > 5 yrs life > 5 yrs life < 5 yrs life < 5 yrs 0.95(0.25)=0.2375 0.05(0.25)=0.0125 0.995(0.75)=0.74625 0.005(0.75)=0.00375 a) P(B) = 0.25 b) P( A B ) = 0.95 c) P( A B ') = 0.995 d) P( A ∩ B ) = P( A B )P(B) = 0.95(0.25) = 0.2375 e) P( A ∩ B ') = P( A B ')P(B') = 0.995(0.75) = 0.74625 f) P(A) = P( A ∩ B ) + P( A ∩ B ') = 0.95(0.25) + 0.995(0.75) = 0.98375 g) 0.95(0.25) + 0.995(0.75) = 0.98375. h) P ( B A' ) = P ( A' B ) P ( B ) P ( A' B ) P ( B ) + P ( A' B ' ) P ( B ' ) 2-16 = 0.05(0.25) = 0.769 0.05(0.25) + 0.005(0.75) 2-97. Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, and poor performers, respectively. a) P(G) = P(G H ) P( H) + P(G M) P( M) + P(G P) P( P) = 0. 95( 0. 40) + 0. 60( 0. 35) + 0. 10( 0. 25) = 0. 615 b) Using the result from part a., P ( G H ) P ( H ) 0. 95( 0. 40) = = 0. 618 P( H G ) = 0. 615 P(G) c) P ( H G ' ) = P ( G ' H ) P ( H ) = 0. 05( 0. 40) = 0. 052 1 − 0. 615 P(G ' ) 2-98. a) P(D)=P(D|G)P(G)+P(D|G’)P(G’)=(.005)(.991)+(.99)(.009)=0.013865 b) P(G|D’)=P(G∩D’)/P(D’)=P(D’|G)P(G)/P(D’)=(.995)(.991)/(1-.013865)=0.9999 2-99. a) P(S) = 0.997(0.60) + 0.9995(0.27) + 0.897(0.13) = 0.9847 b) P(Ch|S) =(0.13)( 0.897)/0.9847 = 0.1184 Section 2-8 2-100. Continuous: a, c, d, f, h, i; Discrete: b, e, and g Supplemental Exercises 2-101. Let Di denote the event that the primary failure mode is type i and let A denote the event that a board passes the test. The sample space is S = {A, A ' D 1, A ' D 2 , A ' D 3 , A ' D 4 , A ' D 5 } . 2-102. a) 20/200 d) b) 135/200 c) 65/200 A 25 B 20 90 2-17 2-103. a) P(A) = 19/100 = 0.19 b) P(A ∩ B) = 15/100 = 0.15 c) P(A ∪ B) = (19 + 95 – 15)/100 = 0.99 d) P(A′∩ B) = 80/100 = 0.80 e) P(A|B) = P(A ∩ B)/P(B) = 0.158 2-104. Let A i denote the event that the ith order is shipped on time. a) By independence, P ( A1 ∩ A2 ∩ A3 ) = P( A1 ) P( A2 ) P ( A3 ) = (0.95) 3 = 0.857 b) Let ' B1 = A 1 ∩ A 2 ∩ A 3 ' B 2 = A1 ∩ A 2 ∩ A 3 ' B 3 = A1 ∩ A 2 ∩ A 3 Then, because the B's are mutually exclusive, P(B1 ∪ B 2 ∪ B 3 ) = P(B1 ) + P(B 2 ) + P(B 3 ) = 3(0.95) 2 (0.05) = 0.135 c) Let ' B1 = A 1 ∩ A '2 ∩ A 3 ' B 2 = A 1 ∩ A 2 ∩ A '3 ' ' B3 = A 1 ∩ A 2 ∩ A 3 ' ' B 4 = A 1 ∩ A 2 ∩ A '3 Because the B's are mutually exclusive, P(B1 ∪ B 2 ∪ B 3 ∪ B 4 ) = P(B1) + P(B 2 ) + P(B 3 ) + P(B 4 ) = 3(0.05) 2 (0.95) + (0.05) 3 = 0.00725 2-105. a) No, P(E1 ∩ E2 ∩ E3) ≠ 0 b) No, E1′ ∩ E2′ is not ∅ c) P(E1′ ∪ E2′ ∪ E3′) = P(E1′) + P(E2′) + P(E3′) – P(E1′∩ E2′) - P(E1′∩ E3′) - P(E2′∩ E3′) + P(E1′ ∩ E2′ ∩ E3′) = 40/240 d) P(E1 ∩ E2 ∩ E3) = 200/240 e) P(E1 ∪ E3) = P(E1) + P(E3) – P(E1 ∩ E3) = 234/240 f) P(E1 ∪ E2 ∪ E3) = 1 – P(E1′ ∩ E2′ ∩ E3′) = 1 - 0 = 1 2-106. (0.20)(0.30) +(0.7)(0.9) = 0.69 2-18 2-107. Let Ai denote the event that the ith bolt selected is not torqued to the proper limit. a) Then, P ( A1 ∩ A2 ∩ A3 ∩ A4 ) = P( A4 A1 ∩ A2 ∩ A3 ) P( A1 ∩ A2 ∩ A3 ) = P( A4 A1 ∩ A2 ∩ A3 ) P( A3 A1 ∩ A2 ) P ( A2 A1 ) P ( A1 ) ⎛ 12 ⎞⎛ 13 ⎞⎛ 14 ⎞⎛ 15 ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 0.282 ⎝ 17 ⎠⎝ 18 ⎠⎝ 19 ⎠⎝ 20 ⎠ 2-108. b) Let B denote the event that at least one of the selected bolts are not properly torqued. Thus, B' is the event that all bolts are properly torqued. Then, ⎛ 15 ⎞ ⎛ 14 ⎞ ⎛ 13 ⎞ ⎛ 12 ⎞ P(B) = 1 - P(B') = 1 − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = 0.718 ⎝ 20 ⎠ ⎝ 19 ⎠ ⎝ 18 ⎠ ⎝ 17 ⎠ Let A,B denote the event that the first, second portion of the circuit operates. Then, P(A) = (0.99)(0.99)+0.9-(0.99)(0.99)(0.9) = 0.998 P(B) = 0.9+0.9-(0.9)(0.9) = 0.99 and P( A ∩ B ) = P(A) P(B) = (0.998) (0.99) = 0.988 2-109. A1 = by telephone, A2 = website; P(A1) = 0.92, P(A2) = 0.95; By independence P(A1 ∪ A2) = P(A1) + P(A2) - P(A1 ∩ A2) = 0.92 + 0.95 - 0.92(0.95) = 0.996 2-110. P(Possess) = 0.95(0.99) +(0.05)(0.90) = 0.9855 2-111. Let D denote the event that a container is incorrectly filled and let H denote the event that a container is filled under high-speed operation. Then, a) P(D) = P( D H )P(H) + P( D H ')P(H') = 0.01(0.30) + 0.001(0.70) = 0.0037 b) P ( H D ) = P ( D H ) P ( H ) = 0.01(0.30) = 0.8108 P( D) 0.0037 2-112. a) P(E’ ∩ T’ ∩ D’) = (0.995)(0.99)(0.999) = 0.984 b) P(E ∪ D) = P(E) + P(D) – P(E ∩ D) = 0.005995 2-113. D = defective copy ⎛ 2 ⎞⎛ 73 ⎞⎛ 72 ⎞ ⎛ 73 ⎞⎛ 2 ⎞⎛ 72 ⎞ ⎛ 73 ⎞⎛ 72 ⎞⎛ 2 ⎞ a) P(D = 1) = ⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟⎜ ⎟ = 0.0778 ⎝ 75 ⎠⎝ 74 ⎠⎝ 73 ⎠ ⎝ 75 ⎠⎝ 74 ⎠⎝ 73 ⎠ ⎝ 75 ⎠⎝ 74 ⎠⎝ 73 ⎠ b) c) ⎛ 2 ⎞⎛ 1 ⎞⎛ 73 ⎞ ⎛ 2 ⎞⎛ 73 ⎞⎛ 1 ⎞ ⎛ 73 ⎞⎛ 2 ⎞⎛ 1 ⎞ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟⎜ ⎟ = 0.00108 ⎝ 75 ⎠⎝ 74 ⎠⎝ 73 ⎠ ⎝ 75 ⎠⎝ 74 ⎠⎝ 73 ⎠ ⎝ 75 ⎠⎝ 74 ⎠⎝ 73 ⎠ Let A represent the event that the two items NOT inspected are not defective. Then, P(A)=(73/75)(72/74)=0.947. P(D = 2) = ⎜ 2-114. The tool fails if any component fails. Let F denote the event that the tool fails. Then, P(F') = 0. 9910 by independence and P(F) = 1 - 0. 9910 = 0.0956 2-115. a) (0.3)(0.99)(0.985) + (0.7)(0.98)(0.997) = 0.9764 b) P ( route1 E ) = P ( E route1) P ( route1) = 0.02485(0.30) = 0.3159 P( E ) 1 − 0.9764 2-19 2-116. a) By independence, 0.15 5 = 7.59 × 10 −5 b) Let A i denote the events that the machine is idle at the time of your ith request. Using independence, the requested probability is ' ' ' ' P( A1A 2 A 3 A 4 A '5 or A1A 2 A 3 A 4 A 5 or A1A 2 A 3 A 4 A 5 or A1A 2 A 3 A 4 A 5 or A1A 2 A 3 A 4 A 5 ) = 0.15 4 (0.85) + 0.15 4 ( 0.85) + 0.15 4 (0.85) + 0.15 4 (0.85 ) + 0.15 4 (0.85) = 5(0.15 4 )(0.85) = 0.0022 c) As in part b,the probability of 3 of the events is ' ' ' ' ' ' ' P ( A1 A2 A3 A4 A5' or A1 A2 A3' A4 A5' or A1 A2 A3 A4 A5 or A1 A2 A3 A4 A5 or A1 A2 A3 A4 A5 or ' ' ' ' A1 A2 A3' A4 A5 or A1' A2 A3 A4 A5 or A1' A2 A3 A4 A5 or A1' A2 A3' A4 A5 or A1' A2 A3 A4 A5 ) = 10(0.15 3 )(0.85 2 ) = 0.0244 So to get the probability of at least 3, add answer parts a.) and b.) to the above to obtain requested probability. Therefore the answer is 0.0000759 + 0.0022 + 0.0244 = 0.0267 2-117. Let A i denote the event that the ith washer selected is thicker than target. ⎛ 30 ⎞⎛ 29 ⎞⎛ 28 ⎞ ⎟⎜ ⎟⎜ ⎟ = 0.207 ⎝ 50 ⎠⎝ 49 ⎠⎝ 8 ⎠ a) ⎜ b) 30/48 = 0.625 c) The requested probability can be written in terms of whether or not the first and second washer selected are thicker than the target. That is, ' ' '' P( A 3 ) = P( A 1A 2 A 3 orA 1A 2 A 3 orA 1A 2 A 3 orA 1A 2 A 3 ) ' ' = P( A 3 A 1 A 2 )P( A 1A 2 ) + P( A 3 A 1 A 2 )P( A 1A 2 ) ' '' '' +P( A 3 A 1 'A 2 )P( A 1A 2 ) + P( A 3 A 1 A 2 )P( A 1A 2 ) ' ' = P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 ) + P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 ) ' ' ' '' ' ' ' +P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 ) + P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 ) = 28 ⎛ 30 29 ⎞ 29 ⎛ 20 30 ⎞ 29 ⎛ 20 30 ⎞ 30 ⎛ 20 19 ⎞ ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟ 48 ⎝ 50 49 ⎠ 48 ⎝ 50 49 ⎠ 48 ⎝ 50 49 ⎠ 48 ⎝ 50 49 ⎠ = 0.60 2-118. a) If n washers are selected, then the probability they are all less than the target is 20 19 20 − n + 1 . ⋅ ⋅.. . 50 49 50 − n + 1 n probability all selected washers are less than target 1 20/50 = 0.4 2 (20/50)(19/49) = 0.155 3 (20/50)(19/49)(18/48) = 0.058 Therefore, the answer is n = 3 b) Then event E that one or more washers is thicker than target is the complement of the event that all are less than target. Therefore, P(E) equals one minus the probability in part a. Therefore, n = 3. 2-20 2-119. a) b) c) d) . e) f) 2-120. 112 + 68 + 246 = 0.453 940 246 P( A ∩ B) = = 0.262 940 514 + 68 + 246 P( A'∪ B) = = 0.881 940 514 P( A'∩ B' ) = = 0.547 940 P( A ∪ B) = P( A ∩ B) 246 / 940 = = 0.783 P(B) 314 / 940 P( B A ) = P ( B ∩ A ) = 246 / 940 = 0. 687 P(A ) 358 / 940 P( A B ) = Let E denote a read error and let S,O,P denote skewed, off-center, and proper alignments, respectively. Then, a) P(E) = P(E|S) P(S) + P(E|O) P (O) + P(E|P) P(P) = 0.01(0.10) + 0.02(0.05) + 0.001(0.85) = 0.00285 b) P(S|E) = P ( E S) P (S) P( E) 2-121. = 0. 01( 0. 10) = 0. 351 0. 00285 Let A i denote the event that the ith row operates. Then, P ( A1 ) = 0. 98, P ( A 2 ) = ( 0. 99) ( 0. 99) = 0. 9801, P ( A 3 ) = 0. 9801, P ( A 4 ) = 0. 98. The probability the circuit does not operate is ' ' ' P ( A1' ) P( A2 ) P ( A3 ) P( A4 ) = (0.02)(0.0199)(0.0199)(0.02) = 1.58 × 10 −7 2-122. a) (0.4)(0.1) + (0.3)(0.1) +(0.2)(0.2) + (0.4)(0.1) = 0.15 b) P(4 or more|provided) = (0.4)(0.1)/0.15 = 0.267 Mind-Expanding Exercises 2-123. Let E denote a read error and let S, O, B, P denote skewed, off-center, both, and proper alignments, respectively. P(E) = P(E|S)P(S) + P(E|O)P(O) + P(E|B)P(B) + P(E|P)P(P) = 0.01(0.10) + 0.02(0.05) + 0.06(0.01) + 0.001(0.84) = 0.00344 2-124. Let n denote the number of washers selected. a) The probability that all are less than the target is 0.4 n , by independence. n 0.4 n 1 0.4 2 0.16 3 0.064 Therefore, n = 3 b) The requested probability is the complement of the probability requested in part a. Therefore, n = 3 2-21 2-125. Let x denote the number of kits produced. Revenue at each demand 50 100 200 100x 100x 100x 0 ≤ x ≤ 50 Mean profit = 100x(0.95)-5x(0.05)-20x -5x 100(50)-5(x-50) 100x 100x 50 ≤ x ≤ 100 Mean profit = [100(50)-5(x-50)](0.4) + 100x(0.55)-5x(0.05)-20x -5x 100(50)-5(x-50) 100(100)-5(x-100) 100x 100 ≤ x ≤ 200 Mean profit = [100(50)-5(x-50)](0.4) + [100(100)-5(x-100)](0.3) + 100x(0.25) - 5x(0.05) - 20x 0 -5x Mean Profit Maximum Profit 74.75 x $ 3737.50 at x=50 32.75 x + 2100 $ 5375 at x=100 1.25 x + 5250 $ 5500 at x=200 Therefore, profit is maximized at 200 kits. However, the difference in profit over 100 kits is small. 0 ≤ x ≤ 50 50 ≤ x ≤ 100 100 ≤ x ≤ 200 2-126. Let E denote the probability that none of the bolts are identified as incorrectly torqued. The requested probability is P(E'). Let X denote the number of bolts in the sample that are incorrect. Then, P(E) = P(E|X=0)P(X=0) + P(E|X=1) P(X=1) + P(E|X=2) P(X=2) + P(E|X=3) P(X=3) + P(E|X=4)P(X=4) and P(X=0) = (15/20)(14/19)(13/18)(12/17) = 0.2817. The remaining probability for x can be determined from the counting methods in Appendix B-1. Then, ( )( ) = ⎜⎝ 4 ! 1! ⎟⎠ ⎜⎝ 3 ! 12! ⎟⎠ = 5 ! 15 ! 4 ! 16 ! = 0.4696 P( X = 1) = ( ) ⎛⎜⎝ 4206! ! ⎞⎟⎠ 4 ! 3 ! 12! 20 ! !1 5 15 13 ⎛ 5 ! ⎞ ⎛ 15 ! ⎞ 20 4 ( )( ) = ⎜⎝ 3 ! 2! ⎟⎠ ⎜⎝ 2! 13 ! ⎟⎠ = 0.2167 P( X = 2) = ⎛ 20 ! ⎞ () ⎟ ⎜ ⎝ 4 ! 16 ! ⎠ 5 15 22 ⎛ 5 ! ⎞ ⎛ 15 ! ⎞ 20 4 ( )( ) = ⎜⎝ 3 ! 2! ⎟⎠ ⎜⎝ 1! 14 ! ⎟⎠ = 0.0309 P( X = 3) = ( ) ⎛⎜⎝ 4206! ! ⎞⎟⎠ !1 5 15 31 ⎛ 5 ! ⎞ ⎛ 15 ! ⎞ 20 4 P(X=4) = (5/20)(4/19)(3/18)(2/17) = 0.0010 and P(E|X=0) = 1, P(E|X=1) = 0.05, P(E|X=2) = 0.05 2 = 0.0025 , P(E|X=3) = 0.05 3 = 1.25 × 10 −4 , P(E|X=4) = 0.05 4 = 6.25 × 10 −6 . Then, P(E) = 1(0.2817) + 0.05(0.4696) + 0.0025(0.2167) + 1.25 × 10 −4 (0.0309) +6.25 × 10 −6 (0.0010) = 0.306 and P(E') = 0.694 2-127. P( A '∩B' ) = 1 − P([ A '∩B' ]' ) = 1 − P( A ∪ B) = 1 − [P( A ) + P(B) − P( A ∩ B)] = 1 − P( A ) − P(B) + P( A )P(B) = [1 − P( A )][1 − P(B)] = P( A ' )P(B' ) 2-22 2-128. The total sample size is ka + a + kb + b = (k + 1)a + (k +1)b. ka + a k (a + b) , P ( B) = P(A ) = ( k + 1) a + ( k + 1) b ( k + 1) a + ( k + 1) b and ka ka P ( A ∩ B) = = ( k + 1) a + ( k + 1) b ( k + 1) ( a + b ) Then , k ( a + b ) ( ka + a ) k ( a + b ) ( k + 1) a ka P ( A ) P ( B) = = = = P ( A ∩ B) 2 2 2 ( k + 1) ( a + b ) [( k + 1) a + ( k + 1) b ] ( k + 1) ( a + b ) Section 2-1.4 on CD S2-1. From the multiplication rule, the answer is 5 × 3 × 4 × 2 = 120 S2-2. From the multiplication rule, 3 ×4 ×3 = 36 S2-3. From the multiplication rule, 3×4×3×4 =1 4 4 S2-4. From equation S2-1, the answer is 10! = 3628800 S2-5. From the multiplication rule and equation S2-1, the answer is 5!5! = 14400 S2-6. From equation S2-3, S2-7. a) From equation S2-4, the number of samples of size five is 7! = 35 sequences are possible 3! 4! ( ) = 5140!! = 416965528 !135 140 5 b) There are 10 ways of selecting one nonconforming chip and there are ( ) = 4130!! = 11358880 !126 130 4 ways of selecting four conforming chips. Therefore, the number of samples that contain exactly one () = 113588800 nonconforming chip is 10 × 4 c) The number of samples that contain at least one nonconforming chip is the total number of samples 130 ( ) minus the number of samples that contain no nonconforming chips ( ) . That is ( ) - ( ) = 5140! ! − 5130! ! = 130721752 ! 135 !125 140 5 130 5 140 5 S2-8. 130 5 a) If the chips are of different types, then every arrangement of 5 locations selected from the 12 results in a 12 12! different layout. Therefore, P5 = = 95040 layouts are possible. 7! b) If the chips are of the same type, then every subset of 5 locations chosen from the 12 results in a different 12! = 792 layouts are possible. layout. Therefore, 12 = 5 5! 7! () 2-23 S2-9. a) b) 7! = 21 sequences are possible. 2!5! 7! = 2520 sequences are possible. 1!1!1!1!1!2! c) 6! = 720 sequences are possible. S2-10. a) Every arrangement of 7 locations selected from the 12 comprises a different design. 12 12! = 3991680 designs are possible. P7 = 5! b) Every subset of 7 locations selected from the 12 comprises a new design. possible. c) First the three locations for the first component are selected in 12! = 792 5!7! 12 ( ) = 3!9!! = 220 ways. Then, the four 12 3 locations for the second component are selected from the nine remaining locations in S2-11. S2-12. designs are ! ( ) = 495! = 126 ! 9 4 ways. From the multiplication rule, the number of designs is 220 × 126 = 27720 a) From the multiplication rule, 10 3 = 1000 prefixes are possible b) From the multiplication rule, 8 × 2 × 10 = 160 are possible c) Every arrangement of three digits selected from the 10 digits results in a possible prefix. 10 ! 10 = 720 prefixes are possible. P3 = 7! a) From the multiplication rule, 2 8 = 256 bytes are possible b) From the multiplication rule, 2 7 = 128 bytes are possible S2-13. a) The total number of samples possible is one tank has high viscosity is ( ) = 424!! = 10626. The number of samples in which exactly !20 24 4 ( )( ) = 165!! × 318!! = 4896 . Therefore, the probability is ! !15 6 1 18 3 4896 = 0.461 10626 b) The number of samples that contain no tank with high viscosity is requested probability is 1 − 3060 = 0.712 . 10626 c) The number of samples that meet the requirements is Therefore, the probability is 2184 = 0.206 10626 2-24 ( ) = 418!! = 3060. Therefore, the !14 18 4 ( )( )( ) = 165!! × 143!! × 214!! = 2184 . ! ! !12 6 1 4 1 14 2 S2-14. () 12 ! a) The total number of samples is 12 = = 220. The number of samples that result in one 3 3! 9! nonconforming part is ( )( ) = 121!! × 10!! = 90. ! 2!8 2 1 10 2 90/220 = 0.409. b) The number of samples with no nonconforming part is nonconforming part is 1 − S2-15. Therefore, the requested probability is 10 ( ) = 3!7!! = 120. The probability of at least one 10 3 120 = 0.455 . 220 54 × = 0.0082 50 49 50! 50 × 49 50 . The number of samples with two defective b) The total number of samples is 2 = = 2!48! 2 5× 4 5× 4 5! 5 × 4 5 2 parts is 2 = . Therefore, the probability is = = 0.0082 . = 50× 49 2!3! 2 50 × 49 2 a) The probability that both parts are defective is () () 2-25 CHAPTER 3 Section 3-1 {0,1,2,...,1000} 3-1. The range of X is 3-2 The range of X is {0,1,2,. ..,50} 3-3. , The range of X is {0,1 2,. ..,99999} 3-4 , The range of X is {0,1 2,3,4,5} 3-5. 3-6 { } The range of X is 1,2,...,491 . Because 490 parts are conforming, a nonconforming part must be selected in 491 selections. , The range of X is {0,1 2,. ..,100} . Although the range actually obtained from lots typically might not exceed 10%. 3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is {0,1,2,.. .} 3-8 The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is {0,1,2,.. .} 3-9. The range of X is 3-10 The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8. ⎧1 3 1 5 6⎫ Therefore the range of X is ⎨ , , , , ⎬ ⎩4 8 2 8 8 ⎭ 3-11 The range of X is {0,1,2, K,10000} 3-12 The range of X is {0,1,2,...,5000} {0,1,2,...,15} Section 3-2 3-13. f X ( 0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3 f X (1.5) = P ( X = 1.5) = 1 / 3 f X ( 2) = 1 / 6 f X (3) = 1 / 6 3-14 a) P(X=1.5) = 1/3 b) P(0.5< X < 2.7) = P(X=1.5) + P(X=2) = 1/6 + 1/3 = 1/2 c) P(X > 3) = 0 d) P ( 0 ≤ X < 2) = P(X=0) + P(X=1.5) = 1/3 + 1/3 = 2/3 e) P(X=0 or X=2) = 1/3 + 1/6 = 1/2 3-15. All probabilities are greater than or equal to zero and sum to one. 3-1 a) P(X ≤ 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(-1 ≤ X ≤ 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X ≤ -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 3-16 All probabilities are greater than or equal to zero and sum to one. a) P(X≤ 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.5714=0.4286 c) P(2<X<6)=P(X=3)=0.1429 d) P(X≤1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1 3-17. Probabilities are nonnegative and sum to one. a) P(X = 4) = 9/25 b) P(X ≤ 1) = 1/25 + 3/25 = 4/25 c) P(2 ≤ X < 4) = 5/25 + 7/25 = 12/25 d) P(X > −10) = 1 3-18 Probabilities are nonnegative and sum to one. a) P(X = 2) = 3/4(1/4)2 = 3/64 b) P(X ≤ 2) = 3/4[1+1/4+(1/4)2] = 63/64 c) P(X > 2) = 1 − P(X ≤ 2) = 1/64 d) P(X ≥ 1) = 1 − P(X ≤ 0) = 1 − (3/4) = 1/4 3-19. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 3-20 P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 3-21. P(X = 0) = 0.023 = 8 x 10-6 P(X = 1) = 3[0.98(0.02)(0.02)]=0.0012 P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576 P(X = 3) = 0.983 = 0.9412 3-22 X = number of wafers that pass P(X=0) = (0.2)3 = 0.008 P(X=1) = 3(0.2)2(0.8) = 0.096 P(X=2) = 3(0.2)(0.8)2 = 0.384 P(X=3) = (0.8)3 = 0.512 3-23 P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 3-24 X = number of components that meet specifications P(X=0) = (0.05)(0.02) = 0.001 P(X=1) = (0.05)(0.98) + (0.95)(0.02) = 0.068 P(X=2) = (0.95)(0.98) = 0.931 3-25. X = number of components that meet specifications P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663 P(X=3) = (0.95)(0.98)(0.99) = 0.92169 3-2 Section 3-3 3-26 x<0 ⎫ ⎧ 0, ⎪1 / 3 0 ≤ x < 1.5⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨2 / 3 1.5 ≤ x < 2⎬ ⎪5 / 6 2 ≤ x < 3 ⎪ ⎪ ⎪ ⎪1 3≤ x ⎪ ⎩ ⎭ f X ( 0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3 where f X (1.5) = P ( X = 1.5) = 1 / 3 f X ( 2) = 1 / 6 f X (3) = 1 / 6 3-27. x < −2 ⎫ ⎧ 0, ⎪1 / 8 − 2 ≤ x < −1⎪ ⎪ ⎪ ⎪3 / 8 − 1 ≤ x < 0 ⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ 0 ≤ x <1 ⎪ ⎪5 / 8 ⎪7 / 8 1≤ x < 2 ⎪ ⎪ ⎪ ⎪ 2≤x ⎪ ⎩1 ⎭ f X (−2) = 1 / 8 f X (−1) = 2 / 8 where f X ( 0) = 2 / 8 f X (1) = 2 / 8 f X (2) = 1 / 8 a) P(X ≤ 1.25) = 7/8 b) P(X ≤ 2.2) = 1 c) P(-1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4 d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8 3-28 ⎧ 0, ⎪ 1 / 25 ⎪ ⎪ 4 / 25 ⎪ F ( x) = ⎨ ⎪ 9 / 25 ⎪16 / 25 ⎪ ⎪1 ⎩ x<0 ⎫ 0 ≤ x < 1⎪ ⎪ 1 ≤ x < 2⎪ ⎪ ⎬ 2 ≤ x < 3⎪ 3 ≤ x < 4⎪ ⎪ 4≤ x ⎪ ⎭ f X (0) = 1 / 25 f X (1) = 3 / 25 f X (2) = 5 / 25 where f X (3) = 7 / 25 f X (4) = 9 / 25 a) P(X < 1.5) = 4/25 b) P(X ≤ 3) = 16/25 c) P(X > 2) = 1 − P(X ≤ 2) = 1 − 9/25 = 16/25 d) P(1 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 1) = 9/25 − 4/25 = 5/25 = 1/5 3-29. x <1 ⎫ ⎧ 0, ⎪ 0.1, 1 ≤ x < 5 ⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ ⎪0.7, 5 ≤ x < 10⎪ ⎪ 1, 10 ≤ x ⎪ ⎩ ⎭ where P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 3-3 3-30 x < 10 ⎫ ⎧ 0, ⎪0.2, 10 ≤ x < 25 ⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ ⎪0.5, 25 ≤ x < 50⎪ ⎪ 1, 50 ≤ x ⎪ ⎩ ⎭ where P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 3-31. x<0 ⎫ ⎧ 0, ⎪0.008, 0 ≤ x < 1⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨0.104, 1 ≤ x < 2 ⎬ ⎪0.488, 2 ≤ x < 3⎪ ⎪ ⎪ ⎪ 3≤ x ⎪ ⎩ 1, ⎭ . where f (0) = 0.2 3 = 0.008, f (1) = 3(0.2)(0.2)(0.8) = 0.096, f (2) = 3(0.2)(0.8)(0.8) = 0.384, f (3) = (0.8) 3 = 0.512, 3-32 x < −0.5 ⎫ ⎧ 0, ⎪0.1, − 0.5 ≤ x < 5⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ ⎪0.4, 5 ≤ x < 15 ⎪ ⎪ 1, 15 ≤ x ⎪ ⎭ ⎩ where P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 3-33. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.5, f(3) = 0.5 a) P(X ≤ 3) = 1 b) P(X ≤ 2) = 0.5 c) P(1 ≤ X ≤ 2) = P(X=1) = 0.5 d) P(X>2) = 1 − P(X≤2) = 0.5 3-34 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.7, f(4) = 0.2, f(7) = 0.1 a) P(X ≤ 4) = 0.9 b) P(X > 7) = 0 c) P(X ≤ 5) = 0.9 d) P(X>4) = 0.1 e) P(X≤2) = 0.7 3-35. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(-10) = 0.25, f(30) = 0.5, f(50) = 0.25 a) P(X≤50) = 1 b) P(X≤40) = 0.75 c) P(40 ≤ X ≤ 60) = P(X=50)=0.25 d) P(X<0) = 0.25 e) P(0≤X<10) = 0 f) P(−10<X<10) = 0 3-4 3-36 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1 a) P(X≤1/18) = 0 b) P(X≤1/4) = 0.9 c) P(X≤5/16) = 0.9 d) P(X>1/4) = 0.1 e) P(X≤1/2) = 1 Section 3-4 3-37 Mean and Variance µ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + 4 f (4) = 0(0.2) + 1(0.2) + 2(0.2) + 3(0.2) + 4(0.2) = 2 V ( X ) = 0 2 f (0) + 12 f (1) + 2 2 f (2) + 3 2 f (3) + 4 2 f (4) − µ 2 = 0(0.2) + 1(0.2) + 4(0.2) + 9(0.2) + 16(0.2) − 2 2 = 2 3- 38 Mean and Variance µ = E ( X ) = 0 f (0) + 1.5 f (1.5) + 2 f (2) + 3 f (3) = 0(1 / 3) + 1.5(1 / 3) + 2(1 / 6) + 3(1 / 6) = 1.333 V ( X ) = 0 2 f (0) + 1.5 2 f (1) + 2 2 f (2) + 3 2 f (3) − µ 2 = 0(1 / 3) + 2.25(1 / 3) + 4(1 / 6) + 9(1 / 6) − 1.333 2 = 1.139 3-39 Determine E(X) and V(X) for random variable in exercise 3-15 . µ = E ( X ) = −2 f (−2) − 1 f (−1) + 0 f (0) + 1 f (1) + 2 f (2) = −2(1 / 8) − 1(2 / 8) + 0(2 / 8) + 1(2 / 8) + 2(1 / 8) = 0 V ( X ) = −2 2 f (−2) − 12 f (−1) + 0 2 f (0) + 12 f (1) + 2 2 f (2) − µ 2 = 4(1 / 8) + 1(2 / 8) + 0(2 / 8) + 1(2 / 8) + 4(1 / 8) − 0 2 = 1.5 3-40 Determine E(X) and V(X) for random variable in exercise 3-15 µ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + 4 f (4) = 0(0.04) + 1(0.12) + 2(0.2) + 3(0.28) + 4(0.36) = 2.8 V ( X ) = 0 2 f (0) + 12 f (1) + 2 2 f (2) + 3 2 f (3) + 4 2 f (4) − µ 2 = 0(0.04) + 1(0.12) + 4(0.2) + 9(0.28) + 16(0.36) − 2.8 2 = 1.36 3-41. Mean and variance for exercise 3-19 µ = E ( X ) = 10 f (10) + 5 f (5) + 1 f (1) = 10(0.3) + 5(0.6) + 1(0.1) = 6.1 million V ( X ) = 10 2 f (10) + 5 2 f (5) + 12 f (1) − µ 2 = 10 2 (0.3) + 5 2 (0.6) + 12 (0.1) − 6.12 = 7.89 million 2 3-5 3-42 Mean and variance for exercise 3-20 µ = E ( X ) = 50 f (50) + 25 f (25) + 10 f (10) = 50(0.5) + 25(0.3) + 10(0.2) = 34.5 million V ( X ) = 50 2 f (50) + 25 2 f (25) + 10 2 f (10) − µ 2 = 50 2 (0.5) + 25 2 (0.3) + 10 2 (0.2) − 34.5 2 = 267.25 million 2 3-43. Mean and variance for random variable in exercise 3-22 µ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) = 0(0.008) + 1(0.096) + 2(0.384) + 3(0.512) = 2.4 V ( X ) = 0 2 f (0) + 12 f (1) + 2 2 f (2) + 3 2 f (3) − µ 2 = 0 2 (0.008) + 1(0.096) + 4(0.384) + 9(0.512) − 2.4 2 = 0.48 3-44 Mean and variance for exercise 3-23 µ = E ( X ) = 15 f (15) + 5 f (5) − 0.5 f (5) = 15(0.6) + 5(0.3) − 0.5(0.1) = 10.45 million V ( X ) = 15 2 f (15) + 5 2 f (5) + (−0.5) 2 f (−0.5) − µ 2 = 15 2 (0.6) + 5 2 (0.3) + (−0.5) 2 (0.1) − 10.45 2 = 33.32 million 2 3-45. Determine x where range is [0,1,2,3,x] and mean is 6. µ = E ( X ) = 6 = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + xf ( x) 6 = 0(0.2) + 1(0.2) + 2(0.2) + 3(0.2) + x(0.2) 6 = 1.2 + 0.2 x 4.8 = 0.2 x x = 24 Section 3-5 3-46 E(X) = (0+100)/2 = 50, V(X) = [(100-0+1)2-1]/12 = 850 3-47. E(X) = (3+1)/2 = 2, V(X) = [(3-1+1)2 -1]/12 = 0.667 3-48 1 ⎛1⎞ 1 ⎛1⎞ 3 ⎛1⎞ 1 E( X ) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = , 8 ⎝3⎠ 4 ⎝ 3⎠ 8 ⎝ 3⎠ 4 2 2 2 2 ⎛1⎞ ⎛1⎞ ⎛ 1⎞ ⎛1⎞ ⎛3⎞ ⎛1⎞ ⎛ 1 ⎞ V ( X ) = ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ = 0.0104 ⎝8⎠ ⎝3⎠ ⎝ 4⎠ ⎝3⎠ ⎝8⎠ ⎝3⎠ ⎝ 4⎠ 3-6 3-49. X=(1/100)Y, Y = 15, 16, 17, 18, 19. E(X) = (1/100) E(Y) = 1 ⎛ 15 + 19 ⎞ ⎜ ⎟ = 0.17 mm 100 ⎝ 2 ⎠ 2 2 ⎛ 1 ⎞ ⎡ (19 − 15 + 1) − 1⎤ 2 V (X ) = ⎜ ⎟⎢ ⎥ = 0.0002 mm 12 ⎝ 100 ⎠ ⎣ ⎦ 3-50 ⎛1⎞ ⎛1⎞ ⎛1⎞ E ( X ) = 2⎜ ⎟ + 3⎜ ⎟ + 4⎜ ⎟ = 3 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ in 100 codes the expected number of letters is 300 2 2⎛1⎞ 2⎛1⎞ 2⎛1⎞ 2 V ( X ) = (2) ⎜ ⎟ + (3) ⎜ ⎟ + (4) ⎜ ⎟ − (3) = 3 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ in 100 codes the variance is 6666.67 3-51. X = 590 + 0.1Y, Y = 0, 1, 2, ..., 9 ⎛0+9⎞ 590 + 0.1⎜ ⎟ = 590.45 mm, ⎝2⎠ ⎡ (9 − 0 + 1) 2 − 1⎤ V ( X ) = (0.1) 2 ⎢ ⎥ = 0.0825 12 ⎣ ⎦ E(X) = 3-52 3-53 mm2 The range of Y is 0, 5, 10, ..., 45, E(X) = (0+9)/2 = 4.5 E(Y) = 0(1/10)+5(1/10)+...+45(1/10) = 5[0(0.1) +1(0.1)+ ... +9(0.1)] = 5E(X) = 5(4.5) = 22.5 V(X) = 8.25, V(Y) = 52(8.25) = 206.25, σY = 14.36 E (cX ) = ∑ cxf ( x) = c∑ xf ( x) = cE ( X ) , x x V (cX ) = ∑ (cx − cµ ) f ( x) = c 2 ∑ ( x − µ ) 2 f ( x) = cV ( X ) 2 x 3-54 x X is a discrete random variable. X is discrete because it is the number of fields out of 28 that has an error. However, X is not uniform because P(X=0) ≠ P(X=1). 3-7 Section 3-6 3-55. A binomial distribution is based on independent trials with two outcomes and a constant probability of success on each trial. a) reasonable b) independence assumption not reasonable c) The probability that the second component fails depends on the failure time of the first component. The binomial distribution is not reasonable. d) not independent trials with constant probability e) probability of a correct answer not constant. f) reasonable g) probability of finding a defect not constant. h) if the fills are independent with a constant probability of an underfill, then the binomial distribution for the number packages underfilled is reasonable. i) because of the bursts, each trial (that consists of sending a bit) is not independent j) not independent trials with constant probability 3-56 0.25 0.20 f(x) 0.15 0.10 0.05 0.00 0 5 10 x a.) E ( X ) = np = 10(0.5) = 5 b.) Values X=0 and X=10 are the least likely, the extreme values 3-57. ⎛10 ⎞ = 5) = ⎜ ⎟0.5 5 (0.5) 5 = 0.2461 ⎜5⎟ ⎝⎠ ⎛10 ⎞ 0 10 ⎛10 ⎞ 1 9 ⎛10 ⎞ 2 8 b) P ( X ≤ 2) = ⎜ ⎟0.5 0.5 + ⎜ ⎟0.5 0.5 + ⎜ ⎟0.5 0.5 ⎜2⎟ ⎜1⎟ ⎜0⎟ ⎝⎠ ⎝⎠ ⎝⎠ 10 10 10 = 0.5 + 10(0.5) + 45(0.5) = 0.0547 ⎛10 ⎞ 10 ⎛10 ⎞ 9 1 0 c) P ( X ≥ 9) = ⎜ ⎟0.5 (0.5) + ⎜ ⎟0.5 (0.5) = 0.0107 ⎜10 ⎟ ⎜9⎟ ⎝⎠ ⎝⎠ a) P ( X d) ⎛10 ⎞ ⎛10 ⎞ P(3 ≤ X < 5) = ⎜ ⎟0.530.57 + ⎜ ⎟0.540.56 ⎜4⎟ ⎜3⎟ ⎝⎠ ⎝⎠ 10 = 120(0.5) + 210(0.5)10 = 0.3223 3-8 3-58 Binom al (10, 0. 01) 0. 9 0. 8 0. 7 pr ob of x 0. 6 0. 5 0. 4 0. 3 0. 2 0. 1 0. 0 0 1 2 3 4 5 6 7 8 9 10 x a) E ( X ) = np = 10(0.01) = 0.1 The value of X that appears to be most likely is 0. 1. b) The value of X that appears to be least likely is 10. 3-59. ⎛10 ⎞ 5 = 5) = ⎜ ⎟0.015 (0.99 ) = 2.40 × 10−8 ⎜5⎟ ⎝⎠ ⎛10 ⎞ ⎛10 ⎞ ⎛10 ⎞ 10 9 8 b) P ( X ≤ 2) = ⎜ ⎟0.010 (0.99 ) + ⎜ ⎟0.011 (0.99 ) + ⎜ ⎟0.012 (0.99 ) ⎜2⎟ ⎜1⎟ ⎜0⎟ ⎝⎠ ⎝⎠ ⎝⎠ = 0.9999 a) P ( X ⎛10 ⎞ ⎛10 ⎞ 1 0 c) P ( X ≥ 9) = ⎜ ⎟0.019 (0.99 ) + ⎜ ⎟0.0110 (0.99 ) = 9.91 × 10 −18 ⎜10 ⎟ ⎜9⎟ ⎝⎠ ⎝⎠ ⎛10 ⎞ ⎛10 ⎞ 7 d ) P (3 ≤ X < 5) = ⎜ ⎟0.013 (0.99 ) + ⎜ ⎟0.014 (0.99) 6 = 1.138 × 10 − 4 ⎜3⎟ ⎜4⎟ ⎝⎠ ⎝⎠ 3-60 n=3 and p=0.5 3 x<0 ⎫ ⎧0 ⎪0.125 0 ≤ x < 1⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨ 0.5 1 ≤ x < 2 ⎬ ⎪0.875 2 ≤ x < 3⎪ ⎪ ⎪ ⎪1 3≤ x ⎪ ⎭ ⎩ where 1 ⎛1⎞ f ( 0) = ⎜ ⎟ = 8 ⎝2⎠ 2 3 ⎛ 1 ⎞⎛ 1 ⎞ f (1) = 3⎜ ⎟⎜ ⎟ = 8 ⎝ 2 ⎠⎝ 2 ⎠ 2 ⎛1⎞ ⎛3⎞ 3 f (2) = 3⎜ ⎟ ⎜ ⎟ = ⎝4⎠ ⎝4⎠ 8 3 1 ⎛1⎞ f (3) = ⎜ ⎟ = 8 ⎝4⎠ 3-9 3-61. n=3 and p=0.25 3 x<0 ⎫ ⎧0 ⎪0.4219 0 ≤ x < 1⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨0.8438 1 ≤ x < 2 ⎬ ⎪0.9844 2 ≤ x < 3⎪ ⎪ ⎪ ⎪1 3≤ x ⎪ ⎭ ⎩ 3-62 Let X denote the number of defective circuits. Then, X has a binomial distribution with n = 40 and p = 0.01. Then, P(X = 0) = 3-63. where 27 ⎛3⎞ f ( 0) = ⎜ ⎟ = 64 ⎝4⎠ 2 27 ⎛ 1 ⎞⎛ 3 ⎞ f (1) = 3⎜ ⎟⎜ ⎟ = 64 ⎝ 4 ⎠⎝ 4 ⎠ 2 ⎛1⎞ ⎛3⎞ 9 f (2) = 3⎜ ⎟ ⎜ ⎟ = ⎝ 4 ⎠ ⎝ 4 ⎠ 64 3 1 ⎛1⎞ f (3) = ⎜ ⎟ = 64 ⎝4⎠ ( )0.01 0.99 40 0 0 40 = 0.6690 . ⎛1000 ⎞ 1 999 = 1) = ⎜ ⎟ ⎜ 1 ⎟0.001 (0.999) = 0.3681 ⎠ ⎝ ⎛1000 ⎞ 999 1 b) P ( X ≥ 1) = 1 − P ( X = 0) = 1 − ⎜ ⎜ 1 ⎟0.001 (0.999 ) = 0.6319 ⎟ ⎝ ⎠ a) P( X ⎛1000 ⎞ ⎛1000 ⎞ 1000 999 0 1 c ) P ( X ≤ 2) = ⎜ ⎜ 0 ⎟0.001 (0.999 ) + ⎜ 1 ⎟0.001 (0.999 ) + ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = 0.9198 d ) E ( X ) = 1000(0.001) = 1 ( )0.001 0.999 1000 2 V ( X ) = 1000(0.001)(0.999) = 0.999 3-64 Let X denote the number of times the line is occupied. Then, X has a binomial distribution with n = 10 and p = 0.4 ⎛ 10⎞ a.) P( X = 3) = ⎜ ⎟ 0.4 3 (0.6) 7 = 0.215 ⎝ 3⎠ ≥ 1) = 1 − P ( X = 0) = 1 − c.) E ( X ) = 10(0.4) = 4 b.) P ( X 3-65. ( )0.4 0.6 10 0 0 10 = 0.994 a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np. ⎛ 50 ⎞ ⎛ 50 ⎞ ⎛ 50 ⎞ 50 49 48 ≤ 2) = ⎜ ⎟0.10 (0.9) + ⎜ ⎟0.11 (0.9) + ⎜ ⎟0.12 (0.9) = 0.112 ⎜2⎟ ⎜1⎟ ⎜0⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎛ 50 ⎞ 50 ⎛ 50 ⎞ 49 1 0 − 48 c) P ( X ≥ 49) = ⎜ ⎟ ⎜⎟ ⎜ 49 ⎟0.1 (0.9) + ⎜ 50 ⎟0.1 (0.9) = 4.51 × 10 ⎝⎠ ⎝⎠ b) P ( X 3-10 2 998 3-66 E(X) = 20 (0.01) = 0.2 V(X) = 20 (0.01) (0.99) = 0.198 µ X + 3σ X = 0.2 + 3 0.198 = 1.53 a) ) X is binomial with n = 20 and p = 0.01 P( X > 1.53) = P( X ≥ 2) = 1 − P( X ≤ 1) = 1− [( )0.01 0.99 + ( )0.01 0.99 ] = 0.0169 20 0 0 20 20 1 1 19 b) X is binomial with n = 20 and p = 0.04 P( X > 1) = 1 − P ( X ≤ 1) = 1− [( )0.04 0.96 20 0 0 20 + ( )0.04 0.96 ] = 0.1897 20 1 1 19 c) Let Y denote the number of times X exceeds 1 in the next five samples. Then, Y is binomial with n = 5 and p = 0.190 from part b. P(Y ≥ 1) = 1 − P (Y = 0) = 1 − [( )0.190 0.810 ] = 0.651 5 0 0 5 The probability is 0.651 that at least one sample from the next five will contain more than one defective. 3-67. Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1. a ) P ( X ≥ 5) = 1 − P ( X ≤ 4) ⎡⎛125 ⎞ 0 ⎛125 ⎞ 2 ⎛125 ⎞ 1 125 124 123 ⎤ ⎢⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ 0 ⎟0.1 (0.9 ) + ⎜ 1 ⎟0.1 (0.9 ) + ⎜ 2 ⎟0.1 (0.9 ) ⎥ ⎠ ⎠ ⎝ ⎠ ⎝ ⎝ ⎥ = 1− ⎢ ⎢ ⎛125 ⎞ ⎥ 125 ⎞ 4 ⎛ ⎢+ ⎜ ⎥ ⎟0.13 (0.9 )122 + ⎜ ⎟0.1 (0.9 )121 ⎜ ⎟ ⎜4⎟ ⎢ ⎝3⎠ ⎥ ⎝ ⎠ ⎣ ⎦ = 0.9961 b) P ( X > 5) = 1 − P ( X ≤ 5) = 0.9886 3-68 Let X denote the number of defective components among those stocked. b.) ( )0.02 0.98 P ( X ≤ 2) = ( )0.02 0.98 c). P ( X ≤ 5) = 0.981 a ). P ( X = 0) = 100 0 0 100 = 0.133 102 0 0 102 + ( )0.02 0.98 102 1 3-11 1 101 + ( )0.02 0.98 102 2 2 100 = 0.666 3-69. Let X denote the number of questions answered correctly. Then, X is binomial with n = 25 and p = 0.25. ⎛ 25 ⎞ ⎛ 25⎞ ⎛ 25 ⎞ 5 4 3 a) P( X ≥ 20) = ⎜ ⎟0.2520 (0.75) + ⎜ ⎟0.2521 (0.75) + ⎜ ⎟0.2522 (0.75) ⎜ 20 ⎟ ⎜ 21⎟ ⎜ 22 ⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎛ 25⎞ ⎛ 25 ⎞ ⎛ 25⎞ 2 1 0 + ⎜ ⎟0.2523 (0.75) + ⎜ ⎟0.2524 (0.75) + ⎜ ⎟0.2525 (0.75) = 9.677 ×10−10 ⎜ 23⎟ ⎜ 24 ⎟ ⎜ 25⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎛ 25⎞ ⎛ 25⎞ ⎛ 25⎞ 25 24 23 b) P( X < 5) = ⎜ ⎟0.250 (0.75) + ⎜ ⎟0.251 (0.75) + ⎜ ⎟0.252 (0.75) ⎜0⎟ ⎜1⎟ ⎜2⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎛ 25⎞ ⎛ 25⎞ 22 21 + ⎜ ⎟0.253 (0.75) + ⎜ ⎟0.254 (0.75) = 0.2137 ⎜3⎟ ⎜4⎟ ⎝⎠ ⎝⎠ 3-70 Let X denote the number of mornings the light is green. b) ( )0.2 0.8 = 0.410 P ( X = 4) = ( )0.2 0.8 = 0.218 c) P ( X > 4) = 1 − P ( X ≤ 4) = 1 − 0.630 = 0.370 a) P ( X = 1) = 5 1 1 20 4 4 4 16 Section 3-7 3-71. a.) b.) c.) d.) P( X P( X P( X P( X = 1) = (1 − 0.5) 0 0.5 = 0.5 = 4) = (1 − 0.5) 3 0.5 = 0.5 4 = 0.0625 = 8) = (1 − 0.5) 7 0.5 = 0.58 = 0.0039 ≤ 2) = P( X = 1) + P( X = 2) = (1 − 0.5) 0 0.5 + (1 − 0.5)1 0.5 = 0.5 + 0.5 2 = 0.75 e.) P( X > 2) = 1 − P ( X ≤ 2) = 1 − 0.75 = 0.25 3-72 E(X) = 2.5 = 1/p giving p = 0.4 P( X = 1) = (1 − 0.4) 0 0.4 = 0.4 3 b.) P ( X = 4) = (1 − 0.4) 0.4 = 0.0864 4 c.) P ( X = 5) = (1 − 0.5) 0.5 = 0.05184 d.) P( X ≤ 3) = P( X = 1) + P( X = 2) + P( X = 3) a.) = (1 − 0.4) 0 0.4 + (1 − 0.4)1 0.4 + (1 − 0.4) 2 0.4 = 0.7840 e.) P( X > 3) = 1 − P( X ≤ 3) = 1 − 0.7840 = 0.2160 3-12 3-73. Let X denote the number of trials to obtain the first successful alignment. Then X is a geometric random variable with p = 0.8 b) P( X = 4) = (1 − 0.8) 3 0.8 = 0.2 3 0.8 = 0.0064 P( X ≤ 4) = P( X = 1) + P( X = 2) + P( X = 3) + P( X = 4) c) = (1 − 0.8) 0 0.8 + (1 − 0.8)1 0.8 + (1 − 0.8) 2 0.8 + (1 − 0.8) 3 0.8 = 0.8 + 0.2(0.8) + 0.2 2 (0.8) + 0.2 3 0.8 = 0.9984 P( X ≥ 4) = 1 − P( X ≤ 3) = 1 − [ P( X = 1) + P( X = 2) + P( X = 3)] a) = 1 − [(1 − 0.8) 0 0.8 + (1 − 0.8)1 0.8 + (1 − 0.8) 2 0.8] = 1 − [0.8 + 0.2(0.8) + 0.2 2 (0.8)] = 1 − 0.992 = 0.008 3-74 Let X denote the number of people who carry the gene. Then X is a negative binomial random variable with r=2 and p = 0.1 a) P( X ≥ 4) = 1 − P( X < 4) = 1 − [ P( X = 2) + P( X = 3)] b) 3-75. ⎡⎛1⎞ ⎤ ⎛ 2⎞ = 1 − ⎢⎜ ⎟(1 − 0.1) 0 0.12 + ⎜ ⎟(1 − 0.1)1 0.12 ⎥ = 1 − (0.01 + 0.018) = 0.972 ⎜⎟ ⎜1⎟ ⎝⎠ ⎣⎝1⎠ ⎦ E ( X ) = r / p = 2 / 0.1 = 20 Let X denote the number of calls needed to obtain a connection. Then, X is a geometric random variable with p = 0.02 a) b) P ( X = 10) = (1 − 0.02) 9 0.02 = 0.98 9 0.02 = 0.0167 P( X > 5) = 1 − P( X ≤ 4) = 1 − [ P( X = 1) + P( X = 2) + P( X = 3) + P( X = 4)] = 1 − [0.02 + 0.98(0.02) + 0.98 2 (0.02) + 0.98 3 (0.02)] = 1 − 0.0776 = 0.9224 c) E(X) = 1/0.02 = 50 3-76 Let X denote the number of mornings needed to obtain a green light. Then X is a geometric random variable with p = 0.20. a) P(X = 4) = (1-0.2)30.2= 0.1024 b) By independence, (0.8)10 = 0.1074. (Also, P(X > 10) = 0.1074) 3-77 p = 0.005 , r = 8 a.) b). P( X = 8) = 0.0058 = 3.91x10 −19 1 µ = E( X ) = = 200 days 0.005 c) Mean number of days until all 8 computers fail. Now we use p=3.91x10-19 µ = E (Y ) = 3-78 1 = 2.56 x1018 days − 91 3.91x10 or 7.01 x1015 years Let Y denote the number of samples needed to exceed 1 in Exercise 3-66. Then Y has a geometric distribution with p = 0.0169. a) P(Y = 10) = (1 − 0.0169)9(0.0169) = 0.0145 b) Y is a geometric random variable with p = 0.1897 from Exercise 3-66. P(Y = 10) = (1 − 0.1897)9(0.1897) = 0.0286 c) E(Y) = 1/0.1897 = 5.27 3-13 3-79. Let X denote the number of trials to obtain the first success. a) E(X) = 1/0.2 = 5 b) Because of the lack of memory property, the expected value is still 5. 3-80 Negative binomial random variable: f(x; p, r) = ⎜ ⎜ ⎛ x − 1⎞ ⎟(1 − p) x − r p r . ⎟ ⎝ r − 1⎠ When r = 1, this reduces to f(x; p, r) = (1−p)x-1p, which is the pdf of a geometric random variable. Also, E(X) = r/p and V(X) = [r(1−p)]/p2 reduce to E(X) = 1/p and V(X) = (1−p)/p2, respectively. 3-81. a) E(X) = 4/0.2 = 20 ⎛19 ⎞ ⎜ ⎟(0.80)16 0.24 = 0.0436 ⎜3⎟ ⎝⎠ ⎛18 ⎞ 15 4 c) P(X=19) = ⎜ ⎟(0.80) 0.2 = 0.0459 ⎜3⎟ ⎝⎠ ⎛ 20 ⎞ 17 4 d) P(X=21) = ⎜ ⎜ 3 ⎟(0.80) 0.2 = 0.0411 ⎟ ⎝⎠ b) P(X=20) = e) The most likely value for X should be near µX. By trying several cases, the most likely value is x = 19. 3-82 Let X denote the number of attempts needed to obtain a calibration that conforms to specifications. Then, X is geometric with p = 0.6. P(X ≤ 3) = P(X=1) + P(X=2) + P(X=3) = 0.6 + 0.4(0.6) + 0.42(0.6) = 0.936. 3-83. Let X denote the number of fills needed to detect three underweight packages. Then X is a negative binomial random variable with p = 0.001 and r = 3. a) E(X) = 3/0.001 = 3000 b) V(X) = [3(0.999)/0.0012] = 2997000. Therefore, σX = 1731.18 3-84 Let X denote the number of transactions until all computers have failed. Then, X is negative binomial random variable with p = 10-8 and r = 3. a) E(X) = 3 x 108 b) V(X) = [3(1−10-80]/(10-16) = 3.0 x 1016 3-14 3-85 Let X denote a geometric random variable with parameter p. Let q = 1-p. ∞ E ( X ) = ∑ x(1 − p ) x −1 p x =1 ∞ = p ∑ xq x −1 = p x =1 = d ⎡ ∞ x⎤ d⎡ 1 ⎤ ⎢∑ q ⎥ = p dq ⎢1 − q ⎥ dq ⎣ x =0 ⎦ ⎦ ⎣ p p 1 = 2= 2 p p (1 − q ) ∞ ( ∞ V ( X ) = ∑ ( x − 1 ) 2 (1 − p ) x −1 p = ∑ px 2 − 2 x + p x =1 x =1 ∞ ∞ = p ∑ x 2 q x −1 − 2∑ xq x −1 + x =1 x =1 ∞ = p ∑ x 2 q x −1 − x =1 2 p2 + ∞ 1 p ∑q 1 p )(1 − p) x −1 x =1 1 p2 ∞ = p ∑ x 2 q x −1 − x =1 1 p2 [ [q(1 + 2q + 3q d = p dq q + 2q 2 + 3q 3 + ... − d = p dq d = p dq = [ ]− q (1− q ) 2 1 p2 2 1 p2 + ...) − 1 p2 = 2 pq(1 − q) −3 + p(1 − q) − 2 − [2(1 − p) + p − 1] = (1 − p) = p 2 p 2 q p2 Section 3-8 3-86 X has a hypergeometric distribution N=100, n=4, K=20 a.) P ( X = 1) = ( )( ) = 20(82160) = 0.4191 ( ) 3921225 20 80 1 3 100 4 b.) P( X = 6) = 0 , the sample size is only 4 c.) P( X = 4) = ( )( ) = 4845(1) = 0.001236 ( ) 3921225 20 80 4 0 100 4 K ⎛ 20 ⎞ = 4⎜ ⎟ = 0.8 N ⎝ 100 ⎠ ⎛ 96 ⎞ ⎛ N −n⎞ V ( X ) = np(1 − p)⎜ ⎟ = 4(0.2)(0.8)⎜ ⎟ = 0.6206 ⎝ 99 ⎠ ⎝ N −1 ⎠ d.) E ( X ) = np = n 3-15 1 p2 x −1 3-87. ( )( ) = (4 ×16 ×15 ×14) / 6 = 0.4623 ( ) (20 ×19 ×18 ×17) / 24 ( )( ) = 1 b) P ( X = 4) = = 0.00021 ( ) (20 ×19 ×18 ×17) / 24 a) c) P ( X = 1) = 4 1 16 3 20 4 4 16 40 20 4 P( X ≤ 2) = P( X = 0) + P( X = 1) + P( X = 2) ( )( ) + ( )( ) + ( )( ) = () () () 4 0 = 16 4 20 4 4 1 16 3 20 4 4 2 16 2 20 4 ⎛ 16×15×14×13 4×16×15×14 6×16×15 ⎞ + + ⎟ ⎜ 24 6 2 ⎠ ⎝ ⎛ 20×19×18×17 ⎞ ⎟ ⎜ 24 ⎠ ⎝ = 0.9866 d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 N=10, n=3 and K=4 0.5 0.4 0.3 P(x) 3-88 0.2 0.1 0.0 0 1 2 3 x 3-16 3-89. ⎫ ⎪ ⎪ 0 ≤ x <1⎪ ⎪ ⎪ ⎪ 1 ≤ x < 2⎬ ⎪ 2 ≤ x < 3⎪ ⎪ ⎪ ⎪ 3≤ x ⎪ ⎭ ⎧ ⎪0, ⎪ ⎪1 / 6, ⎪ ⎪ ⎪ F ( x) = ⎨2 / 3, ⎪ ⎪29 / 30, ⎪ ⎪ ⎪1, ⎪ ⎩ x<0 ⎛ 4 ⎞⎛ 6 ⎞ ⎜ ⎟⎜ ⎟ ⎜ 1 ⎟⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ = 0 .5 , ⎛ 10 ⎞ ⎜ ⎜3⎟ ⎟ ⎝ ⎠ ⎛ 4 ⎞⎛ 6 ⎞ ⎜ ⎟⎜ ⎟ ⎜ 0 ⎟⎜ 3 ⎟ f ( 0 ) = ⎝ ⎠ ⎝ ⎠ = 0 . 1667 , f (1 ) = ⎛ 10 ⎞ ⎜ ⎜3⎟ ⎟ ⎝ ⎠ 3-90 ⎛ 4 ⎞⎛ 6 ⎞ ⎛ 4 ⎞⎛ 6 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ 2 ⎟⎜ 1 ⎟ ⎜ 3 ⎟⎜ 0 ⎟ f ( 2 ) = ⎝ ⎠ ⎝ ⎠ = 0 . 3 , f ( 3 ) = ⎝ ⎠ ⎝ ⎠ = 0 . 0333 ⎛ 10 ⎞ ⎛ 10 ⎞ ⎜ ⎜ ⎜3⎟ ⎟ ⎜3⎟ ⎟ ⎝ ⎠ ⎝ ⎠ Let X denote the number of unacceptable washers in the sample of 10. ( )( ) = () 5 0 70 10 75 10 70! 10!60! 75! 10!65! P( X = 0) = b.) P( X ≥ 1) = 1 − P( X = 0) = 0.5214 c.) ( )( ) = P( X = 1) = () d .) E ( X ) = 10(5 / 75) = 2 / 3 5 1 70 9 75 10 5!70! 9!61! 75! 10!65! = 65 × 64 × 63 × 62 × 61 = 0.4786 75 × 74 × 73 × 72 × 71 a.) 3-91. where = 5 × 65 × 64 × 63 × 62 × 10 = 0.3923 75 × 74 × 73 × 72 × 71 Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high blood pressure. N=800, K=240 n=10 a) n=10 ( )( ) = ( )( ) = 0.1201 P( X = 1) = () 240 560 1 9 800 10 240! 560! 1!239! 9!551! 800! 10!790! b) n=10 P( X > 1) = 1 − P( X ≤ 1) = 1 − [ P( X = 0) + P( X = 1)] P( X = 0) = ( )( ) = ( () 240 560 0 10 800 10 )( 240! 560! 0!240! 10!550! 800! 10!790! ) = 0.0276 P( X > 1) = 1 − P( X ≤ 1) = 1 − [0.0276 + 0.1201] = 0.8523 3-17 3-92 . Let X denote the number of cards in the sample that are defective. a) P( X ≥ 1) = 1 − P( X = 0) P( X = 0) = ( )( ) = () 20 0 120 20 140 20 120! 20!100! 140! 20!120! = 0.0356 P( X ≥ 1) = 1 − 0.0356 = 0.9644 b) P ( X ≥ 1) = 1 − P( X = 0) ( )( ) = P ( X = 0) = () 5 0 135 20 140 20 135! 20!115! 140! 20!120! = 135!120! = 0.4571 115!140! P ( X ≥ 1) = 1 − 0.4571 = 0.5429 3-93. Let X denote the number of blades in the sample that are dull. a) P( X ≥ 1) = 1 − P( X = 0) P( X = 0) = ( )( ) = () 10 0 38 5 48 5 38! 5!33! 48! 5!43! = 38!43! = 0.2931 48!33! P( X ≥ 1) = 1 − P( X = 0) = 0.7069 b) Let Y denote the number of days needed to replace the assembly. P(Y = 3) = 0.29312 (0.7069) = 0.0607 ( )( ) = () ( )( ) = On the second day, P ( X = 0) = () c) On the first day, P ( X = 0) = 2 0 46 5 48 5 6 0 42 5 48 5 46! 5!41! 48! 5!43! = 42! 5!37! 48! 5!43! 46!43! = 0.8005 48!41! = 42!43! = 0.4968 48!37! On the third day, P(X = 0) = 0.2931 from part a. Therefore, P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811. 3-94 Let X denote the count of the numbers in the state's sample that match those in the player's sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6. ( )( ) = ⎛ 40! ⎞ = 2.61 × 10 a) P ( X = 6) = ( ) ⎜ 6!34! ⎟ ⎠ ⎝ ( )( ) = 6 × 34 = 5.31× 10 b) P ( X = 5) = () () ( )( ) = 0.00219 c) P ( X = 4) = () 6 6 6 5 6 4 34 0 40 6 34 1 40 6 34 2 40 6 −1 −7 −5 40 6 d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p = 1 3,838,380 and E(Y) = 1/p = 3,838,380 weeks. This is more than 738 centuries! 3-18 3-95. a) For Exercise 3-86, the finite population correction is 96/99. For Exercise 3-87, the finite population correction is 16/19. Because the finite population correction for Exercise 3-86 is closer to one, the binomial approximation to the distribution of X should be better in Exercise 3-86. b) Assuming X has a binomial distribution with n = 4 and p = 0.2, ( )0.2 0.8 = 0.4096 P ( X = 4) = ( )0.2 0.8 = 0.0016 P ( X = 1) = 4 1 4 4 1 4 3 0 The results from the binomial approximation are close to the probabilities obtained in Exercise 3-86. c) Assume X has a binomial distribution with n = 4 and p = 0.2. Consequently, P(X = 1) and P(X = 4) are the same as computed in part b. of this exercise. This binomial approximation is not as close to the true answer as the results obtained in part b. of this exercise. 3-96 a.) From Exercise 3-92, X is approximately binomial with n = 20 and p = 20/140 = 1/7. P ( X ≥ 1) = 1 − P ( X = 0) = b) ( )( ) ( ) 20 0 1 0 6 20 7 7 = 1 − 0.0458 = 0.9542 finite population correction is 120/139=0.8633 From Exercise 3-92, X is approximately binomial with n = 20 and p = 5/140 =1/28 P ( X ≥ 1) = 1 − P ( X = 0) = ( )( ) ( ) 20 0 1 0 27 20 28 28 finite population correction is 120/139=0.8633 Section 3-9 3-97. e −4 4 0 = e −4 = 0.0183 0! b) P ( X ≤ 2) = P ( X = 0) + P ( X = 1) + P ( X = 2) a) P( X = 0) = e −4 41 e −4 42 + 1! 2! = 0.2381 = e −4 + e −4 4 4 = 0.1954 4! e −4 4 8 = 0.0298 d) P( X = 8) = 8! c) P ( X = 4 ) = 3-98 a) P( X = 0) = e −0.4 = 0.6703 e −0.4 (0.4) e −0.4 (0.4) 2 + = 0.9921 1! 2! e −0.4 (0.4) 4 = 0.000715 c) P ( X = 4 ) = 4! e −0.4 (0.4) 8 = 109 × 10 −8 . d) P( X = 8) = 8! b) P( X ≤ 2) = e −0.4 + 3-19 = 1 − 0.4832 = 0.5168 3-99. P( X = 0) = e − λ = 0.05 . Therefore, λ = −ln(0.05) = 2.996. Consequently, E(X) = V(X) = 2.996. 3-100 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with λ = 10. e −10 105 P( X = 5) = = 0.0378 . 5! e −10 10 e −10 10 2 e −10 103 + + = 0.0103 b) P( X ≤ 3) = e −10 + 1! 2! 3! c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with e −20 2015 = 0.0516 λ = 20. P( Y = 15) = 15! d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with e −5 55 = 0.1755 λ = 5. P(W = 5) = 5! 3-101. a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable with λ = 0.1. P( X = 2) = e −0.1 (0.1) 2 = 0.0045 2! b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable with λ = 1. P(Y = 1) = e−111 = e−1 = 0.3679 1! c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable P(W = 0) = e −2 = 0.1353 P(Y ≥ 2) = 1 − P(Y ≤ 1) = 1 − P(Y = 0) − P(Y = 1) with λ = 2. d) = 1 − e −1 − e −1 = 0.2642 3-102 a) E ( X ) = λ = 0.2 errors per test area b.) P ( X ≤ 2) = e −0.2 + e −0.2 0.2 e −0.2 (0.2) 2 + = 0.9989 1! 2! 99.89% of test areas 3-103. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with λ = 10. P( X = 0) = e −10 = 4.54 × 10−5 b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with λ = 1. P(Y ≥ 1) = 1 − P (Y = 0) = 1 − e −1 = 0.6321 c) The assumptions of a Poisson process require that the probability of a count is constant for all intervals. If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson process are not valid. Separate Poisson random variables might be appropriate for the heavy and light load sections of the highway. 3-20 3-104 3-105. = λ = 0.01 failures per 100 samples. Let Y= the number of failures per day E (Y ) = E (5 X ) = 5E ( X ) = 5λ = 0.05 failures per day. −0.05 = 0.9512 b.)Let W= the number of failures in 500 participants, now λ=0.05 and P (W = 0) = e a.) E ( X ) a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random −0.5 variable with λ = 0.5. P ( X = 0) = e = b) Let Y denote the number of cars with no flaws, 0.6065 ⎛10 ⎞ P (Y = 10) = ⎜ ⎟(0.6065 )10 (0.3935 ) 0 = 0.0067 ⎜10 ⎟ ⎝⎠ c) Let W denote the number of cars with surface flaws. Because the number of flaws has a Poisson distribution, the occurrences of surface flaws in cars are independent events with constant probability. From part a., the probability a car contains surface flaws is 1−0.6065 = 0.3935. Consequently, W is binomial with n = 10 and p = 0.3935. ⎛10 ⎞ P (W = 0) = ⎜ ⎟(0.3935) 0 (0.6065)10 = 0.0067 ⎜0⎟ ⎝⎠ ⎛10 ⎞ P (W = 1) = ⎜ ⎟(0.3935)1 (0.6065) 9 = 0.0437 ⎜1⎟ ⎝⎠ P (W ≤ 1) = 0.0067 + 0.00437 = 0.00504 3-106 a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with λ = 0.16. P( X = 0) = e −0.16 = 0.8521 b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with λ = 0.48. P(Y ≥ 1) = 1 − P (Y = 0) = 1 − e−48 = 0.3812 Supplemental Exercises 3-107. Let X denote the number of totes in the sample that do not conform to purity requirements. Then, X has a hypergeometric distribution with N = 15, n = 3, and K = 2. ⎛ 2 ⎞⎛13 ⎞ ⎜ ⎟⎜ ⎟ ⎜ 0 ⎟⎜ 3 ⎟ 13!12! P( X ≥ 1) = 1 − P( X = 0) = 1 − ⎝ ⎠⎝ ⎠ = 1 − = 0.3714 ⎛15 ⎞ 10!15! ⎜⎟ ⎜3⎟ ⎝⎠ 3-108 Let X denote the number of calls that are answered in 30 seconds or less. Then, X is a binomial random variable with p = 0.75. a) P(X = 9) = ⎛10 ⎞ ⎜ ⎟(0.75)9 (0.25)1 = 0.1877 ⎜9⎟ ⎝⎠ b) P(X ≥ 16) = P(X=16) +P(X=17) + P(X=18) + P(X=19) + P(X=20) ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 20 ⎞ = ⎜ ⎟(0.75) 16 (0.25) 4 + ⎜ ⎟(0.75) 17 (0.25) 3 + ⎜ ⎟(0.75) 18 (0.25) 2 ⎜ 18 ⎟ ⎜ 17 ⎟ ⎜ 16 ⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎛ 20 ⎞ ⎛ 20 ⎞ + ⎜ ⎟(0.75) 19 (0.25) 1 + ⎜ ⎟(0.75) 20 (0.25) 0 = 0.4148 ⎜ 19 ⎟ ⎜ 20 ⎟ ⎝⎠ ⎝⎠ c) E(X) = 20(0.75) = 15 3-21 3-109. Let Y denote the number of calls needed to obtain an answer in less than 30 seconds. a) P (Y = 4) = (1 − 0.75) b) E(Y) = 1/p = 1/0.75 = 4/3 3 0.75 = 0.25 3 0.75 = 0.0117 3-110 Let W denote the number of calls needed to obtain two answers in less than 30 seconds. Then, W has a negative binomial distribution with p = 0.75. ⎛ 5⎞ a) P(W=6) = ⎜ ⎟ (0.25)4 (0.75)2 = 0.0110 ⎝ 1⎠ b) E(W) = r/p = 2/0.75 = 8/3 3-111. a) Let X denote the number of messages sent in one hour. P( X = 5) = e −5 5 5 = 0.1755 5! b) Let Y denote the number of messages sent in 1.5 hours. Then, Y is a Poisson random variable with λ =7.5. P(Y = 10) = e −7.5 (7.5)10 = 0.0858 10! c) Let W denote the number of messages sent in one-half hour. Then, W is a Poisson random variable with λ = 2.5. P (W < 2) = P(W = 0) + P(W = 1) = 0.2873 3-112 X is a negative binomial with r=4 and p=0.0001 E ( X ) = r / p = 4 / 0.0001 = 40000 requests 3-113. X ∼ Poisson(λ = 0.01), X ∼ Poisson(λ = 1) e −1 (1)1 e −1 (1) 2 e −1 (1) 3 + + = 0.9810 P(Y ≤ 3) = e + 1! 2! 3! −1 3-114 Let X denote the number of individuals that recover in one week. Assume the individuals are independent. Then, X is a binomial random variable with n = 20 and p = 0.1. P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.8670 = 0.1330. 3-115 a.) P(X=1) = 0 , P(X=2) = 0.0025, P(X=3) = 0.01, P(X=4) = 0.03, P(X=5) = 0.065 P(X=6) = 0.13, P(X=7) = 0.18, P(X=8) = 0.2225, P(X=9) = 0.2, P(X=10) = 0.16 b.) P(X=1) = 0.0025, P(X=1.5) = 0.01, P(X=2) = 0.03, P(X=2.5) = 0.065, P(X=3) = 0.13 P(X=3.5) = 0.18, P(X=4) = 0.2225, P(X=4.5) = 0.2, P(X=5) = 0.16 3-116 Let X denote the number of assemblies needed to obtain 5 defectives. Then, X is a negative binomial random variable with p = 0.01 and r=5. a) E(X) = r/p = 500. b) V(X) =(5* 0.99/0.012 = 49500 and σX = 222.49 3-117. If n assemblies are checked, then let X denote the number of defective assemblies. If P(X ≥ 1) ≥ 0.95, then P(X=0) ≤ 0.05. Now, ⎛ n⎞ ⎜ ⎟(0.01) 0 (0.99) n = 99 n ⎜ 0⎟ ⎝⎠ n(ln(0.99)) ≤ ln(0.05) P(X=0) = n≥ and 0.99n ≤ 0.05. Therefore, ln(0.05) = 298.07 ln(0.95) This would require n = 299. 3-22 3-118 Require f(1) + f(2) + f(3) + f(4) = 1. Therefore, c(1+2+3+4) = 1. Therefore, c = 0.1. 3-119. Let X denote the number of products that fail during the warranty period. Assume the units are independent. Then, X is a binomial random variable with n = 500 and p = 0.02. a) P(X = 0) = ⎛ 500 ⎞ 0 500 -5 ⎜ ⎟ ⎜ 0 ⎟(0.02) (0.98) = 4.1 x 10 ⎠ ⎝ b) E(X) = 500(0.02) = 10 c) P(X >2) = 1 − P(X ≤ 2) = 0.9995 3-120 3-121. f X (0) = (0.1)(0.7) + (0.3)(0.3) = 0.16 f X (1) = (0.1)(0.7) + (0.4)(0.3) = 0.19 f X (2) = (0.2)(0.7) + (0.2)(0.3) = 0.20 f X (3) = (0.4)(0.7) + (0.1)(0.3) = 0.31 f X (4) = (0.2)(0.7) + (0)(0.3) = 0.14 a) P(X ≤ 3) = 0.2 + 0.4 = 0.6 b) P(X > 2.5) = 0.4 + 0.3 + 0.1 = 0.8 c) P(2.7 < X < 5.1) = 0.4 + 0.3 = 0.7 d) E(X) = 2(0.2) + 3(0.4) + 5(0.3) + 8(0.1) = 3.9 e) V(X) = 22(0.2) + 32(0.4) + 52(0.3) + 82(0.1) − (3.9)2 = 3.09 3-122 x f(x) 3-123. 2 0.2 5.7 0.3 6.5 0.3 8.5 0.2 Let X denote the number of bolts in the sample from supplier 1 and let Y denote the number of bolts in the sample from supplier 2. Then, x is a hypergeometric random variable with N = 100, n = 4, and K = 30. Also, Y is a hypergeometric random variable with N = 100, n = 4, and K = 70. a) P(X=4 or Y=4) = P(X = 4) + P(Y = 4) ⎛ 30 ⎞⎛ 70 ⎞ ⎛ 30 ⎞⎛ 70 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ 4 ⎟⎜ 0 ⎟ ⎜ 0 ⎟⎜ 4 ⎟ = ⎝ ⎠⎝ ⎠ + ⎝ ⎠⎝ ⎠ ⎛100 ⎞ ⎛100 ⎞ ⎜ ⎜ ⎟ ⎜4⎟ ⎟ ⎜4⎟ ⎠ ⎝ ⎠ ⎝ = 0.2408 b) P[(X=3 and Y=1) or (Y=3 and X = 1)]= 3-124 ⎛ 30 ⎛ 70 ⎞ ⎛ 30 ⎞⎛ 70 ⎞ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ ⎜ 3 ⎟⎜ 1 ⎟ ⎜ 1 ⎟⎜ 3 ⎟ = ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ = 0.4913 ⎛100 ⎞ ⎜ ⎟ ⎜4⎟ ⎠ ⎝ Let X denote the number of errors in a sector. Then, X is a Poisson random variable with λ = 0.32768. a) P(X>1) = 1 − P(X≤1) = 1 − e-0.32768 − e-0.32768(0.32768) = 0.0433 b) Let Y denote the number of sectors until an error is found. Then, Y is a geometric random variable and P = P(X ≥ 1) = 1 − P(X=0) = 1 − e-0.32768 = 0.2794 E(Y) = 1/p = 3.58 3-23 3-125. Let X denote the number of orders placed in a week in a city of 800,000 people. Then X is a Poisson random variable with λ = 0.25(8) = 2. a) P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [e-2 + e-2(2) + (e-222)/2!] = 1 − 0.6767 = 0.3233. b) Let Y denote the number of orders in 2 weeks. Then, Y is a Poisson random variable with λ = 4, and P(Y>2) =1- P(Y ≤ 2) = e-4 + (e-441)/1!+ (e-442)/2! =1 - [0.01832+0.07326+0.1465] = 0.7619. 3-126 a.) hypergeometric random variable with N = 500, n = 5, and K = 125 ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎟ ⎟⎜ ⎜ 0 ⎟⎜ 5 ⎟ 6.0164 E10 ⎠= ⎠⎝ = 0.2357 f X (0) = ⎝ 2.5524 E11 ⎛ 500 ⎞ ⎜ ⎟ ⎜5⎟ ⎠ ⎝ ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎟ ⎟⎜ ⎜ 1 ⎟⎜ 4 ⎟ 125(8.10855E8) ⎠= ⎠⎝ = 0.3971 f X (1) = ⎝ 2.5525E11 ⎛ 500 ⎞ ⎜ ⎟ ⎜5⎟ ⎠ ⎝ ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎟ ⎟⎜ ⎜ 2 ⎟⎜ 3 ⎟ 7750(8718875) ⎠= ⎠⎝ = 0.2647 f X (2) = ⎝ 2.5524 E11 ⎛ 500 ⎞ ⎜ ⎟ ⎜5⎟ ⎠ ⎝ ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎟ ⎟⎜ ⎜ 3 ⎟⎜ 2 ⎟ 317750(70125) ⎠= ⎠⎝ = 0.0873 f X (3) = ⎝ 2.5524 E11 ⎛ 500 ⎞ ⎜ ⎟ ⎜5⎟ ⎠ ⎝ ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎟ ⎟⎜ ⎜ 4 ⎟⎜ 1 ⎟ 9691375(375) ⎠= ⎠⎝ = 0.01424 f X (4) = ⎝ 2.5524 E11 ⎛ 500 ⎞ ⎜ ⎟ ⎜5⎟ ⎠ ⎝ ⎛125 ⎞⎛ 375 ⎞ ⎜ ⎟ ⎟⎜ ⎜ 5 ⎟⎜ 0 ⎟ 2.3453E 8 ⎠= ⎠⎝ = 0.00092 f X (5) = ⎝ 500 ⎞ 2.5524 E11 ⎛ ⎜ ⎟ ⎜5⎟ ⎠ ⎝ b.) x f(x) 0 1 2 3 4 5 6 7 8 9 10 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000 3-24 3-127. Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial random variable with n = 30. We are to determine p. If P(X ≥ 1) = 0.9, then P(X = 0) = 0.1. Then ⎛ 30 ⎞ 0 ⎜ ⎟( p ) (1 − p )30 = 0.1 , giving 30ln(1−p)=ln(0.1), ⎜0⎟ ⎝⎠ which results in p = 0.0739. 3-128 Let t denote an interval of time in hours and let X denote the number of messages that arrive in time t. Then, X is a Poisson random variable with λ = 10t. Then, P(X=0) = 0.9 and e-10t = 0.9, resulting in t = 0.0105 hours = 0.63 seconds 3-129. a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with λ = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679. b) Let Y denote the number of flaws in one panel, then P(Y ≥ 1) = 1 − P(Y=0) = 1 − e-0.02 = 0.0198. Let W denote the number of panels that need to be inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and E(W) = 1/0.0198 = 50.51 panels. −0.02 c) P (Y ≥ 1) = 1 − P (Y = 0) = 1 − e = 0.0198 Let V denote the number of panels with 1 or more flaws. Then V is a binomial random variable with n=50 and p=0.0198 ⎛ 50 ⎞ ⎛ 50 ⎞ P (V ≤ 2) = ⎜ ⎟0.0198 0 (.9802) 50 + ⎜ ⎟0.01981 (0.9802) 49 ⎜1⎟ ⎜0⎟ ⎝⎠ ⎝⎠ 50 ⎞ ⎛ + ⎜ ⎟0.0198 2 (0.9802) 48 = 0.9234 ⎜2⎟ ⎝⎠ Mind Expanding Exercises 3-130. Let X follow a hypergeometric distribution with parameters K, n, and N. To solve this problem, we can find the general expectation: n E(Xk) = ∑ i k P( X = i ) i =0 ⎛ K ⎞⎛ N − K ⎞ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ k ⎝ i ⎠⎝ n − i ⎠ = ∑i ⎛N⎞ i =0 ⎜⎟ ⎜n⎟ ⎝⎠ n Using the relationships ⎛K ⎞ ⎛ K − 1⎞ i⎜ ⎟ = K ⎜ ⎜i⎟ ⎜ i −1 ⎟ ⎟ ⎝⎠ ⎝ ⎠ and ⎛N⎞ ⎛ N − 1⎞ n⎜ ⎟ = N ⎜ ⎜n⎟ ⎜ n −1⎟ ⎟ ⎝⎠ ⎝ ⎠ we can substitute into E(XK): 3-25 n E(Xk) = ∑ i k P( X = i ) i =0 ⎛ K ⎞⎛ N − K ⎞ ⎜ ⎟⎜ ⎜ i ⎜ n − i ⎟ ⎟ ⎠ = ∑ i k ⎝ ⎠⎝ ⎛N⎞ i =0 ⎜⎟ ⎜n⎟ ⎝⎠ ⎛ K − 1⎞⎛ N − K ⎞ K⎜ ⎜ i − 1 ⎟⎜ n − i ⎟ ⎟⎜ ⎟ n ⎠⎝ ⎠ = n ∑ i k −1 ⎝ ⎛ N − 1⎞ i =0 N⎜ ⎜ n −1⎟ ⎟ ⎝ ⎠ ⎛ K − 1⎞⎛ N − K ⎞ ⎜ ⎜ j ⎟⎜ n − 1 − j ⎟ ⎟⎜ ⎟ nK n −1 ⎠⎝ ⎠ = ( j + 1) k −1 ⎝ ∑ N j =0 ⎛ N − 1⎞ ⎜ ⎜ n −1⎟ ⎟ ⎝ ⎠ nK = E[( Z + 1) k −1 ] N n Now, Z is also a hypergeometric random variable with parameters n – 1, N – 1, and K – 1. To find the mean of X, E(X), set k = 1: E(X) = = nK nK E[( Z + 1)1−1 ] = N N If we let p = K/N, then E(X) = np. In order to find the variance of X using the formula V(X) = E(X2) – [E(X)}2, the E(X2) must be found. Substituting k = 2 into E(Xk) we get nK nK E[( Z + 1) 2−1 ] = E ( Z + 1) N N nK [E ( Z ) + E (1)] = nK ⎡ (n − 1)( K − 1) + 1⎤ = ⎥ N⎢ N −1 N ⎦ ⎣ E( X 2 ) = 2 nK ⎡ (n − 1)( K − 1) ⎤ ⎛ nK ⎞ nK ⎡ (n − 1)( K − 1) nK ⎤ Therefore, V(X) = +1− + 1⎥ − ⎜ ⎟= ⎢ ⎢ N⎣ N −1 N⎣ N −1 N⎥ ⎦ ⎦ ⎝N⎠ If we let p = K/N, the variance reduces to V(X) = 3-131. ⎛ N −n⎞ ⎜ ⎟np(1 − p ) ⎝ N −1 ⎠ ∞ Show that ∑ (1 − p)i −1 p = 1 using an infinite sum. i =1 ∞ To begin, ∑ (1 − p) i =1 i −1 ∞ p = p ∑ (1 − p)i −1 , by definition of an infinite sum this can be rewritten i =1 as ∞ p ∑ (1 − p)i −1 = i =1 p p = =1 1 − (1 − p ) p 3-26 3-132 E ( X ) = [(a + (a + 1) + ... + b](b − a + 1) a −1 ⎡b ⎤ i − ∑ i⎥ ⎢∑ i =1 ⎦ = ⎣ i =1 (b − a + 1) ⎡ (b 2 − a 2 + b + a ) ⎤ ⎢ ⎥ 2 ⎣ ⎦ = = ⎡ b(b + 1) (a − 1)a ⎤ ⎢2− 2⎥ ⎦ =⎣ (b − a + 1) (b − a + 1) ⎡ (b + a)(b − a + 1) ⎤ ⎢ ⎥ 2 ⎦ =⎣ (b − a + 1) (b + a) 2 b ⎡b 2 (b + a − 1)(b + a) 2 ⎤ i − (b + a)∑ i + ⎥ ∑ [i − ] ⎢∑ 4 i =a ⎣ i=a ⎦ i =a = V (X ) = b + a −1 b + a −1 2 b(b + 1)(2b + 1) (a − 1)a(2a − 1) ⎡ b(b + 1) − (a − 1)a ⎤ (b − a + 1)(b + a) − − (b + a) ⎢ + ⎥ 2 4 6 6 ⎣ ⎦ = b − a +1 2 (b − a + 1) − 1 = 12 b 3-133 Let X denote the number of nonconforming products in the sample. Then, X is approximately binomial with p = 0.01 and n is to be determined. If P ( X ≥ 1) ≥ 0.90 , then P( X = 0) ≤ 0.10 . Now, P(X = 0) = n≤ 3-134 b+ a 2 2 ( )p (1 − p) n 0 0 n = (1 − p ) n . Consequently, (1 − p)n ≤ 0.10 , and ln 0.10 = 229.11 . Therefore, n = 230 is required ln(1 − p) If the lot size is small, 10% of the lot might be insufficient to detect nonconforming product. For example, if the lot size is 10, then a sample of size one has a probability of only 0.2 of detecting a nonconforming product in a lot that is 20% nonconforming. If the lot size is large, 10% of the lot might be a larger sample size than is practical or necessary. For example, if the lot size is 5000, then a sample of 500 is required. Furthermore, the binomial approximation to the hypergeometric distribution can be used to show the following. If 5% of the lot of size 5000 is nonconforming, then the probability of zero nonconforming product in the sample is approximately 7 × 10 −12 . Using a sample of 100, the same probability is still only 0.0059. The sample of size 500 might be much larger than is needed. 3-27 3-135 Let X denote the number of panels with flaws. Then, X is a binomial random variable with n =100 and p is the probability of one or more flaws in a panel. That is, p = 1 − e −0.1 = 0.095. P ( X < 5) = P ( X ≤ 4) = P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) + P ( X = 4) ( )p (1 − p) + ( )p (1 − p) + ( ) p (1 − p ) + ( ) p (1 − p ) = 100 0 0 100 100 3 3 97 100 1 100 4 1 4 99 + ( )p (1 − p) 100 2 2 98 96 = 0.034 3-136 Let X denote the number of rolls produced. Revenue at each demand 1000 2000 3000 0.3x 0.3x 0.3x 0 ≤ x ≤ 1000 mean profit = 0.05x(0.3) + 0.3x(0.7) - 0.1x 0.05x 0.3(1000) + 0.3x 0.3x 1000 ≤ x ≤ 2000 0.05(x-1000) mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x-1000)](0.2) + 0.3x(0.5) - 0.1x 0.05x 0.3(1000) + 0.3(2000) + 0.3x 2000 ≤ x ≤ 3000 0.05(x-1000) 0.05(x-2000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000) + 0.05(x-2000)](0.3) + 0.3x(0.2) - 0.1x 0.05x 0.3(1000) + 0.3(2000) + 0.3(3000)+ 3000 ≤ x 0.05(x-1000) 0.05(x-2000) 0.05(x-3000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000)+0.05(x-2000)]0.3 + [0.3(3000)+0.05(x3000)]0.2 - 0.1x 0 0.05x 0 ≤ x ≤ 1000 1000 ≤ x ≤ 2000 2000 ≤ x ≤ 3000 3000 ≤ x 3-137 Profit 0.125 x 0.075 x + 50 200 -0.05 x + 350 Max. profit $ 125 at x = 1000 $ 200 at x = 2000 $200 at x = 3000 Let X denote the number of acceptable components. Then, X has a binomial distribution with p = 0.98 and n is to be determined such that P( X ≥ 100 ) ≥ 0.95 . n P( X ≥ 100 ) 102 0.666 103 0.848 104 0.942 105 0.981 Therefore, 105 components are needed. 3-28 CHAPTER 4 Section 4-2 ∞ 4-1. ∞ ∫ a) P (1 < X ) = e − x dx = ( −e − x ) 1 = e −1 = 0.3679 1 2 .5 b) P (1 < X < 2.5) = ∫e −x dx = (−e − x ) 2 .5 1 = e −1 − e − 2.5 = 0.2858 1 3 ∫ c) P( X = 3) = e − x dx = 0 3 4 ∫ 4 d) P( X < 4) = e − x dx = (−e − x ) = 1 − e − 4 = 0.9817 0 0 ∞ ∫ e) P(3 ≤ X ) = e − x dx = (−e − x ) ∞ 3 = e− 3 = 0.0498 3 ∞ 4-2. ∞ ∫ a) P( x < X ) = e − x dx = (−e − x ) x = e − x = 0.10 . x Then, x = −ln(0.10) = 2.3 x ∫ x b) P ( X ≤ x) = e − x dx = (−e − x ) = 1 − e − x = 0.10 . 0 0 Then, x = −ln(0.9) = 0.1054 4 4-3 x x2 a) P ( X < 4) = ∫ dx = 8 16 3 4 4 2 − 32 = 0.4375 , because f X ( x) = 0 for x < 3. 16 = 3 5 x x2 b) , P ( X > 3.5) = ∫ dx = 8 16 3 .5 = 3 .5 25 5 c) P ( 4 < X < 5) = 5 x x ∫ 8 dx = 16 4 4 .5 4 2 4. 5 x x d) P ( X < 4.5) = ∫ dx = 8 16 3 3 5 2 − 3 .5 2 = 0.7969 because f X ( x) = 0 for x > 5. 16 = 52 − 42 = 0.5625 16 4 .5 2 − 3 2 = = 0.7031 16 5 3.5 x x x2 x2 52 − 4.52 3.52 − 32 e) P( X > 4.5) + P( X < 3.5) = ∫ dx + ∫ dx = + = + = 0.5 . 8 8 16 4.5 16 3 16 16 4.5 3 5 3.5 4-1 ∞ 4-4 ∫ a) P (1 < X ) = e −( x −4 ) dx = − e −( x −4 ) ∞ = 1 , because f X ( x) = 0 for x < 4. This can also be 4 4 obtained from the fact that f X ( x) is a probability density function for 4 < x. 5 ∫ b) P (2 ≤ X ≤ 5) = e − ( x − 4 ) dx = − e − ( x − 4 ) 5 4 = 1 − e −1 = 0.6321 4 c) P (5 < X ) = 1 − P ( X ≤ 5) . From part b., P ( X ≤ 5) = 0.6321 . Therefore, P (5 < X ) = 0.3679 . 12 ∫ d) P(8 < X < 12) = e − ( x − 4 ) dx = − e − ( x − 4 ) 12 8 = e − 4 − e − 8 = 0.0180 8 x ∫ e) P ( X < x ) = e − ( x − 4 ) dx = − e − ( x − 4 ) x 4 = 1 − e − ( x − 4 ) = 0.90 . 4 Then, x = 4 − ln(0.10) = 6.303 4-5 a) P (0 < X ) = 0.5 , by symmetry. 1 ∫ b) P(0.5 < X ) = 1.5 x 2 dx = 0.5 x 3 1 0.5 = 0.5 − 0.0625 = 0.4375 0.5 0.5 ∫1.5x dx = 0.5 x c) P(−0.5 ≤ X ≤ 0.5) = 2 3 0.5 − 0.5 − 0.5 = 0.125 d) P(X < −2) = 0 e) P(X < 0 or X > −0.5) = 1 1 ∫ f) P( x < X ) = 1.5 x 2 dx = 0.5 x 3 1 x = 0.5 − 0.5 x3 = 0.05 x Then, x = 0.9655 ∞ 4-6. −x −x e 1000 dx = − e 1000 a) P( X > 3000) = ∫ 3000 1000 2000 b) P(1000 < X < 2000) = 1000 c) P ( X < 1000) = ∫ 0 x ∞ = e − 3 = 0.05 3000 −x 1000 − x 2000 e dx = − e 1000 = e −1 − e − 2 = 0.233 ∫ 1000 1000 1000 −x −x e 1000 dx = − e 1000 1000 1000 = 1 − e −1 = 0.6321 0 −x 1000 −x x e dx = − e 1000 = 1 − e − x /1000 = 0.10 . 0 0 1000 d) P ( X < x ) = ∫ Then, e − x/1000 = 0.9 , and x = −1000 ln 0.9 = 105.36. 4-2 50.25 4-7 a) P( X > 50) = ∫ 2.0dx = 2 x 50.25 50 = 0.5 50 50.25 b) P( X > x) = 0.90 = ∫ 2.0dx = 2 x 50.25 x = 100.5 − 2 x x Then, 2x = 99.6 and x = 49.8. 74.8 4-8. a) P( X < 74.8) = ∫ 1.25dx = 1.25x 74.8 74.6 = 0.25 74.6 b) P(X < 74.8 or X > 75.2) = P(X < 74.8) + P(X > 75.2) because the two events are mutually exclusive. The result is 0.25 + 0.25 = 0.50. 75.3 c) P (74.7 < X < 75.3) = ∫ 1.25dx = 1.25x 75.3 74.7 = 1.25(0.6) = 0.750 74.7 4-9 a) P(X < 2.25 or X > 2.75) = P(X < 2.25) + P(X > 2.75) because the two events are mutually exclusive. Then, P(X < 2.25) = 0 and 2.8 P(X > 2.75) = ∫ 2dx = 2(0.05) = 0.10 . 2.75 b) If the probability density function is centered at 2.5 meters, then f X ( x) = 2 for 2.3 < x < 2.8 and all rods will meet specifications. x2 4-10. Because the integral ∫ f ( x)dx is not changed whether or not any of the endpoints x1 and x2 x1 are included in the integral, all the probabilities listed are equal. Section 4-3 4-11. a) P(X<2.8) = P(X ≤ 2.8) because X is a continuous random variable. Then, P(X<2.8) =F(2.8)=0.2(2.8) = 0.56. b) P ( X > 1.5) = 1 − P ( X ≤ 1.5) = 1 − 0.2(1.5) = 0.7 c) P( X < −2) = FX (−2) = 0 d) P( X > 6) = 1 − FX (6) = 0 4-12. a) P( X < 1.8) = P( X ≤ 1.8) = FX (1.8) because X is a continuous random variable. Then, FX (1.8) = 0.25(1.8) + 0.5 = 0.95 b) P ( X > −1.5) = 1 − P ( X ≤ −1.5) = 1 − .125 = 0.875 c) P(X < -2) = 0 d) P(−1 < X < 1) = P(−1 < X ≤ 1) = FX (1) − FX (−1) = .75 − .25 = 0.50 4-3 x 4-13. ∫ Now, f ( x) = e − x for 0 < x and FX ( x) = e − x dx = − e − x x 0 = 1− e −x 0 ⎧ for 0 < x. Then, FX ( x) = ⎨ 0, x ≤ 0 −x ⎩1 − e , x > 0 x 4-14. x x2 Now, f ( x ) = x / 8 for 3 < x < 5 and F X ( x ) = ∫ dx = 8 16 3 x 3 x2 −9 = 16 0, x < 3 ⎧ ⎪2 ⎪x −9 ,3 ≤ x < 5 for 0 < x. Then, F X ( x ) = ⎨ 16 ⎪ 1, x ≥ 5 ⎪ ⎩ x 4-15. ∫ Now, f ( x) = e − ( 4 − x ) for 4 < x and F X ( x ) = e − ( 4 − x ) dx = − e − ( 4 − x ) x 4 = 1 − e −(4− x) 4 for 4 < x. 0, x ≤ 4 ⎧ Then, F X ( x ) = ⎨ −( 4− x ) , x>4 ⎩1 − e 4-16. Now, f ( x) = e − x / 1000 for 0 < x and 1000 x FX ( x) = 1 / 1000 ∫ e − x / 1000 dx = − e − x / 1000 ) x 4 = 1 − e − x / 1000 4 for 0 < x. 0, x ≤ 0 ⎧ Then, F X ( x ) = ⎨ ⎩1 − e − x / 1000 , x>0 P(X>3000) = 1- P(X ≤ 3000) = 1- F(3000) = e-3000/1000 = 0.5 x 4-17. Now, f(x) = 1.25 for 74.6 < x < 75.4 and F ( x) = ∫ 1.25dx = 1.25x − 93.25 74.6 for 74.6 < x < 75.4. Then, 0, x < 74.6 ⎧ ⎪ F ( x) = ⎨1.25 x − 93.25, 74.6 ≤ x < 75.4 ⎪ 1, 75.4 ≤ x ⎩ P ( X > 75) = 1 − P ( X ≤ 75) = 1 − F (75) = 1 − 0.5 = 0.5 because X is a continuous random variable. 4-4 4-18 f ( x) = 2e −2 x , x > 0 4-19. ⎧ ⎪0.2, 0 < x < 4 f ( x) = ⎨ ⎪0.04, 4 ≤ x < 9 ⎩ 4-20. ⎧ ⎪0.25, − 2 < x < 1 f X ( x) = ⎨ ⎪0.5, 1 ≤ x < 1.5 ⎩ 4-21. F ( x) = ∫ 0.5 xdx = x 0 ⎧0, ⎪ ⎪ ⎪ F ( x) = ⎨0.25 x 2 , ⎪ ⎪ ⎪1, ⎩ x 0.5 x 2 20 = 0.25 x 2 for 0 < x < 2. Then, x<0 0≤ x<2 2≤ x Section 4-4 4 4 4-22. x2 E ( X ) = ∫ 0.25 xdx = 0.25 =2 20 0 4 ( x − 2)3 224 V ( X ) = ∫ 0.25( x − 2) dx = 0.25 =+= 3 333 0 0 4 2 4 4 4-23. x3 E ( X ) = ∫ 0.125 x dx = 0.125 = 2.6667 30 0 2 4 4 V ( X ) = ∫ 0.125 x( x − ) dx = 0.125∫ ( x 3 − 16 x 2 + 3 82 3 0 0 = 0.125( x4 − 16 3 4 x3 3 4 + 64 ⋅ 1 x 2 ) = 0.88889 9 2 0 4-5 64 9 x)dx 1 4-24. x4 E ( X ) = ∫ 1.5 x dx = 1.5 4 −1 1 =0 3 −1 1 1 V ( X ) = ∫ 1.5 x ( x − 0) dx = 1.5 ∫ x 4 dx 3 2 −1 −1 5 x 5 = 1 .5 1 −1 = 0 .6 5 x x3 5 3 − 33 = = 4.083 E ( X ) = ∫ x dx = 8 24 3 24 3 5 4-25. 5 ⎛ x 3 8.166 x 2 16.6709 x ⎞ x ⎟dx V ( X ) = ∫ ( x − 4.083) dx = ∫ ⎜ − + ⎟ ⎜8 8 8 8 ⎠ 3 3⎝ 5 2 1 ⎛ x 4 8.166 x 3 16.6709 x 2 =⎜ − + 8⎜ 4 3 2 ⎝ 50.25 4-26. 2 ∫ 2 xdx = x E( X ) = 50.25 49.75 5 ⎞ ⎟ = 0.3264 ⎟ ⎠3 = 50 49.75 50.25 50.25 49.75 49.75 2 2 ∫ 2( x − 50) dx = 2 ∫ ( x − 100 x + 2500)dx V (X ) = = 2( x3 − 100 x2 + 2500 x) 3 2 50.25 49.75 = 0.0208 120 4-27. a.) E ( X ) = ∫x 100 120 600 120 dx = 600 ln x 100 = 109.39 2 x V ( X ) = ∫ ( x − 109.39) 2 100 120 600 dx = 600 ∫ 1 − x2 100 = 600( x − 218.78 ln x − 109.39 2 x −1 ) 2 (109.39 ) x 120 100 b.) Average cost per part = $0.50*109.39 = $54.70 ∞ 4-28. E ( X ) = ∫ x 2 x − 3dx = − 2 x −1 ∞ 1 =2 1 4-6 + (109.39 ) 2 x2 = 33.19 dx ∞ 4-29. ∫ a) E ( X ) = x10e −10( x −5) dx . 5 Using integration by parts with u = x and dv = 10e −10 ( x −5) dx , we obtain E ( X ) = − xe −10 ( x − 5 ) ∞ 5 ∞ + ∫e −10 ( x − 5 ) 5 e −10 ( x −5) dx = 5 − 10 ∞ = 5 .1 5 ∞ ∫ Now, V ( X ) = ( x − 5.1) 2 10e −10( x −5) dx . Using the integration by parts with 5 u = ( x − 5.1) 2 and dv = 10e −10 ( x −5) , we obtain V ( X ) = − ( x − 5.1) 2 e −10( x −5) ∞ 5 ∞ + 2∫ ( x − 5.1)e −10( x −5) dx . 5 From the definition of E(X) the integral above is recognized to equal 0. Therefore, V ( X ) = (5 − 5.1) 2 = 0.01 . ∞ ∫ b) P( X > 5.1) = 10e −10( x −5) dx = − e −10( x −5) ∞ 5.1 = e −10(5.1−5) = 0.3679 5.1 4-30. a) 1210 ∫ x0.1dx = 0.05x E( X ) = 2 1210 1200 = 1205 1200 ( x − 1205) 3 V ( X ) = ∫ ( x − 1205) 0.1dx = 0.1 3 1200 1210 1210 = 8.333 2 Therefore, 1200 σ x = V ( X ) = 2.887 b) Clearly, centering the process at the center of the specifications results in the greatest proportion of cables within specifications. 1205 P(1195 < X < 1205) = P(1200 < X < 1205) = ∫ 0.1dx = 0.1x 1200 Section 4-5 4-31. a) E(X) = (5.5+1.5)/2 = 3.5, V (X ) = (5.5 − 1.5) 2 = 1.333, and σ x = 1.333 = 1.155 . 12 2.5 b) P( X < 2.5) = ∫ 0.25dx = 0.25x 2.5 1.5 = 0.25 1.5 4-7 1205 1200 = 0.5 4-32. a) E(X) = (-1+1)/2 = 0, V (X ) = (1 − (−1)) 2 = 1 / 3, and σ x = 0.577 12 x b) P ( − x < X < x ) = ∫ 1 2 dt = 0.5t −x x = 0.5(2 x) = x −x Therefore, x should equal 0.90. 4-33. a) f(x)= 2.0 for 49.75 < x < 50.25. E(X) = (50.25 + 49.75)/2 = 50.0, V (X ) = (50.25 − 49.75) 2 = 0.0208, and σ x = 0.144 . 12 x b) F ( x ) = ∫ 2.0dx for 49.75 < x < 50.25. Therefore, 49.75 ⎧0, x < 49.75 ⎪ ⎪ ⎪ F ( x) = ⎨2 x − 99.5, 49.75 ≤ x < 50.25 ⎪ ⎪ 50.25 ≤ x ⎪1, ⎩ c) P ( X < 50.1) = F (50.1) = 2(50.1) − 99.5 = 0.7 4-34. a) The distribution of X is f(x) = 10 for 0.95 < x < 1.05. Now, ⎧0, x < 0.95 ⎪ ⎪ ⎪ FX ( x) = ⎨10 x − 9.5, 0.95 ≤ x < 1.05 ⎪ ⎪ 1.05 ≤ x ⎪1, ⎩ b) P( X > 1.02) = 1 − P( X ≤ 1.02) = 1 − FX (1.02) = 0.3 c) If P(X > x)=0.90, then 1 − F(X) = 0.90 and F(X) = 0.10. Therefore, 10x - 9.5 = 0.10 and x = 0.96. (1.05 − 0.95) 2 d) E(X) = (1.05 + 0.95)/2 = 1.00 and V(X) = = 0.00083 12 4-8 4-35 (1.5 + 2.2) = 1.85 min 2 (2.2 − 1.5) 2 V (X ) = = 0.0408 min 2 12 2 2 1 2 dx = ∫ (1 / 0.7)dx = (1 / 0.7) x 1.5 = (1 / 0.7)(0.5) = 0.7143 b) P ( X < 2) = ∫ (2.2 − 1.5) 1.5 1.5 E( X ) = x x 1 x dx = ∫ (1 / 0.7)dx = (1 / 0.7) x 1.5 c.) F ( X ) = ∫ (2.2 − 1.5) 1.5 1.5 for 1.5 < x < 2.2. Therefore, 0, x < 1.5 ⎧ ⎪ F ( x) = ⎨(1 / 0.7) x − 2.14, 1.5 ≤ x < 2.2 ⎪ 1, 2.2 ≤ x ⎩ 4-36 f ( x) = 0.04 for 50< x <75 75 a) P( X > 70) = ∫ 0.04dx = 0.2 x 75 70 = 0.2 70 60 ∫ b) P ( X < 60) = 0.04dx = 0.04 x 50 = 0.4 60 50 75 + 50 = 62.5 seconds 2 (75 − 50) 2 V (X ) = = 52.0833 seconds2 12 c) E ( X ) = 4-37. a) The distribution of X is f(x) = 100 for 0.2050 < x < 0.2150. Therefore, ⎧0, ⎪ ⎪ ⎪ F ( x) = ⎨100x − 20.50, ⎪ ⎪ ⎪1, ⎩ x < 0.2050 0.2050 ≤ x < 0.2150 0.2150 ≤ x b) P ( X > 0.2125) = 1 − F (0.2125) = 1 − [100(0.2125) − 20.50] = 0.25 c) If P(X > x)=0.10, then 1 − F(X) = 0.10 and F(X) = 0.90. Therefore, 100x - 20.50 = 0.90 and x = 0.2140. d) E(X) = (0.2050 + 0.2150)/2 = 0.2100 µm and V(X) = (0.2150 − 0.2050) 2 = 8.33 × 10 −6 µm 2 12 4-9 40 4-38. ∫ a) P( X > 35) = 0.1dx = 0.1x 35 40 = 0.5 35 40 b) P(X > x)=0.90 and P( X > x) = ∫ 0.1dt = 0.1(40 − x) . x Now, 0.1(40-x) = 0.90 and x = 31 c) E(X) = (30 + 40)/2 = 35 and V(X) = (40 − 30) 2 = 8.33 12 Section 4-6 4-39. a) P(Z<1.32) = 0.90658 b) P(Z<3.0) = 0.99865 c) P(Z>1.45) = 1 − 0.92647 = 0.07353 d) P(Z > −2.15) = p(Z < 2.15) = 0.98422 e) P(−2.34 < Z < 1.76) = P(Z<1.76) − P(Z > 2.34) = 0.95116 4-40. a) P(−1 < Z < 1) = P(Z < 1) − P(Z > 1) = 0.84134 − (1 − 0.84134) = 0.68268 b) P(−2 < Z < 2) = P(Z < 2) − [1 − P(Z < 2)] = 0.9545 c) P(−3 < Z < 3) = P(Z < 3) − [1 − P(Z < 3)] = 0.9973 d) P(Z > 3) = 1 − P(Z < 3) = 0.00135 e) P(0 < Z < 1) = P(Z < 1) − P(Z < 0) = 0.84134 − 0.5 = 0.34134 4-41 a) P(Z < 1.28) = 0.90 b) P(Z < 0) = 0.5 c) If P(Z > z) = 0.1, then P(Z < z) = 0.90 and z = 1.28 d) If P(Z > z) = 0.9, then P(Z < z) = 0.10 and z = −1.28 e) P(−1.24 < Z < z) = P(Z < z) − P(Z < −1.24) = P(Z < z) − 0.10749. Therefore, P(Z < z) = 0.8 + 0.10749 = 0.90749 and z = 1.33 4-42. a) Because of the symmetry of the normal distribution, the area in each tail of the distribution must equal 0.025. Therefore the value in Table II that corresponds to 0.975 is 1.96. Thus, z = 1.96. b) Find the value in Table II corresponding to 0.995. z = 2.58. c) Find the value in Table II corresponding to 0.84. z = 1.0 d) Find the value in Table II corresponding to 0.99865. z = 3.0. 4-10 4-43. a) P(X < 13) = P(Z < (13−10)/2) = P(Z < 1.5) = 0.93319 b) P(X > 9) = 1 − P(X < 9) = 1 − P(Z < (9−10)/2) = 1 − P(Z < −0.5) = 0.69146. 6 − 10 14 − 10 ⎞ <Z< c) P(6 < X < 14) = P⎛ ⎜ ⎟ ⎝ 2 ⎠ 2 = P(−2 < Z < 2) = P(Z < 2) −P(Z < − 2)] = 0.9545. 2 − 10 4 − 10 ⎞ <Z< d) P(2 < X < 4) = P⎛ ⎜ ⎟ ⎝ 2 2 ⎠ = P(−4 < Z < −3) = P(Z < −3) − P(Z < −4) = 0.00132 e) P(−2 < X < 8) = P(X < 8) − P(X < −2) = P⎛ Z < ⎜ ⎝ −2 − 10 ⎞ 8 − 10 ⎞ ⎛ ⎟ − P⎜ Z < ⎟ ⎝ 2⎠ 2⎠ = P(Z < −1) − P(Z < −6) = 0.15866. 4-44. x −10 x − 10 ⎞ ⎛ ⎟ = 0.5. Therefore, 2 = 0 and x = 10. 2⎠ ⎝ x − 10 ⎞ x − 10 ⎞ ⎛ ⎛ b) P(X > x) = P⎜ Z > ⎟ = 1 − P⎜ Z < ⎟ 2⎠ 2⎠ ⎝ ⎝ a) P(X > x) = P⎜ Z > = 0.95. x − 10 ⎞ x − 10 Therefore, P⎛ Z < = −1.64. Consequently, x = 6.72. ⎜ ⎟ = 0.05 and 2 ⎝ 2⎠ x − 10 ⎞ ⎛ ⎛ x − 10 ⎞ < Z < 0 ⎟ = P ( Z < 0) − P ⎜ Z < ⎟ 2⎠ ⎝2 ⎠ ⎝ x − 10 ⎞ ⎛ = 0.5 − P⎜ Z < ⎟ = 0.2. 2⎠ ⎝ c) P(x < X < 10) = P⎜ x − 10 ⎞ x − 10 Therefore, P⎛ Z < = −0.52. Consequently, x = 8.96. ⎜ ⎟ = 0.3 and ⎝ 2 ⎠ 2 d) P(10 − x < X < 10 + x) = P(−x/2 < Z < x/2)= 0.95. Therefore, x/2 = 1.96 and x = 3.92 e) P(10 − x < X < 10 + x) = P(−x/2 < Z < x/2) = 0.99. Therefore, x/2 = 2.58 and x = 5.16 4-11 4-45. ⎛ ⎝ a) P(X < 11) = P⎜ Z < 11 − 5 ⎞ ⎟ 4⎠ = P(Z < 1.5) = 0.93319 0 −5⎞ ⎟ 4⎠ ⎛ ⎝ b) P(X > 0) = P⎜ Z > = P(Z > −1.25) = 1 − P(Z < −1.25) = 0.89435 7 −5⎞ ⎛3−5 <Z< ⎟ 4⎠ ⎝4 c) P(3 < X < 7) = P⎜ = P(−0.5 < Z < 0.5) = P(Z < 0.5) − P(Z < −0.5) = 0.38292 9−5⎞ ⎛−2−5 <Z< ⎟ 4⎠ ⎝4 d) P(−2 < X < 9) = P⎜ = P(−1.75 < Z < 1) = P(Z < 1) − P(Z < −1.75)] = 0.80128 8−5⎞ ⎛2−5 <Z< ⎟ 4⎠ ⎝4 e) P(2 < X < 8) = P⎜ =P(−0.75 < Z < 0.75) = P(Z < 0.75) − P(Z < −0.75) = 0.54674 4-46. ⎛ ⎝ a) P(X > x) = P⎜ Z > x − 5⎞ ⎟ = 0.5. 4⎠ Therefore, x = 5. x − 5⎞ ⎛ ⎟ = 0.95. 4⎠ ⎝ x − 5⎞ ⎛ Therefore, P⎜ Z < ⎟ = 0.05 4⎠ ⎝ b) P(X > x) = P⎜ Z > − Therefore, x 4 5 = −1.64, and x = −1.56. ⎛x−5 ⎞ < Z < 1⎟ = 0.2. ⎝4 ⎠ c) P(x < X < 9) = P⎜ − Therefore, P(Z < 1) − P(Z < x 4 5 )= 0.2 where P(Z < 1) = 0.84134. − − Thus P(Z < x 4 5 ) = 0.64134. Consequently, x 4 5 = 0.36 and x = 6.44. 4-12 x − 5⎞ ⎛3−5 <Z< ⎟ = 0.95. 4⎠ ⎝4 x − 5⎞ x − 5⎞ ⎛ ⎛ Therefore, P⎜ Z < ⎟ − P(Z < −0.5) = 0.95 and P⎜ Z < ⎟ − 0.30854 = 0.95. 4⎠ 4⎠ ⎝ ⎝ d) P(3 < X < x) = P⎜ Consequently, x − 5⎞ ⎛ P⎜ Z < ⎟ = 1.25854. Because a probability can not be greater than one, there is 4⎠ ⎝ no solution for x. In fact, P(3 < X) = P(−0.5 < Z) = 0.69146. Therefore, even if x is set to infinity the probability requested cannot equal 0.95. 5 + x − 5⎞ ⎛5− x −5 <Z< ⎟ 4 4 ⎠ ⎝ x⎞ ⎛− x = P⎜ < Z < ⎟ = 0.99 4⎠ ⎝4 e) P(5 − x < X <5 + x) = P⎜ Therefore, x/4 = 2.58 and x = 10.32. 4-47. ⎛ ⎝ a) P(X < 6250) = P⎜ Z < 6250 − 6000 ⎞ ⎟ 100 ⎠ = P(Z < 2.5) = 0.99379 5900 − 6000 ⎞ ⎛ 5800 − 6000 <Z< ⎟ 100 100 ⎝ ⎠ b) P(5800 < X < 5900) = P⎜ =P(−2 < Z < −1) = P(Z <− 1) − P(Z < −2) = 0.13591 x − 6000 ⎞ ⎛ ⎟ = 0.95. 100 ⎠ ⎝ x − 6000 Therefore, 100 = −1.65 and x = 5835. c) P(X > x) = P⎜ Z > 4-48. ⎛ ⎝ a) P(X < 40) = P⎜ Z < 40 − 35 ⎞ ⎟ 2⎠ = P(Z < 2.5) = 0.99379 ⎛ ⎝ b) P(X < 30) = P⎜ Z < 30 − 35 ⎞ ⎟ 2⎠ = P(Z < −2.5) = 0.00621 0.621% are scrapped 4-13 4-49. ⎛ ⎝ a) P(X > 0.62) = P⎜ Z > 0.62 − 0.5 ⎞ ⎟ 0.05 ⎠ = P(Z > 2.4) = 1 − P(Z <2.4) = 0.0082 0.63 − 0.5 ⎞ ⎛ 0.47 − 0.5 <Z< ⎟ 0.05 ⎠ ⎝ 0.05 b) P(0.47 < X < 0.63) = P⎜ = P(−0.6 < Z < 2.6) = P(Z < 2.6) − P(Z < −0.6) = 0.99534 − 0.27425 = 0.72109 ⎛ ⎝ c) P(X < x) = P⎜ Z < x − 0 .5 ⎞ ⎟ = 0.90. 0.05 ⎠ 0. Therefore, x0−055 = 1.28 and x = 0.564. . 4-50. a) P(X < 12) = P(Z < 12−12.4 0.1 ) = P(Z < −4) ≅ 0 12.1 − 12.4 ⎞ b) P(X < 12.1) = P⎛ Z < ⎜ ⎟ = P(Z < −3) = 0.00135 ⎝ 01 . ⎠ and 12.6 − 12.4 ⎞ P(X > 12.6) = P⎛ Z > ⎜ ⎟ = P(Z > 2) = 0.02275. ⎝ 01 . ⎠ Therefore, the proportion of cans scrapped is 0.00135 + 0.02275 = 0.0241, or 2.41% c) P(12.4 − x < X < 12.4 + x) = 0.99. x⎞ ⎛x <Z< ⎟ = 0.99 0 .1 ⎠ ⎝ 0.1 x⎞ ⎛ Consequently, P⎜ Z < ⎟ = 0.995 and x = 0.1(2.58) = 0.258. 0 .1 ⎠ ⎝ Therefore, P⎜ − The limits are ( 12.142, 12.658). 4-51. 45 − 65 ⎞ ⎛ ⎟ = P(Z < -3) = 0.00135 5⎠ ⎝ 65 − 60 ⎞ ⎛ b) P(X > 65) = P⎜ Z > ⎟ = P(Z >1) = 1- P(Z < 1) 5⎠ ⎝ a) P(X <45) = P⎜ Z < = 1 - 0.841345= 0.158655 ⎛ ⎝ c) P(X < x) = P⎜ Z < Therefore, x − 60 ⎞ ⎟ = 0.99. 5⎠ x − 60 5 = 2.33 and x = 72 4-14 4-52. 12 − µ ⎞ a) If P(X > 12) = 0.999, then P⎛ Z > ⎜ ⎟ = 0.999. ⎝ Therefore, 12 − µ = 0.1 0.1 ⎠ −3.09 and µ = 12.309. 12 − µ ⎞ b) If P(X > 12) = 0.999, then P⎛ Z > ⎜ ⎟ = 0.999. ⎝ Therefore, 4-53. 12 − µ = 0. 05 0.05 ⎠ -3.09 and µ = 12.1545. ⎛ ⎝ a) P(X > 0.5) = P⎜ Z > 0.5 − 0.4 ⎞ ⎟ 0.05 ⎠ = P(Z > 2) = 1 − 0.97725 = 0.02275 0.5 − 0.4 ⎞ ⎛ 0 .4 − 0 .4 <Z < ⎟ 0.05 ⎠ ⎝ 0.05 b) P(0.4 < X < 0.5) = P⎜ = P(0 < Z < 2) = P(Z < 2) − P(Z < 0) = 0.47725 ⎛ ⎝ c) P(X > x) = 0.90, then P⎜ Z > x − 0.4 ⎞ ⎟ = 0.90. 0.05 ⎠ x −0.4 Therefore, 0.05 = −1.28 and x = 0.336. 4-54 70 − 60 ⎞ ⎛ ⎟ 4⎠ ⎝ = 1 − P( Z < 2.5) a) P(X > 70) = P⎜ Z > = 1 − 0.99379 = 0.00621 58 − 60 ⎞ ⎛ ⎟ 4⎠ ⎝ = P( Z < −0.5) b) P(X < 58) = P⎜ Z < = 0.308538 c) 1,000,000 bytes * 8 bits/byte = 8,000,000 bits 8,000,000 bits = 133.33 seconds 60,000 bits/sec 4-15 a) P(X > 90.3) + P(X < 89.7) ⎛ ⎝ = P⎜ Z > 89.7 − 90.2 ⎞ 90.3 − 90.2 ⎞ ⎛ ⎟ ⎟ + P⎜ Z < 0 .1 0 .1 ⎠ ⎠ ⎝ = P(Z > 1) + P(Z < −5) = 1 − P(Z < 1) + P(Z < −5) =1 − 0.84134 +0 = 0.15866. Therefore, the answer is 0.15866. b) The process mean should be set at the center of the specifications; that is, at µ = 90.0. 90.3 − 90 ⎞ ⎛ 89.7 − 90 <Z< ⎟ 0 .1 ⎠ ⎝ 0.1 c) P(89.7 < X < 90.3) = P⎜ = P(−3 < Z < 3) = 0.9973. The yield is 100*0.9973 = 99.73% 4-56. 90.3 − 90 ⎞ ⎛ 89.7 − 90 <Z< ⎟ 0 .1 ⎠ ⎝ 0.1 a) P(89.7 < X < 90.3) = P⎜ = P(−3 < Z < 3) = 0.9973. P(X=10) = (0.9973)10 = 0.9733 b) Let Y represent the number of cases out of the sample of 10 that are between 89.7 and 90.3 ml. Then Y follows a binomial distribution with n=10 and p=0.9973. Thus, E(Y)= 9.973 or 10. 4-57. 80 − 100 ⎞ ⎛ 50 − 100 <Z< ⎟ 20 ⎠ ⎝ 20 a) P(50 < X < 80) = P⎜ = P(−2.5 < Z < -1) = P(Z < −1) − P(Z < −2.5) = 0.15245. ⎛ ⎝ b) P(X > x) = 0.10. Therefore, P⎜ Z > x −100 x − 100 ⎞ ⎟ = 0.10 and 20 = 1.28. 20 ⎠ Therefore, x = 126. hours 4-16 4-58. ⎛ ⎝ a) P(X < 5000) = P⎜ Z < 5000 − 7000 ⎞ ⎟ 600 ⎠ = P(Z < −3.33) = 0.00043. ⎛ ⎝ b) P(X > x) = 0.95. Therefore, P⎜ Z > x − 7000 ⎞ ⎟ = 0.95 and 600 ⎠ x −7000 600 = −1.64. Consequently, x = 6016. ⎛ ⎝ c) P(X > 7000) = P⎜ Z > 7000 − 7000 ⎞ ⎟ = P( Z > 0) = 0.5 600 ⎠ P(three lasers operating after 7000 hours) = (1/2)3 =1/8 4-59. ⎛ ⎝ a) P(X > 0.0026) = P⎜ Z > 0.0026 − 0.002 ⎞ ⎟ 0.0004 ⎠ = P(Z > 1.5) = 1-P(Z < 1.5) = 0.06681. 0.0026 − 0.002 ⎞ ⎛ 0.0014 − 0.002 <Z< ⎟ 0.0004 0.0004 ⎝ ⎠ b) P(0.0014 < X < 0.0026) = P⎜ = P(−1.5 < Z < 1.5) = 0.86638. 0.0026 − 0.002 ⎞ ⎛ 0.0014 − 0.002 <Z< ⎟ σ σ ⎠ ⎝ 0.0006 ⎞ − 0.0006 ⎛ = P⎜ <Z< ⎟. σ σ⎠ ⎝ 0.0006 ⎞ ⎛ Therefore, P⎜ Z < ⎟ = 0.9975. Therefore, 0.0006 = 2.81 and σ = 0.000214. σ σ⎠ ⎝ c) P(0.0014 < X < 0.0026) = P⎜ 4-60. ⎛ ⎝ a) P(X > 13) = P⎜ Z > 13 − 12 ⎞ ⎟ 0 .5 ⎠ = P(Z > 2) = 0.02275 ⎛ ⎝ b) If P(X < 13) = 0.999, then P⎜ Z < 13 − 12 ⎞ ⎟ = 0.999. σ⎠ Therefore, 1/ σ = 3.09 and σ = 1/3.09 = 0.324. ⎛ ⎝ c) If P(X < 13) = 0.999, then P⎜ Z < Therefore, 13 − µ 0 .5 13 − µ ⎞ ⎟ = 0.999. 0.5 ⎠ = 3.09 and µ = 11.455 4-17 Section 4-7 4-61. a) E(X) = 200(0.4) = 80, V(X) = 200(0.4)(0.6) = 48 and σ ⎛ Then, P( X ≤ 70) ≅ P⎜ Z ≤ ⎜ ⎝ X = 48 . 70 − 80 ⎞ ⎟ = P( Z ≤ −1.44) = 0.074934 ⎟ 48 ⎠ b) ⎛ 70 − 80 90 − 80 ⎞ ⎟ = P (−1.44 < Z ≤ 1.44) P (70 < X ≤ 90) ≅ P⎜ <Z≤ ⎟ ⎜ 48 ⎠ ⎝ 48 = 0.925066 − 0.074934 = 0.85013 4-62. a) P ( X < 4) = ( )0.1 0.9 + ( )0.1 0.9 + ( )0.1 0.9 + ( )0.1 0.9 100 0 0 100 2 100 2 100 1 98 1 100 3 99 3 97 = 0.0078 b) E(X) = 10, V(X) = 100(0.1)(0.9) = 9 and σ X = 3 . Then, P( X < 4) ≅ P( Z < 4 −10 3 ) = P( Z < −2) = 0.023 c) P(8 < X < 12) ≅ P( 8 −310 < Z < 4-63. 12 −10 3 ) = P(−0.67 < Z < 0.67) = 0.497 Let X denote the number of defective chips in the lot. Then, E(X) = 1000(0.02) = 20, V(X) = 1000(0.02)(0.98) = 19.6. 25 − 20 ⎞ ⎟ = P ( Z > 1.13) = 1 − P( Z ≤ 1.13) = 0.129 ⎟ 19.6 ⎠ ⎝ 10 b) P(20 < X < 30) ≅ P(0 < Z < 19.6 ) = P(0 < Z < 2.26) ⎛ a) P ( X > 25) ≅ P⎜ Z > ⎜ = P ( Z ≤ 2.26) − P ( Z < 0) = 0.98809 − 0.5 = 0.488 4-64. Let X denote the number of defective electrical connectors. Then, E(X) = 25(100/1000) = 2.5, V(X) = 25(0.1)(0.9) = 2.25. a) P(X=0)=0.925=0.0718 .5 b) P( X ≤ 0) ≅ P( Z < 0−225 ) = P( Z < −1.67) = 0.047 2. The approximation is smaller than the binomial. It is not satisfactory since np<5. c) Then, E(X) = 25(100/500) = 5, V(X) = 25(0.2)(0.8) = 4. P(X=0)=0.825=0.00377 P( X ≤ 0) ≅ P( Z < 0 −5 4 ) = P( Z < −2.5) = 0.006 Normal approximation is now closer to the binomial; however, it is still not satisfactory since np = 5 is not > 5. 4-18 4-65. Let X denote the number of original components that fail during the useful life of the product. Then, X is a binomial random variable with p = 0.001 and n = 5000. Also, E(X) = 5000 (0.001) = 5 and V(X) = 5000(0.001)(0.999) = 4.995. ⎛ 10 − 5 ⎞ ⎟ = P( Z ≥ 2.24) = 1 − P( Z < 2.24) = 1 − 0.987 = 0.013 . P( X ≥ 10) ≅ P⎜ Z ≥ ⎟ ⎜ 4.995 ⎠ ⎝ 4-66. Let X denote the number of particles in 10 cm2 of dust. Then, X is a Poisson random variable with λ = 10(1000) = 10,000 . Also, E(X) = λ = 10,000 = V(X) P ( X > 10,000 ) ≅ P ( Z > 4-67 10 , 000 −10 , 000 10 , 000 ) = P ( Z > 0 ) = 0 .5 Let X denote the number of errors on a web site. Then, X is a binomial random variable with p = 0.05 and n = 100. Also, E(X) = 100 (0.05) = 5 and V(X) = 100(0.05)(0.95) = 4.75 ⎛ 1− 5 ⎞ ⎟ = P( Z ≥ −1.84) = 1 − P( Z < −1.84) = 1 − 0.03288 = 0.96712 P( X ≥ 1) ≅ P⎜ Z ≥ ⎜ ⎟ 4.75 ⎠ ⎝ 4-68. Let X denote the number of particles in 10 cm2 of dust. Then, X is a Poisson random variable with λ = 10,000 . Also, E(X) = λ = 10,000 =V(X) a) 20 , 000 −10 , 000 10 , 000 P ( X ≥ 20,000) ≅ P ( Z ≥ ) = P ( Z ≥ 100) = 1 − P ( Z < 100) = 1 − 1 = 0 b.) P ( X < 9,900 ) ≅ P ( Z < ⎛ ⎜ ⎝ 9 , 900 −10 , 000 10 , 000 c.) If P(X > x) = 0.01, then P⎜ Z > Therefore, 4-69 x −10 , 000 100 ) = P ( Z < −1) = 0.1587 x − 10,000 ⎞ ⎟ = 0.01. 10,000 ⎟ ⎠ = 2.33 and x = 10,233 Let X denote the number of hits to a web site. Then, X is a Poisson random variable with a of mean 10,000 per day. E(X) = λ = 10,000 and V(X) = 10,000 a) ⎛ 10,200 − 10,000 ⎞ ⎟ = P( Z ≥ 2) = 1 − P( Z < 2) P( X ≥ 10,200) ≅ P⎜ Z ≥ ⎟ ⎜ 10,000 ⎠ ⎝ = 1 − 0.9772 = 0.0228 Expected value of hits days with more than 10,200 hits per day is (0.0228)*365=8.32 days per year 4-19 4-69 b.) Let Y denote the number of days per year with over 10,200 hits to a web site. Then, Y is a binomial random variable with n=365 and p=0.0228. E(Y) = 8.32 and V(Y) = 365(0.0228)(0.9772)=8.13 15 − 8.32 ⎞ ⎛ P(Y > 15) ≅ P⎜ Z ≥ ⎟ = P( Z ≥ 2.34) = 1 − P( Z < 2.34) 8.13 ⎠ ⎝ = 1 − 0.9904 = 0.0096 4-70 E(X) = 1000(0.2) = 200 and V(X) = 1000(0.2)(0.8) = 160 − 200 a) P( X ≥ 225) ≅ 1 − P( Z ≤ 225160 ) = 1 − P( Z ≤ 1.98) = 1 − 0.97615 = 0.02385 b) − 200 P(175 ≤ X ≤ 225) ≅ P( 175160 ≤ Z ≤ 225 − 200 160 ) = P(−1.98 ≤ Z ≤ 1.98) = 0.97615 − 0.02385 = 0.9523 ⎛ x − 200 ⎞ ⎟ = 0.01. c) If P(X > x) = 0.01, then P⎜ Z > ⎜ ⎟ 160 ⎠ ⎝ Therefore, 4-71 x − 200 160 = 2.33 and x = 229.5 X is the number of minor errors on a test pattern of 1000 pages of text. X is a Poisson random variable with a mean of 0.4 per page a. ) The number of errors per page is a random variable because it will be different for each page. e −0.4 0.4 0 = 0.670 0! P ( X ≥ 1) = 1 − P ( X = 0) = 1 − 0.670 = 0.330 b.) P( X = 0) = The mean number of pages with one or more errors is 1000(0.330)=330 pages c.) Let Y be the number of pages with errors. ⎛ ⎞ 350 − 330 ⎟ = P ( Z ≥ 1.35) = 1 − P ( Z < 1.35) P (Y > 350) ≅ P⎜ Z ≥ ⎜ 1000(0.330)(0.670) ⎟ ⎝ ⎠ = 1 − 0.9115 = 0.0885 4-20 Section 4-9 0 4-72. ∫ a) P( X ≤ 0) = λe − λx dx = 0 0 ∞ ∞ 2 1 2 ∫ b) P ( X ≥ 2) = 2e − 2 x dx = − e − 2 x 1 ∫ c) P( X ≤ 1) = 2e − 2 x dx = − e − 2 x = e − 4 = 0.0183 = 1 − e − 2 = 0.8647 0 0 2 2 1 1 ∫ d) P (1 < X < 2) = 2e − 2 x dx = − e − 2 x x e) P( X ≤ x) = 2e − 2t dt = − e − 2t x 0 0 ∫ 4-73. = e − 2 − e − 4 = 0.1170 = 1 − e − 2 x = 0.05 and x = 0.0256 If E(X) = 10, then λ = 01 . . ∞ ∞ 10 10 ∫ a) P( X > 10) = 0.1e − 0.1x dx = − e − 0.1x ∞ b) P ( X > 20) = − e − 0.1x 20 ∞ c) P ( X > 30) = − e − 0.1x = e −1 = 0.3679 = e − 2 = 0.1353 = e − 3 = 0.0498 30 x x 0 0 ∫ − 0.1t dt = − e − 0.1t d) P( X < x) = 0.1e 4-74. = 1 − e − 0.1x = 0.95 and x = 29.96. Let X denote the time until the first count. Then, X is an exponential random variable with λ = 2 counts per minute. ∞ ∞ 0.5 0.5 −2x −2x ∫ 2e dx = − e a) P( X > 0.5) = 1/ 6 1/ 6 0 = e −1 = 0.3679 0 −2x −2x ∫ 2e dx = − e b) P( X < 10 ) = 60 c) P (1 < X < 2) = − e − 2 x 2 = 1 − e −1/ 3 = 0.2835 = e − 2 − e − 4 = 0.1170 1 4-75. a) E(X) = 1/λ = 1/3 = 0.333 minutes b) V(X) = 1/λ2 = 1/32 = 0.111, σ = 0.3333 c) P( X < x) = 3e − 3t dt = − e − 3t x x 0 0 ∫ = 1 − e − 3 x = 0.95 , x = 0.9986 4-21 4-76. The time to failure (in hours) for a laser in a cytometry machine is modeled by an exponential distribution with 0.00004. ∞ −.0.00004 x dx = − e − 0.00004 x ∫ 0.00004e ∞ 20000 20000 ∞ 30000 30000 0 a) P( X > 20,000) = −.0.00004 x dx = − e − 0.00004 x ∫ 0.00004e b) P( X < 30,000) = = e − 0.8 = 0.4493 = 1 − e −1.2 = 0.6988 c) 30000 P (20,000 < X < 30,000) = ∫ 0.00004e −.0.00004 x dx 20000 = − e − 0.00004 x 30000 = e − 0.8 − e −1.2 = 0.1481 20000 4-77. Let X denote the time until the first call. Then, X is exponential and λ = E (1X ) = 115 calls/minute. ∞ a) P ( X > 30) = ∫ 1 15 − x e 15 dx = − e − x 15 ∞ = e − 2 = 0.1353 30 30 b) The probability of at least one call in a 10-minute interval equals one minus the probability of zero calls in a 10-minute interval and that is P(X > 10). P ( X > 10) = − e − x 15 ∞ = e − 2 / 3 = 0.5134 . 10 Therefore, the answer is 1- 0.5134 = 0.4866. Alternatively, the requested probability is equal to P(X < 10) = 0.4866. c) P (5 < X < 10) = − e − x 15 10 = e −1 / 3 − e − 2 / 3 = 0.2031 5 d) P(X < x) = 0.90 and P ( X < x) = − e − t 15 x = 1 − e − x / 15 = 0.90 . Therefore, x = 34.54 0 minutes. 4-78. Let X be the life of regulator. Then, X is an exponential random variable with λ = 1 / E( X ) = 1 / 6 a) Because the Poisson process from which the exponential distribution is derived is memoryless, this probability is 6 P(X < 6) = ∫ 0 1 6 e − x / 6 dx = − e − x / 6 6 = 1 − e −1 = 0.6321 0 b) Because the failure times are memoryless, the mean time until the next failure is E(X) = 6 years. 4-22 4-79. Let X denote the time to failure (in hours) of fans in a personal computer. Then, X is an exponential random variable and λ = 1 / E ( X ) = 0.0003 . ∞ ∞ ∫ 0.0003e a) P(X > 10,000) = − x 0.0003 dx = − e − x 0.0003 10 , 000 10 , 000 7 , 000 7 , 000 b) P(X < 7,000) = ∫ 0.0003e − x 0.0003 dx = − e − x 0.0003 0 4-80. = e − 3 = 0.0498 = 1 − e − 2.1 = 0.8775 0 Let X denote the time until a message is received. Then, X is an exponential random variable and λ = 1 / E ( X ) = 1 / 2 . ∞ a) P(X > 2) = ∫ 1 2 e − x / 2 dx = − e − x / 2 ∞ = e −1 = 0.3679 2 2 b) The same as part a. c) E(X) = 2 hours. 4-81. Let X denote the time until the arrival of a taxi. Then, X is an exponential random variable with λ = 1 / E ( X ) = 0.1 arrivals/ minute. ∞ − 0.1x − 0.1 x ∫ 0.1e dx = − e ∞ 60 60 10 10 0 a) P(X > 60) = 0 − 0.1 x − 0.1 x ∫ 0.1e dx = − e b) P(X < 10) = ∞ a) P(X > x) = ∞ x 4-82. = e − 6 = 0.0025 = 1 − e −1 = 0.6321 x − 0.1t − 0.1t ∫ 0.1e dt = − e = e − 0.1x = 0.1 and x = 23.03 minutes. b) P(X < x) = 0.9 implies that P(X > x) = 0.1. Therefore, this answer is the same as part a. x c) P(X < x) = − e −0.1t = 1 − e −0.1x = 0.5 and x = 6.93 minutes. 0 4-83. Let X denote the distance between major cracks. Then, X is an exponential random variable with λ = 1 / E ( X ) = 0.2 cracks/mile. ∞ a) P(X > 10) = ∫ 0.2e − 0.2 x dx = − e − 0.2 x 10 ∞ = e − 2 = 0.1353 10 b) Let Y denote the number of cracks in 10 miles of highway. Because the distance between cracks is exponential, Y is a Poisson random variable with λ = 10(0.2) = 2 cracks per 10 miles. P(Y = 2) = e −2 22 = 0.2707 2! c) σ X = 1 / λ = 5 miles. 4-23 15 4-84. a) P (12 < X < 15) = 0.2e − 0.2 x dx = − e − 0.2 x 15 12 ∞ 12 ∫ b) P(X > 5) = − e − 0.2 x = e − 2.4 − e − 3 = 0.0409 = e −1 = 0.3679 . By independence of the intervals in a Poisson 5 process, the answer is 0.3679 2 = 0.1353 . Alternatively, the answer is P(X > 10) = e −2 = 0.1353 . The probability does depend on whether or not the lengths of highway are consecutive. c) By the memoryless property, this answer is P(X > 10) = 0.1353 from part b. 4-85. Let X denote the lifetime of an assembly. Then, X is an exponential random variable with λ = 1 / E ( X ) = 1 / 400 failures per hour. 100 a) P(X < 100) = ∫ 1 400 e − x / 400 dx = − e − x / 400 100 = 1 − e − 0.25 = 0.2212 0 0 b) P(X > 500) = − e − x / 400 ∞ = e − 5 / 4 = 0.2865 500 c) From the memoryless property of the exponential, this answer is the same as part a., P(X < 100) = 0.2212. 4-86. a) Let U denote the number of assemblies out of 10 that fail before 100 hours. By the memoryless property of a Poisson process, U has a binomial distribution with n = 10 and p =0.2212 (from Exercise 4-85a). Then, P (U ≥ 1) = 1 − P (U = 0) = 1 − ( )0.2212 (1 − 0.2212) 10 0 0 10 = 0.9179 b) Let V denote the number of assemblies out of 10 that fail before 800 hours. Then, V is a binomial random variable with n = 10 and p = P(X < 800), where X denotes the lifetime of an assembly. 800 Now, P(X < 800) = ∫ 1 400 0 Therefore, P(V = 10) = 4-87. e − x / 400dx = −e − x / 400 ( )0.8647 10 10 800 = 1 − e− 2 = 0.8647 . 0 10 (1 − 0.8647 ) 0 = 0.2337 . Let X denote the number of calls in 3 hours. Because the time between calls is an exponential random variable, the number of calls in 3 hours is a Poisson random variable. Now, the mean time between calls is 0.5 hours and λ = 1 / 0.5 = 2 calls per hour = 6 calls in 3 hours. −6 0 −6 1 −6 2 −6 3 P( X ≥ 4) = 1 − P( X ≤ 3) = 1 − ⎡ e 6 + e 6 + e 6 + e 6 ⎤ = 0.8488 ⎢ 0! 1! 2! 3! ⎥ ⎣ ⎦ 4-88. Let Y denote the number of arrivals in one hour. If the time between arrivals is exponential, then the count of arrivals is a Poisson random variable and λ = 1 arrival per hour. −1 1 −1 2 −1 3 ⎡ −1 0 ⎤ P(Y > 3) = 1 − P (Y ≤ 3) = 1 − e 1 + e 1 + e 1 + e 1 = 0.01899 ⎢ ⎥ ⎣ 0! 1! 4-24 2! 3! ⎦ 4-89. a) From Exercise 4-88, P(Y > 3) = 0.01899. Let W denote the number of one-hour intervals out of 30 that contain more than 3 arrivals. By the memoryless property of a Poisson process, W is a binomial random variable with n = 30 and p = 0.01899. P(W = 0) = 30 0.01899 0 (1 − 0.01899 ) 30 = 0.5626 0 b) Let X denote the time between arrivals. Then, X is an exponential random variable with () ∞ ∞ x x λ = 1 arrivals per hour. P(X > x) = 0.1 and P( X > x) = ∫ 1e −1t dt =− e −1t = e −1x = 0.1 . Therefore, x = 2.3 hours. 4-90. Let X denote the number of calls in 30 minutes. Because the time between calls is an exponential random variable, X is a Poisson random variable with λ = 1 / E( X ) = 0.1 calls per minute = 3 calls per 30 minutes. [ −3 0 −3 1 −3 2 −3 3 a) P(X > 3) = 1 − P( X ≤ 3) = 1 − e 0!3 + e 1!3 + e 2!3 + e 3!3 = 0.3528 b) P(X = 0) = e 0!3 = 0.04979 −3 0 c) Let Y denote the time between calls in minutes. Then, P (Y ≥ x) = 0.01 and ∞ ∞ x x P(Y ≥ x) = ∫ 0.1e − 0.1t dt = −e − 0.1t = e − 0.1x . Therefore, e −0.1x = 0.01 and x = 46.05 minutes. ∞ a) From Exercise 4-90, P(Y > 120) = ∞ 120 4-91. 120 − 0.1 y − 0.1 y ∫ 0.1e dy = −e = e −12 = 6.14 × 10− 6 . b) Because the calls are a Poisson process, the number of calls in disjoint intervals are independent. From Exercise 4-90 part b., the probability of no calls in one-half hour is 4 e −3 = 0.04979 . Therefore, the answer is e − 3 = e −12 = 6.14 × 10− 6 . Alternatively, the answer is the probability of no calls in two hours. From part a. of this exercise, this is e−12 . c) Because a Poisson process is memoryless, probabilities do not depend on whether or not intervals are consecutive. Therefore, parts a. and b. have the same answer. 4-92. a) P ( X > θ ) = ∞ ∫ θe 1 −x /θ dx = −e − x / θ θ b) P ( X > 2θ ) = −e − x / θ c) P ( X > 3θ ) = −e − x / θ ∞ 2θ ∞ 3θ ∞ θ = e −1 = 0.3679 = e − 2 = 0.1353 = e − 3 = 0.0498 d) The results do not depend on θ . 4-25 4-93. X is an exponential random variable with λ = 0.2 flaws per meter. a) E(X) = 1/ λ = 5 meters. ∞ ∞ 10 b) P(X > 10) = 10 − 0.2 x − 0.2 x ∫ 0.2e dx = −e = e − 2 = 0.1353 c) No, see Exercise 4-91 part c. d) P(X < x) = 0.90. Then, P(X < x) = − e − 0.2t x = 1 − e − 0.2 x . Therefore, 1 − e −0.2 x = 0.9 0 and x = 11.51. ∞ 4-94. P(X > 8) = ∫ 0.2e − 0.2 x dx = −e −8 / 5 = 0.2019 8 The distance between successive flaws is either less than 8 meters or not. The distances are independent and P(X > 8) = 0.2019. Let Y denote the number of flaws until the distance exceeds 8 meters. Then, Y is a geometric random variable with p = 0.2019. a) P(Y = 5) = (1 − 0.2019) 4 0.2019 = 0.0819 . b) E(Y) = 1/0.2019 = 4.95. ∞ 4-95. E ( X ) = ∫ xλe − λx dx. Use integration by parts with u = x and dv = λe − λx . 0 Then, E ( X ) = − xe − λx ∞ 0 ∞ V(X) = ∫ ( x − λ ) λe 12 − λx ∞ + ∫e − λx dx = − e − λx λ 0 ∞ = 1/ λ 0 dx. Use integration by parts with u 1 = (x − λ )2 and 0 dv = λe − λx . Then, V ( X ) = −( x − λ ) e 12 − λx ∞ ∞ + 2 ∫ ( x − λ )e 1 0 0 − λx ∞ 1 dx = ( λ ) + λ ∫ ( x − λ )λe −λx dx 12 2 0 The last integral is seen to be zero from the definition of E(X). Therefore, V(X) = ( λ1 ) 2 Section 4-10 4-96 a) The time until the tenth call is an Erlang random variable with λ = 5 calls per minute and r = 10. b) E(X) = 10/5 = 2 minutes. V(X) = 10/25 = 0.4 minutes. c) Because a Poisson process is memoryless, the mean time is 1/5=0.2 minutes or 12 seconds 4-26 . 4-97. Let Y denote the number of calls in one minute. Then, Y is a Poisson random variable minute. with λ = 5 calls per e −5 5 4 = 0.1755 4! a) P(Y = 4) = b) P(Y > 2) = 1 - P(Y ≤ 2) = 1 − e −5 50 e −5 51 e −5 52 − − = 0.8754 . 0! 1! 2! Let W denote the number of one minute intervals out of 10 that contain more than 2 calls. Because the calls are a Poisson process, W is a binomial random variable with n = 10 and p = 0.8754. Therefore, P(W = 10) = 10 0.875410 (1 − 0.8754 ) 0 = 0.2643 . 10 () 4-98. Let X denote the pounds of material to obtain 15 particles. Then, X has an Erlang distribution with r = 15 and λ = 0.01 . a) E(X) = b) V(X) = 4-99 r λ = 15 = 1500 pounds. 0.01 15 = 150,000 and σ X = 150,000 = 387.3 pounds. 0.012 Let X denote the time between failures of a laser. X is exponential with a mean of 25,000. a.) Expected time until the second failure E ( X ) = r / λ = 2 / 0.00004 = 50,000 hours b.) N=no of failures in 50000 hours 50000 =2 25000 2 e −2 (2) k P( N ≤ 2) = ∑ = 0.6767 k! k =0 E(N ) = 4-100 Let X denote the time until 5 messages arrive at a node. Then, X has an Erlang distribution with r = 5 and λ = 30 messages per minute. a) E(X) = 5/30 = 1/6 minute = 10 seconds. b)V(X) = 5 30 2 = 1 / 180 minute 2 = 1/3 second and σ X = 0.0745 minute = 4.472 seconds. c) Let Y denote the number of messages that arrive in 10 seconds. Then, Y is a Poisson random variable with λ = 30 messages per minute = 5 messages per 10 seconds. [ −5 0 −5 1 −5 2 −5 3 −5 4 P(Y ≥ 5) = 1 − P(Y ≤ 4) = 1 − e 0!5 + e 1!5 + e 2!5 + e 3!5 + e 4!5 = 0.5595 d) Let Y denote the number of messages that arrive in 5 seconds. Then, Y is a Poisson random variable with λ = 2.5 messages per 5 seconds. P (Y ≥ 5) = 1 − P (Y ≤ 4) = 1 − 0.8912 = 0.1088 4-27 4-101. Let X denote the number of bits until five errors occur. Then, X has an Erlang distribution with r = 5 and λ = 10 −5 error per bit. a) E(X) = b) V(X) = r = 5 × 105 bits. λ r λ 2 = 5 × 1010 and σ X = 5 × 1010 = 223607 bits. c) Let Y denote the number of errors in 105 bits. Then, Y is a Poisson random variable with λ = 1 / 105 = 10 −5 error per bit = 1 error per 105 bits. [ −1 0 −1 1 −1 2 P(Y ≥ 3) = 1 − P(Y ≤ 2) = 1 − e 0!1 + e 1!1 + e 2!1 = 0.0803 4-102 λ = 1 / 20 = 0.05 r = 100 a) E ( X ) = r / λ = 100 / .05 = 5 minutes b) 4 min - 2.5 min=1.5 min c)Let Y be the number of calls before 15 seconds λ = 0.25 * 20 = 5 [ −5 0 −5 1 −5 2 P(Y > 3) = 1 − P( X ≤ 2) = 1 − e 0!5 + e 1!5 + e 2!5 = 1 − .1247 = 0.8753 4-103. a) Let X denote the number of customers that arrive in 10 minutes. Then, X is a Poisson random variable with λ = 0.2 arrivals per minute = 2 arrivals per 10 minutes. [ −2 0 −2 1 −2 2 −2 3 P( X > 3) = 1 − P( X ≤ 3) = 1 − e 0!2 + e 1!2 + e 2!2 + e 3!2 = 0.1429 b) Let Y denote the number of customers that arrive in 15 minutes. Then, Y is a Poisson random variable with λ = 3 arrivals per 15 minutes. [ −3 0 −3 1 −3 2 −3 3 −3 4 P( X ≥ 5) = 1 − P( X ≤ 4) = 1 − e 0!3 + e 1!3 + e 2!3 + e 3!3 + e 4!3 = 0.1847 4-104. Let X denote the time in days until the fourth problem. Then, X has an Erlang distribution with r = 4 and λ = 1 / 30 problem per day. a) E(X) = 4 30 −1 = 120 days. b) Let Y denote the number of problems in 120 days. Then, Y is a Poisson random variable with λ = 4 problems per 120 days. [ −4 0 −4 1 −4 2 −4 3 P(Y < 4) = e 0!4 + e 1!4 + e 2!4 + e 3!4 = 0.4335 4-105. a) Γ (6) = 5!= 120 b) Γ ( 5 ) = 3 Γ ( 3 ) = 2 2 2 c) 31 22 Γ ( 1 ) = 3 π 1 / 2 = 1.32934 2 4 Γ( 9 ) = 7 Γ( 7 ) = 7 5 3 1 Γ( 1 ) = 105 π 1 / 2 = 11.6317 2 2 2 2222 2 16 ∞ 4-106 Γ(r ) = ∫ x r −1e − x dx . Use integration by parts with u = x r −1 and dv =e-x. Then, 0 r −1 − x Γ(r ) = − x e ∞ 0 ∞ + (r − 1) ∫ x r − 2e − x dx = (r − 1)Γ(r − 1) . 0 4-28 ∞ 4-107 ∫ 0 ∞ f ( x; λ , r )dx = ∫ λr x r −1e − λx Γ(r ) 0 ∞ λx , dx . Let y = then the integral is ∫ λ y r −1e− y 0 Γ( r ) dy λ . From the definition of Γ(r ) , this integral is recognized to equal 1. 4-108. If X is a chi-square random variable, then X is a special case of a gamma random variable. Now, E(X) = r λ = r ( 7 / 2) ( 7 / 2) = 14 . = 7 and V ( X ) = 2 = λ (1 / 2) (1 / 2) 2 Section 4-11 4-109. β=0.2 and δ=100 hours E ( X ) = 100Γ(1 + V ( X ) = 100 2 Γ(1 + 1 0.2 ) = 100 × 5!= 12,000 2 0.2 ) − 100 2 [Γ(1 + 1 0.2 )]2 = 3.61 × 1010 −100 4-110. a) P( X < 10000) = FX (10000) = 1 − e −50 b) P( X > 5000) = 1 − FX (5000) = e 0.2 0.2 = 1 − e −2.512 = 0.9189 = 0.1123 4-111. Let X denote lifetime of a bearing. β=2 and δ=10000 hours a) b) P ( X > 8000) = 1 − F X (8000) = e 2 ⎛ 8000 ⎞ −⎜ ⎟ ⎝ 10000 ⎠ = e − 0.8 = 0.5273 2 E ( X ) = 10000Γ(1 + 1 ) = 10000Γ(1.5) 2 = 10000(0.5)Γ(0.5) = 5000 π = 8862.3 = 8862.3 hours c) Let Y denote the number of bearings out of 10 that last at least 8000 hours. Then, Y is a binomial random variable with n = 10 and p = 0.5273. P (Y = 10) = 10 0.527310 (1 − 0.5273) 0 = 0.00166 . 10 () 1 4-112 a.) E ( X ) = δΓ(1 + β ) = 900Γ(1 + 1 / 3) = 900Γ(4 / 3) = 900(0.89298). = 803.68 hours b.) [ 2 2 V ( X ) = δ 2 Γ(1 + β ) − δ 2 Γ(1 + β ) = 900 2 Γ(1 + 2 ) − 900 2 [Γ(1 + 1 )] 3 3 2 2 = 900 2 (0.90274) - 900 2 (0.89298) 2 = 85314.64 hours 2 c.) P ( X < 500) = F X (500) = 1 − e ⎛ 500 ⎞ −⎜ ⎟ ⎝ 900 ⎠ 3 4-29 = 0.1576 4-113. Let X denote the lifetime. a) E ( X ) = δΓ(1 + 015 ) = δΓ(3) = 2δ = 600. Then δ = 300 . Now, . P(X > 500) = b) P(X < 400) = 4-114 e − ( 500 ) 0.5 300 1− e = 0.2750 400 −( 300 ) 0.5 = 0.6848 Let X denote the lifetime a) E ( X ) = 700Γ(1 + 1 ) = 620.4 2 b) V ( X ) = 7002 Γ(2) − 7002 [Γ(1.5)]2 = 7002 (1) − 700 2 (0.25π ) = 105,154.9 c) P(X > 620.4) = e − ( 620 .4 )2 700 = 0 . 4559 4-115 a.)β=2, δ=500 E ( X ) = 500Γ(1 + 1 ) = 500Γ(1.5) 2 = 500(0.5)Γ(0.5) = 250 π = 443.11 = 443.11 hours b.) V ( X ) = 500 2 Γ (1 + 1) − 500 2 [Γ(1 + 1 2 )] 2 = 500 2 Γ(2) − 500 2 [Γ(1.5)] 2 = 53650.5 c.) P(X < 250) = F(250)= 1 − e 4-116 250 − ( 500 ) 25 = 1 − 0.7788 = 0.2212 If X is a Weibull random variable with β=1 and δ=1000, the distribution of X is the exponential distribution with λ=.001. 0 ⎛x⎞ 1 ⎛ 1 ⎞⎛ x ⎞ −⎜ 1000 ⎟ f ( x) = ⎜ ⎟⎜ ⎟ e ⎝ ⎠ for x > 0 ⎝ 1000 ⎠⎝ 1000 ⎠ = 0.001e − 0.001x for x > 0 The mean of X is E(X) = 1/λ = 1000. 4-30 Section 4-11 4-117 X is a lognormal distribution with θ=5 and ω2=9 a. ) ⎛ ln(13330) − 5 ⎞ P ( X < 13300) = P (e W < 13300) = P (W < ln(13300)) = Φ⎜ ⎟ 3 ⎝ ⎠ = Φ (1.50) = 0.9332 b.) Find the value for which P(X≤x)=0.95 ⎛ ln( x) − 5 ⎞ P( X ≤ x) = P(e W ≤ x) = P(W < ln( x)) = Φ⎜ ⎟ = 0.95 3 ⎝ ⎠ ln( x) − 5 = 1.65 x = e 1.65 ( 3) + 5 = 20952 .2 3 θ +ω 2 / 2 = e 5+ 9 / 2 = e 9.5 = 13359.7 c.) µ = E ( X ) = e V ( X ) = e 2θ +ω (e ω − 1) = e 10 + 9 (e 9 − 1) = e 19 (e 9 − 1) = 1.45 x1012 2 4-118 2 a.) X is a lognormal distribution with θ=-2 and ω2=9 P(500 < X < 1000) = P(500 < eW < 1000) = P (ln(500) < W < ln(1000)) ⎛ ln(500) + 2 ⎞ ⎛ ln(1000) + 2 ⎞ = Φ⎜ ⎟ = Φ (2.97) − Φ (2.74) = 0.0016 ⎟ − Φ⎜ 3 3 ⎝ ⎠ ⎝ ⎠ ⎛ ln( x) + 2 ⎞ ⎟ = 0. 1 3 ⎝ ⎠ b.) P ( X < x ) = P (e W ≤ x ) = P (W < ln( x )) = Φ⎜ ln( x) + 2 = −1.28 3 c.) µ = E ( X ) = e x = e −1.28 ( 3) − 2 = 0.0029 θ +ω 2 / 2 = e −2 + 9 / 2 = e 2.5 = 12 .1825 V ( X ) = e 2θ +ω ( e ω − 1) = e −4 + 9 ( e 9 − 1) = e 5 ( e 9 − 1) = 1, 202 , 455 .87 2 2 4-31 4-119 a.) X is a lognormal distribution with θ=2 and ω2=4 ⎛ ln(500) − 2 ⎞ P ( X < 500) = P (e W < 500) = P (W < ln(500)) = Φ⎜ ⎟ 2 ⎝ ⎠ = Φ (2.11) = 0.9826 b.) P( X < 15000 | X > 1000) = P(1000 < X < 1500) P( X > 1000) ⎡ ⎛ ln(1500) − 2 ⎞ ⎛ ln(1000) − 2 ⎞⎤ ⎟ − Φ⎜ ⎟⎥ ⎢Φ ⎜ 2 2 ⎝ ⎠ ⎠⎦ ⎣⎝ = ⎡ ⎛ ln(1000) − 2 ⎞⎤ ⎟⎥ ⎢1 − Φ⎜ 2 ⎝ ⎠⎦ ⎣ = Φ (2.66) − Φ (2.45) 0.9961 − 0.9929 = 0.0032 / 0.007 = 0.45 = (1 − Φ(2.45) ) (1 − 0.9929) c.) The product has degraded over the first 1000 hours, so the probability of it lasting another 500 hours is very low. 4-120 X is a lognormal distribution with θ=0.5 and ω2=1 a) ⎛ ln(10) − 0.5 ⎞ P ( X > 10) = P (e W > 10) = P (W > ln(10)) = 1 − Φ⎜ ⎟ 1 ⎝ ⎠ = 1 − Φ (1.80) = 1 − 0.96407 = 0.03593 b.) ⎛ ln( x) − 0.5 ⎞ P ( X ≤ x) = P (e W ≤ x) = P (W < ln( x)) = Φ⎜ ⎟ = 0.50 1 ⎝ ⎠ ln( x) − 0.5 = 0 x = e 0 (1 ) + 0 . 5 = 1 . 65 seconds 1 c.) µ = E ( X ) = e θ +ω 2 / 2 = e 0.5 +1 / 2 = e 1 = 2 .7183 V ( X ) = e 2θ + ω ( e ω − 1) = e 1 + 1 ( e 1 − 1) = e 2 ( e 1 − 1) = 12 . 6965 2 2 4-32 4-121 Find the values of θand ω2 given that E(X) = 100 and V(X) = 85,000 100 = e θ +ω 2 85000 = e 2θ +ω (e ω − 1) 2 /2 ω let x = e θ and y = e 2 2 then (1) 100 = x y and (2) 85000= x y( y −1) = x y − x y 2 2 2 2 Square (1) 10000 = x 2 y and substitute into (2) 85000 = 10000 ( y − 1) y = 9 .5 Substitute y into (1) and solve for x x = 100 = 32.444 9 .5 θ = ln(32.444) = 3.48 and ω 2 = ln(9.5) = 2.25 4-122 a.) Find the values of θand ω2 given that E(X) = 10000 and σ= 20,000 10000 = e θ +ω 2 20000 2 = e 2θ +ω (e ω − 1) 2 /2 ω let x = e θ and y = e 2 2 then (1) 10000 = x y and (2) 20000 = x y( y −1) = x y − x y 2 2 22 2 Square (1) 10000 2 = x 2 y and substitute into (2) 20000 y=5 2 = 10000 2 ( y − 1) Substitute y into (1) and solve for x x = 10000 = 4472.1360 5 θ = ln(4472.1360) = 8.4056 and ω 2 = ln(5) = 1.6094 b.) ⎛ ln(10000) − 8.4056 ⎞ P ( X > 10000) = P (eW > 10000) = P (W > ln(10000)) = 1 − Φ⎜ ⎟ 1.2686 ⎝ ⎠ = 1 − Φ (0.63) = 1 − 0.7357 = 0.2643 ⎛ ln( x) − 8.4056 ⎞ ⎟ = 0.1 1.2686 ⎝ ⎠ c.) P ( X > x) = P (e W > x) = P(W > ln( x)) = Φ⎜ ln( x) − 8.4056 = −1.28 x = e − 1 . 280 ( 1 . 2686 1.2686 4-33 ) + 8 . 4056 = 881 . 65 hours 4-123 Let X ~N(µ, σ2), then Y = eX follows a lognormal distribution with mean µ and variance σ2. By ⎛ log y − µ ⎞ ⎟. σ ⎝ ⎠ definition, FY(y) = P(Y ≤ y) = P(eX < y) = P(X < log y) = FX(log y) = Φ⎜ Since Y = eX and X ~ N(µ, σ2), we can show that fY (Y ) = 1 f X (log y ) y ⎛ log y − µ ⎞ ⎟ 2σ ⎠ −⎜ ∂FY ( y ) ∂FX (log y ) 1 1 1 Finally, fY(y) = = f X (log y ) = ⋅ e⎝ = ∂y y y σ 2π ∂y Supplemental Exercises 2.5 ⎛ ⎞ x2 − x ⎟ = 0.0625 P( X < 2.5) = ∫ (0.5 x − 1)dx = ⎜ 0.5 ⎜ ⎟ 2 ⎝ ⎠2 2 2.5 4-124 a) 4 4 P( X > 3) = ∫ (0.5 x − 1)dx = 0.5 x2 − x = 0.75 2 b) 3 3 3.5 P(2.5 < X < 3.5) = ∫ (0.5 x − 1)dx = 0.5 x2 − x 2 c) 2.5 x 4-125 x F ( x) = ∫ (0.5t − 1)dt = 0.5 t2 − t = 2 2 2 x2 4 3.5 = 0.5 2.5 − x + 1 . Then, ⎧0, x<2 ⎪ ⎪2 ⎪ F ( x) = ⎨ x4 − x + 1, 2 ≤ x < 4 ⎪ ⎪ 4≤ x ⎪1, ⎩ 4 4-126 E ( X ) = ∫ x(0.5 x − 1)dx = 0.5 x3 − 3 2 4 4 = 2 32 3 − 8 − ( 4 − 2) = 3 10 3 4 2 x2 2 2 V ( X ) = ∫ ( x − 10 ) 2 (0.5 x − 1)dx = ∫ ( x 2 − 20 x + 100 )(0.5 x − 1)dx 3 3 9 4 = ∫ (0.5 x 3 − 13 x 2 + 110 x − 100 )dx = 3 9 9 2 = 0.2222 4-34 x4 8 − 13 x 3 + 55 x 2 − 100 x 9 9 9 4 2 2 . 4-127. Let X denote the time between calls. Then, λ = 1 / E ( X ) = 0.1 calls per minute. 5 5 0 0 ∫ a) P( X < 5) = 0.1e − 0.1x dx = −e − 0.1x b) P (5 < X < 15) = −e − 0.1 x 15 = 1 − e − 0.5 = 0.3935 = e − 0.5 e −1.5 = 0.3834 5 x ∫ c) P(X < x) = 0.9. Then, P( X < x) = 0.1e − 0.1t dt = 1 − e − 0.1x = 0.9 . Now, x = 23.03 0 minutes. 4-128 a) This answer is the same as part a. of Exercise 4-127. 5 5 0 0 P( X < 5) = ∫ 0.1e − 0.1x dx = −e − 0.1x = 1 − e − 0.5 = 0.3935 b) This is the probability that there are no calls over a period of 5 minutes. Because a Poisson process is memoryless, it does not matter whether or not the intervals are consecutive. ∞ ∞ 5 5 P( X > 5) = ∫ 0.1e −0.1x dx = −e −0.1x = e −0.5 = 0.6065 4-129. a) Let Y denote the number of calls in 30 minutes. Then, Y is a Poisson random variable with λ = 3 . P(Y ≤ 2) = e −3 3 0 e −3 31 e −3 3 2 + + = 0.423 . 0! 1! 2! b) Let W denote the time until the fifth call. Then, W has an Erlang distribution with λ = 0.1 and r = 5. E(W) = 5/0.1 = 50 minutes. 4-130 Let X denote the lifetime. Then λ = 1 / E ( X ) = 1 / 6 . 3 P ( X < 3) = ∫ 0 1 6 e − x / 6 dx = −e − x / 6 3 = 1 − e − 0.5 = 0.3935 . 0 4-131. Let W denote the number of CPUs that fail within the next three years. Then, W is a binomial random variable with n = 10 and p = 0.3935 (from Exercise 4-130). Then, P (W ≥ 1) = 1 − P (W = 0) = 1 − 10 0.3935 0 (1 − 0.3935)10 = 0.9933 . 0 () 4-35 4-132 X is a lognormal distribution with θ=0 and ω2=4 a.) P(10 < X < 50) = P(10 < e W < 50) = P(ln(10) < W > ln(50)) ⎛ ln(50) − 0 ⎞ ⎛ ln(10) − 0 ⎞ = Φ⎜ ⎟ − Φ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ = Φ (1.96) − Φ (1.15) = 0.975002 − 0.874928 = 0.10007 ⎛ ln( x) − 0 ⎞ b.) P ( X < x ) = P (e W < x) = P (W < ln( x )) = Φ⎜ ⎟ = 0.05 2 ⎝ ⎠ ln( x) − 0 = −1.64 x = e − 1 . 64 ( 2 ) = 0 . 0376 2 c.) µ = E ( X ) = e θ +ω 2 / 2 = e 0 + 4 / 2 = e 2 = 7 .389 V ( X ) = e 2 θ + ω ( e ω − 1) = e 0 + 4 ( e 4 − 1) = e 4 ( e 4 − 1) = 2926 . 40 2 2 4-133 a. ) Find the values of θand ω2 given that E(X) = 50 and V(X) = 4000 50 = e θ +ω 2 4000 = e 2θ +ω (e ω − 1) 2 /2 ω let x = e θ and y = e Square (1) for x x = 2 then (1) 50 = x y and (2) 4000= x y( y −1) = x y − x y 2 2 50 and substitute into (2) y 2 2 ⎛ 50 ⎞ ⎛ 50 ⎞ 4000= ⎜ ⎟ y2 − ⎜ ⎟ y = 2500 y −1) ( ⎜ y⎟ ⎜ y⎟ ⎝⎠ ⎝⎠ y = 2.6 substitute y back in to (1) and solve for x x = 50 = 31 2. 6 θ = ln(31) = 3.43 and ω 2 = ln(2.6) = 0.96 b .) ⎛ ln(150) − 3.43 ⎞ P ( X < 150) = P (e W < 150) = P (W < ln(150)) = Φ⎜ ⎟ 0.98 ⎝ ⎠ = Φ (1.61) = 0.946301 4-36 22 2 4-134 Let X denote the number of fibers visible in a grid cell. Then, X has a Poisson distribution and λ = 100 fibers per cm2 = 80,000 fibers per sample = 0.5 fibers per grid cell. e −0.5 0.50 = 0.3935 . a) P ( X ≥ 1) = 1 − P( X ) = 1 − 0! b) Let W denote the number of grid cells examined until 10 contain fibers. If the number of fibers have a Poisson distribution, then the number of fibers in each grid cell are independent. Therefore, W has a negative binomial distribution with p = 0.3935. Consequently, E(W) = 10/0.3935 = 25.41 cells. c) V(W) = 10(1 − 0.3935) . Therefore, σ W = 6.25 cells. 0.39352 4-135. Let X denote the height of a plant. 2.25 − 2.5 ⎞ ⎟ = P(Z > -0.5) = 1 - P(Z ≤ -0.5) = 0.6915 0 .5 ⎠ 3 .0 − 2 .5 ⎞ ⎛ 2.0 − 2.5 b) P(2.0 < X < 3.0) = P⎜ <Z< ⎟ =P(-1 < Z < 1) = 0.683 0 .5 ⎠ ⎝ 0 .5 x − 2 .5 ⎞ x − 2.5 ⎛ = -1.28. c.)P(X > x) = 0.90 = P⎜ Z > ⎟ = 0.90 and 0 .5 ⎠ 0.5 ⎝ ⎛ ⎝ a) P(X>2.25) = P⎜ Z > Therefore, x = 1.86. 4-136 a) P(X > 3.5) from part a. of Exercise 4-135 is 0.023. b) Yes, because the probability of a plant growing to a height of 3.5 centimeters or more without irrigation is low. 4-137. Let X denote the thickness. 5 .5 − 5 ⎞ ⎟ = P(Z > 2.5) = 0. 0062 0 .2 ⎠ 5 .5 − 5 ⎞ ⎛ 4 .5 − 5 b) P(4.5 < X < 5.5) = P⎜ <Z< ⎟ = P (-2.5 < Z < 2.5) = 0.9876 0 .2 ⎠ ⎝ 0.2 ⎛ ⎝ a) P(X > 5.5) = P⎜ Z > Therefore, the proportion that do not meet specifications is 1 − P(4.5 < X < 5.5) = 0.012. ⎛ ⎝ c) If P(X < x) = 0.95, then P⎜ Z > 4-138 x−5⎞ x −5 = 1.65 and x = 5.33. ⎟ = 0.95. Therefore, 0 .2 ⎠ 0.2 Let X denote the dot diameter. If P(0.0014 < X < 0.0026) = 0.9973, then P( 0.0014− 0.002 < Z < σ Therefore, 0.0006 σ 0.0026− 0.002 σ 0006 ) = P( −0.σ < Z < = 3 and σ = 0.0002 . 4-37 0.0006 σ ) = 0.9973 . 4-139. If P(0.002-x < X < 0.002+x), then P(-x/0.0004 < Z < x/0.0004) = 0.9973. Therefore, x/0.0004 = 3 and x = 0.0012. The specifications are from 0.0008 to 0.0032. 4-140 Let X denote the life. a) P( X < 5800) = P( Z < 5800 − 7000 600 d) If P(X > x) = 0.9, then P(Z < x = 6232 hours. ) = P( Z < −2) = 1 − P( Z ≤ 2) = 0.023 x − 7000 ) 600 4-141. If P(X > 10,000) = 0.99, then P(Z > µ = 11,398 . 4-142 = -1.28. Consequently, x − 7000 = -1.28 and 600 10 , 000 − µ 10, 000 − µ ) = 0.99. Therefore, = -2.33 and 600 600 The probability a product lasts more than 10000 hours is [ P( X > 10000)]3 , by independence. If [ P( X > 10000)]3 = 0.99, then P(X > 10000) = 0.9967. Then, P(X > 10000) = P( Z > µ = 11,632 hours. 4-143 10000− µ 600 ) = 0.9967 . Therefore, 10000 − µ 600 X is an exponential distribution with E(X) = 7000 hours 5800 a.) P ( X < 5800) = ∫ 0 ∞ ⎛ 5800 ⎞ x −⎜ ⎟ − 1 e 7000 dx = 1 − e ⎝ 7000 ⎠ = 0.5633 7000 x x − − 1 e 7000 dx =0.9 Therefore, e 7000 = 0.9 b.) P ( X > x ) = ∫ 7000 x and x = −7000 ln(0.9) = 737.5 hours 4-38 = -2.72 and 4-144 Find the values of θand ω2 given that E(X) = 7000 and σ= 600 7000 = e θ +ω 2 /2 ω let x = e θ and y = e (2) 600 2 600 2 = e 2θ +ω (e ω − 1) 2 2 2 then (1) 7000 = x y and = x 2 y ( y − 1) = x 2 y 2 − x 2 y Square (1) 7000 2 = x 2 y and substitute into (2) 600 2 = 7000 2 ( y − 1) y = 1 .0073 Substitute y into (1) and solve for x x = 7000 = 6974.6 1.0073 θ = ln(6974 .6) = 8.850 and ω 2 = ln(1.0073) = 0.0073 a.) ⎛ ln(5800) − 8.85 ⎞ P ( X < 5800) = P (e W < 5800) = P (W < ln(5800)) = Φ⎜ ⎟ 0.0854 ⎝ ⎠ = Φ ( −2.16) = 0.015 ⎛ ln( x ) − 8.85 ⎞ b.) P ( X > x) = P (e W > x) = P (W > ln( x )) = Φ ⎜ ⎟ = 0 .9 ⎝ 0.0854 ⎠ ln( x) − 8.85 = −1.28 x = e −1.28 ( 0.0854 ) +8.85 = 6252 .20 hours 0.0854 4-145. a) Using the normal approximation to the binomial with n = 50*36*36=64,800, and p = 0.0001 we have: E(X)=64800(0.0001)=6.48 ⎞ ⎛ X − np 1 − 6.48 ⎟ ≥ P( X ≥ 1) ≅ P⎜ ⎜ np(1 − p) 64800(0.0001)(0.9999) ⎟ ⎠ ⎝ = P(Z > − 2.15 ) = 1 − 0.01578 = 0.98422 b) ⎛ X − np ⎞ 4 − 6.48 ⎟ P( X ≥ 4) ≅ P⎜ ≥ ⎜ np(1 − p) 64800(0.0001)(0.9999) ⎟ ⎝ ⎠ = P( Z ≥ −0.97) = 1 − 0.166023 = 0.8340 4-39 4-146 Using the normal approximation to the binomial with X being the number of people who will be seated. X ~Bin(200, 0.9). ⎞ ⎟ = P( Z ≤ 1.18) = 0.8810 200(0.9)(0.1) ⎟ ⎠ ⎛ X − np ≥ ⎜ np(1 − p ) ⎝ 185 − 180 a) P(X ≤ 185) = P⎜ b) P ( X < 185) ⎛ X − np ≥ = P ( X ≤ 184) = P⎜ ⎜ np (1 − p ) ⎝ ⎞ ⎟ = P ( Z ≤ 0.94) = 0.8264 200(0.9)(0.1) ⎟ ⎠ 184 − 180 c) P(X ≤ 185) ≅ 0.95, Successively trying various values of n: The number of reservations taken could be reduced to about 198. Probability P(Z < n Zo Z0 ) 190 3.39 0.99965 195 2.27 0.988396 198 1.61 0.946301 Mind-Expanding Exercises 4-147. a) P(X > x) implies that there are r - 1 or less counts in an interval of length x. Let Y denote the number of counts in an interval of length x. Then, Y is a Poisson random variable with parameter λx . Then, P ( X > x ) = P (Y ≤ r − 1) = r −1 ∑e λ i =0 b) P ( X ≤ x ) = 1 − r −1 c) f X ( x) = . ∑e λ i =0 d dx i − x ( λx ) i! i − x ( λx ) i! FX ( x ) = λ e − λx r −1 (λx )i i =0 i! ∑ −e − λx (λx )i = λe −λx (λx )r −1 ∑ λi r −1 i =0 (r − 1)! i! 4-148. Let X denote the diameter of the maximum diameter bearing. Then, P(X > 1.6) = 1 P ( X ≤ 1.6) . Also, X ≤ 1.6 if and only if all the diameters are less than 1.6. Let Y denote the diameter of a bearing. Then, by independence . −1. P( X ≤ 1.6) = [ P(Y ≤ 1.6)]10 = [P( Z ≤ 106.0255 )] = 0.99996710 = 0.99967 10 Then, P(X > 1.575) = 0.0033. 4-40 4-149. a) Quality loss = Ek ( X − m) 2 = kE ( X − m) 2 = kσ 2 , by the definition of the variance. b) Quality loss = Ek ( X − m) 2 = kE ( X − µ + µ − m) 2 = kE[( X − µ ) 2 + ( µ − m) 2 + 2( µ − m)( X − µ )] = kE ( X − µ ) 2 + k ( µ − m) 2 + 2k ( µ − m) E ( X − µ ). The last term equals zero by the definition of the mean. Therefore, quality loss = kσ 2 + k ( µ − m) 2 . 4-150. Let X denote the event that an amplifier fails before 60,000 hours. Let A denote the event that an amplifier mean is 20,000 hours. Then A' is the event that the mean of an amplifier is 50,000 hours. Now, P(E) = P(E|A)P(A) + P(E|A')P(A') and 60 , 000 ∫ P( E | A) = 1 20 , 000 e− x / 20, 000 dx = −e− x / 20, 000 60 , 000 0 0 P ( E | A' ) = −e − x / 50, 000 = 1 − e− 3 = 0.9502 60 , 000 = 1 −e − 6 / 5 = 0.6988 . 0 Therefore, P(E) = 0.9502(0.10) + 0.6988(0.90) = 0.7239 4-151. P ( X < t1 + t2 X > t1 ) = P ( t1 < X < t1 + t 2 ) P ( X > t1 ) from the definition of conditional probability. Now, P( X > t1 ) = −e − λx ∞ t1 + t 2 t1 + t 2 t1 P(t1 < X < t1 + t2 ) = t1 − λx − λx ∫ λe dx = −e =e − λt1 − e − λ ( t1 + t 2 ) =e − λt1 t1 Therefore, P( X < t1 + t2 X > t1 ) = e − λt1 (1 − e − λt 2 ) = 1 − e − λt 2 = P ( X < t 2 ) e − λt 1 4-41 4-152. a) 1 − P ( µ 0 − 6σ < X < µ 0 + 6σ ) = 1 − P (−6 < Z < 6) = 1.97 × 10 −9 = 0.00197 ppm b) 1 − P ( µ 0 − 6σ < X < µ 0 + 6σ ) = 1 − P (−7.5 < X − ( µ 0 +1.5σ ) σ < 4.5) = 3.4 × 10 −6 = 3.4 ppm c) 1 − P ( µ 0 − 3σ < X < µ 0 + 3σ ) = 1 − P (−3 < Z < 3) = .0027 = 2,700 ppm d) 1 − P ( µ 0 − 3σ < X < µ 0 + 3σ ) = 1 − P (−4.5 < X − ( µ 0 +1.5σ ) σ < 1.5) = 0.0668106 = 66,810.6 ppm e) If the process is centered six standard deviations away from the specification limits and the process shift, there will be significantly less product loss. If the process is centered only three standard deviations away from the specifications and the process shifts, there could be a great loss of product. Section 4-8 on CD S4-1. E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5 a) b) c) P( X ≤ 2) = P( X ≤ 2.5) ≅ P( Z ≤ P( X ≤ 2) ≅ P( Z ≤ P ( X ≤ 2) = 2 − 2.5 4.5 ( )0.1 0.9 50 0 0 50 2.5−5 4.5 ) = P( Z ≤ −1.18) = 0.119 ) = P( Z ≤ −0.24) = 0.206 + ( )0.1 0.9 50 1 1 49 + ( )0.1 0.9 50 2 2 48 = 0.118 The probability computed using the continuity correction is closer. d) S4-2. 9.5 − 5 ⎞ ⎛ P( X < 10) = P( X ≤ 9.5) ≅ P⎜ Z ≤ ⎟ = P( Z ≤ 2.12) = 0.983 4.5 ⎠ ⎝ E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5 a) P( X ≥ 2) = P( X ≥ 1.5) ≅ P( Z ≤ 1.54−55 ) = P( Z ≥ −1.65) = 0.951 . b) P( X ≥ 2) ≅ P( Z ≥ c) P ( X ≥ 2) = 1 − P ( X < 2) = 1 − 2−5 4.5 ) = P( Z ≥ −1.414) = 0.921 ( )0.1 0.9 − ( )0.1 0.9 50 0 0 50 50 1 The probability computed using the continuity correction is closer. d) 1 49 = 0.966 P ( X > 6) = P ( X ≥ 7) = P ( X ≥ 6.5) ≅ P ( Z ≥ 0.707 ) = 0.24 4-42 S4-3. E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5 a) P(2 ≤ X ≤ 5) = P(1.5 ≤ X ≤ 5.5) ≅ P( 1.54−55 ≤ Z ≤ . 5.5− 5 4.5 ) = P(−1.65 ≤ Z ≤ 0.236) = P ( Z ≤ 0.24) − P( Z ≤ −1.65) = 0.5948 − (1 − 0.95053) = 0.5453 b) − P (2 ≤ X ≤ 5) ≅ P ( 24.5 ≤ Z ≤ 5 5−5 4.5 ) = P (−1.414 ≤ Z ≤ 0) = 0.5 − P ( Z ≤ −1.414) = 0.5 − (1 − 0.921) = 0.421 The exact probability is 0.582 S4-4. E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5 a) 5 P( X = 10) = P(9.5 ≤ X ≤ 10.5) ≅ P( 9.54−55 ≤ Z ≤ 10.4.−5 ) . 5 = P(2.121 ≤ Z ≤ 2.593) = 0.012 b) P ( X = 5) = P(4.5 ≤ X ≤ 5.5) ≅ P( 4.5−55 ≤ Z ≤ 4. 5.5 −5 4.5 ) = P(−0.24 ≤ Z ≤ 0.24) = 0.1897 S4-5 Let X be the number of chips in the lot that are defective. Then E(X)=1000(0.02)=20 and V(X)=1000(0.02)(0.98)=19.6 a) P(20 X ≤ 30)=P(19.5 ≤ X ≤ 30.5) = ⎛ 19.5 − 20 30.5 − 20 ⎞ P⎜ ≤Z≤ ⎟ = P (−.11 ≤ Z ≤ 2.37) = 0.9911 − 0.4562 = 0.5349 ⎜ ⎟ 19.6 ⎠ ⎝ 19.6 b) P(X=20)=P(19.5 ≤ X ≤ 20.5) =P(-0.11 ≤ Z ≤ 0.11)=0.5438-0.4562=0.0876. c) The answer should be close to the mean. Substituting values close to the mean, we find x=20 gives the maximum probability. 4-43 CHAPTER 5 Section 5-1 5-1. First, f(x,y) ≥ 0. Let R denote the range of (X,Y). f ( x, y ) = 1 + 1 + 1 + 1 + 1 = 1 Then, 4 8 4 4 8 R 5-2 a) P(X < 2.5, Y < 3) = f(1.5,2)+f(1,1) = 1/8+1/4=3/8 b) P(X < 2.5) = f (1.5, 2) + f (1.5, 3) +f(1,1)= 1/8 + 1/4+1/4 = 5/8 c) P(Y < 3) = f (1.5, 2)+f(1,1) = 1/8+1/4=3/8 d) P(X > 1.8, Y > 4.7) = f ( 3, 5) = 1/8 5-3. E(X) = 1(1/4)+ 1.5(3/8) + 2.5(1/4) + 3(1/8) = 1.8125 E(Y) = 1(1/4)+2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875 5-4 a) marginal distribution of X x 1 1.5 2.5 3 b) fY 1.5 ( y ) = f(x) 1/4 3/8 1/4 1/8 f XY (1.5, y ) and f X (1.5) = 3/8. Then, f X (1.5) y 2 3 c) f X 2 ( x ) = fY 1.5 ( y ) (1/8)/(3/8)=1/3 (1/4)/(3/8)=2/3 f XY ( x,2) and fY (2) = 1/8. Then, f Y ( 2) x f X 2 ( y) 1.5 (1/8)/(1/8)=1 d) E(Y|X=1.5) = 2(1/3)+3(2/3) =2 1/3 e) Since fY|1.5(y)≠fY(y), X and Y are not independent 5-5 Let R denote the range of (X,Y). Because f ( x, y ) = c(2 + 3 + 4 + 3 + 4 + 5 + 4 + 5 + 6) = 1 , 36c = 1, and c = 1/36 R 5-6. 1 a) P ( X = 1, Y < 4) = f XY (1,1) + f XY (1,2) + f XY (1,3) = 36 ( 2 + 3 + 4) = 1 / 4 b) P(X = 1) is the same as part a. = 1/4 1 c) P (Y = 2) = f XY (1,2) + f XY ( 2,2) + f XY (3,2) = 36 (3 + 4 + 5) = 1 / 3 d) P ( X < 2, Y < 2) = f XY (1,1) = 1 36 ( 2) = 1 / 18 5-1 5-7. E ( X ) = 1[ f XY (1,1) + f XY (1,2) + f XY (1,3)] + 2[ f XY (2,1) + f XY (2,2) + f XY (2,3)] + 3[ f XY (3,1) + f XY (3,2) + f XY (3,3)] 9 = (1 × 36 ) + (2 × 12 ) + (3 × 15 ) = 13 / 6 = 2.167 36 36 V ( X ) = (1 − 13 ) 2 6 E (Y ) = 2.167 + (2 − 13 ) 2 6 9 36 12 36 + (3 − 13 ) 2 6 15 36 = 0.639 V (Y ) = 0.639 5-8 a) marginal distribution of X x f X ( x) = f XY ( x,1) + f XY ( x,2) + f XY ( x,3) 1 1/4 2 1/3 3 5/12 b) f Y X ( y ) = f XY (1, y ) f X (1) y f Y X ( y) 1 2 3 (2/36)/(1/4)=2/9 (3/36)/(1/4)=1/3 (4/36)/(1/4)=4/9 c) f X Y ( x ) = f XY ( x,2) and f Y ( 2) = f XY (1,2) + f XY ( 2,2) + f XY (3,2) = f Y ( 2) x f X Y ( x) 1 2 3 (3/36)/(1/3)=1/4 (4/36)/(1/3)=1/3 (5/36)/(1/3)=5/12 d) E(Y|X=1) = 1(2/9)+2(1/3)+3(4/9) = 20/9 e) Since fXY(x,y) ≠fX(x)fY(y), X and Y are not independent. 5-9. f ( x, y ) ≥ 0 and f ( x, y ) = 1 R 5-10. a) P ( X < 0.5, Y < 1.5) = f XY ( −1,−2) + f XY ( −0.5,−1) = 1 + 8 b) P ( X < 0.5) = f XY ( −1,−2) + f XY ( −0.5,−1) = 3 8 c) P (Y < 1.5) = f XY ( −1,−2) + f XY ( −0.5,−1) + f XY (0.5,1) = d) P ( X > 0.25, Y < 4.5) = f XY (0.5,1) + f XY (1,2) = 5-2 1 4 5 8 7 8 = 3 8 12 36 = 1/ 3 5-11. E ( X ) = −1( 1 ) − 0.5( 1 ) + 0.5( 1 ) + 1( 1 ) = 8 4 2 8 E (Y ) = −2( 1 ) − 1( 1 ) + 1( 1 ) + 2( 1 ) = 8 4 2 8 5-12 1 8 1 4 a) marginal distribution of X x f X ( x) -1 -0.5 0.5 1 1/8 ¼ ½ 1/8 b) f Y X ( y ) = f XY (1, y ) f X (1) y f Y X ( y) 2 1/8/(1/8)=1 c) f X Y ( x ) = f XY ( x,1) f Y (1) x f X Y ( x) 0.5 ½/(1/2)=1 d) E(Y|X=1) = 0.5 e) no, X and Y are not independent 5-13. Because X and Y denote the number of printers in each category, X ≥ 0, Y ≥ 0 and X + Y = 4 5-14. a) The range of (X,Y) is 3 y2 1 0 1 3 2 x Let H = 3, M = 2, and L = 1 denote the events that a bit has high, moderate, and low distortion, respectively. Then, 5-3 x,y 0,0 0,1 0,2 0,3 1,0 1,1 1,2 2,0 2,1 3,0 fxy(x,y) 0.85738 0.1083 0.00456 0.000064 0.027075 0.00228 0.000048 0.000285 0.000012 0.000001 b) x 0 1 2 3 c) f Y 1 ( y ) = fx(x) 0.970299 0.029835 0.000297 0.000001 f XY (1, y ) , fx(1) = 0.29835 f X (1) y 0 1 2 fY|1(x) 0.09075 0.00764 0.000161 E(X)=0(0.970299)+1(0.029403)+2(0.000297)+3*(0.000001)=0.03 (or np=3*.01). f (1, y ) d) fY 1 ( y ) = XY , fx(1) = 0.029403 f X (1) y 0 1 2 fY|1(x) 0.920824 0.077543 0.001632 e) E(Y|X=1)=0(.920824)+1(0.077543)+2(0.001632)=0.080807 f) No, X and Y are not independent since, for example, fY(0)≠fY|1(0). 5-4 5-15 a) The range of (X,Y) is X ≥ 0, Y ≥ 0 and X + Y ≤ 4 . X is the number of pages with moderate graphic content and Y is the number of pages with high graphic output out of 4. x=0 x=1 x=2 x=3 x=4 -05 5.35x10 0 0 0 0 0.00184 0.00092 0 0 0 0.02031 0.02066 0.00499 0 0 0.08727 0.13542 0.06656 0.01035 0 0.12436 0.26181 0.19635 0.06212 0.00699 y=4 y=3 y=2 y=1 y=0 b.) x=0 f(x) x=1 0.2338 x=2 0.4188 x=3 0.2679 x=4 0.0725 0.0070 c.) E(X)= 4 xi f ( xi ) = 0(0.2338) + 1(0.4188) + 2(0.2679) + 3(0.7248) = 4(0.0070) = 1.2 0 d.) f Y 3 ( y ) = f XY (3, y ) , fx(3) = 0.0725 f X (3) y 0 1 2 3 4 fY|3(y) 0.857 0.143 0 0 0 e) E(Y|X=3) = 0(0.857)+1(0.143) = 0.143 f) V(Y|X=3) = 02(0.857)+12(0.143)- 0.1432= 0.123 g) no, X and Y are not independent 5-5 5-16 a) The range of (X,Y) is X ≥ 0, Y ≥ 0 and X + Y ≤ 4 . X is the number of defective items found with inspection device 1 and Y is the number of defective items found with inspection device 2. x=0 x=1 x=2 x=3 x=4 -19 -16 -14 -12 -11 1.94x10 1.10x10 2.35x10 2.22x10 7.88x10 -16 -13 -11 -9 -7 2.59x10 1.47x10 3.12x10 2.95x10 1.05x10 -13 -11 -8 -6 -5 1.29x10 7.31x10 1.56x10 1.47x10 5.22x10 -11 -8 -6 -4 2.86x10 1.62x10 3.45x10 3.26x10 0.0116 2.37x10-9 1.35x10-6 2.86x10-4 0.0271 0.961 y=0 y=1 y=2 y=3 y=4 f (x, y) = 4 ( 0 . 993 ) x ( 0 . 007 ) 4 − x x 4 ( 0 . 997 ) y ( 0 . 003 ) 4 − y y For x=1,2,3,4 and y=1,2,3,4 b.) x=0 f (x, y) = f(x) x=1 x=2 4 ( 0 . 993 ) x ( 0 . 007 ) 4 − x x 2.40 x 10-9 1.36 x 10-6 x=3 for 2.899 x 10-4 c.)since x is binomial E(X)= n(p) = 4*(0.993)=3.972 d.) f Y |2 ( y ) = f XY (2, y ) = f ( y ) , fx(2) = 0.0725 f X ( 2) y 0 1 2 3 4 fY|1(y)=f(y ) 8.1 x 10-11 1.08 x 10-7 5.37 x 10-5 0.0119 0.988 e) E(Y|X=2) = E(Y)= n(p)= 4(0.997)=3.988 f) V(Y|X=2) = V(Y)=n(p)(1-p)=4(0.997)(0.003)=0.0120 g) yes, X and Y are independent. 5-6 x=4 x = 1,2,3,4 0.0274 0.972 Section 5-2 5-17. a) P( X = 2) = f XYZ (2,1,1) + f XYZ (2,1,2) + f XYZ (2,2,1) + f XYZ (2,2,2) = 0.5 b) P( X = 1, Y = 2) = f XYZ (1,2,1) + f XYZ (1,2,2) = 0.35 c) c) P( Z < 1.5) = f XYZ (1,1,1) + f XYZ (1,2,1) + f XYZ (2,1,1) + f XYZ (2,2,1) = 0.5 d) P ( X = 1 or Z = 2) = P ( X = 1) + P ( Z = 2) − P ( X = 1, Z = 2) = 0.5 + 0.5 − 0.3 = 0.7 e) E(X) = 1(0.5) + 2(0.5) = 1.5 5-18 a) b) c) 5-19. P ( X = 1, Y = 1) 0.05 + 0.10 = = 0 .3 P(Y = 1) 0.15 + 0.2 + 0.1 + 0.05 P ( X = 1, Y = 1, Z = 2) 0 .1 P ( X = 1, Y = 1 | Z = 2) == = = 0 .2 P ( Z = 2) 0.1 + 0.2 + 0.15 + 0.05 P( X = 1, Y = 1, Z = 2) 0.10 = = 0 .4 P( X = 1 | Y = 1, Z = 2) = 0.10 + 0.15 P(Y = 1, Z = 2) P ( X = 1 | Y = 1) = f X YZ ( x) = f XYZ ( x,1,2) and f YZ (1,2) = f XYZ (1,1,2) + f XYZ (2,1,2) = 0.25 f YZ (1,2) x 1 2 5-20 f X YZ ( x) 0.10/0.25=0.4 0.15/0.25=0.6 a.) percentage of slabs classified as high = p1 = 0.05 percentage of slabs classified as medium = p2 = 0.85 percentage of slabs classified as low = p3 = 0.10 b.) X is the number of voids independently classified as high X ≥ 0 Y is the number of voids independently classified as medium Y ≥ 0 Z is the number of with a low number of voids and Z ≥ 0 And X+Y+Z = 20 c.) p1 is the percentage of slabs classified as high. d) E(X)=np1 = 20(0.05) = 1 V(X)=np1 (1-p1)= 20(0.05)(0.95) = 0.95 5-7 5-21. a) P ( X = 1, Y = 17, Z = 3) = 0 Because the point (1,17,3) ≠ 20 is not in the range of (X,Y,Z). b) P ( X ≤ 1, Y = 17, Z = 3) = P ( X = 0, Y = 17, Z = 3) + P ( X = 1, Y = 17, Z = 3) = 20! 0.05 0 0.8517 0.10 3 + 0 = 0.07195 0!17!3! Because the point (1,17,3) ≠ 20 is not in the range of (X,Y,Z). () 20 c) Because X is binomial, P ( X ≤ 1) = 0 0.05 0 0.95 20 + d.) Because X is binomial E(X) = np = 20(0.05) = 1 5-22 ( )0.05 0.95 20 1 1 19 = 0.7358 a) The probability is 0 since x+y+z>20 P ( X = 2, Z = 3 | Y = 17) = P ( X = 2, Z = 3, Y = 17) . P (Y = 17) Because Y is binomial, P (Y = 17) = ( )0.85 20 17 17 0.15 3 = 0.2428 . 20! 0.05 2 0.8517 0.10 3 =0 2!3!17! 0.2054 P ( X = 2, Y = 17) b) P ( X = 2 | Y = 17) = . Now, because x+y+z = 20, P (Y = 17) 20! P(X=2, Y=17) = P(X=2, Y=17, Z=1) = 0.05 2 0.8517 0.10 1 = 0.0540 2!17!1! P( X = 2, Y = 17) 0.0540 P( X = 2 | Y = 17) = = = 0.2224 P(Y = 17) 0.2428 Then, P ( X = 2, Z = 3, Y = 17) = c) E ( X | Y = 17) = 0 P ( X = 0, Y = 17) P ( X = 1, Y = 17) +1 P (Y = 17) P (Y = 17) +2 E ( X | Y = 17) = 0 P ( X = 2, Y = 17) P ( X = 3, Y = 17) +3 P(Y = 17) P (Y = 17) 0.07195 0.1079 0.05396 0.008994 +1 +2 +3 0.2428 0.2428 0.2428 0.2428 =1 5-23. a) The range consists of nonnegative integers with x+y+z = 4. b) Because the samples are selected without replacement, the trials are not independent and the joint distribution is not multinomial. 5-8 5-24 f XY ( x , 2 ) fY (2) P ( X = x | Y = 2) = ( )( )( ) + ( )( )( ) + ( )( )( ) = 0.1098 + 0.1758 + 0.0440 = 0.3296 P (Y = 2) = () () () ( )( )( ) = 0.1098 P ( X = 0 and Y = 2) = () ( )( )( ) = 0.1758 P ( X = 1 and Y = 2) = () ( )( )( ) = 0.0440 P ( X = 2 and Y = 2) = () 4 0 5 2 15 4 6 2 4 1 4 0 4 1 4 1 5 2 15 4 5 2 15 4 5 2 15 4 5-25. 6 0 6 1 6 1 f XY ( x,2) 0 1 2 5 2 15 4 6 2 5 2 15 4 x 4 2 6 1 0.1098/0.3296=0.3331 0.1758/0.3296=0.5334 0.0440/0.3296=0.1335 P(X=x, Y=y, Z=z) is the number of subsets of size 4 that contain x printers with graphics enhancements, y printers with extra memory, and z printers with both features divided by the number of subsets of size 4. From the results on the CD material on counting techniques, it can be shown that ( )( )( ) P ( X = x, Y = y , Z = z ) = for x+y+z = 4. () ( )( )( ) = 0.1758 a) P ( X = 1, Y = 2, Z = 1) = () ( )( )( ) = 0.2198 b) P ( X = 1, Y = 1) = P ( X = 1, Y = 1, Z = 2) = () 4 x 5 y 6 z 15 4 4 1 5 2 15 4 6 1 4 1 5 1 15 4 6 2 c) The marginal distribution of X is hypergeometric with N = 15, n = 4, K = 4. Therefore, E(X) = nK/N = 16/15 and V(X) = 4(4/15)(11/15)[11/14] = 0.6146. 5-9 5-26 a) P ( X = 1, Y = 2 | Z = 1) = P ( X = 1, Y = 2, Z = 1) / P ( Z = 1) = [( )(( )() )] [( ( )( ) )] = 0.4762 4 1 5 2 15 4 6 1 6 1 9 3 15 4 b) P ( X = 2 | Y = 2 ) = P ( X = 2 , Y = 2 ) / P (Y = 2 ) [ 45 6 2 = ( 2 )(15 )( 0 ) (4 ) [( ()( ) )] = 0 .1334 5 2 10 2 15 4 c) Because X+Y+Z = 4, if Y = 0 and Z = 3, then X = 1. Because X must equal 1, f X YZ (1) = 1 . 5-27. a)The probability distribution is multinomial because the result of each trial (a dropped oven) results in either a major, minor or no defect with probability 0.6, 0.3 and 0.1 respectively. Also, the trials are independent b.) Let X, Y, and Z denote the number of ovens in the sample of four with major, minor, and no defects, respectively. 4! 0.6 2 0.3 2 0.10 = 0.1944 2!2!0! 4! P ( X = 0, Y = 0, Z = 4) = 0.6 0 0.3 0 0.14 = 0.0001 0!0!4! P ( X = 2, Y = 2, Z = 0) = c. ) 5-28 a.) fXY ( x, y) = fXYZ ( x, y, z) where R is the set of values for z such that x+y+z = 4. That is, R R consists of the single value z = 4-x-y and f XY ( x, y ) = 4! 0.6 x 0.3 y 0.14 − x − y x! y!(4 − x − y )! for x + y ≤ 4. b.) E ( X ) = np1 = 4(0.6) = 2.4 c.) E (Y ) = np 2 = 4(0.3) = 1.2 5-29 a.) P ( X = 2 | Y = 2) = P (Y = 2) = 4 2 P ( X = 2, Y = 2) 0.1944 = = 0.7347 P (Y = 2) 0.2646 0.3 2 0.7 4 = 0.2646 from the binomial marginal distribution of Y b) Not possible, x+y+z=4, the probability is zero. 5-10 c.) P ( X | Y = 2) = P ( X = 0 | Y = 2), P ( X = 1 | Y = 2), P ( X = 2 | Y = 2) P ( X = 0, Y = 2) = P (Y = 2) P ( X = 1, Y = 2) P ( X = 1 | Y = 2) = = P (Y = 2) P ( X = 2, Y = 2) P ( X = 2 | Y = 2) = = P (Y = 2) P ( X = 0 | Y = 2) = 4! 0.6 0 0.3 2 0.12 0.2646 = 0.0204 0!2!2! 4! 0.610.3 2 0.11 0.2646 = 0.2449 1!2!1! 4! 0.6 2 0.3 2 0.10 0.2646 = 0.7347 2!2!0! d.) E(X|Y=2)=0(0.0204)+1(0.2449)+2(0.7347) = 1.7143 5-30 Let X,Y, and Z denote the number of bits with high, moderate, and low distortion. Then, the joint distribution of X, Y, and Z is multinomial with n =3 and p1 = 0.01, p2 = 0.04, and p3 = 0.95 . a) P ( X = 2, Y = 1) = P ( X = 2, Y = 1, Z = 0) 3! 0.0120.0410.950 = 1.2 × 10 − 5 2!1!0! 3! b) P ( X = 0, Y = 0, Z = 3) = 0.0100.0400.953 = 0.8574 0!0!3! = 5-31 a., X has a binomial distribution with n = 3 and p = 0.01. Then, E(X) = 3(0.01) = 0.03 and V(X) = 3(0.01)(0.99) = 0.0297. b. first find P ( X | Y = 2) P (Y = 2) = P ( X = 1, Y = 2, Z = 0) + P ( X = 0, Y = 2, Z = 1) 3! 3! 0.01(0.04) 2 0.95 0 + 0.010 (0.04) 2 0.951 = 0.0046 1!2!0! 0!2!1! 3! P ( X = 0, Y = 2) P ( X = 0 | Y = 2) = = 0.010 0.04 2 0.951 0.004608 = 0.98958 0!2!1! P (Y = 2) = P ( X = 1 | Y = 2) = P ( X = 1, Y = 2) 3! = 0.0110.04 2 0.95 0 P (Y = 2) 1!2!1! 0.004608 = 0.01042 E ( X | Y = 2) = 0(0.98958) + 1(0.01042) = 0.01042 V ( X | Y = 2) = E ( X 2 ) − ( E ( X )) 2 = 0.01042 − (0.01042) 2 = 0.01031 5-11 5-32 a.) Let X,Y, and Z denote the risk of new competitors as no risk, moderate risk, and very high risk. Then, the joint distribution of X, Y, and Z is multinomial with n =12 and p1 = 0.13, p 2 = 0.72, and p 3 = 0.15 . X, Y and Z ≥ 0 and x+y+z=12 b.) P ( X = 1, Y = 3, Z = 1) = 0, not possible since x+y+z≠12 c) P ( Z ≤ 2) = 12 0 0.15 0 0.8512 + 12 1 0.151 0.8511 + 12 2 0.15 2 0.8510 = 0.1422 + 0.3012 + 0.2924 = 0.7358 5-33 a.) P ( Z = 2 | Y = 1, X = 10) = 0 b.) first get P( X = 10) = P( X = 10, Y = 2, Z = 0) + P( X = 10, Y = 1, Z = 1) + P( X = 10, Y = 0, Z = 2) 12! 12! 12! = 0.1310 0.72 2 0.15 0 + 0.1310 0.7210.151 + 0.1310 0.72 0 0.15 2 10!2!0! 10!1!1! 10!0!2! −8 −8 −9 −8 = 4.72 x10 + 1.97 x10 + 2.04 x10 = 6.89 x10 P ( Z = 0, Y = 2, X = 10) P ( Z = 1, Y = 1, X = 10) P ( Z ≤ 1 | X = 10) = + P ( X = 10) P ( X = 10) 12! 12! = 0.1310 0.72 2 0.15 0 6.89 x10 −8 + 0.1310 0.721 0.151 6.89 x10 −8 10!2!0! 10!1!1! = 0.9698 c.) P(Y ≤ 1, Z ≤ 1 | X = 10) = P ( Z = 1, Y = 1, X = 10) P( X = 10) 12! 0.13100.7210.151 6.89 x10 −8 10!1!1! = 0.2852 = d. E ( Z | X = 10) = (0(4.72 x10 −8 ) + 1(1.97 x10 −8 ) + 2(2.04 x10 −9 )) / 6.89 x10 −8 = 2.378 x10 −8 5-12 Section 5-3 33 5-34 3 3 2 xydxdy = c y x2 Determine c such that c 00 dy = c(4.5 0 0 y2 2 3 )= 0 81 4 c. Therefore, c = 4/81. 32 5-35. a) P ( X < 2, Y < 3) = 3 4 81 4 81 xydxdy = 4 (2) ydy = 81 (2)( 9 ) = 0.4444 2 00 0 b) P(X < 2.5) = P(X < 2.5, Y < 3) because the range of Y is from 0 to 3. 3 2.5 P( X < 2.5, Y < 3) = 3 xydxdy = 4 81 4 81 4 (3.125) ydy = 81 (3.125) 9 = 0.6944 2 00 0 2.5 3 c) P (1 < Y < 2.5) = 2.5 xydxdy = 4 81 4 81 (4.5) ydy = 18 81 10 5-35 2.5 =0.5833 1 1 d) 2.5 3 P( X > 1.8,1 < Y < 2.5) = 2.5 xydxdy = 4 81 4 81 (2.88) ydy = 1 1.8 33 e) E ( X ) = x 2 ydxdy = 4 81 4 81 9 ydy = 4 9 0 40 f) P ( X < 0, Y < 4) = y2 2 3 =2 0 4 0 3 3 4 4 f XY ( x, y )dy = x 81 ydy = 81 x(4.5) = 0 b) fY 1.5 ( y ) = for 0 < x < 3 . 0 f XY (1.5, y ) = f X (1.5) 3 c) E(Y|X=1.5) = 0 4 81 2 9 y (1.5) (1.5) 2 = 9 y for 0 < y < 3. 3 2 22 2y3 y y dy = y dy = 9 90 27 2 0 f XY ( x,2) = f Y ( 2) 4 81 2 9 x ( 2) ( 2) 3 =6 0 2 1 ydy = y 2 9 9 d.) P (Y < 2 | X = 1.5) = f Y |1.5 ( y ) = e) f X 2 ( x ) = 2x 9 2 =9x 2 = 0 for 0 < x < 3. 5-13 2 (2.88) ( 2.52 −1) =0.3733 xydxdy = 0 ydy = 0 4 81 00 a) f X ( x) = 4 81 1 3 00 5-36 y2 2 4 4 −0 = 9 9 5-37. 3 x+2 3 ( x + y )dydx = xy + c 0 = dx x 0 x 3 x+2 y2 2 [x( x + 2) + ( x+ 2)2 2 x2 2 − x2 − dx 0 3 [ = c (4 x + 2 )dx = 2 x 2 + 2 x 3 0 = 24c 0 Therefore, c = 1/24. 5-38 a) P(X < 1, Y < 2) equals the integral of f XY ( x, y ) over the following region. y 2 0 x 12 0 Then, P( X < 1, Y < 2) = 1 12 11 ( x + y )dydx = xy + 24 0 x 24 0 12 x + 2x − 24 x3 2 y2 2 2 dx = x 2 13 2 x + 2 − 3 x dx = 2 24 0 1 = 0.10417 0 b) P(1 < X < 2) equals the integral of f XY ( x, y ) over the following region. y 2 0 1 x 2 0 1 P (1 < X < 2) = 24 2 x+2 1 2 1 ( x + y )dydx = xy + 24 1 x 3 = y2 2 x+2 dx x 2 1 1 (4 x + 2)dx = 2x 2 + 2x = 1 . 6 24 0 24 1 5-14 c) P(Y > 1) is the integral of f XY ( x , y ) over the following region. 1 11 P (Y > 1) = 1 − P (Y ≤ 1) = 1 − 1 y2 1 ( xy + ) ( x + y )dydx = 1 − 24 0 2x x 1 24 0 1 1 132 1 x2 1 1 3 = 1− x + − x dx = 1 − +−x 24 0 22 24 2 2 2 = 1 − 0.02083 = 0.9792 d) P(X < 2, Y < 2) is the integral of fXY ( x, y) over the following region. y 2 0 x 2 0 3 x+2 1 E( X ) = 24 0 3 1 x( x + y )dydx = x2 y + 24 0 x 3 = 1 1 4x3 (4 x 2 + 2 x)dx = + x2 24 0 24 3 xy 2 2 x+2 dx x 3 = 0 15 8 e) 3 x+2 E( X ) = 1 24 0 3 x( x + y )dydx = x 1 x2 y + 24 0 3 = 1 13 (3 x 2 + 2 x)dx = x + x2 24 0 24 5-15 xy 2 2 3 = 0 x+2 dx x 15 8 1 0 5-39. a) f X ( x) is the integral of f XY ( x, y ) over the interval from x to x+2. That is, 1 24 f X ( x) = b) f Y 1 ( y ) = x+2 1 xy + 24 ( x + y )dy = x f XY (1, y ) f X (1) = 1 (1+ y ) 24 11 + 6 12 = 1+ y 6 y2 2 x+2 = x x1 + for 0 < x < 3. 6 12 for 1 < y < 3. See the following graph, y f 2 Y|1 (y) defined over this line segment 1 0 x 12 0 3 c) E(Y|X=1) = 1 3 1+ y 1 1 y2 y3 y dy = ( y + y 2 )dy = + 6 61 62 3 3 d.) P (Y > 2 | X = 1) = 2 e.) f X 2 ( x) = f XY ( x , 2 ) . fY ( 2) 3 = 2.111 1 3 1+ y y2 1 1 dy = (1 + y )dy = y + 6 62 6 2 3 =0.5833 2 Here f Y ( y ) is determined by integrating over x. There are three regions of integration. For 0 < y ≤ 2 the integration is from 0 to y. For 2 < y ≤ 3 the integration is from y-2 to y. For 3 < y < 5 the integration is from y to 3. Because the condition is y=2, only the first integration is needed. y fY ( y) = 1 1 ( x + y )dx = 24 0 24 x2 2 y + xy = y2 16 for 0 < y ≤ 2 . 0 y f X|2 (x) defined over this line segment 2 1 0 12 x 0 1 ( x + 2) x+2 24 Therefore, fY (2) = 1 / 4 and f X 2 ( x) = = for 0 < x < 2 1/ 4 6 5-16 3x 5-40 c 00 x 3 3 y2 x3 x 4 81 xydyd x = c x dx = c dx = c. Therefore, c = 8/81 20 2 8 8 0 0 1x 5-41. 1 8 8 x3 81 1 a) P(X<1,Y<2)= xydyd x = dx = =. 81 0 0 81 0 2 81 8 81 2x 2 x2 8 8 8 x4 xydyd x = x dx = b) P(1<X<2) = 81 1 0 81 1 2 81 8 2 1 8 (2 4 − 1) 5 = = . 81 8 27 c) 3x 3 3 3 8 8 x2 −1 8x x 8 x4 x2 P (Y > 1) = xydyd x = x dx = − dx = − 81 1 1 81 1 2 81 1 2 2 81 8 4 = 8 81 4 2 4 2 3 3 11 − − − 8 4 84 2x = 1 = 0.01235 81 2 8 8 x3 8 24 16 d) P(X<2,Y<2) = xydyd x = dx = =. 81 0 0 81 0 2 81 8 81 e.) 3x 3x 3 3x 3 3 8 8 8 x2 2 8 x4 2 E( X ) = x( xy )dyd x = x ydyd x = x dx = dx 81 0 0 81 0 0 81 0 2 81 0 2 = 8 35 12 = 81 10 5 f) 3x 8 8 8 x3 E (Y ) = y ( xy )dyd x = xy 2 dyd x = x dx 81 0 0 81 0 0 81 0 3 3 = 8 x4 8 35 8 dx = = 81 0 3 81 15 5 5-17 3 1 x 5-42 a.) f ( x) = 8 4x 3 xydy = 81 0 81 0< x<3 8 (1) y f (1, y ) 81 b.) f Y | x =1 ( y ) = = = 2y f (1) 4(1) 3 81 1 c.) E (Y | X = 1) = 2 ydy = y 2 1 0 0 < y <1 =1 0 d.) P (Y > 2 | X = 1) = 0 this isn’t possible since the values of y are 0< y < x. e.) f ( y ) = 3 4y 8 xydx = , therefore 81 0 9 8 x ( 2) f ( x,2) 81 2x f X |Y = 2 ( x) = = = 4( 2) f ( 2) 9 9 5-43. 0< x<2 Solve for c ∞x ∞ e − 2 x − 3 y dyd x = c 00 ∞ c −2x c −2 x e 1 − e −3 x d x = e − e −5 x d x = 30 30 ( ) c11 1 − = c. c = 10 32 5 10 5-44 a) 1x P( X < 1, Y < 2) = 10 1 e − 2 x −3 y 00 1 10 e − 5 x e − 2 x = − 3 5 2 2x P(1 < X < 2) = 10 e 10 b) = 0.77893 0 2 − 2 x −3 y 10 − 2 x dyd x = e − e −5 x d x 31 2 10 e − 5 x e − 2 x = − 3 5 2 = 0.19057 1 c) ∞x 1∞ e − 2 x −3 y dyd x = P(Y > 3) = 10 1 10 − 2 x 10 − 2 x dyd x = e (1 − e − 3 x )dy = e − e − 5 x dy 30 30 33 10 e −5 x e −9 e − 2 x = − 3 5 2 10 − 2 x −9 e (e − e −3 x )dy 33 ∞ = 3.059 x10 −7 3 5-18 d) 2x P ( X < 2, Y < 2) = 10 2 e − 2 x −3 y 00 10 − 2 x 10 e −10 e − 4 dyd x = e (1 − e −3 x )dx = − 30 35 2 = 0.9695 ∞x xe − 2 x − 3 y dyd x = 7 10 ye − 2 x − 3 y dyd x = e) E(X) = 10 1 5 00 ∞x f) E(Y) = 10 00 x 5-45. a) f ( x ) = 10 e − 2 x −3 y 0 b) f Y \ X =1 ( y ) = 10e −2 z 10 dy = (1 − e − 3 x ) = (e − 2 x − e − 5 x ) for 0 < x 3 3 f X ,Y (1, y ) f X (1) = 10e −2 − 3 y 10 − 2 (e − e − 5 ) 3 = 3.157e − 3 y 0 < y < 1 1 c) E(Y|X=1 )=3.157 ye -3 y dy=0.2809 0 d) f X |Y = 2 ( x) = f X ,Y (x,2 ) fY ( 2 ) = 10e −2 x − 6 = 2e − 2 x + 4 for 2 < x, −10 5e where f(y) = 5e-5y for 0 < y ∞∞ 5-46 ∞ e − 2 x e −3 y dydx = c 0x 5-47. ∞ c − 2 x −3 x c 1 e (e )dx = e −5 x dx = c 30 30 15 c = 15 a) 12 1 e − 2 x − 3 y dyd x = 5 e − 2 x (e −3 x − e − 6 )d x P ( X < 1, Y < 2) = 15 0x 0 1 1 5 = 5 e − 5 x dx − 5e − 6 e − 2 x dx = 1 − e −5 + e − 6 (e − 2 − 1) = 0.9879 2 0 0 2∞ 2 e − 2 x − 3 y dyd x = 5 e −5 x dyd x =(e −5 − e 10 ) = 0.0067 b) P(1 < X < 2) = 15 1x 1 c) 3∞ P (Y > 3) = 15 ∞∞ e 03 − 2 x −3 y dydx + ∞ 3 e − 2 x −3 y 3x −9 dydx = 5 e e 0 3 5 = − e −15 + e −9 = 0.000308 2 2 5-19 −2 x dx + 5 e −5 x dx 3 2 0 d) 22 2 e − 2 x −3 y dyd x = 5 e − 2 x (e −3 x − e −6 )d x = P ( X < 2, Y < 2) = 15 0x 0 2 2 5 = 5 e −5 x dx − 5e −6 e − 2 x dx = 1 − e −10 + e − 6 e − 4 − 1 = 0.9939 2 0 0 ∞∞ ( ∞ xe − 2 x −3 y dyd x = 5 xe −5 x dx = e) E(X) = 15 0x 0 ) ( 1 = 0 .0 4 52 f) ∞∞ ∞ ye − 2 x −3 y dyd x = E (Y ) = 15 0x =− 358 += 10 6 15 ∞ 5-48 ∞ −3 5 5 ye −5 y dy + 3 ye −3 y dy 20 20 a.) f ( x) = 15 e − 2 x − 3 y dy = x b) f X (1) = 5e −5 f Y | X =1 ( y ) = 15 − 2 z − 3 x (e ) = 5e − 5 x for x > 0 3 f XY (1, y ) = 15e −2 − 3 y 15e −2 − 3 y = 3e 3− 3 y for 1 <y 5e − 5 ∞ 3 ye 3−3 y dy = − y e 3−3 y c) E (Y | X = 1) = ∞ 1 + ∞ 1 e 3 − 3 y dy = 4 / 3 1 2 d) 1 3e 3−3 y dy = 1 − e −3 = 0.9502 for 0 < y , f Y (2) = f X |Y = 2 ( y ) = 15e −2 x − 6 = 2e − 2 x 15 − 6 e 2 5-20 15 − 6 e 2 ) 5-49. The graph of the range of (X, Y) is y 5 4 3 2 1 0 1 2 1 x +1 x 4 3 4 x +1 cdydx + cdydx = 1 1 x −1 00 1 4 = c ( x + 1)dx + 2c dx 0 1 3 2 = c + 6c = 7.5c = 1 Therefore, c = 1/7.5=2/15 0.50.5 5-50 1 7.5 a) P ( X < 0.5, Y < 0.5) = dydx = 1 30 00 0.5x +1 1 7.5 b) P ( X < 0.5) = 0.5 dydx = 1 7.5 00 ( x + 1)dx = 2 15 (5) = 8 1 12 0 c) 1 x +1 4 x +1 x 7.5 E( X ) = dydx + 00 x 7.5 1 x −1 dydx 4 1 = ( x 2 + x)dx + 1 7.5 2 7.5 ( x)dx = 1 0 d) 1 x +1 E (Y ) = 1 7.5 4 x +1 ydydx + 1 7.5 1 = 1 7.5 ydydx 1 x −1 00 ( x +1) 2 2 4 dx + 715 . 0 ( x +1) 2 − ( x −1) 2 2 1 = 4 2 1 15 dx 1 ( x + 2 x + 1)dx + 0 1 1 = 15 ( 7 ) + 15 (30 ) = 3 1 15 4 xdx 1 97 45 5-21 12 15 (5) + 6 2 7.5 (7.5) = 19 9 5-51. a. ) x +1 f ( x) = 0 1 x +1 dy = 7 .5 7 .5 0 < x < 1, for x +1 f ( x) = x + 1 − ( x − 1) 1 2 dy = = for 1 < x < 4 7 .5 7 .5 7 .5 x −1 b. ) f Y | X =1 ( y ) = f XY (1, y ) 1 / 7.5 = = 0 .5 f X (1) 2 / 7 .5 f Y | X =1 ( y ) = 0.5 for 0 < y < 2 2 y y2 c. ) E (Y | X = 1) = dy = 2 4 0 2 =1 0 0.5 0.5 d.) P (Y < 0.5 | X = 1) = 0.5dy = 0.5 y 0 5-52 = 0.25 0 Let X, Y, and Z denote the time until a problem on line 1, 2, and 3, respectively. a) P ( X > 40, Y > 40, Z > 40) = [ P ( X > 40)] 3 because the random variables are independent with the same distribution. Now, ∞ 1 40 P( X > 40) = e− x / 40 dx = −e − x / 40 (e ) = e−1 and the answer is 40 40 −1 3 ∞ = e − 3 = 0.0498 . b) P ( 20 < X < 40,20 < Y < 40,20 < Z < 40) = [ P (20 < X < 40)]3 and P (20 < X < 40) = −e − x / 40 40 = e − 0.5 − e −1 = 0.2387 . 20 3 The answer is 0.2387 = 0.0136. c.) The joint density is not needed because the process is represented by three independent exponential distributions. Therefore, the probabilities may be multiplied. 5-22 5-53 a.) µ=3.2 λ=1/3.2 ∞∞ P ( X > 5, Y > 5 ) = (1 / 10 .24 ) e − x y − 3 .2 3 . 2 ∞ dydx = 3 .2 e 55 =e − 5 3 .2 e − 5 3 .2 e − x y − 3.2 3.2 ∞ dydx = 3.2 e 10 10 10 3.2 e − 10 3.2 e − 5 3 .2 dx = 0 .0439 ∞∞ − x 3 .2 5 P( X > 10, Y > 10) = (1 / 10.24) =e − − x 3.2 e − 10 3.2 dx 10 = 0.0019 b.) Let X denote the number of orders in a 5-minute interval. Then X is a Poisson random variable with λ=5/3.2 = 1.5625. e −1.5625 (1.5625) 2 P ( X = 2) = = 0.256 2! For both systems, P ( X = 2) P (Y = 2) = 0.256 2 = 0.0655 c. ) The joint probability distribution is not necessary because the two processes are independent and we can just multiply the probabilities. a) fY|X=x(y), for x = 2, 4, 6, 8 5 4 f(y2) 5-54 3 2 1 0 0 1 2 3 y 5-23 4 b) P(Y < 2 | X = 2) = c) d) ∞ E (Y | X = 2) = E (Y | X = x) = e) Use fX(x) = 0 ∞ 0 2 0 2e − 2 y dy = 0.9817 2 ye − 2 y dy = 1 / 2 (using integration by parts) xye − xy dy = 1 / x (using integration by parts) 1 1 f ( x, y ) − xy = , fY | X ( x, y ) = xe , and the relationship fY | X ( x, y ) = XY b − a 10 f X ( x) f XY ( x, y ) xe − xy f XY ( x, y ) = and 1 / 10 10 − xy −10 y −10 y 10 xe −e 1 − 10 ye f) fY(y) = dx = (using integration by parts) 2 0 10 10 y Therefore, xe − xy = Section 5-4 0.5 1 1 5-55. a) P ( X < 0.5) = 0.5 0.5 1 (4 xy )dydx = (2 x)dx = x 2 (8 xyz )dzdydx = 000 0.5 = 0.25 0 0 00 b) 0.5 0.5 1 P ( X < 0.5, Y < 0.5) = (8 xyz )dzdydx 0 00 0.5 0.5 = 0.5 (4 xy )dydx = (0.5 x)dx = 00 0 x2 4 0.5 = 0.0625 0 c) P(Z < 2) = 1, because the range of Z is from 0 to 1. d) P(X < 0.5 or Z < 2) = P(X < 0.5) + P(Z < 2) - P(X < 0.5, Z < 2). Now, P(Z < 2) =1 and P(X < 0.5, Z < 2) = P(X < 0.5). Therefore, the answer is 1. 111 1 2 (8 x yz )dzdydx = (2 x 2 )dx = e) E ( X ) = 000 5-56 0 1 2 x3 30 = 2/3 a) P( X < 0.5 Y = 0.5) is the integral of the conditional density f X Y ( x) . Now, f X 0.5 ( x) = f XY ( x,0.5) fY (0.5) 1 f XY ( x,0.5) = (8 xyz )dz = 4 xy for 0 < x < 1 and and 0 11 0 < y < 1. Also, fY ( y ) = (8 xyz )dzdx = 2 y for 0 < y < 1. 00 Therefore, f X 0.5 ( x) = 2x = 2 x for 0 < x < 1. 1 0.5 Then, P ( X < 0.5 Y = 0.5) = 2 xdx = 0.25 . 0 5-24 b) P( X < 0.5, Y < 0.5 Z = 0.8) is the integral of the conditional density of X and Y. Now, f Z ( z ) = 2 z for f XY Z ( x, y ) = 0 < z < 1 as in part a. and f XYZ ( x, y, z ) 8 xy(0.8) = = 4 xy for 0 < x < 1 and 0 < y < 1. fZ ( z) 2(0.8) 0.50.5 Then, P ( X < 0.5, Y < 0.5 Z = 0.8) = 0.5 (4 xy )dydx = ( x / 2)dx = 00 1 16 = 0.0625 0 1 5-57. a) fYZ ( y, z ) = (8 xyz )dx = 4 yz for 0 < y < 1 and 0 < z < 1. 0 Then, f X YZ ( x) = f XYZ ( x, y, z ) 8 x(0.5)(0.8) = = 2 x for 0 < x < 1. fYZ ( y, z ) 4(0.5)(0.8) 0.5 b) Therefore, P ( X < 0.5 Y = 0.5, Z = 0.8) = 2 xdx = 0.25 0 4 5-58 a) 0 x2 + y2 ≤4 cdzdydx = the volume of a cylinder with a base of radius 2 and a height of 4 = (π 2 2 )4 = 16π . Therefore, c = 1 16π b) P ( X 2 + Y 2 ≤ 1) equals the volume of a cylinder of radius =8π) times c. Therefore, the answer is 2 and a height of 4 ( 8π = 1 / 2. 16π c) P(Z < 2) equals half the volume of the region where f XYZ ( x, y, z ) is positive times 1/c. Therefore, the answer is 0.5. 2 4− x2 4 2 x c d) E ( X ) = dzdydx = − 2− 4 − x 2 0 2 4− x 2 1 c − 4− x −2 2 substitution, u = 4 − x 2 , du = -2x dx, and E ( X ) = 5-59. 4− y −2 1 c 4 u du = −4 2 c3 2 3 (4 − x 2 ) 2 = 0. −2 4 f ( x,1) a) f X 1 ( x) = XY and fY (1) Also, f Y ( y ) = (8 x 4 − x 2 )dx . Using 1 c dx = 4 xy f XY ( x, y ) = 1 c dz = 4 = c 1 4π for x 2 + y 2 < 4 . 0 2 4 dzdx = 8 4 − y 2 for -2 < y < 2. c 1 c − 4− y 2 0 Then, f ( x) = Xy 4/c 8 c 4− y 2 evaluated at y = 1. That is, f X 1 ( x) = − 3<x< 3. 1 1 23 Therefore, P ( X < 1 | Y < 1) = −3 5-25 dx = 1+ 3 = 0.7887 23 1 23 for b) 2 f ( x, y,1) f XY 1( x, y) = XYZ and f Z ( z) = f Z (1) −2 4 − x2 − 4 − x2 1 dydx = c 2 −2 4 − x 2 dx 2 c Because f Z ( z ) is a density over the range 0 < z < 4 that does not depend on Z, f Z ( z ) =1/4 for 1/ c 1 = for x 2 + y 2 < 4 . 1 / 4 4π area in x 2 + y 2 < 1 P ( x 2 + y 2 < 1 | Z = 1) = = 1/ 4 . 4π Then, f XY 1 ( x, y ) = 0 < z < 4. Then, 5-60 f XYZ ( x, y, z ) and from part 5-59 a., f XY ( x, y ) = f XY ( x, y ) f Z xy ( z ) = Therefore, f Z 5-61 xy ( z ) = 1 16π 1 4π 1 4π for x 2 + y 2 < 4 . = 1 / 4 for 0 < z < 4. Determine c such that f ( xyz ) = c is a joint density probability over the region x>0, y>0 and z>0 with x+y+z<1 1 1− x 1− x − y f ( xyz ) = c 1 1− x dzdydx = 00 0 1 c(1 − x − y )dydx = 00 0 1− x y2 c( y − xy − ) 20 dx 1 (1 − x) 2 (1 − x ) 1 x2 x3 = c (1 − x) − x(1 − x) − dx = c dx = c x − + 2 2 2 2 6 0 0 2 1 1 =c . 6 Therefore, c = 6. 1 1− x 1− x − y 5-62 a.) P ( X < 0.5, Y < 0.5, Z < 0.5) = 6 dzdydx 00 The conditions x>0.5, y>0.5, 0 z>0.5 and x+y+z<1 make a space that is a cube with a volume of 0.125. Therefore the probability of P ( X < 0.5, Y < 0.5, Z < 0.5) = 6(0.125) = 0.75 b.) 0 . 50 . 5 P ( X < 0 .5, Y < 0 .5) = 0 .5 00 0.5 = 0 (6 y − 6 xy − 3 y ) 2 6 (1 − x − y ) dydx = 0.5 0 dx 0 9 9 3 − 3 x dx = x − x2 4 4 2 0 .5 = 3/ 4 0 c.) 0.51− x 1− x − y P( X < 0.5) = 6 dzdydx = 00 0.5 = 6( 0 2 0.51 1− x 0 0 5-26 1− x y2 6(1 − x − y )dydx = 6( y − xy − ) 20 0 0 x 1 − x + )dx = x 3 − 3 x 2 + 3 x 2 2 ( 0.5 ) 0.5 0 = 0.875 1 0 d. ) 1 1− x 1− x − y E( X ) = 6 1 1− x xdzdydx = 00 0 0 y2 6 x(1 − x − y )dydx = 6x( y − xy − ) 20 0 0 1 x3 x 3x 4 3x 2 = 6( − x 2 + )dx = − 2x 3 + 2 2 4 2 0 5-63 1− x 1 1 = 0 .2 5 0 a.) 1− x 1− x − y 1− x y2 dzdy = 6(1 − x − y )dy = 6 y − xy − 2 0 0 f ( x) = 6 0 1− x 0 2 = 6( x 1 − x + ) = 3( x − 1) 2 for 0 < x < 1 2 2 b.) 1− x − y f ( x, y ) = 6 dz = 6(1 − x − y ) 0 for x > 0 , y > 0 and x + y < 1 c.) f ( x | y = 0.5, z = 0.5) = f ( x, y = 0.5, z = 0,5) 6 = = 1 For, x = 0 f ( y = 0.5, z = 0.5) 6 d. ) The marginal f Y ( y ) is similar to f X ( x) and f Y ( y ) = 3(1 − y ) 2 for 0 < y < 1. f X |Y ( x | 0.5) = 5-64 f ( x,0.5) 6(0.5 − x) = = 4(1 − 2 x) for x < 0.5 3(0.25) f Y (0.5) Let X denote the production yield on a day. Then, P ( X > 1400) = P ( Z > 1400 −1500 10000 ) = P( Z > −1) = 0.84134 . a) Let Y denote the number of days out of five such that the yield exceeds 1400. Then, by independence, Y has a binomial distribution with n = 5 and p = 0.8413. Therefore, the answer is P (Y = 5) = 5 0.84135 (1 − 0.8413) 0 = 0.4215 . 5 b) As in part a., the answer is () P (Y ≥ 4) = P (Y = 4) + P (Y = 5) = ( )0.8413 (1 − 0.8413) 5 4 4 1 5-27 + 0.4215 = 0.8190 5-65. a) Let X denote the weight of a brick. Then, P ( X > 2.75) = P ( Z > 2.75 − 3 0.25 ) = P( Z > −1) = 0.84134 . Let Y denote the number of bricks in the sample of 20 that exceed 2.75 pounds. Then, by independence, Y has a binomial distribution with n = 20 and p = 0.84134. Therefore, the answer is P (Y = 20) = 20 0.84134 20 = 0.032 . 20 b) Let A denote the event that the heaviest brick in the sample exceeds 3.75 pounds. Then, P(A) = 1 - P(A') and A' is the event that all bricks weigh less than 3.75 pounds. As in part a., P(X < 3.75) = P(Z < 3) and P ( A) = 1 − [ P ( Z < 3)] 20 = 1 − 0.99865 20 = 0.0267 . () 5-66 a) Let X denote the grams of luminescent ink. Then, P ( X < 1.14) = P ( Z < 1.14.−1.2 ) = P ( Z < −2) = 0.022750 . 03 Let Y denote the number of bulbs in the sample of 25 that have less than 1.14 grams. Then, by independence, Y has a binomial distribution with n = 25 and p = 0.022750. Therefore, the answer is 25 P (Y ≥ 1) = 1 − P (Y = 0) = 0 0.02275 0 (0.97725) 25 = 1 − 0.5625 = 0.4375 . b) () P (Y ≤ 5) = P (Y = 0) + P (Y = 1) + P (Y = 2) + ( P (Y = 3) + P (Y = 4) + P(Y = 5) ( )0.02275 (0.97725) + ( )0.02275 (0.97725) 25 0 0 25 25 3 = 3 22 ( )0.02275 (0.97725) + ( )0.02275 (0.97725) 25 1 1 24 25 4 + 4 21 ( )0.02275 + ( )0.02275 + 25 2 2 (0.97725) 23 25 5 5 (0.97725) 20 . = 0.5625 + 0.3274 + 0.09146 + 0.01632 + 0.002090 + 0.0002043 = 0.99997 ≅ 1 c.) P (Y = 0) = ( )0.02275 25 0 0 (0.97725) 25 = 0.5625 d.) The lamps are normally and independently distributed, therefore, the probabilities can be multiplied. Section 5-5 5-67. E(X) = 1(3/8)+2(1/2)+4(1/8)=15/8 = 1.875 E(Y) = 3(1/8)+4(1/4)+5(1/2)+6(1/8)=37/8 = 4.625 E(XY) = [1 × 3 × (1/8)] + [1 × 4 × (1/4)] + [2 × 5 × (1/2)] + [4 × 6 × (1/8)] = 75/8 = 9.375 σ XY = E ( XY ) − E ( X ) E (Y ) = 9.375 − (1.875)(4.625) = 0.703125 V(X) = 12(3/8)+ 22(1/2) +42(1/8)-(15/8)2 = 0.8594 V(Y) = 32(1/8)+ 42(1/4)+ 52(1/2) +62(1/8)-(37/8)2 = 0.7344 ρ XY = σ XY 0.703125 = = 0.8851 σ XσY (0.8594)(0.7344) 5-28 5-68 E ( X ) = −1(1 / 8) + (−0.5)(1 / 4) + 0.5(1 / 2) + 1(1 / 8) = 0.125 E (Y ) = −2(1 / 8) + (−1)(1 / 4) + 1(1 / 2) + 2(1 / 8) = 0.25 E(XY) = [-1× −2 × (1/8)] + [-0.5 × -1× (1/4)] + [0.5 × 1× (1/2)] + [1× 2 × (1/8)] = 0.875 V(X) = 0.4219 V(Y) = 1.6875 σ XY = 0.875 − (0.125)(0.25) = 0.8438 ρ XY = σσ σ = 0.8438 XY X Y =1 0.4219 1.6875 5-69. 3 3 c( x + y ) = 36c, c = 1 / 36 x =1 y =1 13 13 E( X ) = E (Y ) = 6 6 16 16 E( X 2 ) = E (Y 2 ) = 3 3 −1 36 ρ= = −0.0435 23 23 36 36 5-70 14 E ( XY ) = 3 σ xy V ( X ) = V (Y ) = 14 13 = − 3 6 = −1 36 23 36 E ( X ) = 0(0.01) + 1(0.99) = 0.99 E (Y ) = 0(0.02) + 1(0.98) = 0.98 E(XY) = [0 × 0 × (0.002)] + [0 × 1× (0.0098)] + [1× 0 × (0.0198)] + [1× 1× (0.9702)] = 0.9702 V(X) = 0.99-0.992=0.0099 V(Y) = 0.98-0.982=0.0196 σ XY = 0.9702 − (0.99)(0.98) = 0 ρ XY = σσ σ = 0 XY X 5-71 2 Y =0 0.0099 0.0196 E(X1) = np1 = 20(1/3)=6.67 E(X2) = np2=20(1/3)=6.67 V(X1) = np1(1-p1)=20(1/3)(2/3)=4.44 V(X2) = np2(1-p2)=20(1/3)(2/3)=4.44 E(X1X2) =n(n-1)p1p2 =20(19)(1/3)(1/3)=42.22 σ XY = 42.22 − 6.67 2 = −2.267 and ρ XY = The sign is negative. 5-29 − 2.267 (4.44)(4.44) = − 0 .5 1 5-72 From Exercise 5-40, c=8/81. From Exercise 5-41, E(X) = 12/5, and E(Y) = 8/5 3x E ( XY ) = 3x 3 8 36 =4 81 18 = σ xy = 4 − 12 5 8 = 0.16 5 E( X 2 ) = 6 E (Y 2 ) = 3 V ( x) = 0.24, V (Y ) = 0.44 0.16 ρ= = 0.4924 0.24 0.44 5-73. Similarly to 5-49, c = 2 / 19 1 x +1 E( X ) = E (Y ) = 3 8 8 8 x3 2 8 x5 xy ( xy )dyd x = x 2 y 2 dyd x = x dx = dx 81 0 0 81 0 0 81 0 3 81 0 3 2 19 0 2 19 5 x +1 xdydx + 2 xdydx = 2.614 19 1 x −1 ydydx + 2 19 0 1 x +1 00 Now, E ( XY ) = 2 19 5 x +1 ydydx = 2.649 1 x −1 1 x +1 xydydx + 00 5 x +1 2 xydydx = 8.7763 19 1 x −1 σ xy = 8.7763 − (2.614)(2.649) = 1.85181 E ( X 2 ) = 8.7632 E (Y 2 ) = 9.11403 V ( x) = 1.930, V (Y ) = 2.0968 1.852 ρ= = 0.9206 1.930 2.062 5-30 5-74 1 x +1 4 x +1 x 7.5 E( X ) = dydx + 00 x 7.5 1 x −1 1 = 4 2 1 7.5 dydx ( x + x)dx + ( x)dx = 12 ( 5 ) + 15 6 2 7.5 0 2 7.5 (7.5) = 19 9 1 2 E(X )=222,222.2 V(X)=222222.2-(333.33)2=111,113.31 E(Y2)=1,055,556 V(Y)=361,117.11 ∞∞ E ( XY ) = 6 × 10 −6 xye −.001x −.002 y dydx = 388,888.9 0x σ xy = 388,888.9 − (333.33)(833.33) = 111,115.01 ρ= 5-75. 111,115.01 111113.31 361117.11 = 0.5547 a) E(X) = 1 E(Y) = 1 ∞∞ xye − x − y dxdy E ( XY ) = 00 ∞ ∞ −x = xe dx ye − y dy 0 0 = E ( X ) E (Y ) Therefore, σ XY = ρ XY = 0 . 5-76. Suppose the correlation between X and Y is ρ. for constants a, b, c, and d, what is the correlation between the random variables U = aX+b and V = cY+d? Now, E(U) = a E(X) + b and E(V) = c E(Y) + d. Also, U - E(U) = a[ X - E(X) ] and V - E(V) = c[ Y - E(Y) ]. Then, σ UV = E{[U − E (U )][V − E (V )]} = acE{[ X − E ( X )][Y − E (Y )]} = acσ XY 2 2 2 2 Also, σ U = E[U − E (U )]2 = a 2 E[ X − E ( X )]2 = a 2σ X and σ V = c 2σ Y . ρ UV = acσ XY 2 a 2σ X 2 c 2σ Y = ρ XY if a and c are of the same sign - ρ XY if a and c differ in sign 5-31 Then, 5-77 E ( X ) = −1(1 / 4) + 1(1 / 4) = 0 E (Y ) = −1(1 / 4) + 1(1 / 4) = 0 E(XY) = [-1× 0 × (1/4)] + [-1× 0 × (1/4)] + [1× 0 × (1/4)] + [0 × 1× (1/4)] = 0 V(X) = 1/2 V(Y) = 1/2 σ XY = 0 − (0)(0) = 0 0 ρ XY = σσ σ = =0 1/ 2 1/ 2 The correlation is zero, but X and Y are not independent, since, for example, if y=0, X must be –1 or 1. XY X 5-78 Y If X and Y are independent, then f XY ( x, y ) = f X ( x) f Y ( y ) and the range of (X, Y) is rectangular. Therefore, E ( XY ) = xyf X ( x) fY ( y )dxdy = xf X ( x)dx yfY ( y )dy = E ( X ) E (Y ) hence σXY=0 Section 5-6 5-79 a.) 0.04 0.03 0.02 z(0) 0.01 10 0.00 -2 0 -1 0 y 1 2 3 -10 4 5-32 x b.) 0.07 0.06 0.05 0.04 z(.8) 0.03 0.02 0.01 10 0.00 -2 0 0 -1 y 0 1 2 3 x x -10 4 c) 0.07 0.06 0.05 0.04 z(-.8) 0.03 0.02 0.01 10 0.00 -2 -1 y 0 1 2 3 -10 4 5-33 5-80 Because ρ = 0 and X and Y are normally distributed, X and Y are independent. Therefore, P(2.95 < X < 3.05, 7.60 < Y < 7.80) = P(2.95 < X < 3.05) P(7.60 < Y < 7.80) = P ( 2.095− 3 < Z < .04 3.05 − 3 0.04 ) P ( 7.60.− 7.70 < Z < 0 08 7.80 − 7.70 0.08 ) = 0.7887 2 = 0.6220 5-81. Because ρ = 0 and X and Y are normally distributed, X and Y are independent. Therefore, µX = 0.1mm σX=0.00031mm µY = 0.23mm σY=0.00017mm Probability X is within specification limits is 0.099535 − 0.1 0.100465 − 0.1 <Z< 0.00031 0.00031 = P(−1.5 < Z < 1.5) = P ( Z < 1.5) − P( Z < −1.5) = 0.8664 P (0.099535 < X < 0.100465) = P Probability that Y is within specification limits is 0.22966 − 0.23 0.23034 − 0.23 <Z < 0.00017 0.00017 = P (−2 < Z < 2) = P ( Z < 2) − P ( Z < −2) = 0.9545 P (0.22966 < X < 0.23034) = P Probability that a randomly selected lamp is within specification limits is (0.8664)(.9594)=0.8270 5-82 a) By completing the square in the numerator of the exponent of the bivariate normal PDF, the joint PDF can be written as 1 − 1 f Y |X = x 2 σY ( y −( µY + ρ x−µ x 1 e 2π σ x 1 Also, fx(x) = 2π σ x e − − 2 σY ( y − (µY + ρ − 2π σ x = e 2 σY σY ( x − µ x )) σX 2 1 − σx − 2(1− ρ 2 ) e f XY ( x, y ) 2πσ x σ y 1 − ρ = f X ( x) 1 2 2 By definition, 2 1 1 + (1− ρ 2 ) 2 σx 1 fY | X = x = 2 2 (1− ρ 2 ) e 2 f XY ( x , y ) 2πσ x σ y 1 − ρ = = f X ( x) x−µx σY ( x − µ x )) σX ( y − ( µY + ρ σY ( x −µ x )) σX e 2 2 (1−ρ 2 ) 2π σ y 1 − ρ 2 5-34 x −µ x σx 2 2(1− ρ 2 ) 2 + (1−ρ 2 ) x −µ x σx 2 x−µ x σx 2 Now fY|X=x is in the form of a normal distribution. b) E(Y|X=x) = µ y + ρ σy σx ( x − µ x ) . This answer can be seen from part 5-82a. Since the PDF is in the form of a normal distribution, then the mean can be obtained from the exponent. c) V(Y|X=x) = σ 2 (1 − ρ2 ) . This answer can be seen from part 5-82a. Since the PDF is in the form y of a normal distribution, then the variance can be obtained from the exponent. 5-83 ∞∞ ∞∞ f XY ( x, y )dxdy = − ∞− ∞ 2π 1 σσ e X ( x−µ X )2 −1 2 σ X 2 + ( y − µY ) 2 σ Y Y 2 dxdy = −∞−∞ ∞ 1 2π σe −∞ ( x−µ X )2 −1 2 σ X ∞ 2 1 2π dx X σe −∞ ( y − µY ) 2 −1 2 σ Y 2 dy Y and each of the last two integrals is recognized as the integral of a normal probability density function from −∞ to ∞. That is, each integral equals one. Since f XY( x, y ) = f ( x) f ( y ) then X and Y are independent. 5-84 − 1 Let f XY ( x, y ) = X −µ X σX 2 − 2 2ρ ( X −µ X )(Y − µ X ) Y − µY + σ X σY σY 2(1− ρ 2 ) e 2πσ xσ y 1 − ρ2 Completing the square in the numerator of the exponent we get: X −µ X σX 2 − 2 ρ ( X − µ X )(Y − µ X ) σ XσY + 2 2 Y − µY = σY Y − µY σY −ρ X −µ X 2 2 + (1 − ρ ) σX X −µ X σX But, Y − µY −ρ σY X −µ X = σX 1 σY (Y − µ Y ) − ρ 1 σY ( X −µ x ) = σY σX (Y − ( µY + ρ σY ( X − µ x )) σX Substituting into fXY(x,y), we get 1 ∞ ∞ −∞ −∞ f XY ( x, y ) = − 1 2πσ xσ y 1 − ρ 2 2 σY 2 y − ( µY + ρ σY x−µx ( x − µ x )) + (1− ρ 2 ) σX σx 2(1− ρ 2 ) e y −( µ y + ρ = ∞ −∞ 1 2πσ x − e 1 x−µx 2 σx − 2 dx × ∞ −∞ 1 2πσ y 1 − ρ 2 e 2 dydx σy ( x − µ x )) σx 2 2 2σ x (1− ρ 2 ) dy The integrand in the second integral above is in the form of a normally distributed random variable. By definition of the integral over this function, the second integral is equal to 1: 5-35 y −( µ y + ρ 1 ∞ e 2πσ x −∞ = − − 1 ∞ e 2πσ x −∞ 1 x−µx 2 σx − 2 dx × 1 x−µx 2 σx 1 ∞ −∞ σy ( x − µ x )) σx 2 2 2σ x (1− ρ 2 ) e 2πσ y 1 − ρ 2 dy 2 dx × 1 The remaining integral is also the integral of a normally distributed random variable and therefore, it also integrates to 1, by definition. Therefore, ∞∞ f ( x, y ) − ∞ − ∞ XY =1 5-85 f X ( x) = 1 πσ σ 2 −∞ X Y 1− ρ 2 − 0.5 ( x − µ X = ( x−µ X )2 − 0 .5 ∞ 1 2π σe 1− ρ 2 σ e )2 σ 1− ρ 2 2 X − 2 ρ ( x − µ X )( y − µ Y ) σσ X − 0 .5 ∞ 2 X X −∞ 2π σ 1 Y 1− ρ 2 e 1− ρ 2 + ( y − µY ) 2 σ Y ( y − µY ) σ Y − 2 Y dy ρ ( x−µ X ) σ 2 − X ρ ( x−µ X ) σ X 2 dy 2 − 0 .5 = 1 e 2πσ X ( x−µ X )2 σ − 0 .5 ∞ 2 X −∞ 2π σ 1− ρ 2 1 Y 1− ρ 2 e ( y − µY ) σ Y − ρ ( x−µ X ) σ X dy The last integral is recognized as the integral of a normal probability density with mean µ Y + σ ρ ( x− µ σ Y X X ) 2 and variance σ Y (1 − ρ 2 ) . Therefore, the last integral equals one and the requested result is obtained. 5-36 5-86 2 2 E ( X ) = µ X , E (Y ) = µY , V ( X ) = σ X , and V (Y ) = σ Y . Also, − 0.5 ( x−µ X )2 1− ρ 2 ∞∞ σ 2 X − 2 ρ ( x−µ X ( x − µ X )( y − µ Y )e E ( X − µ X )(Y − µ Y ) = 2πσ X σ Y (1 − ρ 2 ) 1 / 2 − ∞− ∞ Let u = x−µ X σX and v = y − µY σY −0.5 1− ρ 2 σ dxdy u 2 − 2 ρuv + v 2 uve 2 1/ 2 − ∞− ∞ 2π (1 − ρ ) − 0. 5 1− ρ 2 ∞∞ = X . Then, ∞∞ E ( X − µ X )(Y − µ Y ) = )( y − µ Y ) ( y − µ Y ) 2 + 2 Y Y σσ σ X σ Y dudv [u − ρv ]2 + (1− ρ 2 ) v 2 uve 2π (1 − ρ 2 ) 1 / 2 − ∞− ∞ σ X σ Y dudv The integral with respect to u is recognized as a constant times the mean of a normal random variable with mean ρv and variance 1 − ρ 2 . Therefore, ∞ E ( X − µ X )(Y − µY ) = −∞ ∞ v 2 2π e −0.5v ρvσ X σ Y dv = ρσ X σ Y −∞ v2 2π 2 e −0.5v dv . The last integral is recognized as the variance of a normal random variable with mean 0 and variance 1. Therefore, E( X − µ X )( Y − µ Y ) = ρσ X σ Y and the correlation between X and Y is ρ . Section 5-7 5-87. a) E(2X + 3Y) = 2(0) + 3(10) = 30 b) V(2X + 3Y) = 4V(X) + 9V(Y) = 97 c) 2X + 3Y is normally distributed with mean 30 and variance 97. Therefore, P(2 X + 3Y < 30) = P( Z < 30 − 30 97 d) P(2 X + 3Y < 40) = P( Z < 5-88 ) = P( Z < 0) = 0.5 40 −30 97 ) = P( Z < 1.02) = 0.8461 Y = 10X and E(Y) =10E(X) = 50mm. V(Y)=102V(X)=25mm2 5-37 5-89 a) Let T denote the total thickness. Then, T = X + Y and E(T) = 4 mm, V(T) = 0.12 + 0.12 = 0.02mm 2 , and σ T = 0.1414 mm. b) 4 .3 − 4 = P ( Z > 2.12) 0.1414 = 1 − P ( Z < 2.12) = 1 − 0.983 = 0.0170 P (T > 4.3) = P Z > 5-90 a) X∼N(0.1, 0.00031) and Y∼N(0.23, 0.00017) Let T denote the total thickness. Then, T = X + Y and E(T) = 0.33 mm, V(T) = 0.000312 + 0.00017 2 = 1.25 x10 −7 mm 2 , and σ T = 0.000354 mm. 0.2337 − 0.33 = P (Z < −272) ≅ 0 0.000354 P (T < 0.2337 ) = P Z < b) P ( T > 0 . 2405 5-91. )=P = P ( Z > − 253 ) = 1 − P ( Z < 253 ) ≅ 1 Let D denote the width of the casing minus the width of the door. Then, D is normally distributed. 5 1 a) E(D) = 1/8 V(D) = ( 1 ) 2 + ( 16 ) 2 = 256 8 b) P ( D > 1 ) = P ( Z > 4 c ) P ( D < 0) = P ( Z < 5-92 0 . 2405 − 0 . 33 0 . 000345 Z> 11 − 48 5 0− 1 8 5 ) = P ( Z > 0.89) = 0.187 256 ) = P ( Z < −0.89) = 0.187 256 D=A-B-C a) E(D) = 10 - 2 - 2 = 6 mm V ( D ) = 0.12 + 0.052 + 0.052 = 0.015mm 2 σ D = 0.1225mm b) P(D < 5.9) = P(Z < 5-93. 5 .9 − 6 ) = P(Z < -0.82) = 0.206. 0.1225 a) Let X denote the average fill-volume of 100 cans. σ X = 0.5 2 100 = 0.05 . 12 − 12.1 = P ( Z < −2) = 0.023 0.05 12 − µ c) P( X < 12) = 0.005 implies that P Z < = 0.005. 0.05 12 − µ Then 0.05 = -2.58 and µ = 12.129 . b) E( X ) = 12.1 and P ( X < 12) = P Z < d.) P( X < 12) = 0.005 implies that P Z < Then 12 −12.1 σ / 100 = -2.58 and σ = 0.388 . 5-38 12 − 12.1 σ / 100 = 0.005. e.) P( X < 12) = 0.01 implies that P Z < Then 5-94 12 −12.1 0.5 / n 12 − 12.1 0 .5 / n = -2.33 and n = 135.72 ≅ 136 . = 0.01. Let X denote the average thickness of 10 wafers. Then, E( X ) = 10 and V( X ) = 0.1. a) P (9 < X < 11) = P ( 9 −10 < Z < 11−10 ) = P ( −3.16 < Z < 3.16) = 0.998 . 0.1 0.1 The answer is 1 − 0.998 = 0.002 b) P( X > 11) = 0.01 and σ X = Therefore, P ( X > 11) = P ( Z > 1 n 11−10 1 n . ) = 0 .0 1 , 11− 10 1 = 2.33 and n = 5.43 which is n rounded up to 6. c.) P( X > 11) = 0.0005 and σ X = σ Therefore, P ( X > 11) = P ( Z > 11−10 σ 10 . ) = 0.0005 , 11−10 σ 10 = 3.29 10 σ = 10 / 3.29 = 0.9612 5-95. X~N(160, 900) a) Let Y = 25X, E(Y)=25E(X)=4000, V(Y) = 252(900)=562500 P(Y>4300) = P Z> 4300 − 4000 562500 = P ( Z > 0.4) = 1 − P ( Z < 0.4) = 1 − 0.6554 = 0.3446 b.) c) P( Y > x) = 0.0001 implies that P Z > Then x − 4000 750 x − 4000 562500 = 0.0001. = 3.72 and x = 6790 . Supplemental Exercises 5-96 ( )( )( )( )( ) The sum of x and 5-97. y f ( x, y ) = 1 , 1 + 1 + 1 + 1 + 1 = 1 4 8 8 4 4 f XY ( x, y ) ≥ 0 a) P( X < 0.5, Y < 1.5) = f XY (0,1) + f XY (0,0) = 1 / 8 + 1 / 4 = 3 / 8 . b) P( X ≤ 1) = f XY (0,0) + f XY (0,1) + f XY (1,0) + f XY (1,1) = 3 / 4 c) P(Y < 1.5) = f XY (0,0) + f XY (0,1) + f XY (1,0) + f XY (1,1) = 3 / 4 d) P( X > 0.5, Y < 1.5) = f XY (1,0) + f XY (1,1) = 3 / 8 e) E(X) = 0(3/8) + 1(3/8) + 2(1/4) = 7/8. V(X) = 02(3/8) + 12(3/8) + 22(1/4) - 7/82 =39/64 E(Y) = 1(3/8) + 0(3/8) + 2(1/4) = 7/8. . V(Y) = 12(3/8) + 02(3/8) + 22(1/4) - 7/82 =39/64 5-39 5-98 a) f X ( x ) = f XY ( x, y ) and f X (0) = 3 / 8, f X (1) = 3 / 8, f X (2) = 1/ 4 . y b) fY 1 ( y ) = f XY (1, y ) and fY 1 (0) = f X (1) c) E (Y | X = 1) = 1/8 3/8 = 1 / 3, fY 1 (1) = 1/ 4 3/8 = 2/3. yf XY (1, y ) =0(1/ 3) + 1(2 / 3) = 2 / 3 x =1 d) Because the range of (X, Y) is not rectangular, X and Y are not independent. e.) E(XY) = 1.25, E(X) = E(Y)= 0.875 V(X) = V(Y) = 0.6094 COV(X,Y)=E(XY)-E(X)E(Y)= 1.25-0.8752=0.4844 0.4844 = 0.7949 0.6094 0.6094 20! a. ) P ( X = 2, Y = 4, Z = 14) = 0.10 2 0.20 4 0.7014 = 0.0631 2!4!14! b.) P ( X = 0) = 0.10 0 0.90 20 = 0.1216 c.) E ( X ) = np1 = 20(0.10) = 2 V ( X ) = np1 (1 − p1 ) = 20(0.10)(0.9) = 1.8 f ( x, z ) d.) f X | Z = z ( X | Z = 19) XZ f Z ( z) 20! f XZ ( xz ) = 0.1 x 0.2 20− x − z 0.7 z x! z!(20 − x − z )! 20! f Z ( z) = 0.3 20 − z 0.7 z z! (20 − z )! ρ XY = 5-99 f XZ ( x, z ) (20 − z )! 0.1 x 0.2 20 − x − z (20 − z )! 1 f X | Z = z ( X | Z = 19) = = 20 − z f Z ( z) x! (20 − x − z )! 0.3 x! (20 − x − z )! 3 x Therefore, X is a binomial random variable with n=20-z and p=1/3. When z=19, 2 1 and f X |19 (1) = . 3 3 2 1 1 e.) E ( X | Z = 19) = 0 +1 = 3 3 3 f X |19 (0) = 5-100 Let X, Y, and Z denote the number of bolts rated high, moderate, and low. Then, X, Y, and Z have a multinomial distribution. a) P ( X = 12, Y = 6, Z = 2) = 20! 0.6120.360.12 = 0.0560 . 12!6!2! b) Because X, Y, and Z are multinomial, the marginal distribution of Z is binomial with n = 20 and p = 0.1. c) E(Z) = np = 20(0.1) = 2. 5-40 2 3 20 − x − z 5-101. a) f Z |16 ( z ) = f XZ (16, z ) 20! and f XZ ( x, z ) = 0.6 x 0.3 ( 20 − x − z ) 0.1 z for f X (16) x! z! (20 − x − z )! x + z ≤ 20 and 0 ≤ x,0 ≤ z . Then, f Z 16 ( z ) = 20! 16!z!( 4 − z )! 0.616 0.3( 4 − z ) 0.1z 20! 16!4! 16 0.6 0.4 4 = ()() 4! 0.3 4 − z 0.1 z z!( 4 − z )! 0.4 0.4 for 0 ≤ z ≤ 4 . That is the distribution of Z given X = 16 is binomial with n = 4 and p = 0.25. b) From part a., E(Z) = 4 (0.25) = 1. c) Because the conditional distribution of Z given X = 16 does not equal the marginal distribution of Z, X and Z are not independent. 5-102 Let X, Y, and Z denote the number of calls answered in two rings or less, three or four rings, and five rings or more, respectively. a) P ( X = 8, Y = 1, Z = 1) = 10! 0.780.2510.051 = 0.0649 8! ! ! 11 b) Let W denote the number of calls answered in four rings or less. Then, W is a binomial random variable with n = 10 and p = 0.95. Therefore, P(W = 10) = 10 0.95100.050 = 0.5987 . 10 c) E(W) = 10(0.95) = 9.5. () 5-103 a) f Z 8 ( z ) = 10! f XZ (8, z ) 0.70 x 0.25(10 − x − z ) 0.05 z for and f XZ ( x, z ) = x! z!(10 − x − z )! f X (8) x + z ≤ 10 and 0 ≤ x,0 ≤ z . Then, f Z 8 ( z) = 10! 8!z!( 2 − z )! 0.70 8 0.25 ( 2 − z ) 0.05 z 10! 8!2! 8 0.70 0.30 2 = ()() 2! 0.25 2 − z 0.05 z z !( 2 − z )! 0.30 0.30 for 0 ≤ z ≤ 2 . That is Z is binomial with n =2 and p = 0.05/0.30 = 1/6. b) E(Z) given X = 8 is 2(1/6) = 1/3. c) Because the conditional distribution of Z given X = 8 does not equal the marginal distribution of Z, X and Z are not independent. 32 3 cx 2 ydydx = cx 2 5-104 00 0 y2 2 2 0 3 dx = 2c x3 3 = 18c . Therefore, c = 1/18. 0 5-41 11 1 1 18 5-105. a) P ( X < 1, Y < 1) = x 2 ydydx = 1 18 00 2.5 1 18 x 2 ydydx = 00 1 18 dx = 0 0 32 3 1 18 c) P (1 < Y < 2.5) = 0 2 y2 2 x2 x 2 ydydx = 01 1 18 x2 1 x3 93 2 y2 2 dx = 1 0 1 1 x3 36 3 dx = 0 2.5 2 b) P ( X < 2.5) = 1 y2 2 x2 = 0 1 108 2.5 = 0.5787 0 3 1 x3 12 3 = 0 3 4 d) 3 1.5 P ( X > 2,1 < Y < 1.5) = 2 32 3 00 95 432 = x 3 2dx = 00 1.5 5 x3 144 3 3 9 4 dx = 1 2 1 18 3 1 x4 94 x 2 8 dx = 3 0 3 x 2 y 2 dydx = y2 2 = 0.2199 0 1 18 x2 2 1 18 x ydydx = 32 1 18 x ydydx = 3 1 18 f) E (Y ) = 2 1 = e) E ( X ) = 3 1 18 4 x3 27 3 0 3 = 0 4 3 2 1 18 5-106 a) f X ( x) = x 2 ydy = 1 x 2 for 0 < x < 3 9 0 b) f Y X ( y ) = f XY (1, y ) = f X (1) 1 18 1 9 y = y for 0 < y < 2. 2 2 1 f XY ( x,1) 18 x c) f X 1 ( x) = = and fY ( y ) = fY (1) fY (1) Therefore, f X 1 ( x) = 3 1 18 x 2 ydx = y 2 for 0 < y < 2 . 0 x2 1 2 = x for 0 < x < 3. 1/ 2 9 1 18 5-107. The region x2 + y 2 ≤ 1 and 0 < z < 4 is a cylinder of radius 1 ( and base area π ) and height 4. Therefore, the volume of the cylinder is 4 π and f XYZ ( x, y, z) = 1 4π for x2 + y 2 ≤ 1 and 0 < z < 4. a) The region X 2 + Y 2 ≤ 0.5 is a cylinder of radius 0.5 and height 4. Therefore, P( X 2 + Y 2 ≤ 0.5) = 4 ( 0.5π ) 4π = 1/ 2 . b) The region X 2 + Y 2 ≤ 0.5 and 0 < z < 2 is a cylinder of radius 0.5 and height 2. Therefore, P( X 2 + Y 2 ≤ 0.5, Z < 2) = 2 ( 0.5π ) 4π = 1/ 4 5-42 c) f XY 1 ( x, y ) = f XYZ ( x, y,1) and f Z ( z ) = f Z (1) for 0 < z < 4. Then, f XY 1 ( x, y ) = 4 1− x 2 0 − 1− x 1 / 4π = 1/ 4 dydx = 1 / 4 for x 2 + y 2 ≤ 1 . 1 π 4 1 4π f X ( x) = d) 1 4π x 2 + y 2 ≤1 1 2π dydz = 2 1 − x 2 dz = π 1 − x 2 for -1 < x < 1 0 2 4 f XYZ (0,0, z ) 2 2 1 and f XY ( x, y ) = 4π dz = 1 / π for x + y ≤ 1 . Then, f XY (0,0) 0 1 / 4π f Z 0,0 ( z ) = = 1 / 4 for 0 < z < 4 and µ Z 0, 0 = 2 . 1/ π f ( x, y, z ) 1 / 4π f Z xy ( z ) = XYZ = = 1 / 4 for 0 < z < 4. Then, E(Z) given X = x and Y = y is f XY ( x, y ) 1/ π 5-108 a) f Z 0, 0 ( z ) = b) 4 z 4 dz = 2 . 0 1 5-109. 1 f XY ( x, y ) = c for 0 < x < 1 and 0 < y < 1. Then, cdxdy = 1 and c = 1. Because 0 0 f XY ( x, y ) is constant, P( X − Y < 0.5) is the area of the shaded region below 1 0.5 0 That is, 0.5 1 P( X − Y < 0.5) = 3/4. 5-110 a) Let X1, X 2 ,..., X 6 denote the lifetimes of the six components, respectively. Because of independence, P( X1 > 5000, X 2 > 5000,..., X 6 > 5000 ) = P( X1 > 5000)P( X 2 > 5000 )... P( X 6 > 5000) If X is exponentially distributed with mean θ , then λ = ∞ 1 P( X > x) = θ e − t / θ dt = −e − t / θ ∞ e −5 / 8 e −0.5 e −0.5 e −0.25 e −0.25 e −0.2 = e −2.325 = 0.0978 . 5-43 and = e − x / θ . Therefore, the answer is x x 1 θ b) The probability that at least one component lifetime exceeds 25,000 hours is the same as 1 minus the probability that none of the component lifetimes exceed 25,000 hours. Thus, 1-P(Xa<25,000, X2<25,000, …, X6<25,000)=1-P(X1<25,000)…P(X6<25,000) =1-(1-e-25/8)(1-e-2.5)(1-e-2.5)(1-e-1.25)(1-e-1.25)(1-e-1)=1-.2592=0.7408 5-111. Let X, Y, and Z denote the number of problems that result in functional, minor, and no defects, respectively. a) P ( X = 2, Y = 5) = P ( X = 2, Y = 5, Z = 3) = 210!!3! 0.2 2 0.5 5 0.3 3 = 0.085 !5 b) Z is binomial with n = 10 and p = 0.3. c) E(Z) = 10(0.3) = 3. 5-112 a) Let X denote the mean weight of the 25 bricks in the sample. Then, E( X ) = 3 and 25 σ X = 0.25 = 0.05 . Then, P( X < 2.95) = P(Z < 2 .095 − 3 ) = P (Z < -1) = 0.159. . 05 b) P( X > x) = P( Z > x−3 x−3 ) = 0.99 . So, = -2.33 and x=2.8835. .05 0.05 5-113. a.) 18.25 5.25 17.75 4.75 cdydx = 0.25c, c = 4. The area of a panel is XY and P(XY > 90) is Because the shaded area times 4 below, 5.25 4.75 18.25 17.25 18.25 18.25 5.25 18.25 4dydx = 4 5.25 − 90 dx = 4(5.25 x − 90 ln x x That is, 17.75 17.75 90 / x ) = 0.499 17.75 b. The perimeter of a panel is 2X+2Y and we want P(2X+2Y >46) 18.25 5.25 18.25 4dydx = 4 5.25 − (23 − x)dx 17.75 23− x 17.75 18.25 = 4 (−17.75 + x)dx = 4(−17.75 x + 17.75 5-44 x2 2 18.25 ) = 0 .5 17.75 5-114 a)Let X denote the weight of a piece of candy and X∼N(0.1, 0.01). Each package has 16 candies, then P is the total weight of the package with 16 pieces and E( P ) = 16(0.1)=1.6 ounces and V(P) = 162(0.012)=0.0256 ounces2 6 b) P ( P < 1.6) = P ( Z < 1.0.−1.6 ) = P ( Z < 0) = 0.5 . 16 c) Let Y equal the total weight of the package with 17 pieces, E(Y) = 17(0.1)=1.7 ounces and V(Y) = 172(0.012)=0.0289 ounces2 −1.7 P(Y < 1.6) = P( Z < 1.06.0289 ) = P( Z < −0.59) = 0.2776 . 5-115. Let X denote the average time to locate 10 parts. Then, E( X ) =45 and σ X = a) P ( X > 60) = P ( Z > 60 − 45 30 / 10 30 10 ) = P ( Z > 1.58) = 0.057 b) Let Y denote the total time to locate 10 parts. Then, Y > 600 if and only if X > 60. Therefore, the answer is the same as part a. 5-116 a) Let Y denote the weight of an assembly. Then, E(Y) = 4 + 5.5 + 10 + 8 = 27.5 and V(Y)= 0.4 2 + 0.5 2 + 0.2 2 + 0.5 2 = 0.7 . P (Y > 29.5) = P( Z > 29.5 − 27.5 0.7 ) = P ( Z > 2.39) = 0.0084 b) Let X denote the mean weight of 8 independent assemblies. Then, E( X ) = 27.5 and 27.5 V( X ) = 0.7/8 = 0.0875. Also, P ( X > 29) = P ( Z > 290−0875 ) = P ( Z > 5.07) = 0 . . 5-117 0.07 0.06 0.05 0.04 z(-.8) 0.03 0.02 0.01 10 0.00 -2 0 -1 y 0 1 2 3 -10 4 5-45 x 5-118 1 f XY ( x , y ) = e 1 .2π f XY ( x , y ) = f XY ( x , y ) = −1 {( x −1) 2 −1 .6 ( x −1)( y − 2 ) + ( y − 2 ) 2 } 0.72 1 2π .36 e −1 {( x −1) 2 −1.6 ( x −1)( y − 2 ) + ( y − 2 ) 2 } 2 ( 0 .36 ) −1 1 2π 1 − .8 2 e 2 (1−.8 2 ) {( x −1) 2 − 2 (. 8 )( x −1)( y − 2 ) + ( y − 2 ) 2 } E ( X ) = 1 , E (Y ) = 2 V ( X ) = 1 V (Y ) = 1 and ρ = 0.8 5-119 Let T denote the total thickness. Then, T = X1 + X2 and a.) E(T) = 0.5+1=1.5 mm V(T)=V(X1) +V(X2) + 2Cov(X1X2)=0.01+0.04+2(0.014)=0.078mm2 where Cov(XY)=ρσXσY=0.7(0.1)(0.2)=0.014 b.) P (T < 1) = P Z < 1 − 1 .5 0.078 = P ( Z < −1.79) = 0.0367 c.) Let P denote the total thickness. Then, P = 2X1 +3 X2 and E(P) =2(0.5)+3(1)=4 mm V(P)=4V(X1) +9V(X2) + 2(2)(3)Cov(X1X2)=4(0.01)+9(0.04)+2(2)(3)(0.014)=0.568mm2 where Cov(XY)=ρσXσY=0.7(0.1)(0.2)=0.014 5-120 Let T denote the total thickness. Then, T = X1 + X2 + X3 and a) E(T) = 0.5+1+1.5 =3 mm V(T)=V(X1) +V(X2) +V(X3)+2Cov(X1X2)+ 2Cov(X2X3)+ 2Cov(X1X3)=0.01+0.04+0.09+2(0.014)+2(0.03)+ 2(0.009)=0.246mm2 where Cov(XY)=ρσXσY b.) P (T < 1.5) = P Z < 1 .5 − 3 = P ( Z < −6.10) ≅ 0 0.246 5-46 5-121 Let X and Y denote the percentage returns for security one and two respectively. If ½ of the total dollars is invested in each then ½X+ ½Y is the percentage return. E(½X+ ½Y)= 0.05 (or 5 if given in terms of percent) V(½X+ ½Y)=1/4 V(X)+1/4V(Y)+2(1/2)(1/2)Cov(X,Y) where Cov(XY)=ρσXσY=-0.5(2)(4)=-4 V(½X+ ½Y)=1/4(4)+1/4(6)-2=3 Also, E(X)=5 and V(X) = 4. Therefore, the strategy that splits between the securities has a lower standard deviation of percentage return than investing 2million in the first security. Mind-Expanding Exercises 5-122. By the independence, P( X 1 ∈ A1 , X 2 ∈ A2 ,..., X p ∈ A p ) = ... A1 = A2 f X 1 ( x1 ) f X 2 ( x 2 )... f X p ( x p )dx1 dx 2 ...dx p Ap f X 1 ( x1 )dx1 A1 f X 2 ( x 2 )dx 2 ... A2 f X p ( x p )dx p Ap = P( X 1 ∈ A1 ) P( X 2 ∈ A2 )...P( X p ∈ A p ) 5-123 E (Y ) = c1µ1 + c2 µ 2 + ... + c p µ p . Also, V (Y ) = = [c x + c x + ... + c x − (c µ + c µ + ... + c µ )] [c ( x µ ) + ... + c ( x − µ )] f ( x ) f ( x )... f 2 11 22 p 1 p 11 p 2 2 p p 2 1 1 p p X1 1 X2 2 Xp f X 1 ( x1 ) f X 2 ( x2 )... f X p ( x p )dx1dx2 ...dx p ( x p )dx1dx2 ...dx p Now, the cross-term c1c ( x1 − µ1 )( x2 − µ 2 ) f X 1 ( x1 ) f X 2 ( x2 )... f X p ( x p )dx1dx2 ...dx p 2 [ = c1c2 ( x1 − µ1 ) f X 1 ( x1 )dx1 [ ( x − µ 2 ) f X 2 ( x2 )dx2 = 0 2 from the definition of the mean. Therefore, each cross-term in the last integral for V(Y) is zero and V (Y ) = [ c (x − µ ) 2 1 1 1 2 [ f X 1 ( x1 )dx1 ... c 2 ( x p − µ p ) 2 f X p ( x p )dx p p = c12V ( X 1 ) + ... + c 2V ( X p ). p 5-47 a b a b 0 0 f XY ( x, y )dydx = 5-124 0 0 cdydx = cab . Therefore, c = 1/ab. Then, b f X ( x) = cdy = 1 a a for 0 < x < a, and f Y ( y) = cdx = 1 b 0 0 for 0 < y < b. Therefore, f XY (x,y)=f X (x)f Y (y) for all x and y and X and Y are independent. b 5-125 f X ( x) = b b g ( x)h( y )dy = g ( x) h( y )dy = kg ( x) where k = 0 h( y )dy. Also, 0 0 a f Y ( y ) = lh( y ) where l = g ( x)dx. Because f XY (x,y) is a probability density 0 a b a g ( x)h( y )dydx = function, 0 0 b h( y )dy = 1. Therefore, kl = 1 and g ( x)dx 0 0 f XY (x,y)=f X (x)f Y (y) for all x and y. Section 5-8 on CD S5-1. S5-2. f Y ( y) = 1 4 at y = 3, 5, 7, 9 from Theorem S5-1. Because X ≥ 0 , the transformation is one-to-one; that is ( )p (1 − p) ( y ) = ( )(0.25) (0.75) f Y ( y) = f X ( y ) = If p = 0.25, S5-3. a) fY fY ( y ) = f X y 3− y y y − 10 2 1 y − 10 = 2 72 y 2 − 10 y 1 E (Y ) = dy = 72 72 10 ( y3 3 Because y = -2 ln x, e −2 x= y . From Theorem S5-2, for y = 0, 1, 4, 9. for 10 ≤ y ≤ 22 − 102y 2 ) 22 = 18 10 −y y S5-4. and for y = 0, 1, 4, 9. y 3 22 b) 3− y y 3 y = x2 = x . Then, f Y ( y ) = f X (e 2 ) − 1 e 2 −y 2 −y =1e2 2 −y for 0 ≤ e 2 ≤ 1 or y ≥ 0 , which is an exponential distribution (which equals a chi-square distribution with k = 2 degrees of freedom). 5-48 S5-5. a) Let Q = R. Then, p = i 2r q=r J= p q i= and r=q ∂i ∂p ∂r ∂p ∂i ∂q ∂r ∂q = 1 2 ( pq ) −1 / 2 0 p q f PQ ( p, q ) = f IR ( p q for 0 ≤ That is, for − 1 p 1 / 2 q −3 / 2 1 2 = 2 ( pq ) −1 / 2 1 , q ) 1 ( pq ) −1 / 2 = 2 2 () p q 1 2 ( pq ) −1 / 2 = q −1 ≤ 1, 0 ≤ q ≤ 1 0 ≤ p ≤ q, 0 < q ≤ 1 . 1 f P ( p ) = q −1dq = − ln p for 0 < p ≤ 1. p 1 b) E ( P ) = − p ln p dp . Let u = ln p and dv = p dp. Then, du = 1/p and 0 2 p 2 v= S5-6. a) If . Therefore, E ( P) = − (ln p ) y = x 2 , then x = y p2 2 1 1 p 2 + 0 dp = p2 4 0 for x ≥ 0 and y ≥ 0 . Thus, 1 = 0 1 4 f Y ( y) = f X ( y ) 1 y 2 −1 2 = e− y for 2y y > 0. b) If 0. y = x 1 / 2 , then x = y 2 c) If y = ln x, then x = e y for −∞ < y < ∞. for x≥0 and y ≥ 0 . Thus, fY ( y ) = f X ( y 2 )2 y = 2 ye− y y x ≥ 0 . Thus, f Y ( y ) = f X (e y )e y = e y e − e = e y − e ∞ S5-7. ∞ av 2 e −bv dv a) Now, must equal one. Let u = bv, then 1 = 0 the definition of the gamma function the last expression is b) If w= mv 2 2 , then v= f W ( w) = f V = for b 3 m −3 / 2 2 w1 / 2 e −b 2w m w ≥ 0. 5-49 for y > y for ∞ a ( u ) 2 e −u du = a3 u 2 e −u du. b b b0 0 a 2a b3 Γ (3) = 3 . Therefore, a = 2 b3 b 2w for v ≥ 0, w ≥ 0 . m b 3 2 w −b 2mw 2 w dv = e (2mw) −1 / 2 m dw 2m () 2 . From S5-8. If y = e x , then x = ln y for 1 ≤ x ≤ 2 and e 1 ≤ y ≤ e 2 . Thus, fY ( y ) = f X (ln y ) 1≤ ln y ≤ 2 . That is, fY ( y ) = 1 y for 11 = yy for e ≤ y ≤ e2 . ∞ S5-9. Now P (Y ≤ a ) = P ( X ≥ u (a )) = f X ( x)dx . By changing the variable of integration from x to y u(a) −∞ by using x = u(y), we obtain P(Y ≤ a ) = f X (u ( y ))u ' ( y )dy because as x tends to ∞, y = h(x) tends a a to - ∞. Then, P (Y ≤ a ) = f X (u ( y ))(−u ' ( y ))dy . Because h(x) is decreasing, u'( y) is negative. −∞ Therefore, | u'( y) | = - u'( y) and Theorem S5-1 holds in this case also. S5-10. If y = ( x − 2) 2 , then x = 2 − y for 0≤ x≤2 and x = 2+ y for 2 ≤ x ≤ 4 . Thus, fY ( y ) = f X (2 − y ) | − 1 y −1 / 2 | + f X (2 + y ) | 1 y −1 / 2 | 2 2 = 2− y 16 y + 2+ y 16 y = ( 1 ) y −1 / 2 for 0 ≤ y ≤ 4 4 S5-11. a) Let a = s1s 2 and y = s1 . Then, s1 = y, s 2 = a y and ∂s ∂s 0 1 ∂a ∂y a −3 / 2 = − 1 . Then, 1 J= = −y ∂s ∂s y y 2 ∂a ∂y f AY (a, y ) = f S S ( y, a )( 1 ) = 2 y ( 8ay )( 1 ) = 4ay for y y y 1 1 2 2 12 0 ≤ y ≤ 1 and 0 ≤ a ≤ 4 y . 1 b) f A (a) = a a dy = − ln( a ) 44 a /4 4y for 0 < a ≤ 4. 5-50 0 ≤ y ≤ 1 and 0 ≤ a ≤ 4 . That is, for y S5-12. i = s and v = rs Let r = v/i and s = i. Then, ∂i J = ∂r ∂v ∂r ∂i ∂s = 0 1 = s ∂v s r ∂s f RS (r , s ) = f IV ( s, rs ) s = e − rs s for rs ≥ 0 f RS (r , s) = se − rs and 1 ≤ s ≤ 2 . That is, for 1 ≤ s ≤ 2 and r ≥ 0 . 2 Then, f R (r ) = se − rs ds . Let u = s and dv = e − rs ds. Then, du = ds and v = 1 − e − rs r Then, 2 2 2 e−r −2e−2r e−rs −e−rs −e−rs f R (r) = −s + ds= −2 r11r r r1 e−r −2e−2r e−r −e−2r + r r2 e−r (r +1) −e−2r (2r +1) = r2 = for r > 0. Section 5-9 on CD e tx 1 = m x =1 m m S5-13 . a) b) E (e tX ) = M (t ) = m (e t ) x = x =1 1t e (1 − e tm )(1 − e t ) −1 m (e t ) m +1 − e t e t (1 − e tm ) = m(e t − 1) m(1 − e t ) and dM(t ) 1 t = e (1 − etm )(1 − et )−1 + et (−metm )(1 − et )−1 + et (1 − etm )(−1)(1 − et )−2 (−et ) dt m { } dM (t ) (1 − e tm )e t et = 1 − e tm − me tm + dt m(1 − e t ) 1− et = et 1 − e tm − me tm + me ( m +1) t m(1 − e t ) 2 { } Using L’Hospital’s rule, dM (t ) et − me tm − m 2 e tm + m(m + 1)e ( m +1)t lim = lim lim t →0 t →0 m t →0 dt − 2(1 − e t )e t = lim t →0 et − m 2 e tm − m 3 e tm + m(m + 1) 2 e ( m +1)t lim m t →0 − 2(1 − e t )e t − 2e t (−e t ) 1 m(m + 1) 2 − m 2 − m 3 m 2 + m m + 1 =× = = m 2 2m 2 5-51 Therefore, E(X) = d 2 M (t ) d 2 =2 dt 2 dt = 1 m m +1 . 2 m e tx x =1 11 = mm d2 (tx ) 2 1 + tx + + 2 2 x =1 dt m m ( x 2 + term sin volvingpow ersoft ) x =1 Thus, d 2 M (t ) 1 m(m + 1)(m + 2) (m + 1)(2m + 2) = = 6 6 dt 2 t =0 m Then, 2m 2 + 3m + 1 (m + 1) 2 4m 2 + 6m + 2 − 3m 2 − 6m − 3 − = 6 4 12 2 m −1 = 12 V (X ) = ∞ S5-14. a) E (e tX ) = e tx x =0 b) ∞ t t ( λe t ) x e −λ λx = e −λ = e −λ e λe = e λ ( e −1) x! x! x =0 t dM (t ) = λe t e λ ( e −1) dt dM (t ) = λ = E( X ) dt t = 0 t t d 2 M (t ) = λ 2 e 2t e λ ( e −1) + λe t e λ ( e −1) dt 2 d 2 M (t ) = λ2 + λ 2 dt t =0 V ( X ) = λ 2 + λ − λ2 = λ 5-52 ∞ S5-15 a) etx (1 − p)x−1 p = E(etX ) = x =1 b) p∞t [e (1 − p)]x 1 − p x=1 et (1 − p ) p pet = = 1 − (1 − p )et 1 − p 1 − (1 − p )et dM (t ) = pe t (1 − (1 − p ) e t ) − 2 (1 − p ) e t + pe t (1 − (1 − p ) e t ) −1 dt = p (1 − p )e 2t (1 − (1 − p )et )−2 + pet (1 − (1 − p )et )−1 dM(t) 1− p 1 = +1= = E(X ) dt t =0 p p d 2M(t) = p(1 − p)e2t 2(1− (1 − p)et )−3(1− p)et + p(1− p)(1 − (1 − p)et )−2 2e2t dt2 + pet (1− (1− p)et )−2 (1− p)et + pet (1− (1− p)et )−1 d 2 M (t ) 2(1 − p ) 2 2(1 − p ) 1 − p 2(1 − p ) 2 + 3 p (1 − p ) + p 2 = + + +1 = dt 2 t = 0 p2 p p p2 2 + 2 p2 − 4 p + 3 p − 3 p2 + p2 2 − p = p2 p2 2 − p 1 1− p V (X ) = − 2= 2 p2 p p = S5-16. M Y (t ) = Ee tY = Ee t ( X 1 + X 2 ) = Ee tX 1 Ee tX 2 = (1 − 2t ) − k1 / 2 (1 − 2t ) − k 2 / 2 = (1 − 2t ) − ( k1 + k 2 ) / 2 Therefore, Y has a chi-square distribution with k 1 + k 2 degrees of freedom. ∞ S5-17. a) ∞ E (e tX ) = e tx 4 xe − 2 x dx = 4 xe ( t − 2) x dx 0 0 Using integration by parts with u = x and dv = e ( t − 2 ) x dx and du = dx, (t − 2) x v= e t−2 xe (t − 2) x 4 t−2 we obtain ∞ 0 ∞ e (t −2 ) x xe ( t − 2) x − dx = 4 t−2 t−2 0 This integral only exists for t < 2. In that case, b) dM (t ) = −8(t − 2) −3 dt c) d 2 M (t ) = 24(t − 2) − 4 2 dt and ∞ 0 e (t −2 ) x − (t − 2) 2 E (e tX ) = ∞ 0 4 for t < 2 (t − 2) 2 dM (t ) = −8(−2) −3 = 1 = E ( X ) dt t = 0 and 24 3 d 2 M (t ) 3 1 = = . Therefore, V ( X ) = − 12 = 2 16 2 2 2 dt t =0 5-53 β S5-18. a) tX E (e ) = α b) β etx etx etβ − etα dx = = β −α t(β − α ) α t(β − α ) β e tβ − αe tα dM (t ) e tβ − e tα = + dt t (β − α ) − ( β − α )t 2 = ( βt − 1)e tβ − (αt − 1)e tα t 2 (β − α ) Using L’Hospital’s rule, dM (t ) (βt − 1)βe tβ + βe tβ − (αt − 1)αe tα − αe tα lim = t →0 dt 2t (β − α ) dM (t ) β 2 (βt − 1)e tβ + β 2 e tβ + β 2 e tβ − α 2 (αt − 1)e tα − α 2 e tα − α 2 e tα lim = t →0 dt 2(β − α ) dM (t ) β 2 − α 2 (β + α ) = = = E( X ) t →0 dt 2(β − α ) 2 lim b d 2 M (t ) d 2 1 tx 1 d 2 e tb − e ta =2 ( ) e dx = b − a dt 2 t dt 2 dt a b − a tb + (tb) 2 (tb) 3 (ta ) 2 (ta ) 3 + − ta − − + ... 2 3! 2 3! = t = 1 d2 b − a dt 2 = 1 d2 (b 2 − a 2 )t (b 3 − a 3 )t 2 b−a+ + + ... 2 3! b − a dt 2 b 3 − a 3 b 2 + ba + a 2 = = 3(b − a ) 3 Thus, b 2 + ba + a 2 (b + a ) 2 b 2 − 2ab + a 2 (b − a ) 2 V(X)= − = = 3 4 12 12 5-54 ∞ S5-19. a) ∞ M (t ) = e tx λe − λx dx = λ e ( t − λ ) x dx 0 0 (t −λ ) x ∞ =λ b) e t −λ −λ 1 = = (1 − t t − λ 1− λ = 0 t λ )−1 for t < λ dM (t ) 1 t −2 = (−1)(1 − λ ) ( −1 ) = λ t2 dt λ (1 − λ ) dM (t ) dt 1 = λ t =0 d 2 M (t ) 2 =2 2 t dt λ (1 − λ ) 3 d 2 M (t ) dt 2 λ2 1 λ ∞ a) t =0 2 V (X ) = S5-20. − 2 λ M (t ) = e tx Γ (r ) 0 2 = 2 = λ 1 λ2 (λx) r −1 e − λx dx = λr Γ(r ) ∞ x r −1 e ( t − λ ) x dx 0 Let u = (λ-t)x. Then, M (t ) = ∞ λr Γ( r ) u λ −t 0 r −1 e −u du 1 λr Γ(r ) t = = = (1 − λ ) − r from r tr λ − t Γ(r )(λ − t ) (1 − λ ) the definition of the gamma function for t < λ. b) M ' ( t ) = − r (1 − t ) − r − 1 ( − 1 ) λ r M ' (t ) t = 0 = M ' ' (t ) = λ V (X ) = = E( X ) r (r + 1) λ2 M ' ' (t ) t = 0 = λ t (1 − λ ) r − 2 r (r + 1) λ2 r (r + 1) 2 λ − r λ 2 = r λ2 5-55 n S5-21. a) t E (e ) = ∏ E (e tX i ) = (1 − λ ) tY −n i =1 b) From Exercise S5-20, Y has a gamma distribution with parameter λ and n. 2 S5-22. a) M Y (t ) = e 2 2 µ1t +σ 12 t2 + µ 2t +σ 2 t2 2 =e 2 2 ( µ1 + µ 2 ) t + (σ 1 +σ 2 ) t2 2 b) Y has a normal distribution with mean µ 1 + µ 2 and variances σ 1 + σ 2 2 S5-23. Because a chi-square distribution is a special case of the gamma distribution with λ = 1 k and r = , from 2 2 Exercise S5-20. M (t ) = (1 − 2t ) − k / 2 k − k −1 − k −1 (1 − 2t ) 2 (−2) = k (1 − 2t ) 2 2 = k = E( X ) M ' (t ) = − M ' (t ) t =0 M ' ' (t ) = 2k ( k + 1)(1 − 2t ) 2 − k −2 2 M ' ' (t ) t = 0 = 2k ( k + 1) = k 2 + 2k 2 V ( X ) = k 2 + 2k − k 2 = 2k 2 S5-24. a) = 1 and M ( r ) (0) = µ r' b) From Exercise S5-20, c) r M (t ) = M (0) + M ' (0)t + M ' (0) t2! + ... + M ( r ) (0) tr! + ... by Taylor’s expansion. Now, M(0) µ 1' = r λ and ' µ2 = and the result is obtained. r M (t ) = 1 + λ t + r ( r +1) λ2 r ( r +1) t 2 2! λ2 + ... which agrees with Exercise S5-20. Section 5-10 on CD S5-25. Use Chebychev's inequality with c = 4. Then, S5-26. E(X) = 5 and σ X = 2.887 . Then, The actual probability is P( X − 10 > 4) ≤ 116 . P( X − 5 > 2σ X ) ≤ 1 4 . P( X − 5 > 2σ X ) = P( X − 5 > 5.77) = 0 . 5-56 P( X − 20 > 2σ ) ≤ S5-27. E(X) = 20 and V(X) = 400. Then, 1 4 and P( X − 20 > 3σ ) ≤ 1 9 The actual probabilities are P ( X − 20 > 2σ ) = 1 − P ( X − 20 < 40) 60 = 1 − 0.05e − 0.05 x dx = 1 − − e −0.05 x 60 = 0.0498 0 0 P ( X − 20 > 3σ ) = 1 − P ( X − 20 < 60) 80 = 1 − 0.05e − 0.05 x dx = 1 − − e −0.05 x 80 = 0.0183 0 0 S5-28. E(X) = 4 and σ X = 2 P ( X − 4 ≥ 4) ≤ 1 4 1 9 and P( X − 4 ≥ 6) ≤ . The actual probabilities are 7 P ( X − 4 ≥ 4) = 1 − P ( X − 4 < 4) = 1 − e−2 2x x! x =1 9 e− 2 2 x x! P ( X − 4 ≥ 6) = 1 − P ( X − 4 < 6) = 1 − = 1 − 0.8636 = 0.1364 = 0.000046 x =1 S5-29. Let X denote the average of 500 diameters. Then, a) σX = 0.01 500 = 4.47 x10 −4 . 1 P ( X − µ ≥ 4σ X ) ≤ 16 and P ( X − µ < 0.0018) ≥ 15 . Therefore, the bound is 16 0.0018. If P ( X − µ < x) = P( −x 4.47 x10 − −4 S5-30. a) E(Y) = <Z< 15 16 , then x 4.47 x10 − 4 − P( σ x < X X −µ σX < σx ) = 0.9375. Then, ) = 0.9375. and X x 4.47 x10 − 4 = 1.86. Therefore, x = 8.31x10 −4 . P( X − µ ≥ cσ ) b) Because Y ≤ 1 , ( X − µ )2 ≥ ( X − µ )2Y If X − µ ≥ cσ , then Y = 1 and ( X − µ ) 2 Y ≥ c 2σ 2 Y If X − µ < cσ , then Y = 0 and ( X − µ )2 Y = c 2σ 2Y . c) Because ( X − µ ) 2 ≥ c 2σ 2Y , E[( X − µ ) 2 ] ≥ c 2σ 2 E (Y ) . d) From part a., E(Y) = 1 c2 P( X − µ ≥ cσ ) . From part c., σ 2 ≥ c 2σ 2 P( X − µ ≥ cσ ) . Therefore, ≥ P ( X − µ ≥ cσ ) . 5-57 CHAPTER 6 Section 6-1 6-1. Sample average: n xi i =1 x= n = 592.035 = 74.0044 mm 8 Sample variance: 8 xi = 592.035 i =1 8 xi2 = 43813.18031 i =1 2 n xi n 2 i x− s2 = (592.035)2 43813.18031 − i =1 n i =1 8 = n −1 8 −1 0.0001569 = = 0.000022414 (mm) 2 7 Sample standard deviation: s = 0.000022414 = 0.00473 mm The sample standard deviation could also be found using n s= ( x i − x )2 i =1 n −1 8 where ( xi − x)2 = 0.0001569 i =1 Dot Diagram: .. ...: . -------+---------+---------+---------+---------+---------diameter 73.9920 74.0000 74.0080 74.0160 74.0240 74.0320 There appears to be a possible outlier in the data set. 6-1 6-2. Sample average: 19 xi i =1 x= = 19 272.82 = 14.359 min 19 Sample variance: 19 xi = 272.82 i =1 19 xi2 =10333.8964 i =1 2 n xi n i =1 2 i x− s2 = n i =1 10333.8964 − = n −1 6416.49 = = 356.47 (min) 2 18 19 − 1 Sample standard deviation: s = 356.47 = 18.88 min The sample standard deviation could also be found using n s= ( x i − x )2 i =1 n −1 where 19 ( x i − x )2 = 6416.49 i =1 6-2 (272.82)2 19 6-3. Sample average: x= 84817 = 7068.1 yards 12 Sample variance: 12 xi = 84817 i =1 19 xi2 =600057949 i =1 2 n xi n 2 i x− s2 = (84817 )2 600057949 − i =1 i =1 n 12 = n −1 564324.92 = = 51302.265 11 12 − 1 ( yards )2 Sample standard deviation: s = 51302.265 = 226.5 yards The sample standard deviation could also be found using n s= ( x i − x )2 i =1 n −1 where 12 ( x i − x )2 = 564324.92 i =1 Dot Diagram: (rounding was used to create the dot diagram) . . : .. .. : : -+---------+---------+---------+---------+---------+-----C1 6750 6900 7050 7200 7350 7500 6-3 6-4. Sample mean: 18 xi x= i =1 = 18 2272 = 126.22 kN 18 Sample variance: 18 xi = 2272 i =1 18 xi2 =298392 i =1 2 n xi n 2 i x− s2 = i =1 i =1 n 298392 − = (2272)2 18 n −1 18 − 1 11615.11 = = 683.24 (kN) 2 17 Sample standard deviation: s = 683.24 = 26.14 kN The sample standard deviation could also be found using n s= ( x i − x )2 i =1 n −1 where 18 ( x i − x )2 = 11615.11 i =1 Dot Diagram: . : :: . :: . . :. . +---------+---------+---------+---------+---------+-------yield 90 105 120 135 150 165 6-5. Sample average: 6-4 x= 351.8 = 43.975 8 Sample variance: 8 xi = 351.8 i =1 19 xi2 =16528.403 i =1 2 n xi n 2 i x− s2 = (351.8)2 16528.043 − i =1 i =1 n 8 = n −1 1057.998 = = 151.143 7 8 −1 Sample standard deviation: s = 151.143 = 12.294 The sample standard deviation could also be found using n s= ( x i − x )2 i =1 n −1 where 8 (xi − x )2 = 1057.998 i =1 Dot Diagram: . . .. . .. . +---------+---------+---------+---------+---------+------24.0 32.0 40.0 48.0 56.0 64.0 6-5 6-6. Sample average: 35 xi x= i =1 = 35 28368 = 810.514 watts / m 2 35 Sample variance: 19 xi = 28368 i =1 19 xi2 = 23552500 i =1 2 n xi n 2 i x− s2 = i =1 i =1 n 23552500 − = n −1 = 16465.61 ( watts / m 2 ) 2 (28368)2 35 − 1 35 = 559830.743 34 Sample standard deviation: s = 16465.61 = 128.32 watts / m 2 The sample standard deviation could also be found using n s= ( x i − x )2 i =1 n −1 where 35 (xi − x )2 = 559830.743 i =1 6-7. µ= 6905 = 5.44 ; The value 5.44 is the population mean since the actual physical population 1270 of all flight times during the operation is available. 6-6 6-8 a.) Sample average: n xi x= i =1 = n 19.56 = 2.173 mm 9 b.) Sample variance: 9 xi = 19.56 i =1 9 xi2 =45.953 i =1 2 n xi n 2 i x− s2 = i =1 i =1 n n −1 = 0.4303 (mm) 2 45.953 − = (19.56)2 9 = 9 −1 3.443 8 Sample standard deviation: s = 0.4303 = 0.6560 mm c.) Dot Diagram . . . .. .. .. -------+---------+---------+---------+---------+---------crack length 1.40 1.75 2.10 2.45 2.80 3.15 6-9. Dot Diagram (rounding of the data is used to create the dot diagram) x .. . : .: ... . . .: :. . .:. .:: ::: -----+---------+---------+---------+---------+---------+-x 500 600 700 800 900 1000 The sample mean is the point at which the data would balance if it were on a scale. 6-7 a. Dot Diagram of CRT data in exercise 6-5 (Data were rounded for the plot) Dotplot for Exp 1-Exp 2 Exp 2 Exp 1 10 20 30 40 50 60 70 The data are centered a lot lower in the second experiment. The lower CRT resolution reduces the visual accommodation. n xi 6-11. a) x= i =1 = n 57.47 = 7.184 8 2 n xi n 2 i x− b) s2 = i =1 412.853 − n i =1 = n −1 s = 0.000427 = 0.02066 (57.47 )2 8 = 8 −1 0.00299 = 0.000427 7 c) Examples: repeatability of the test equipment, time lag between samples, during which the pH of the solution could change, and operator skill in drawing the sample or using the instrument. n xi 6-12 sample mean x= i =1 = n 748.0 = 83.11 drag counts 9 2 n xi n 2 i x− sample variance s2 = i =1 i =1 n −1 n 62572 − = (748.0)2 9 9 −1 404.89 = 50.61 drag counts 2 8 sample standard deviation s = 50.61 = 7.11 drag counts = Dot Diagram . ....... . ---+---------+---------+---------+---------+---------+---drag counts 75.0 80.0 85.0 90.0 95.0 100.0 6-8 6-13. a) x = 65.86 °F s = 12.16 °F b) Dot Diagram :: . . . . .. .: .: . .:..: .. :: .... .. -+---------+---------+---------+---------+---------+-----temp 30 40 50 60 70 80 c) Removing the smallest observation (31), the sample mean and standard deviation become x = 66.86 °F s = 10.74 °F Section 6-3 6-14 Stem-and-leaf display of octane rating Leaf Unit = 0.10 N = 83 83|4 represents 83.4 1 83|4 3 84|33 4 85|3 7 86|777 13 87|456789 24 88|23334556679 34 89|0233678899 (13) 90|0111344456789 36 91|00011122256688 22 92|22236777 14 93|023347 8 94|2247 4 95| 4 96|15 2 97| 2 98|8 1 99| 1 100|3 6-15 a.) Stem-and-leaf display for cycles to failure: unit = 100 1 1 5 10 22 33 (15) 22 11 5 2 1|2 represents 1200 0T|3 0F| 0S|7777 0o|88899 1*|000000011111 1T|22222223333 1F|444445555555555 1S|66667777777 1o|888899 2*|011 2T|22 b) No, only 5 out of 70 coupons survived beyond 2000 cycles. 6-9 6-16 Stem-and-leaf display of percentage of cotton N = 64 Leaf Unit = 0.10 32|1 represents 32.1% 1 6 9 17 24 (14) 26 17 12 9 5 3 6-17. Stem-and-leaf display for Problem 2-4.yield: unit = 1 1 1 7 21 38 (11) 41 27 19 7 1 6-18 32|1 32|56789 33|114 33|56666688 34|0111223 34|55666667777779 35|001112344 35|56789 36|234 36|6888 37|13 37|689 7o|8 8*| 8T|223333 8F|44444444555555 8S|66666666667777777 8o|88888999999 9*|00000000001111 9T|22233333 9F|444444445555 9S|666677 9o|8 Descriptive Statistics Variable Octane Rating 6-19. 6-21. N 83 Median Q1 90.400 88.600 Q3 92.200 Descriptive Statistics Variable cycles 6-20 1|2 represents 12 N 70 Median 1436.5 Q1 1097.8 Q3 1735.0 N Median Q1 90 89.250 86.100 Q3 93.125 median: ~ = 34.700 % x mode: 34.7 % sample average: x = 34.798 % Descriptive Statistics Variable yield 6-10 6-22 a.) sample mean: x = 65.811 inches standard deviation s = 2.106 inches b.) Stem-and-leaf display of female engineering student heights Leaf Unit = 0.10 61|0 represents 61.0 inches 1 3 5 9 17 (4) 16 8 3 1 N = 37 61|0 62|00 63|00 64|0000 65|00000000 66|0000 67|00000000 68|00000 69|00 70|0 x c.) median: ~ = 66.000 inches 6-23 Stem-and-leaf display for Problem 6-23. Strength: unit = 1.0 1|2 represents 12 1 1 2 4 6 11 22 28 39 47 (12) 41 32 22 12 8 5 532|9 533| 534|2 535|47 536|6 537|5678 538|12345778888 539|016999 540|11166677889 541|123666688 542|0011222357899 543|011112556 544|00012455678 545|2334457899 546|23569 547|357 548|11257 6-11 6-24 Stem-and-leaf of concentration N = 60 Leaf Unit = 1.0 2|2 represents 29 Note: Minitab has dropped the value to the right of the decimal to make this display. 1 2 3 8 12 20 (13) 27 22 13 7 3 1 2|9 3|1 3|9 4|22223 4|5689 5|01223444 5|5666777899999 6|11244 6|556677789 7|022333 7|6777 8|01 8|9 The data have a symmetrical bell-shaped distribution, and therefore may be normally distributed. n xi Sample Mean x = i =1 = n 3592.0 = 59.87 60 Sample Standard Deviation 60 60 x i = 3592 .0 x i2 = 224257 and i =1 i =1 2 n xi n 2 i x− s2 = i =1 n i =1 n −1 = 156 .20 224257 − = (3592 .0 )2 60 60 − 1 and s = 156 .20 = 12 .50 Sample Median Variable concentration ~ = 59.45 x N 60 Median 59.45 6-12 = 9215 .93 59 6-25 Stem-and-leaf display for Problem 6-25. Yard: unit = 1.0 Note: Minitab has dropped the value to the right of the decimal to make this display. 1 5 8 16 20 33 46 (15) 39 31 12 4 1 22 | 6 23 | 2334 23 | 677 24 | 00112444 24 | 5578 25 | 0111122334444 25 | 5555556677899 26 | 000011123334444 26 | 56677888 27 | 0000112222233333444 27 | 66788999 28 | 003 28 | 5 100 n xi Sample Mean x = i =1 xi = n i =1 100 = 26030 .2 = 260 .3 yards 100 Sample Standard Deviation 100 100 x i = 26030 .2 x i2 = 6793512 and i =1 i =1 2 n xi n 2 i x− s2 = i =1 ((26030 .2 ) 17798 .42 100 = 100 − 1 99 2 i =1 6793512 − n n −1 = 179 .782 yards 2 = and s = 179 .782 = 13 .41 yards Sample Median Variable yards N 100 Median 260.85 6-13 6-26 Stem-and-leaf of speed (in megahertz) N = 120 Leaf Unit = 1.0 63|4 represents 634 megahertz 2 7 16 35 48 (17) 55 36 24 17 5 3 1 1 63|47 64|24899 65|223566899 66|0000001233455788899 67|0022455567899 68|00001111233333458 69|0000112345555677889 70|011223444556 71|0057889 72|000012234447 73|59 74|68 75| 76|3 35/120= 29% exceed 700 megahertz. 120 xi Sample Mean x = i =1 = 120 82413 = 686.78 mhz 120 Sample Standard Deviation 120 120 x i = 82413 i =1 i =1 2 n xi n 2 i x− i =1 n n −1 = 658 .85 mhz 2 s2 = x i2 =56677591 and i =1 56677591 − = (82413 )2 120 − 1 and s = 658 .85 = 25 .67 mhz Sample Median Variable speed ~ = 683.0 mhz x N 120 Median 683.00 6-14 120 = 78402 .925 119 6-27 a.)Stem-and-leaf display of Problem 6-27. Rating: unit = 0.10 1|2 represents 1.2 1 2 5 7 9 12 18 (7) 15 8 4 3 2 83|0 84|0 85|000 86|00 87|00 88|000 89|000000 90|0000000 91|0000000 92|0000 93|0 94|0 95|00 b.) Sample Mean 40 n xi xi i =1 x= = n i =1 = 40 3578 = 89 .45 40 Sample Standard Deviation 40 40 xi = 3578 xi2 = 320366 and i =1 i =1 2 n xi n 2 i x− s2 = i =1 i =1 n n −1 320366 − = (3578 )2 40 40 − 1 = 313 .9 39 = 8 .05 and s = 8 .05 = 2 .8 Sample Median Variable rating N 40 Median 90.000 c.) 22/40 or 55% of the taste testers considered this particular Pinot Noir truly exceptional. 6-15 6-28 a.) Stem-and-leaf diagram of NbOCl3 N = 27 Leaf Unit = 100 0|4 represents 40 gram-mole/liter x 10-3 6 7 (9) 11 7 7 3 0|444444 0|5 1|001122233 1|5679 2| 2|5677 3|124 27 xi b.) sample mean x = i =1 27 = 41553 = 1539 gram − mole/liter x10 − 3 27 Sample Standard Deviation 27 27 x i = 41553 x i2 =87792869 and i =1 i =1 2 n xi n x i2 − s2 = i =1 (41553 )2 23842802 n 27 = = n −1 27 − 1 26 -3 917030 .85 = 957 .62 gram - mole/liter x 10 i =1 and s = ~ Sample Median x = 1256 Variable NbOCl3 6-29 87792869 − = 917030 .85 gram − mole/liter x10 −3 N 40 Median 1256 a.)Stem-and-leaf display for Problem 6-29. Height: unit = 0.10 1|2 represents 1.2 Female Students| Male Students 0|61 1 00|62 3 00|63 5 0000|64 9 00000000|65 17 2 65|00 0000|66 (4) 3 66|0 00000000|67 16 7 67|0000 00000|68 8 17 68|0000000000 00|69 3 (15) 69|000000000000000 0|70 1 18 70|0000000 11 71|00000 6 72|00 4 73|00 2 74|0 1 75|0 b.) The male engineering students are taller than the female engineering students. Also there is a slightly wider range in the heights of the male students. 6-16 Section 6-4 Frequency Tabulation for Exercise 6-14.Octane Data -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below 81.75 0 .0000 0 .0000 1 81.75 84.25 83.0 1 .0120 1 .0120 2 84.25 86.75 85.5 6 .0723 7 .0843 3 86.75 89.25 88.0 19 .2289 26 .3133 4 89.25 91.75 90.5 33 .3976 59 .7108 5 91.75 94.25 93.0 18 .2169 77 .9277 6 94.25 96.75 95.5 4 .0482 81 .9759 7 96.75 99.25 98.0 1 .0120 82 .9880 8 99.25 101.75 100.5 1 .0120 83 1.0000 above 101.75 0 .0000 83 1.0000 -------------------------------------------------------------------------------Mean = 90.534 Standard Deviation = 2.888 Median = 90.400 30 Frequency 6-30 20 10 0 83.0 85.5 88.0 90.5 93.0 octane data 6-17 95.5 98.0 100.5 6-31. Frequency Tabulation for Exercise 6-15.Cycles -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below .000 0 .0000 0 .0000 1 .000 266.667 133.333 0 .0000 0 .0000 2 266.667 533.333 400.000 1 .0143 1 .0143 3 533.333 800.000 666.667 4 .0571 5 .0714 4 800.000 1066.667 933.333 11 .1571 16 .2286 5 1066.667 1333.333 1200.000 17 .2429 33 .4714 6 1333.333 1600.000 1466.667 15 .2143 48 .6857 7 1600.000 1866.667 1733.333 12 .1714 60 .8571 8 1866.667 2133.333 2000.000 8 .1143 68 .9714 9 2133.333 2400.000 2266.667 2 .0286 70 1.0000 above 2400.000 0 .0000 70 1.0000 -------------------------------------------------------------------------------Mean = 1403.66 Standard Deviation = 402.385 Median = 1436.5 Frequency 15 10 5 0 500 750 1000 1250 1500 1750 number of cycles to failure 6-18 2000 2250 6-32 Frequency Tabulation for Exercise 6-16.Cotton content -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below 31.0 0 .0000 0 .0000 1 31.0 32.0 31.5 0 .0000 0 .0000 2 32.0 33.0 32.5 6 .0938 6 .0938 3 33.0 34.0 33.5 11 .1719 17 .2656 4 34.0 35.0 34.5 21 .3281 38 .5938 5 35.0 36.0 35.5 14 .2188 52 .8125 6 36.0 37.0 36.5 7 .1094 59 .9219 7 37.0 38.0 37.5 5 .0781 64 1.0000 8 38.0 39.0 38.5 0 .0000 64 1.0000 above 39.0 0 .0000 64 1.0000 -------------------------------------------------------------------------------Mean = 34.798 Standard Deviation = 1.364 Median = 34.700 Frequency 20 10 0 31.5 32.5 33.5 34.5 35.5 36.5 cotton percentage 6-19 37.5 38.5 6-33. Frequency Tabulation for Exercise 6-17.Yield -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below 77.000 0 .0000 0 .0000 1 77.000 79.400 78.200 1 .0111 1 .0111 2 79.400 81.800 80.600 0 .0000 1 .0111 3 81.800 84.200 83.000 11 .1222 12 .1333 4 84.200 86.600 85.400 18 .2000 30 .3333 5 86.600 89.000 87.800 13 .1444 43 .4778 6 89.000 91.400 90.200 19 .2111 62 .6889 7 91.400 93.800 92.600 9 .1000 71 .7889 8 93.800 96.200 95.000 13 .1444 84 .9333 9 96.200 98.600 97.400 6 .0667 90 1.0000 10 98.600 101.000 99.800 0 .0000 90 1.0000 above 101.000 0 .0000 90 1.0000 -------------------------------------------------------------------------------Mean = 89.3756 Standard Deviation = 4.31591 Median = 89.25 Frequency 15 10 5 0 78 80 82 84 86 88 90 92 yield 6-20 94 96 98 Frequency Tabulation for Exercise 6-14.Octane Data -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below 83.000 0 .0000 0 .0000 1 83.000 84.125 83.5625 1 .0120 1 .0120 2 84.125 85.250 84.6875 2 .0241 3 .0361 3 85.250 86.375 85.8125 1 .0120 4 .0482 4 86.375 87.500 86.9375 5 .0602 9 .1084 5 87.500 88.625 88.0625 13 .1566 22 .2651 6 88.625 89.750 89.1875 8 .0964 30 .3614 7 89.750 90.875 90.3125 16 .1928 46 .5542 8 90.875 92.000 91.4375 15 .1807 61 .7349 9 92.000 93.125 92.5625 9 .1084 70 .8434 10 93.125 94.250 93.6875 7 .0843 77 .9277 11 94.250 95.375 94.8125 2 .0241 79 .9518 12 95.375 96.500 95.9375 2 .0241 81 .9759 13 96.500 97.625 97.0625 0 .0000 81 .9759 14 97.625 98.750 98.1875 0 .0000 81 .9759 15 98.750 99.875 99.3125 1 .0120 82 .9880 16 99.875 101.000 100.4375 1 .0120 83 1.0000 above 101.000 0 .0000 83 1.0000 -------------------------------------------------------------------------------Mean = 90.534 Standard Deviation = 2.888 Median = 90.400 20 Frequency 6-34 10 0 84 86 88 90 92 94 96 98 100 octane The histograms have the same shape. Not much information is gained by doubling the number of bins. 6-21 6-35 Frequency Tabulation for Problem 6-22. Height Data -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below 60.500 0 .0000 0 .0000 1 60.500 61.500 61.000 1 .0270 1 .0270 2 61.500 62.500 62.000 2 .0541 3 .0811 3 62.500 63.500 63.000 2 .0541 5 .1351 4 63.500 64.500 64.000 4 .1081 9 .2432 5 64.500 65.500 65.000 8 .2162 17 .4595 6 65.500 66.500 66.000 4 .1081 21 .5676 7 66.500 67.500 67.000 8 .2162 29 .7838 8 67.500 68.500 68.000 5 .1351 34 .9189 9 68.500 69.500 69.000 2 .0541 36 .9730 10 69.500 70.500 70.000 1 .0270 37 1.0000 above 70.500 0 .0000 37 1.0000 -------------------------------------------------------------------------------Mean = 65.811 Standard Deviation = 2.106 Median = 66.0 8 7 Frequency 6 5 4 3 2 1 0 61 62 63 64 65 66 67 68 69 70 height The histogram for the spot weld shear strength data shows that the data appear to be normally distributed (the same shape that appears in the stem-leaf-diagram). 20 Frequency 6-36 10 0 5320 5340 5360 5380 5400 5420 5440 5460 5480 5500 shear strength 6-22 6-37 Frequency Tabulation for exercise 6-24. Concentration data -------------------------------------------------------------------------------Lower Upper Relative Cumulative Cum. Rel. Class Limit Limit Midpoint Frequency Frequency Frequency Frequency -------------------------------------------------------------------------------at or below 29.000 0 .0000 0 .0000 1 29.0000 37.000 33.000 2 .0333 2 .0333 2 37.0000 45.000 41.000 6 .1000 8 .1333 3 45.0000 53.000 49.000 8 .1333 16 .2667 4 53.0000 61.000 57.000 17 .2833 33 .5500 5 61.0000 69.000 65.000 13 .2167 46 .7667 6 69.0000 77.000 73.000 8 .1333 54 .9000 7 77.0000 85.000 81.000 5 .0833 59 .9833 8 85.0000 93.000 89.000 1 .0167 60 1.0000 above 93.0000 0 .0800 60 1.0000 -------------------------------------------------------------------------------Mean = 59.87 Standard Deviation = 12.50 Median = 59.45 Frequency 20 10 0 25 35 45 55 65 75 85 95 concentration Yes, the histogram shows the same shape as the stem-and-leaf display. Yes, the histogram of the distance data shows the same shape as the stem-and-leaf display in exercise 6-25. 20 Frequency 6-38 10 0 220 228 236 244 252 260 268 276 284 292 distance 6-23 6-39 Histogram for the speed data in exercise 6-26. Yes, the histogram of the speed data shows the same shape as the stem-and-leaf display in exercise 6-26 Frequency 20 10 0 620 720 770 speed (megahertz) Yes, the histogram of the wine rating data shows the same shape as the stem-and-leaf display in exercise 6-27. 7 6 5 Frequency 6-40 670 4 3 2 1 0 85 90 rating 6-24 95 6-41 Pareto Chart for Automobile Defects 81 64.8 80.2 86.4 96.3 91.4 100 100.0 80 72.8 scores 48.6 60 63.0 percent of total 32.4 40 37.0 16.2 20 0 contour holes/slots lubrication pits trim assembly dents deburr 0 Roughly 63% of defects are described by parts out of contour and parts under trimmed. Section 6-5 Descriptive Statistics of O-ring joint temperature data Variable Temp N 36 Mean 65.86 Variable Temp Minimum 31.00 Median 67.50 Maximum 84.00 TrMean 66.66 Q1 58.50 StDev 12.16 SE Mean 2.03 Q3 75.00 a.) Lower Quartile: Q1=58.50 Upper Quartile: Q3=75.00 b.) Median = 67.50 c.) Data with lowest point removed Variable Temp N 35 Variable Temp Mean 66.86 Minimum 40.00 Median 68.00 Maximum 84.00 TrMean 67.35 Q1 60.00 StDev 10.74 SE Mean 1.82 Q3 75.00 The mean and median have increased and the standard deviation and difference between the upper and lower quartile has decreased. d.)Box Plot - The box plot indicates that there is an outlier in the data. 90 80 70 Temp 6-42 60 50 40 30 6-25 6-43. Descriptive Statistics Variable PMC Variable PMC N 20 Min 2.000 Mean 4.000 Max 5.200 Median 4.100 Q1 3.150 Tr Mean 4.044 Q3 4.800 StDev 0.931 SE Mean 0.208 a) Sample Mean: 4 b) Sample Variance: 0.867 Sample Standard Deviation: 0.931 c) PMC 5 4 3 2 Descriptive Statistics Variable time N 8 Variable time Mean 2.415 Minimum 1.750 Median 2.440 Maximum 3.150 a.)Sample Mean: 2.415 Sample Standard Deviation: 0.543 b.) Box Plot – There are no outliers in the data. 3.0 time 6-44 2.5 2.0 6-26 TrMean 2.415 Q1 1.912 StDev 0.534 Q3 2.973 SE Mean 0.189 6-45. Descriptive Statistics Variable Temperat Variable Temperat N 9 Min 948.00 Mean 952.44 Max 957.00 Median 953.00 Q1 949.50 Tr Mean 952.44 Q3 955.00 StDev 3.09 SE Mean 1.03 a) Sample Mean: 952.44 Sample Variance: 9.55 Sample Standard Deviation: 3.09 b) Median: 953; Any increase in the largest temperature measurement will not affect the median. c) 957 956 Temperatur 955 954 953 952 951 950 949 948 Descriptive statistics Variable drag coefficients N Variable drag coefficients 9 Mean 83.11 Minimum 74.00 Median 82.00 Maximum 100.00 a.) Upper quartile: Q1 =79.50 Lower Quartile: Q3=84.50 b.) 100 drag coefficients 6-46 90 80 6-27 TrMean 83.11 Q1 79.50 StDev 7.11 Q3 84.50 SE Mean 2.37 c.) Variable drag coefficients N 8 Mean 81.00 Variable drag coefficients Median 81.50 Minimum 74.00 TrMean 81.00 Maximum 85.00 StDev 3.46 Q1 79.25 SE Mean 1.22 Q3 83.75 drag coefficients 85 80 75 Removing the largest observation (100) lowers the mean and median. Removing this “outlier also greatly reduces the variability as seen by the smaller standard deviation and the smaller difference between the upper and lower quartiles. 6-47. Descriptive Statistics Variable temperat Variable temperat N 24 Min 43.000 Mean 48.125 Max 52.000 Median 49.000 Q1 46.000 Tr Mean 48.182 Q3 50.000 StDev 2.692 SE Mean 0.549 a) Sample Mean: 48.125 Sample Median: 49 b) Sample Variance: 7.246 Sample Standard Deviation: 2.692 c) 52 51 temperatur 50 49 48 47 46 45 44 43 The data appear to be slightly skewed. 4-48 The golf course yardage data appear to be skewed. Also, there is an outlying data point above 7500 yards. 6-28 7500 7400 yardage 7300 7200 7100 7000 6900 6800 6-49 Boxplot for problem 6-49 100 Octane 95 90 85 This plot conveys the same basic information as the stem and leaf plot but in a different format. The outliers that were separated from the main portion of the stem and leaf plot are shown here separated from the whiskers. 6-50 The box plot shows that the data are symmetrical about the mean. It also shows that there are no outliers in the data. These are the same interpretations seen in the stem-leaf-diagram. 6-29 shear strength 5490 5440 5390 5340 6-51 Boxplot for problem 6-51 70 Height in inches 69 68 67 66 65 64 63 62 61 This plot, as the stem and leaf one, indicates that the data fall mostly in one region and that the measurements toward the ends of the range are more rare. 6-52 The box plot and the stem-leaf-diagram show that the data are very symmetrical about the mean. It also shows that there are no outliers in the data. 6-30 90 concentration 80 70 60 50 40 30 6-53 Boxplot for problem 6-53 285 Distance in yards 275 265 255 245 235 225 The plot indicates that most balls will fall somewhere in the 250-275 range. In general, the population is grouped more toward the high end of the region. This same type of information could have been obtained from the stem and leaf graph of problem 6-25. 6-54 The box plot shows that the data are not symmetrical about the mean. The data are skewed to the right and have a longer right tail (at the lower values). It also shows that there is an outlier in the data. These are the same interpretations seen in the stem-leaf-diagram. 6-31 rating 95 90 85 6-55 Boxplot for problem 6-55 75 70 Height 65 60 Female students Male students We can see that the two distributions seem to be centered at different values. 6-32 6-56 The box plot shows that there is a difference between the two formulations. Formulation 2 has a higher mean cold start ignition time and a larger variability in the values of the start times. The first formulation has a lower mean cold start ignition time and is more consistent. Care should be taken, though since these box plots for formula 1 and formula 2 are made using 8 and 10 data points respectively. More data should be collected on each formulation to get a better determination. 3.5 seconds 3.0 2.5 2.0 formulation 1 formulation 2 Section 6-6 Time Series Plot 13 12 11 10 Res_Time 6-57. 9 8 7 6 5 4 Index 5 10 15 Computer response time appears random. No trends or patterns are obvious. 6-33 20 6-58 a.) Stem-leaf-plot of viscosity N = 40 Leaf Unit = 0.10 2 12 16 16 16 16 16 16 16 17 (4) 19 7 42 43 43 44 44 45 45 46 46 47 47 48 48 89 0000112223 5566 2 5999 000001113334 5666689 The stem-leaf-plot shows that there are two “different” sets of data. One set of data is centered about 43 and the second set is centered about 48. The time series plot shows that the data starts out at the higher level and then drops down to the lower viscosity level at point 24. Each plot gives us a different set of information. b.) If the specifications on the product viscosity are 48.0±2, then there is a problem with the process performance after data point 24. An investigation needs to take place to find out why the location of the process has dropped from around 48.0 to 43.0. The most recent product is not within specification limits. 49 48 viscosity 47 46 45 44 43 Index 6-59. 10 20 30 a) 6-34 40 260 250 240 230 Force 220 210 200 190 180 170 In d e x 10 20 30 b) Stem-and-leaf display for Problem 2-23.Force: unit = 1 3 6 14 18 (5) 17 14 10 3 40 1|2 represents 12 17|558 18|357 19|00445589 20|1399 21|00238 22|005 23|5678 24|1555899 25|158 In the time series plot there appears to be a downward trend beginning after time 30. The stem and leaf plot does not reveal this. 6-60 18 17 17 17 17 17 16 16 16 16 | | | | | | | | | | 1 888 6667 44444444444445555 2233333 000011111 8899 567 3 1 18 concentration 1 4 8 25 (7) 18 9 5 2 1 17 16 Index 10 20 30 40 6-35 50 6-61 a) Sunspots 150 100 50 0 Index 10 20 30 40 50 60 70 80 90 100 b) Stem-and-leaf display for Problem 2-25.Sunspots: unit = 1 17 29 39 50 50 38 33 23 20 16 10 8 7 4 1|2 represents 12 0|01224445677777888 1|001234456667 2|0113344488 3|00145567789 4|011234567788 5|04579 6|0223466778 7|147 8|2356 9|024668 10|13 11|8 12|245 13|128 HI|154 The data appears to decrease between 1790 and 1835, the stem and leaf plot indicates skewed data. a.) Time Series Plot 16 miles flown 6-62 11 6 Index 10 20 30 40 50 60 70 80 Each year the miles flown peaks during the summer hours. The number of miles flown increased over the years 1964 to 1970. b.) Stem-and-leaf of miles fl N = 84 6-36 Leaf Unit = 0.10 1 10 22 33 (18) 33 24 13 6 2 1 6 7 8 9 10 11 12 13 14 15 16 7 246678889 013334677889 01223466899 022334456667888889 012345566 11222345779 1245678 0179 1 2 When grouped together, the yearly cycles in the data are not seen. The data in the stem-leafdiagram appear to be nearly normally distributed. Section 6-7 6-63 Normal Probability Plot for 6-1 Piston Ring Diameter ML Estimates - 95% CI 99 ML Estimates Mean 74.0044 StDev 95 0.0043570 90 Percent 80 70 60 50 40 30 20 10 5 1 73.99 74.00 74.01 74.02 Data The pattern of the data indicates that the sample may not come from a normally distributed population or that the largest observation is an outlier. Note the slight bending downward of the sample data at both ends of the graph. 6-37 6-64 Normal Probability Plot for time Exercise 6-2 Data 99 ML Estimates Mean 14.3589 StDev 95 18.3769 90 Percent 80 70 60 50 40 30 20 10 5 1 -40 -20 0 20 40 60 Data It appears that the data do not come from a normal distribution. Very few of the data points fall on the line. 6-65 There is no evidence to doubt that data are normally distributed Normal Probability Plot for 6-5 Visual Accomodation Data ML Estimates - 95% CI 99 ML Estimates Mean 90 Percent 80 70 60 50 40 30 20 10 5 1 0 10 20 30 40 Data 6-38 50 60 70 80 43.975 StDev 95 11.5000 6-66 Normal Probability Plot for temperature Data from exercise 6-13 99 ML Estimates Mean 65.8611 StDev 95 11.9888 90 Percent 80 70 60 50 40 30 20 10 5 1 30 40 50 60 70 80 90 100 Data The data appear to be normally distributed. Although, there are some departures from the line at the ends of the distribution. 6-67 Normal Probability Plot for 6-14 Octane Rating ML Estimates - 95% CI ML Estimates Mean Percent 95 90 80 70 60 50 40 30 20 10 5 1 80 90 100 Data 6-39 90.5256 StDev 99 2.88743 There are a few points outside the confidence limits, indicating that the sample is not perfectly normal. These deviations seem to be fairly small though. 6-68 Normal Probability Plot for cycles to failure Data from exercise 6-15 ML Estimates Mean 1282.78 StDev 99 539.634 Percent 95 90 80 70 60 50 40 30 20 10 5 1 0 1000 2000 3000 Data The data appear to be normally distributed. Although, there are some departures from the line at the ends of the distribution. 6-69 Normal Probability Plot for 6-27 Wine Quality Rating ML Estimates - 95% CI 99 ML Estimates Mean 89.45 StDev 95 2.80134 90 Percent 80 70 60 50 40 30 20 10 5 1 82 87 92 97 Data The data seem to be normally distributed. Notice that there are clusters of observations because of the discrete nature of the ratings. 6-40 6-70 Normal Probability Plot for concentration Data from exercise 6-24 99 ML Estimates Mean 59.8667 StDev 95 12.3932 90 Percent 80 70 60 50 40 30 20 10 5 1 25 35 45 55 65 75 85 95 Data The data appear to be normally distributed. Nearly all of the data points fall very close to, or on the line. 6-71 Normal Probability Plot for 6-22...6-29 ML Estimates - 95% CI 99 Female Students 95 Male Students 90 Percent 80 70 60 50 40 30 20 10 5 1 60 65 70 75 Data Both populations seem to be normally distributed, moreover, the lines seem to be roughly parallel indicating that the populations may have the same variance and differ only in the value of their mean. 6-72 Yes, it is possible to obtain an estimate of the mean from the 50th percentile value of the normal probability plot. The fiftieth percentile point is the point at which the sample mean should equal the population mean and 50% of the data would be above the value and 50% below. An estimate of the standard deviation would be to subtract the 50th percentile from the 64th percentile These values are based on the values from the ztable that could be used to estimate the standard deviation. 6-41 Supplemental Exercises 6-73. a) Sample Mean = 65.083 The sample mean value is close enough to the target value to accept the solution as conforming. There is a slight difference due to inherent variability. b) s2 = 1.86869 s = 1.367 A major source of variability would include measurement to measurement error. A low variance is desirable since it may indicate consistency from measurement to measurement. 6 6-74 2 6 2 i x = 10,433 a) xi i =1 = 62,001 n=6 i =1 2 6 xi 6 2 i x− s2 = i =1 n i =1 62,001 6 = 19.9Ω 2 6 −1 10,433 − = n −1 s = 19.9Ω 2 = 4.46Ω 6 2 6 2 i x = 353 b) = 1521 xi i =1 n=6 i =1 2 6 xi 6 2 i x− s2 = i =1 n i =1 1,521 6 = 19.9Ω 2 6 −1 353 − = n −1 s = 19.9Ω 2 = 4.46Ω Shifting the data from the sample by a constant amount has no effect on the sample variance or standard deviation. 6 2 6 xi2 = 1043300 c) xi i =1 n=6 2 6 xi 6 2 i x− s2 = = 6200100 i =1 i =1 i =1 n −1 n 6200100 6 = 1990Ω 2 6 −1 1043300 − = s = 1990Ω 2 = 44.61Ω Yes, the rescaling is by a factor of 10. Therefore, s2 and s would be rescaled by multiplying s2 by 102 (resulting in 1990Ω2) and s by 10 (44.6Ω). 6-42 6-75 a) Sample 1 Range = 4 Sample 2 Range = 4 Yes, the two appear to exhibit the same variability b) Sample 1 s = 1.604 Sample 2 s = 1.852 No, sample 2 has a larger standard deviation. c) The sample range is a relatively crude measure of the sample variability as compared to the sample standard deviation since the standard deviation uses the information from every data point in the sample whereas the range uses the information contained in only two data points - the minimum and maximum. 6-76 a.) It appears that the data may shift up and then down over the 80 points. 17 viscosity 16 15 14 13 Index 10 20 30 40 50 60 70 80 b.)It appears that the mean of the second set of 40 data points may be slightly higher than the first set of 40. c.) Descriptive Statistics: viscosity 1, viscosity 2 Variable Viscosity1 Viscosity2 N 40 40 Mean 14.875 14.923 Median 14.900 14.850 TrMean 14.875 14.914 There is a slight difference in the mean levels and the standard deviations. 6-77 6-43 StDev 0.948 1.023 SE Mean 0.150 0.162 Comparative boxplots for viscosity data 17 16 Viscosity 15 14 13 6-76 First half 6-76 Second half Both sets of data appear to have the same mean although the first half seem to be concentrated a little more tightly. Two data points appear as outliers in the first half of the data. 6-78 15 sales 10 5 0 Index 10 20 30 40 50 60 70 80 90 There appears to be a cyclic variation in the data with the high value of the cycle generally increasing. The high values are during the winter holiday months. b) We might draw another cycle, with the peak similar to the last year’s data (1969) at about 12.7 thousand bottles. 6-79 a)Stem-and-leaf display for Problem 2-35: unit = 1 6-44 1|2 represents 12 1 8 18 (7) 15 12 7 5 3 0T|3 0F|4444555 0S|6666777777 0o|8888999 1*|111 1T|22233 1F|45 1S|77 1o|899 b) Sample Average = 9.325 Sample Standard Deviation = 4.4858 c) 20 springs 15 10 5 Index 10 20 30 40 The time series plot indicates there was an increase in the average number of nonconforming springs made during the 40 days. In particular, the increase occurs during the last 10 days. 6-80 a.) Stem-and-leaf of errors Leaf Unit = 0.10 3 10 10 6 1 0 1 2 3 4 N = 20 000 0000000 0000 00000 0 b.) Sample Average = 1.700 Sample Standard Deviation = 1.174 c.) 6-45 4 errors 3 2 1 0 Index 5 10 15 20 The time series plot indicates a slight decrease in the number of errors for strings 16 - 20. 6-81 Normal Probability Plot for 6-76 First h...6-76 Second ML Estimates - 95% CI 99 6-76 First half 95 6-76 Second half 90 Percent 80 70 60 50 40 30 20 10 5 1 12 13 14 15 16 17 18 Data Both sets of data appear to be normally distributed and with roughly the same mean value. The difference in slopes for the two lines indicates that a change in variance might have occurred. This could have been the result of a change in processing conditions, the quality of the raw material or some other factor. 6-46 6-82 Normal Probability Plot for Temperature 99 ML Estimates Mean 48.125 StDev 95 2.63490 90 Percent 80 70 60 50 40 30 20 10 5 1 40 45 50 55 Data There appears to be no evidence that the data are not normally distributed. There are some repeat points in the data that cause some points to fall off the line. 6-83 Normal Probability Plot for 6-44...6-56 ML Estimates - 95% CI 6-44 99 6-56 95 90 Percent 80 70 60 50 40 30 20 10 5 1 0.5 1.5 2.5 3.5 4.5 Data Although we do not have sufficient data points to really see a pattern, there seem to be no significant deviations from normality for either sample. The large difference in slopes indicates that the variances of the populations are very different. 6-47 8-84 a.) Normal Probability Plot for No. of Cycles 99 ML Estimates Mean 1051.88 StDev 95 1146.43 90 Percent 80 70 60 50 40 30 20 10 5 1 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 Data The data do not appear to be normally distributed. There is a curve in the line. b.) Normal Probability Plot for y* 99 ML Estimates Mean 2.75410 StDev 95 0.499916 90 Percent 80 70 60 50 40 30 20 10 5 1 1 2 3 4 Data After the transformation y*=log(y), the normal probability plot shows no evidence that the data are not normally distributed. 6-85 6-48 Boxplot for Exercise 6-85 1100 1000 900 800 700 600 Trial 1 - - Trial 3 Trial 4 Trial 5 There is a difference in the variability of the measurements in the trials. Trial 1 has the most variability in the measurements. Trial 3 has a small amount of variability in the main group of measurements, but there are four outliers. Trial 5 appears to have the least variability without any outliers. All of the trials except Trial 1 appear to be centered around 850. Trial 1 has a higher mean value All five trials appear to have measurements that are greater than the “true” value of 734.5. The difference in the measurements in Trial 1 may indicate a “start-up” effect in the data. There could be some bias in the measurements that is centering the data above the “true” value. a.) Descriptive Statistics Variable N Mean density 29 5.4541 Median 5.4600 TrMean 5.4611 StDev 0.4072 b.) There does appear to be a low outlier in the data. Normal Probability Plot for density data ML Estimates - 95% CI 99 ML Estimates Mean 5.38517 StDev 95 0.379670 90 Goodness of Fit 80 Percent 6-86 Trial 2 AD* 70 60 50 40 30 20 10 5 1 4.5 5.5 Data 6-49 6.5 1.837 SE Mean 0.0756 c.)Due to the very low data point at 4.07, the mean may be lower than it should be. Therefore, the median would be a better estimate of the density of the earth. The median is not affected by outliers. Mind Expanding Exercises 9 2 9 2 i x = 62572 6-87 = 559504 xi i =1 n=9 i =1 2 9 xi 9 2 i x− s2 = i =1 i =1 559504 9 = 50.61 9 −1 62572 − n = n −1 s = 50.61 = 7.11 Subtract 30 and multiply by 10 9 2 9 2 i x = 2579200 = 22848400 xi i =1 n=9 i =1 2 9 xi 9 i =1 2 i x− s2 = n i =1 n −1 s = 5061.1 = 71.14 22848400 9 = 5061.1 9 −1 2579200 − = Yes, the rescaling is by a factor of 10. Therefore, s2 and s would be rescaled by multiplying s2 by 102 (resulting in 5061.1) and s by 10 (71.14). Subtracting 30 from each value has no affect on the variance or standard deviation. This is because n n ( xi − a) 2 = 6-88 V (aX + b) = a 2V ( X ) . i =1 ( xi − x ) + n( x − a ) 2 ; The sum written in this form shows that the quantity is i =1 minimized when a = x . n 6-89 Of the two quantities (xi − x )2 and i =1 given that x≠µ. This is because n (xi − µ )2 n , the quantity i =1 (xi − x )2 will be smaller i =1 x is based on the values of the xi ’s. different for this sample. 6-50 The value of µ may be quite yi = a + bxi 6-90 n n i =1 x= 2 y= and n n i =1 and n −1 sx = sx n i =1 n −1 Therefore, x = 835.00 °F 6-91 n = a + bx 2 (a + bxi − a + bx ) 2 sy = xi i =1 = n n 2 na + b (xi − x )2 i =1 sx = n (a + bxi ) xi n (bxi − bx ) 2 = i =1 n −1 b2 = ( xi − x ) 2 i =1 n −1 = b2 sx 2 s y = bs x s x = 10.5 °F The results in °C: y = −32 + 5 / 9 x = −32 + 5 / 9(835.00) = 431.89 °C 2 2 s y = b 2 s x = (5 / 9) 2 (10.5) 2 = 34.028 °C 6-92 Using the results found in Exercise 6-90 with a = − x and b = 1/s, the mean and standard deviation s of the zi are z = 0 and sZ = 1. 6-93. Yes, in this case, since no upper bound on the last electronic component is available, use a measure of central location that is not dependent on this value. That measure is the median. Sample Median = x ( 4) + x (5) 2 = 63 + 75 = 69 hours 2 6-51 n +1 n xi + x n +1 xi 6-94 a) i =1 x n +1 = x n +1 x n +1 i =1 = n +1 n +1 nx n + x n +1 = n +1 x n = x n + n +1 n +1 n +1 2 n xi + x n +1 n b) ns 2 n +1 2 1 = x +x 2 n +1 i =1 − n +1 i =1 2 n n xi n 2 i = x +x 2 n +1 2 x n +1 i =1 − − n +1 i =1 xi i =1 − n +1 2 x n+1 n +1 2 n n xi n2 n x+ x n +1 − 2x n +1 x n − n +1 n +1 i =1 2 i = i =1 n 2 i = x− ( 2 ( n 2 i x+ xi ) 2 n i =1 n 2 i = x− + n +1 i =1 = xi ) ( i =1 2 n = (n − 1) s + xi ) − n ( n 2 x n +1 − 2 x n +1 x n n +1 ( [ xi ) 2 − n (n + 1)( 2 + n +1 ( xi ) 2 n +1 x i ) − n( 2 2 n(n + 1) + n 2 x n +1 − 2 x n x n n +1 [ nx 2 n 2 + x n +1 − 2 x n x n n +1 n +1 n 2 = (n − 1) s n + x n +1 − 2 x n x n + x n2 n +1 2 n 2 2 = (n − 1) s n + xn+1 − xn n +1 [ ( ( + ) ) 6-52 n 2 x n +1 − 2 x n x n n +1 [ xi ) 2 n(n + 1) xi ) 2 = (n − 1) s n + + n 2 x n +1 − 2 x n x n n +1 [ c) xn = 65.811 inches xn+1 = 64 2 s n = 4.435 n = 37 s n = 2.106 37(65.81) + 64 = 65.76 x n +1 = 37 + 1 37 (37 − 1)4.435 + (64 − 65.811) 2 37 + 1 s n +1 = 37 = 2.098 6-95. The trimmed mean is pulled toward the median by eliminating outliers. a) 10% Trimmed Mean = 89.29 b) 20% Trimmed Mean = 89.19 Difference is very small c) No, the differences are very small, due to a very large data set with no significant outliers. 6-96. If nT/100 is not an integer, calculate the two surrounding integer values and interpolate between the two. For example, if nT/100 = 2/3, one could calculate the mean after trimming 2 and 3 observations from each end and then interpolate between these two means. 6-53 CHAPTER 7 Section 7-2 2n 7-1. () E X1 = E Xi i =1 2n = 2n 1 E 2n Xi = i =1 1 (2nµ ) = µ 2n n () E X2 = E Xi i =1 n = 1 E n () The variances are V X 1 = n Xi = i =1 σ2 1 (nµ ) = µ , n () and V X 2 = 2n σ2 n X 1 and X 2 are unbiased estimators of µ. ; compare the MSE (variance in this case), ˆ MSE (Θ1 ) σ 2 / 2n n 1 =2 = = ˆ 2n 2 σ /n MSE (Θ 2 ) Since both estimators are unbiased, examination of the variances would conclude that X1 is the “better” estimator with the smaller variance. 7-2. () 1 ˆ E Θ1 = [E ( X 1 ) + E ( X 2 ) + 7 + E ( X 7 )] = 1 1 (7 E ( X )) = (7 µ ) = µ 7 7 () 1 1 ˆ E Θ 2 = [E (2 X 1 ) + E ( X 6 ) + E ( X 7 )] = [2 µ − µ + µ ] = µ 2 2 ˆ ˆ a) Both Θ1 and Θ 2 are unbiased estimates of µ since the expected values of these statistics are equivalent to the true mean, µ. X + X 2 + ... + X 7 1 ˆ b) V Θ1 = V 1 = 2 (V ( X 1 ) + V ( X 2 ) + 7 7 () + V ( X 7 )) = 1 1 (7σ 2 ) = σ 2 49 7 2 σ ˆ V (Θ1 ) = 7 () 2 X1 − X 6 + X 4 1 1 ˆ = 2 (V (2 X 1 ) + V ( X 6 ) + V ( X 4 ) ) = (4V ( X 1 ) + V ( X 6 ) + V ( X 4 )) V Θ2 = V 2 4 2 1 = 4σ 2 + σ 2 + σ 2 4 1 = (6σ 2 ) 4 ( ) 3σ 2 ˆ V (Θ 2 ) = 2 Since both estimators are unbiased, the variances can be compared to decide which is the better ˆ ˆˆ estimator. The variance of Θ is smaller than that of Θ , Θ is the better estimator. 1 7-3. 2 1 ˆ ˆ Since both Θ1 and Θ 2 are unbiased, the variances of the estimators can be examined to determine which ˆ is the “better” estimator. The variance of θ2 is smaller than that of θ1 thus θ2 may be the better estimator. ˆ MSE (Θ1 ) V (Θ1 ) 10 Relative Efficiency = = = = 2 .5 ˆ ˆ MSE (Θ 2 ) V (Θ 2 ) 4 7-4. Since both estimators are unbiased: 7-1 Relative Efficiency = ˆ ˆ MSE (Θ1 ) V (Θ1 ) σ 2 / 7 2 = = = 2 ˆ ˆ MSE (Θ 2 ) V (Θ 2 ) 3σ / 2 21 7-5. ˆ ˆ MSE (Θ1 ) V (Θ1 ) 10 = = = 2 .5 ˆ ) V (Θ ) 4 ˆ MSE (Θ 2 2 7-6. ˆ E (Θ1 ) = θ ˆ E (Θ 2 ) = θ / 2 ˆ Bias = E (Θ 2 ) − θ = θ θ −θ = − 2 2 ˆ V (Θ1 ) = 10 ˆ V (Θ 2 ) = 4 ˆ For unbiasedness, use Θ1 since it is the only unbiased estimator. As for minimum variance and efficiency we have: ˆ (V (Θ1 ) + Bias 2 )1 Relative Efficiency = where, Bias for θ1 is 0. ˆ ) + Bias 2 ) (V (Θ 2 2 Thus, Relative Efficiency = (10 + 0) 4+ −θ 2 2 = 40 (16 + θ ) 2 ˆ If the relative efficiency is less than or equal to 1, Θ1 is the better estimator. ˆ Use Θ1 , when 40 (16 + θ2 ) ≤1 40 ≤ (16 + θ2 ) 24 ≤ θ 2 θ ≤ −4.899 or θ ≥ 4.899 ˆ If −4.899 < θ < 4.899 then use Θ 2 . ˆ ˆ ˆ For unbiasedness, use Θ1 . For efficiency, use Θ1 when θ ≤ −4.899 or θ ≥ 4.899 and use Θ 2 when −4.899 < θ < 4.899 . 7-7. ˆ E (Θ1 ) = θ No bias ˆ ˆ V (Θ1 ) = 12 = MSE (Θ1 ) ˆ E (Θ 2 ) = θ No bias ˆ ˆ V (Θ 2 ) = 10 = MSE (Θ 2 ) ˆ ˆ E (Θ 3 ) ≠ θ Bias MSE (Θ 3 ) = 6 [note that this includes (bias2)] To compare the three estimators, calculate the relative efficiencies: ˆ MSE (Θ1 ) 12 ˆ = = 1.2 , since rel. eff. > 1 use Θ 2 as the estimator for θ ˆ MSE (Θ 2 ) 10 ˆ MSE (Θ1 ) 12 = =2, ˆ) 6 MSE (Θ 3 ˆ since rel. eff. > 1 use Θ 3 as the estimator for θ ˆ MSE (Θ 2 ) 10 = = 1.8 , ˆ MSE (Θ 3 ) 6 ˆ since rel. eff. > 1 use Θ 3 as the estimator for θ Conclusion: ˆ ˆ Θ 3 is the most efficient estimator with bias, but it is biased. Θ 2 is the best “unbiased” estimator. 7-2 7-8. n1 = 20, n2 = 10, n3 = 8 Show that S2 is unbiased: 2 2 20S12 + 10S 2 + 8S3 38 () E S2 = E (( )( ) ( )) 1 2 2 E 20 S12 + E 10 S 2 + E 8S3 38 1 2 2 = 20σ 12 + 10σ 2 + 8σ 3 38 1 = 38σ 2 38 =σ2 = ( ) ( ) ∴ S2 is an unbiased estimator of σ2 . n 7-9. (X 2 i −X ) i =1 Show that is a biased estimator of σ2 : n a) n = = − X) 2 i i =1 E = (X n n 1 E n (X i =1 n 1 n − nX ) 2 i () () E X i2 − nE X 2 i =1 n 1 n (µ 2 ) +σ 2 − n µ2 + ( =σ2 − ∴ (X n i =1 1 nµ 2 + nσ 2 − nµ 2 − σ 2 n 1 = ((n − 1)σ 2 ) n = σ2 i σ2 −X n b) Bias = E ) n ) 2 is a biased estimator of σ2 . (X 2 i − nX n ) 2 −σ 2 = σ 2 − σ2 n −σ 2 = − c) Bias decreases as n increases. 7-3 σ2 n 7-10 () 2 a) Show that X 2 is a biased estimator of µ. Using E X2 = V( X) + [ E ( X)] () E X2 = = = = = 1 n2 n2 n2 Xi i =1 n 2 n Xi + E i =1 nσ 2 + n2 n2 1 E V 1 1 2 n 1 Xi i =1 2 n µ i =1 (nσ 2 + (nµ ) 2 ) (nσ 2 + n2µ 2 ) () E X2 = σ2 n + µ2 ∴ X 2 is a biased estimator of µ.2 () b) Bias = E X 2 − µ 2 = σ2 + µ2 − µ2 = n c) Bias decreases as n increases. 7-11 σ2 n a.) The average of the 26 observations provided can be used as an estimator of the mean pull force since we know it is unbiased. This value is 75.615 pounds. b.) The median of the sample can be used as an estimate of the point that divides the population into a “weak” and “strong” half. This estimate is 75.2 pounds. c.) Our estimate of the population variance is the sample variance or 2.738 square pounds. Similarly, our estimate of the population standard deviation is the sample standard deviation or 1.655 pounds. d.) The standard error of the mean pull force, estimated from the data provided is 0.325 pounds. This value is the standard deviation, not of the pull force, but of the mean pull force of the population. e.) Only one connector in the sample has a pull force measurement under 73 pounds. Our point estimate for the proportion requested is then 1/26 = 0.0385 7-4 7-12. Descriptive Statistics Variable N Oxide Thickness 24 Mean 423.33 Median 424.00 TrMean 423.36 StDev 9.08 SE Mean 1.85 a) The mean oxide thickness, as estimated by Minitab from the sample, is 423.33 Angstroms. b) Standard deviation for the population can be estimated by the sample standard deviation, or 9.08 Angstroms. c) The standard error of the mean is 1.85 Angstroms. d) Our estimate for the median is 424 Angstroms. e) Seven of the measurements exceed 430 Angstroms, so our estimate of the proportion requested is 7/24 = 0.2917 7.13 a) 1 1 1 1 1 θ E ( X ) = x (1 + θ x)dx = xdx + x 2 dx 2 2 2 −1 −1 −1 = 0+ θ 3 = θ 3 X be the sample average of the observations in the random sample. We know that E ( X ) = µ , the mean of the distribution. However, the mean of the distribution is /3, so θˆ = 3 X is an unbiased estimator of . b) Let 7.14 1 1 E ( X ) = np = p n n p ⋅ (1 − p ) ˆ b.) We know that the variance of p is so its standard error must be n a.) ˆ E ( p ) = E ( X n) = p ⋅ (1 − p ) . To n estimate this parameter we would substitute our estimate of p into it. 7.15 a.) b.) E ( X1 − X 2 ) = E ( X1) − E ( X 2 ) = µ1 − µ 2 s.e. = V ( X 1 − X 2 ) = V ( X 1 ) + V ( X 2 ) + 2COV ( X 1 , X 2 ) = σ 12 n1 + σ 22 n2 This standard error could be estimated by using the estimates for the standard deviations of populations 1 and 2. 7-5 7-16 2 E (S p ) = E (n1 − 1) ⋅ S12 + (n 2 − 1) ⋅ S 2 2 1 2 2 (n1 − 1) E (S1 ) + (n 2 − 1) ⋅ E ( S 2 ) = = n1 + n2 − 2 n1 + n 2 − 2 [ [ = 7-17 n + n2 − 2 2 1 2 2 (n1 − 1) ⋅ σ 1 + (n2 − 1) ⋅ σ 2 ) = 1 σ =σ2 n1 + n 2 − 2 n1 + n 2 − 2 a.) ˆ E(µ) = E(αX1 + (1 − α ) X 2 ) = αE( X1) + (1 − α )E( X 2 ) = αµ + (1 − α )µ = µ b.) ˆ s.e.( µ ) = V (α X 1 + (1 − α ) X 2 ) = α 2V ( X 1 ) + (1 − α ) 2 V ( X 2 ) = α2 =σ1 c.) σ 12 n1 + (1 − α ) 2 σ 22 = α2 n2 σ 12 n1 + (1 − α ) 2 a σ 12 n2 α 2 n 2 + (1 − α ) 2 an 1 n1 n 2 The value of alpha that minimizes the standard error is: α= an1 n2 + an1 b.) With a = 4 and n1=2n2, the value of alpha to choose is 8/9. The arbitrary value of α=0.5 is too small and will result in a larger standard error. With α=8/9 the standard error is ˆ s.e.( µ ) = σ 1 (8 / 9) 2 n 2 + (1 / 9) 2 8 n 2 = 2n 2 2 0 .667 σ 1 n2 If α=0.5 the standard error is ˆ s.e.( µ ) = σ 1 7-18 a.) E( ( 0 .5 ) 2 n 2 + ( 0 .5 ) 2 8 n 2 2n 2 2 = 1 .0607 σ 1 n2 X1 X 2 1 1 1 1 − )= E( X 1 ) − E ( X 2 ) = n1 p1 − n 2 p 2 = p1 − p 2 = E ( p1 − p 2 ) n1 n2 n1 n2 n1 n2 p1 (1 − p1 ) p 2 (1 − p 2 ) + n1 n2 b.) c.) An estimate of the standard error could be obtained substituting for p2 in the equation shown in (b). d.) Our estimate of the difference in proportions is 0.01 e.) The estimated standard error is 0.0413 7-6 X1 for n1 p1 and X2 n2 Section 7-3 n 7-19. e − λ λ xi e − nλ λ L (λ ) = ∏ = n xi ! i =1 e − λ λx f ( x) = x! n i =1 xi ∏x ! i i =1 n ln L(λ ) = −nλ ln e + n x i ln λ − i =1 d ln L(λ ) 1 = −n + λ dλ ln x i ! i =1 n xi ≡ 0 i =1 n xi i =1 −n+ =0 λ n x i = nλ i =1 n xi ˆ λ= i =1 n n n 7-20. f ( x) = λe − λ ( x −θ ) for x ≥ θ −λ n L(λ ) = ∏ λe − λ ( x −θ ) = λne −λ ( x −θ ) i =1 = λn e x − nθ i =1 i =1 n ln L(λ , θ ) = − n ln λ − λ x i − λnθ i =1 d ln L(λ , θ ) nn = − − x i − nθ ≡ 0 λ i =1 dλ n n −= x i − nθ λ i =1 ˆ λ =n/ n x i − nθ i =1 ˆ λ= 1 x −θ n − Let λ = 1 then L(θ ) = e i =1 ( xi−θ ) for xi ≥ 0 n − xi −nθ L(θ ) = e i=1 = e nθ − nx LnL(θ ) = nθ − nx θ cannot be estimated using ML equations since dLnL(θ ) = 0 . Therefore, θ is estimated using Min( X 1 , X 2 , d (θ ) , Xn) . ˆ Ln(θ) is maximized at xmin and θ = xmin b.) Example: Consider traffic flow and let the time that has elapsed between one car passing a fixed point and the instant that the next car begins to pass that point be considered time headway. This headway can be modeled by the shifted exponential distribution. Example in Reliability: Consider a process where failures are of interest. Say that a sample or population is put into operation at x = 0, but no failures will occur until θ period of operation. Failures will occur after the time θ. 7-7 7-21. f ( x) = p (1 − p ) x −1 n L( p ) = ∏ p (1 − p ) xi −1 i =1 n xi − n = p n (1 − p ) i =1 n ln L( p ) = n ln p + xi − n ln(1 − p ) i =1 n x −n ∂ ln L( p) n i =1 i =− p ∂p 1− p ≡0 n (1 − p )n − p xi − n i =1 0= p (1 − p ) n n − np − p xi + pn i =1 0= p (1 − p ) n 0=n− p xi i =1 n ˆ p= n xi i =1 7-22. f ( x) = (θ + 1) xθ n θ θ θ L(θ ) = ∏ (θ + 1) xi = (θ + 1) x1 × (θ + 1) x2 × ... i =1 n = (θ + 1) n ∏ xiθ i =1 ln L(θ ) = n ln(θ + 1) + θ ln x1 + θ ln x2 + ... n = n ln(θ + 1) + θ ln xi i =1 n n ∂ ln L (θ ) = + ln xi = 0 ∂θ θ + 1 i =1 n n =− ln xi θ +1 i =1 n ˆ θ= n −1 − ln xi i =1 7-8 7-23. a) β −1 n β xi L( β , δ ) = ∏ δ i =1 δ n =e i =1 ln L( β , δ ) = ln β −1 n βx ∏ δ δi i =1 δ n δ e β xi − β xi − [ ( ) ]− xi β −1 xi δ β δ β δ i =1 β = n ln( δ ) + ( β − 1) ln ( )− ( ) xi xi β δ δ b) ∂ ln L( β , δ ) n =+ ∂β β ln ( )− ln( )( ) xi xi xi β δ δ δ β xi ∂ ln L( β , δ ) n n = − − ( β − 1) + β β +1 ∂δ δ δ δ ∂ ln L( β , δ ) Upon setting equal to zero, we obtain ∂δ β δ n= xi β and n n β + and + ln xi − n ln δ = ln xi − 1 β = δ= 1/ β β ∂ ln L( β , δ ) equal to zero and substituting for δ, we obtain ∂β Upon setting β xi n n β 1 δβ β ln xiβ ln xi xiβ xiβ (ln xi − ln δ ) n xiβ xi n = + n xiβ β ln xi n xiβ ln xi − xiβ 1 β ln xi n c) Numerical iteration is required. 7-24 7-25 θˆ = 1 n n θ (θ + 1) xi = i =1 E( X ) = (θ + 1) n a−0 1 n = Xi = X 2 n i =1 n xi θ i =1 , therefore: ˆ a = 2X The expected value of this estimate is the true parameter so it must be unbiased. This estimate is reasonable in one sense because it is unbiased. However, there are obvious problems. Consider ˆ the sample x1=1, x2=2 and x3=10. Now x =4.37 and a = 2 x = 8.667 . This is an unreasonable estimate of a, because clearly a ≥ 10. 7-9 7-26. ˆ a) a cannot be unbiased since it will always be less than a. b) bias = na a (n + 1) a − =− → 0. n +1 n +1 n + 1 n →∞ c) 2 X y a d) P(Y≤y)=P(X1, …,Xn ≤ y)=(P(X1≤y))n= bias=E(Y)-a = 7-27 n . Thus, f(y) is as given. Thus, an a −a =− . n +1 n +1 For any n>1 n(n+2) > 3n so the variance of better than the first. ˆ a2 ˆ is less than that of a1 . It is in this sense that the second estimator is 7-28 n x e− x θ L(θ ) = ∏ i θ2 i ln L(θ ) = ln( xi ) − i =1 ∂ ln L(θ ) 1 = ∂θ θ2 xi − xi θ − 2n ln θ 2n θ setting the last equation equal to zero and solving for theta, we find: n xi ˆ = i =1 θ 2n 7-29 a) E ( X 2 ) = 2θ = ˆ Θ= 1n 2 Xi 2n i =1 n 1n 2 X i so n i =1 xi e − x 2θ b) L(θ ) = ∏ i =1 2 i θ ∂ ln L(θ ) 1 = ∂θ 2θ 2 xi 2 − ln L(θ ) = ln( xi ) − xi 2 − n ln θ 2θ n θ setting the last equation equal to zero, we obtain the maximum likelihood estimate ˆ Θ= 1n 2 Xi 2n i =1 which is the same result we obtained in part (a) 7-10 c) a f ( x)dx = 0.5 = 1 − e − a / 2θ 2 0 a = − 2θ ln(0.5) We can estimate the median (a) by substituting our estimate for theta into the equation for a. 1 x2 ) = 2c a) c (1 + θx ) dx = 1 = (cx + cθ 2 −1 −1 1 7-30 so the constant c should equal 0.5 b) E( X ) = θ = 3⋅ 1n θ Xi = n i =1 3 1n Xi n i =1 c) n 1 θ E (θˆ) = E 3 ⋅ X i = E (3 X ) = 3E ( X ) = 3 = θ n i =1 3 d) n1 L(θ ) = ∏ (1 + θX i ) i =1 2 n 1 ln L(θ ) = n ln( ) + ln(1 + θX i ) 2 i =1 ∂ ln L(θ ) n Xi = ∂θ i =1(1 + θX i ) by inspection, the value of θ that maximizes the likelihood is max (Xi) 7-11 7-31 a)Using the results from Example 7-12 we obtain that the estimate of the mean is 423.33 and the estimate of the variance is 82.4464 b) 0 -200000 log L -400000 -600000 7 8 9 Std. Dev. 10 11 12 13 410 14 415 420 425 430 435 Mean 15 The function seems to have a ridge and its curvature is not too pronounced. The maximum value for std deviation is at 9.08, although it is difficult to see on the graph. 7-32 When n is increased to 40, the graph will look the same although the curvature will be more pronounced. As n increases it will be easier to determine the where the maximum value for the standard deviation is on the graph. Section 7-5 7-33. P(1.009 ≤ X ≤ 1.012) = P 1.009−1.01 0.003 / 9 ≤ X −µ σ/ n ≤ 1.012−1.01 0.003 / 9 = P(−1 ≤ Z ≤ 2) = P( Z ≤ 2) − P( Z ≤ −1) = 0.9772 − 0.1586 = 0.8186 7-34. X i ~ N (100,102 ) µ X = 100 σ X = n = 25 σ n = 10 =2 25 P[(100 − 1.8(2)) ≤ X ≤ (100 + 2)] = P(96.4 ≤ X ≤ 102) =P 96.4 −100 2 ≤ X −µ σ/ n ≤ 102 −100 2 = P ( −1.8 ≤ Z ≤ 1) = P ( Z ≤ 1) − P ( Z ≤ −1.8) = 0.8413 − 0.0359 = 0.8054 7-12 7-35. µ X = 75.5 psi σ X = σ = 3.5 = 1.429 6 n P ( X ≥ 75 . 75 ) = P ( X −µ σ/ n ≥ 75 .75 − 75 . 5 1 . 429 ) = P ( Z ≥ 0 . 175 ) = 1 − P ( Z ≤ 0 . 175 ) = 1 − 0 .56945 = 0 . 43055 7-36. n=6 σX = σX 7-37. n = 49 σ = 3 .5 n 6 = 1.429 is reduced by 0.929 psi σX = σ n = 0.5 = 3.5 49 Assuming a normal distribution, σ 50 µ X = 2500 σ X = = = 22.361 n 5 P (2499 ≤ X ≤ 2510) = P ( 2499− 2500 22.361 ≤ σX/− µn ≤ 2510− 2500 22.361 ) = P (−0.045 ≤ Z ≤ 0.45) = P ( Z ≤ 0.45) − P ( Z ≤ −0.045) = 0.6736 − 0.4821 = 0.1915 7-38. 7-39. σX = σ 50 = = 22.361 psi = standard error of X 5 n σ 2 = 25 σ σX = n 2 n= σ σX 2 5 1.5 = 11.11 ~ 12 = 7-13 7-40. Let Y = X − 6 a + b (0 + 1) = = µX = 2 2 µX = µX 2 σX = 2 σX = (b − a )2 = 12 σ2 n = 1 12 12 1 2 1 12 1 144 = 1 σ X = 12 µ Y = 1 − 6 = −5 1 2 2 2 1 σ Y = 144 1 Y = X − 6 ~ N (−5 1 , 144 ) , approximately, using the central limit theorem. 2 7-41. n = 36 µX = a + b (3 + 1) = =2 2 2 σX = (b − a + 1)2 − 1 = 12 µ X = 2, σ X = z= 2/3 36 = (3 − 1 + 1) 2 − 1 = 8= 12 12 2 3 2/3 6 X −µ σ/ n Using the central limit theorem: P(2.1 < X < 2.5) = P 2.1− 2 2/3 6 <Z< 2. 5 − 2 2/3 6 = P(0.7348 < Z < 3.6742) = P( Z < 3.6742) − P( Z < 0.7348) = 1 − 0.7688 = 0.2312 7-42. µ X = 8.2 minutes σ X = 15 minutes . n = 49 σX = σX = n 15 . = 0.2143 49 µ X = µ X = 8.2 mins Using the central limit theorem, X is approximately normally distributed. 10 − 8.2 a) P ( X < 10) = P ( Z < ) = P ( Z < 8.4) = 1 0.2143 5 − 8. 2 b) P (5 < X < 10) = P ( 0.2143 < Z < 10 −8.2 ) 0.2143 = P ( Z < 8.4) − P ( Z < −14.932) = 1 − 0 = 1 c) P ( X < 6) = P ( Z < 6 − 8.2 ) = P ( Z < −10.27) = 0 0.2143 7-14 7-43. n1 = 16 n2 = 9 µ1 = 75 σ1 = 8 µ 2 = 70 σ 2 = 12 X 1 − X 2 ~ N (µ X − µ X , σ 2 + σ 2 ) 1 X1 2 2 ~ N ( µ1 − µ 2 , σ1 n1 + 82 ~ N (75 − 70, 16 + X2 2 σ2 n2 12 2 9 ) ) ~ N (5,20) a) P ( X 1 − X 2 > 4) P( Z > 4 −5 ) 20 = P ( Z > −0.2236) = 1 − P ( Z ≤ −0.2236) = 1 − 0.4115 = 0.5885 b) P (3.5 ≤ X 1 − X 2 ≤ 5.5) − P ( 3.5205 ≤ Z ≤ 5.5 −5 20 ) = P ( Z ≤ 0.1118) − P ( Z ≤ −0.3354) = 0.5445 − 0.3687 = 0.1759 7-44. If µ B = µ A , then X B − X A is approximately normal with mean 0 and variance Then, P ( X B − X A > 3.5) = P ( Z > 3.5 − 0 20.48 2 σB 25 + 2 σA 25 = 20.48 . ) = P ( Z > 0.773) = 0.2196 The probability that X B exceeds X A by 3.5 or more is not that unusual when µ B and µ A are equal. Therefore, there is not strong evidence that µ B is greater than µ A . 7-45. Assume approximate normal distributions. 2 ( X high − X low ) ~ N (60 − 55, 4 + 16 42 ) 16 ~ N (5, 2) P ( X high − X low ≥ 2) = P ( Z ≥ 2 −5 ) 2 = 1 − P ( Z ≤ −2.12) = 1 − 0.0170 = 0.983 Supplemental Exercises 1 7-46. f ( x1 , x 2 , x3 , x 4 , x5 ) = 7-47. f ( x1 , x 2 ,..., x n ) = ∏ λe −λxi 2π σ − e 5 ( x −µ )2 i 2σ 2 i =1 n for x1 > 0, x 2 > 0,..., x n > 0 i =1 7-48. f ( x1 , x 2 , x 3 , x 4 ) = 1 for 0 ≤ x1 ≤ 1,0 ≤ x 2 ≤ 1,0 ≤ x 3 ≤ 1,0 ≤ x 4 ≤ 1 7-49. 1.5 2 2 2 X 1 − X 2 ~ N (100 − 105, +) 25 30 ~ N (−5,0.2233) 7-15 7-50. σ X1 − X 2 = 0.2233 = 0.4726 7-51. X ~ N (50,144) P ( 47 ≤ X ≤ 53) = P 47 − 50 12 / 36 ≤Z≤ 53− 50 12 / 36 = P ( −1.5 ≤ Z ≤ 1.5) = P ( Z ≤ 1.5) − P ( Z ≤ −1.5) = 0.9332 − 0.0668 = 0.8664 7-52. No, because Central Limit Theorem states that with large samples (n ≥ 30), X is approximately normally distributed. 7-53. Assume X is approximately normally distributed. P ( X > 4985) = 1 − P ( X ≤ 4985) = 1 − P ( Z ≤ 4985 − 5500 ) 100 / 9 = 1 − P ( Z ≤ −15.45) = 1 − 0 = 1 7-54. t= X −µ s/ n = 52 − 50 2 / 16 = 5.6569 t .05,15 = 1.753 . Since 5.33 >> t.05,15 , the results are very unusual. 7-55. P ( X ≤ 37) = P ( Z ≤ −5.36) = 0 7-56. Binomial with p equal to the proportion of defective chips and n = 100. 7-57. E (aX 1 + (1 − a ) X 2 = aµ + (1 − a ) µ = µ V ( X ) = V [aX 1 + (1 − a ) X 2 ] = a 2V ( X 1 ) + (1 − a) 2V ( X 2 ) 2 2 1 2 = a 2 ( σ ) + (1 − 2a + a 2 )( σ ) n n = a 2σ 2 σ 2 2aσ 2 a 2σ 2 + − + n2 n2 n2 n1 2 = (n2a 2 + n1 − 2n1a + n1a 2 )( σ ) n1n2 2 ∂V ( X ) = ( σ )(2n2 a − 2n1 + 2n1a) ≡ 0 n1n2 ∂a 0 = 2n2a − 2n1 + 2n1a 2a (n2 + n1 ) = 2n1 a (n2 + n1 ) = n1 a= n1 n2 + n1 7-16 7-58 L(θ ) = 2θ n −x i =1 θ n 1 3 ln L(θ ) = n ln e i n ∏ xi 2 i =1 n 1 2θ +2 3 n ln xi − i =1 xi i =1 θ ∂ ln L(θ ) − 3n n xi = + 2 θ ∂θ i =1θ Making the last equation equal to zero and solving for theta, we obtain: n ˆ Θ= xi i =1 3n as the maximum likelihood estimate. 7-59 n L(θ ) = θ n ∏ xiθ −1 i =1 n ln L(θ ) = n lnθ + (θ − 1) ln( xi ) i =1 ∂ ln L(θ ) n n = + ln( xi ) ∂θ θ i =1 making the last equation equal to zero and solving for theta, we obtain the maximum likelihood estimate. ˆ Θ= −n n ln( xi ) i =1 7-17 7-60 L(θ ) = 1n θ 1−θ x n∏ i θ i =1 ln L(θ ) = − n ln θ + 1−θ n θ ln( xi ) i =1 ∂ ln L(θ ) n 1n =− − ln( xi ) θ θ 2 i =1 ∂θ making the last equation equal to zero and solving for the parameter of interest, we obtain the maximum likelihood estimate. ˆ Θ =− 1n ln( xi ) n i =1 1 ˆ E (θ ) = E − n = 1 n n ln( xi ) = i =1 n θ= i =1 1 E− n n ln( xi ) = − i =1 n E[ln( xi )] i =1 nθ =θ n 1−θ 1 E (ln( X i )) = (ln x ) x θ dx 1−θ let 0 1 then, 1 n E (ln( X )) = −θ x 1−θ θ dx = −θ 0 7-18 u = ln x and dv = x θ dx Mind-Expanding Exercises P ( X 1 = 0, X 2 = 0) = M ( M − 1) N ( N − 1) P ( X 1 = 0, X 2 = 1) = 7-61. M (N − M ) N ( N − 1) ( N − M )M N ( N − 1) ( N − M )( N − M − 1) P ( X 1 = 1, X 2 = 1) = N ( N − 1) P ( X 1 = 1, X 2 = 0) = P ( X 1 = 0) = M / N P ( X 1 = 1) = N − M N P ( X 2 = 0) = P ( X 2 = 0 | X 1 = 0) P ( X 1 = 0) + P ( X 2 = 0 | X 1 = 1) P ( X 1 = 1) M −1 M M N −M M × + × = N −1 N N −1 N N P ( X 2 = 1) = P ( X 2 = 1 | X 1 = 0) P ( X 1 = 0) + P ( X 2 = 1 | X 1 = 1) P ( X 1 = 1) = = N − M M N − M −1 N − M N −M × + × = N −1 N N −1 N N Because P ( X 2 = 0 | X 1 = 0) = 7-62 M −1 N −1 is not equal to P ( X 2 = 0) = M , X1 N and X 2 are not independent. a) cn = Γ[(n − 1) / 2] Γ(n / 2) 2 /(n − 1) b.) When n = 10, cn = 1.0281. When n = 25, cn = 1.0105. So S is a pretty good estimator for the standard deviation even when relatively small sample sizes are used. 7-19 7-63 (a) Z i = Yi − X i ; so Z i is N(0, 2σ 2 ). Let σ *2 = 2σ 2 . The likelihood function is n L(σ * ) = ∏ 1 2 i =1 − σ * 2π e 1 2σ *2 1 ( zi2 ) n 2 − zi 1 2σ *2 i=1 = e (σ *2 2π ) n / 2 n 1 ln[ L(σ *2 )] = − (2πσ *2 ) − 2 2σ *2 The log-likelihood is n zi2 i =1 Finding the maximum likelihood estimator: d ln[ L(σ *2 )] n 1 =− + 2 dσ * 2σ * 2σ *4 n zi2 = 0 i =1 n nσ *2 = zi2 i =1 ˆ σ *2 = 1 n n zi2 = i =1 1 n n ( yi − xi ) 2 i =1 But σ *2 = 2σ 2 , so the MLE is 1 n 1 ˆ σ2 = 2n n ˆ 2σ 2 = ( yi − xi ) 2 i =1 n ( yi − xi ) 2 i =1 b) ˆ E (σ 2 ) = E 1 2n 1 = 2n 1 2n i =1 E (Yi − X i ) 2 i =1 n E (Yi 2 − 2Yi X i + X i2 ) i =1 n [ E (Yi 2 ) − E (2Yi X i ) + E ( X i2 )] = = (Yi − X i ) 2 n = 1 2n 1 = 2n n i =1 n [σ 2 − 0 + σ 2 ] i =1 2nσ 2 2n =σ2 So the estimator is unbiased. 7-20 7-64. 1 from Chebyshev's inequality. c2 1 Then, P | X − µ |< cσ ≥ 1 − 2 . Given an ε, n and c can be chosen sufficiently large that the last probability n c is near 1 and c σ is equal to ε. P | X − µ |≥ cσ n ≤ n 7-65 ( )( ) P(X (1) > t ) = P (X > t for i = 1,..., n ) = [1 − F (t )] Then, P(X (1) ≤ t ) = 1 − [1 − F (t )] P X ( n) ≤ t = P X i ≤ t for i = 1,..., n = [ F (t )]n n i n ∂ F X (t ) = n[1 − F (t )] n −1 f (t ) ∂t (1) ∂ f X ( n ) (t ) = F X ( n ) (t ) = n[ F (t )] n −1 f (t ) ∂t f X (1) (t ) = 7-66 7-67. ( P (X ) = 1) = 1 − F (0) = 1 − [ F (0)] P X (1) = 0 = F X (1) (0 ) = 1 − [1 − F (0 )] n = 1 − p n because F(0) = 1 - p. (n) n t −µ σ t −µ − n −1 σ { [ ]} f X ( n ) (t ) = n Φ 7-68. = 1 − (1 − p ) [ ]. From Exercise 7-65, 1 (t ) = n{ − Φ [ ]} 1 e 2π σ P ( X ≤ t ) = F (t ) = Φ f X (1) n X (n) t −µ σ n −1 1 2π σ − e (t − µ ) 2 2σ 2 (t − µ ) 2 2σ 2 P ( X ≤ t ) = 1 − e − λt . From Exercise 7-65, F X (1) (t ) = 1 − e − nλt f X (1) (t ) = nλe − nλt F X ( n ) (t ) = [1 − e − λt ] n f X ( n ) (t ) = n[1 − e − λt ] n −1 λe − λt 7-21 7-69. ))( (( ) P F X ( n ) ≤ t = P X ( n ) ≤ F −1 (t ) = t n from Exercise 7-65 for 0 ≤ t ≤ 1 . 1 If Y = F ( X ( n) ) , then fY ( y ) = ny n −1 ,0 ≤ y ≤ 1 . Then, E (Y ) = ny n dy = ( ) P (F (X (1) ) > t ) = P X (1) < F −1 (t ) = 1 − (1 − t ) n 0 n n +1 from Exercise 7-65 for 0 ≤ t ≤ 1 . If Y = F ( X (1) ) , then fY ( y ) = n(1 − t ) n −1,0 ≤ y ≤ 1 . 1 Then, E (Y ) = yn(1 − y )n −1 dy = 0 E[ F ( X ( n ) )] = n −1 7-70. E (V ) = k 1 where integration by parts is used. Therefore, n +1 n 1 and E[ F ( X (1) )] = n +1 n +1 [ E ( X i2+1 ) + E ( X i2 ) − 2 E ( X i X i +1 )] i =1 n −1 =k (σ 2 + µ 2 + σ 2 + µ 2 − 2 µ 2 ) Therefore, k = i =1 1 2( n −1) = k (n − 1)2σ 2 7-71 a.)The traditional estimate of the standard deviation, S, is 3.26. The mean of the sample is 13.43 so the values of X i − X corresponding to the given observations are 3.43, 1.43, 4.43, 0.57, 4.57, 1.57 and 2.57. The median of these new quantities is 2.57 so the new estimate of the standard deviation is 3.81; slightly larger than the value obtained with the traditional estimator. b.) Making the first observation in the original sample equal to 50 produces the following results. The traditional estimator, S, is equal to 13.91. The new estimator remains unchanged. 7-72 a.) Tr = X 1 + X1 + X 2 − X1 + X1 + X 2 − X1 + X 3 − X 2 + ... + X 1 + X 2 − X 1 + X 3 − X 2 + ... + X r − X r −1 + ( n − r )( X 1 + X 2 − X 1 + X 3 − X 2 + ... + X r − X r −1 ) 1 . nλ Then, X2 − X1 is the minimum lifetime of (n-1) items from the memoryless property of the exponential and 1 E ( X 2 − X1 ) = . (n − 1)λ Because X1 is the minimum lifetime of n items, E ( X 1 ) = Similarly, E ( X k − X k −1 ) = E (Tr ) = 1 . Then, (n − k + 1)λ n n −1 n − r +1 r T 1 + + ... + = and E r = = µ (n − r + 1)λ λ nλ (n − 1)λ r λ b.) V (Tr / r ) = 1 /(λ2r ) is related to the variance of the Erlang distribution V ( X ) = r / λ2 . They are related by the value (1/r2) . The censored variance is (1/r2) times the uncensored variance. 7-22 Section 7.3.3 on CD S7-1 From Example S7-2 the posterior distribution for µ is normal with mean (σ 2 / n) µ 0 + σ 02 x σ 02 + σ 2 / n and σ /(σ 2 / n) . σ σ 2 /n 2 0 variance 2 0 + The Bayes estimator for µ goes to the MLE as n increases. This follows since σx σ σ /n 2 2 goes to 0, and the estimator approaches 0 2 (the σ 02 ’s cancel). Thus, in the limit ˆ µ = x. 0 S7-2 f (x | µ) = a) Because is f ( x, µ ) = 1 e 2π σ 1 e (b − a ) 2π σ − − ( x−µ )2 2σ 2 ( x−µ )2 2σ 2 and f (µ ) = 1 for a ≤ µ ≤ b , the joint distribution b−a for -∞ < x < ∞ and a ≤ µ ≤ b . Then, 2 b ( x− µ ) − 1 1 2 f ( x) = e 2σ dµ and this integral is recognized as a normal probability. Therefore, b − a a 2π σ 1 [Φ(bσ− x ) − Φ( aσ− x )] where Φ(x) is the standard normal cumulative distribution function. f ( x) = b−a Then, f ( µ | x) = − f ( x, µ ) = f ( x) b) The Bayes estimator is ~ µ= e 2σ 2 − − 2π σ [Φ ( bσ x ) − Φ ( aσ x )] − b ( x− µ )2 ( x−µ )2 µ e σ dµ . − − 2π σ [Φ( bσ x ) − Φ( aσ x )] a 22 Let v = (x - µ). Then, dv = - dµ and ~ µ= x−a ( x − v )e v2 2σ 2 − − x[Φ( xσ a ) − Φ ( xσ b )] [Φ( σ ) − Φ( σ )] Let w = dv − − 2π σ [Φ( bσ x ) − Φ( aσ x )] x −b = − b− x v2 2σ 2 a− x . Then, x−a − x −b ve 2σ 2 ( x −b) 2 dw = [ 22v2 ]dv = [ σv2 ]dv σ − − 2π [Φ(bσ x ) − Φ( aσ x )] 2σ 2 = x+ ( x −a)2 − ( x −b) 2 −e σ σe b−x − 2π Φ( σ ) − Φ( aσ x ) 2σ 2 dv − 2π σ [Φ ( σ ) − Φ ( aσ x )] σe− wdw − v2 2σ 2 b− x ( x−a)2 ~ µ = x− − 22 7-23 and S7-3. a) e − λ λx f ( x) = x! for x = 0, 1, 2, and (m + 1)m +1 λm+ x e f ( x, λ ) = f (λ ) = m + 1 m +1 λ0 λm e − ( m +1) λλ 0 Γ(m + 1) for λ > 0. Then, − λ − ( m +1) λλ 0 λm+1Γ(m + 1) x! 0 . This last density is recognized to be a gamma density as a function of λ. Therefore, the posterior m +1 distribution of λ is a gamma distribution with parameters m + x + 1 and 1 + . λ0 b) The mean of the posterior distribution can be obtained from the results for the gamma distribution to be m + x +1 1+ S7-4 λ0 = x = 4.85 a) = x = 5.05 ~ µ= 9 25 (4) + 1(4.85) 9 25 +1 = 4.625 The Bayes estimate appears to underestimate the mean. a) From Example S7-2, ˆ b.) µ S7-6. m + x +1 m + λ0 + 1 = λ0 a) From Example S7-2, the Bayes estimate is ˆ b.) µ S7-5. m +1 ~ µ= 1 (0.01)(5.03) + ( 25 )(5.05) = 5.046 1 0.01 + 25 The Bayes estimate is very close to the MLE of the mean. f ( x | λ ) = λ e − λx , x ≥ 0 and f (λ ) = 0.01e −0.01λ . Then, f ( x1 , x 2 , λ ) = λ 2 e − λ ( x1 + x2 ) 0.01e −0.01λ = 0.01λ 2 e − λ ( x1 + x2 + 0.01) . As a function of λ, this is recognized as a gamma density with parameters 3 and x1 + x 2 + 0.01 . Therefore, the posterior mean for λ is ~ λ= 3 3 = = 0.00133 . x1 + x2 + 0.01 2 x + 0.01 1000 0.00133e −.00133 x dx =0.736. b) Using the Bayes estimate for λ, P(X<1000)= 0 7-24 CHAPTER 8 Section 8-2 8-1 a.) The confidence level for x − 2.14σ / n ≤ µ ≤ x + 2.14σ / n is determined by the by the value of z0 which is 2.14. From Table II, we find Φ(2.14) = P(Z<2.14) = 0.9838 and the confidence level is 100(1-.032354) = 96.76%. b.) The confidence level for x − 2.49σ / n ≤ µ ≤ x + 2.49σ / n is determined by the by the value of z0 which is 2.14. From Table II, we find Φ(2.49) = P(Z<2.49) = 0.9936 and the confidence level is is 100(1-.012774) = 98.72%. c.) The confidence level for x − 1.85σ / n ≤ µ ≤ x + 1.85σ / n is determined by the by the value of z0 which is 2.14. From Table II, we find Φ(1.85) = P(Z<1.85) = 0.9678 and the confidence level is 93.56%. 8-2 a.) A zα = 2.33 would give result in a 98% two-sided confidence interval. b.) A zα = 1.29 would give result in a 80% two-sided confidence interval. c.) A zα = 1.15 would give result in a 75% two-sided confidence interval. 8-3 a.) A zα = 1.29 would give result in a 90% one-sided confidence interval. b.) A zα = 1.65 would give result in a 95% one-sided confidence interval. c.) A zα = 2.33 would give result in a 99% one-sided confidence interval. 8-4 a.) 95% CI for µ, n = 10, σ = 20 x = 1000, z = 1.96 x − zσ / n ≤ µ ≤ x + zσ / n 1000 − 1.96(20 / 10 ) ≤ µ ≤ 1000 + 1.96(20 / 10 ) b.) .95% CI for 987.6 ≤ µ ≤ 1012.4 µ , n = 25, σ = 20 x = 1000, z = 1.96 x − z σ / n ≤ µ ≤ x + zσ / n 1000 − 1.96(20 / 25 ) ≤ µ ≤ 1000 + 1.96(20 / 25 ) 992.2 ≤ µ ≤ 1007.8 c.) 99% CI for µ, n = 10, σ = 20 x = 1000, z = 2.58 x − z σ / n ≤ µ ≤ x + zσ / n 1000 − 2.58(20 / 10 ) ≤ µ ≤ 1000 + 2.58(20 / 10 ) d.) 99% CI for 983.7 ≤ µ ≤ 1016.3 µ , n = 25, σ = 20 x = 1000, z = 2.58 x − zσ / n ≤ µ ≤ x + zσ / n 1000 − 2.58(20 / 25 ) ≤ µ ≤ 1000 + 2.58(20 / 25 ) 989.7 ≤ µ ≤ 1010.3 8-1 8-5 Find n for the length of the 95% CI to be 40. Za/2 = 1.96 1/2 length = (1.96)(20) / n = 20 39.2 = 20 n n= 39.2 20 2 = 3.84 Therefore, n = 4. 8-6 Interval (1): 3124.9 ≤ µ ≤ 3215.7 and Interval (2):: 3110.5 ≤ µ ≤ 3230.1 Interval (1): half-length =90.8/2=45.4 Interval (2): half-length =119.6/2=59.8 x1 = 3124.9 + 45.4 = 3170.3 x 2 = 3110.5 + 59.8 = 3170.3 The sample means are the same. a.) b.) Interval (1): 3124.9 ≤ µ ≤ 3215.7 was calculated with 95% Confidence because it has a smaller half-length, and therefore a smaller confidence interval. The 99% confidence level will make the interval larger. 8-7 a.) The 99% CI on the mean calcium concentration would be longer. b). No, that is not the correct interpretation of a confidence interval. The probability that µ is between 0.49 and 0.82 is either 0 or 1. c). Yes, this is the correct interpretation of a confidence interval. The upper and lower limits of the confidence limits are random variables. 8-8 95% Two-sided CI on the breaking strength of yarn: where x = 98 , σ = 2 , n=9 and z0.025 = 1.96 x − z 0 . 025 σ / n ≤ µ ≤ x + z 0 . 025 σ / n 98 − 1 . 96 ( 2 ) / 9 ≤ µ ≤ 98 + 1 . 96 ( 2 ) / 96 . 7 ≤ µ ≤ 99 . 3 8-9 95% Two-sided CI on the true mean yield: where x = 90.480 , σ = 3 , n=5 and z0.025 = 1.96 x − z 0 . 025 σ / n ≤ µ ≤ x + z 0 . 025 σ / n 90 . 480 − 1 . 96 ( 3 ) / 5 ≤ µ ≤ 90 . 480 + 1 . 96 ( 3 ) / 87 . 85 ≤ µ ≤ 93 . 11 8-10 9 5 99% Two-sided CI on the diameter cable harness holes: where x =1.5045 , σ = 0.01 , n=10 and z0.005 = 2.58 x − z 0 . 005 σ / n ≤ µ ≤ x + z 0 . 005 σ / n 1 . 5045 − 2 . 58 ( 0 . 01 ) / 10 ≤ µ ≤ 1 . 5045 + 2 . 58 ( 0 . 01 ) / 10 1 . 4963 ≤ µ ≤ 1 . 5127 8-2 8-11 a.) 99% Two-sided CI on the true mean piston ring diameter For α = 0.01, zα/2 = z0.005 = 2.58 , and x = 74.036, σ = 0.001, n=15 σ σ x − z0.005 ≤ µ ≤ x + z0.005 n n 74.036 − 2.58 0.001 ≤ µ ≤ 74.036 + 2.58 15 0.001 15 74.0353 ≤ µ ≤ 74.0367 b.) 95% One-sided CI on the true mean piston ring diameter For α = 0.05, zα = z0.05 =1.65 and x = 74.036, σ = 0.001, n=15 σ ≤µ n 0.001 74.036 − 1.65 ≤µ 15 x − z0.05 74.0356 ≤ µ 8-12 a.) 95% Two-sided CI on the true mean life of a 75-watt light bulb For α = 0.05, zα/2 = z0.025 = 1.96 , and x = 1014, σ =25 , n=20 x − z 0.025 1014 − 1.96 σ n 25 ≤ µ ≤ x + z 0.025 σ ≤ µ ≤ 1014 + 1.96 20 1003 ≤ µ ≤ 1025 n 25 20 b.) 95% One-sided CI on the true mean piston ring diameter For α = 0.05, zα = z0.05 =1.65 and x = 1014, σ =25 , n=20 x − z 0.05 1014 − 1.65 σ n ≤µ 25 ≤µ 20 1005 ≤ µ 8-3 8-13 a) 95% two sided CI on the mean compressive strength zα/2 = z0.025 = 1.96, and x = 3250, σ2 = 1000, n=12 σ σ x − z0.025 ≤ µ ≤ x + z0.025 n n 3250 − 1.96 31.62 ≤ µ ≤ 3250 + 1.96 31.62 12 3232.11 ≤ µ ≤ 3267.89 12 b.) 99% Two-sided CI on the true mean compressive strength zα/2 = z0.005 = 2.58 x − z0.005 3250 − 2.58 8-14 σ n ≤ µ ≤ x + z0.005 31.62 σ n ≤ µ ≤ 3250 + 2.58 12 3226.4 ≤ µ ≤ 3273.6 31.62 12 95% Confident that the error of estimating the true mean life of a 75-watt light bulb is less than 5 hours. For α = 0.05, zα/2 = z0.025 = 1.96 , and σ =25 , E=5 zσ n = a/2 E 2 1.96(25) = 5 2 = 96.04 Always round up to the next number, therefore n=97 8-15 Set the width to 6 hours with σ = 25, z0.025 = 1.96 solve for n. 1/2 width = (1.96)( 25) / n = 3 49 = 3 n n= 49 3 2 = 266.78 Therefore, n=267. 8-16 99% Confident that the error of estimating the true compressive strength is less than 15 psi. For α = 0.01, zα/2 = z0.005 = 2.58 , and σ =31.62 , E=15 zσ n = a/2 E 2 2.58(31.62) = 15 2 = 29.6 ≅ 30 Therefore, n=30 8-4 8-17 To decrease the length of the CI by one half, the sample size must be increased by 4 times (22). zα / 2σ / n = 0.5l Now, to decrease by half, divide both sides by 2. ( zα / 2σ / n ) / 2 = (l / 2) / 2 ( zα / 2σ / 2 n ) = l / 4 ( zα / 2σ / 22 n ) = l / 4 Therefore, the sample size must be increased by 22. 8-18 If n is doubled in Eq 8-7: zα / 2σ 2n = zα / 2σ 1.414 n x − zα / 2 = σ zα / 2σ 1.414 n ≤ µ ≤ x + zα / 2 n = σ n z α / 2σ 1 1.414 n The interval is reduced by 0.293 29.3% If n is increased by a factor of 4 Eq 8-7: zα / 2σ 4n = zα / 2σ 2n = z α / 2σ 2n = 1 zα / 2σ 2 n The interval is reduced by 0.5 or ½. Section 8-3 t 0.025,15 = 2.131 t 0.05,10 = 1.812 t 0.005, 25 = 2.787 8-19 t 0.001,30 = 3.385 t 0.025,12 = 2.179 t 0.025, 24 = 2.064 c.) t 0.005,13 = 3.012 d.) 8-21 a.) b.) 8-20 t 0.0005,15 = 4.073 a.) t 0.05,14 = 1.761 b.) t 0.01,19 = 2.539 c.) t 0.001, 24 = 3.467 8-5 t 0.10, 20 = 1.325 8-22 95% confidence interval on mean tire life n = 16 x = 60,139.7 s = 3645.94 t 0.025,15 = 2.131 x − t 0.025,15 n 3645.94 60139.7 − 2.131 8-23 s s ≤ µ ≤ x + t 0.025,15 n ≤ µ ≤ 60139.7 + 2.131 16 58197.33 ≤ µ ≤ 62082.07 3645.94 16 99% lower confidence bound on mean Izod impact strength n = 20 x = 1.25 s = 0.25 t 0.01,19 = 2.539 s x − t 0.01,19 1.25 − 2.539 8-24 n ≤µ 0.25 ≤µ 20 1.108 ≤ µ 99% confidence interval on mean current required Assume that the data are normally distributed and that the variance is unknown. n = 10 x = 317.2 s = 15.7 t 0.005,9 = 3.250 x − t 0.005,9 317.2 − 3.250 s n 15.7 ≤ µ ≤ x + t 0.005,9 s n ≤ µ ≤ 317.2 + 3.250 10 301.06 ≤ µ ≤ 333.34 8-6 15.7 10 a.) The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the level of polyunsaturated fatty acid is normally distributed. Norm al Probability Plot for 8-25 ML Estimates - 95% CI 99 95 90 80 Percent 8-25 70 60 50 40 30 20 10 5 1 16 17 18 Data b.) 99% CI on the mean level of polyunsaturated fatty acid. For α = 0.01, tα/2,n-1 = t0.005,5 = 4.032 x − t 0.005,5 16.98 − 4.032 s n 0.319 ≤ µ ≤ x + t 0.005,5 s n ≤ µ ≤ 16.98 + 4.032 6 16.455 ≤ µ ≤ 17.505 0.319 6 The 99% confidence for the mean polyunsaturated fat is (16.455, 17.505). to be somewhere in this region. 8-7 We would look for the true mean a.) The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the compressive strength is normally distributed. N o r m a l P r o b a b i li t y P l o t f o r S t r e n g t h ML Estimates - 95% CI 99 M L Estimates Mean 2259.92 StDev 95 34.0550 90 80 Percent 8-26 70 60 50 40 30 20 10 5 1 2150 2250 2 350 Data b.) 95% two-sided confidence interval on mean comprehensive strength n = 12 x = 2259.9 s = 35.6 t 0.025,11 = 2.201 x − t 0.025,11 2259.9 − 2.201 s n 35.6 ≤ µ ≤ x + t 0.025,11 n ≤ µ ≤ 2259.9 + 2.201 12 2237.3 ≤ µ ≤ 2282.5 c.) 95% lower-confidence bound on mean strength x − t 0.05,11 2259.9 − 1.796 s s n 35.6 ≤µ ≤µ 12 2241.4 ≤ µ 8-8 35.6 12 8-27 a.) According to the normal probability plot there does not seem to be a severe deviation from normality for this data. This is evident by the fact that the data appears to fall along a straight line. N orm al P rob ab ility P lot for 8 -2 7 M L E stim a te s - 9 5 % C I 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 8 .1 5 8 .2 0 8 .2 5 8 .3 0 D ata b.) 95% two-sided confidence interval on mean rod diameter For α = 0.05 and n = 15, tα/2,n-1 = t0.025,14 = 2.145 s x − t 0.025,14 8.23 − 2.145 ≤ µ ≤ x + t 0.025,14 n s n 0.025 0.025 ≤ µ ≤ 8.23 + 2.145 15 15 8.216 ≤ µ ≤ 8.244 8-28 95% lower confidence bound on mean rod diameter t0.05,14 = 1.761 s x −t 0.05 ,14 8.23 − 1.761 n ≤µ 0.025 ≤µ 15 8.219 ≤ µ The lower bound of the one sided confidence interval is lower than the lower bound of the two-sided confidence interval even though the level of significance is the same. This is because all of the alpha value is in the left tail (or in the lower bound). 8-9 8-29 95% lower bound confidence for the mean wall thickness given x = 4.05 s = 0.08 n = 25 tα,n-1 = t0.05,24 = 1.711 x − t 0.05, 24 4.05 − 1.711 s ≤µ n 0.08 ≤µ 25 4.023 ≤ µ It may be assumed that the mean wall thickness will most likely be greater than 4.023 mm. a.) The data appear to be normally distributed. Therefore, there is no evidence to support that the percentage of enrichment is not normally distributed. Normal Probability Plot for Percent Enrichment ML Estimates - 95% CI 99 ML Estimates Mean 2.90167 StDev 95 0.0951169 90 80 Percent 8-30 70 60 50 40 30 20 10 5 1 2.6 2.7 2.8 2.9 3.0 3.1 3.2 Data b.) 99% two-sided confidence interval on mean percentage enrichment For α = 0.01 and n = 12, tα/2,n-1 = t0.005,12 = 3.106, x = 2.9017 x − t 0.005,11 2.902 − 3.106 s n 0.0993 ≤ µ ≤ x + t 0.005,11 s n ≤ µ ≤ 2.902 + 3.106 12 2.813 ≤ µ ≤ 2.991 8-10 s = 0.0993 0.0993 12 8-31 x = 1.10 s = 0.015 n = 25 95% CI on the mean volume of syrup dispensed For α = 0.05 and n = 25, tα/2,n-1 = t0.025,24 = 2.064 s x − t 0.025, 24 1.10 − 2.064 n 0.015 s n 0.015 ≤ µ ≤ 1.10 + 2.064 25 1.094 ≤ µ ≤ 1.106 25 90% CI on the mean frequency of a beam subjected to loads x = 231.67, s = 1.53, n = 5, tα/2, n -1 = t.05, 4 = 2.132 x − t 0.05, 4 231.67 − 2.132 s n 1.53 ≤ µ ≤ x + t 0.05, 4 s n ≤ µ ≤ 231.67 − 2.132 5 230.2 ≤ µ ≤ 233.1 1.53 5 By examining the normal probability plot, it appears that the data are normally distributed. There does not appear to be enough evidence to reject the hypothesis that the frequencies are normally distributed. N o rm a l P ro b ab ility P lo t fo r fre q u e n c ie s M L E s tim a te s - 9 5 % C I 99 M L E stim a te s M ean 90 80 70 60 50 40 30 20 10 5 1 226 231 236 D a ta 8-11 2 3 1 .6 7 S tD e v 95 Percent 8-32 ≤ µ ≤ x + t 0.025, 24 1 .3 6 9 4 4 Section 8-4 8-34 χ 02.05,10 = 18.31 χ 02.025,15 = 27.49 χ 02.01,12 = 26.22 χ 02.005, 25 = 46.93 8-33 χ 02.95, 20 = 10.85 χ 02.99,18 = 7.01 a.) χ 0.05, 24 2 = 36.42 b.) c.) 8-35 χ 02.99,9 = 2.09 χ 2 0 . 95 , 19 = 10 . 12 and χ 02.05,19 = 30.14 99% lower confidence bound for σ2 For α = 0.01 and n = 15, 2 χ α , n −1 = χ 02.01,14 = 29.14 14(0.008) 2 ≤σ2 29.14 0.00003075 ≤ σ 2 8-36 95% two sided confidence interval for σ n = 10 s = 4.8 2 χ α / 2,n −1 = χ 02.025,9 = 19.02 and χ 12−α / 2,n −1 = χ 02.975,9 = 2.70 9(4.8) 2 9(4.8) 2 2 ≤σ ≤ 19.02 2.70 2 10.90 ≤ σ ≤ 76.80 3.30 < σ < 8.76 8-37 95% lower confidence bound for σ2 given n = 16, s2 = (3645.94)2 For α = 0.05 and n = 16, χ2 ,n −1 = χ2.05,15 = 25.00 α 0 15(3645.94) 2 ≤σ2 25 7,975,727.09 ≤ σ 2 8-38 99% two-sided confidence interval on σ2 for Izod impact test data n = 20 2 s = 0.25 χ 0.005,19 = 38.58 and χ 02.995,19 = 6.84 19(0.25) 2 19(0.25) 2 ≤σ 2 ≤ 38.58 6.84 2 0.03078 ≤ σ ≤ 0.1736 0.1754 < σ < 0.4167 8-12 χ 02.995,16 = 5.14 8-39 95% confidence interval for σ: given n = 51, s = 0.37 First find the confidence interval for σ2 : 2 2 For α = 0.05 and n = 51, χα / 2 , n −1 = χ2.025,50 = 71.42 and χ1− α / 2 ,n −1 = χ2.975,50 = 32.36 0 0 50(0.37) 2 50(0.37) 2 ≤σ 2 ≤ 71.42 32.36 0.096 ≤ σ2 0.2115 Taking the square root of the endpoints of this interval we obtain, 0.31 < σ < 0.46 8-40 99% two-sided confidence interval for σ for the hole diameter data. (Exercise 8-35) For α = 0.01 and n = 15, 2 χ α / 2,n −1 = χ 02.005,14 = 31.32 and χ 12−α / 2,n −1 = χ 02.995,14 = 4.07 14(0.008) 2 14(0.008) 2 ≤σ2 ≤ 31.32 4.07 2 0.00002861 ≤ σ ≤ 0.0002201 0.005349 < σ < 0.01484 8-41 90% lower confidence bound on σ (the standard deviation the sugar content) given n = 10, s2 = 23.04 For α = 0.1 and n = 10, χ α , n −1 2 2 = χ 0.1,9 = 14.68 9(23.04) ≤σ2 14.68 14.13 ≤ σ2 Take the square root of the endpoints of this interval to find the confidence interval for σ: 3.8 < σ Section 8-5 8-42 95% Confidence Interval on the death rate from lung cancer. ˆ p= 823 = 0.823 1000 ˆ p − zα / 2 0.823 − 1.96 8-43 n = 1000 zα / 2 = 1.96 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + zα / 2 n ˆ ˆ p (1 − p ) n 0.823(0.177) 0.823(0.177) ≤ p ≤ 0.823 + 1.96 1000 1000 0.7993 ≤ p ≤ 0.8467 E = 0.03, α = 0.05, zα/2 = z0.025 = 1.96 and p = 0.823 as the initial estimate of p, z n = α /2 E 2 1.96 ˆ ˆ p (1 − p ) = 0.03 2 0.823(1 − 0.823) = 621.79 , n ≅ 622. 8-13 8-44 a.) 95% Confidence Interval on the true proportion of helmets showing damage. ˆ p= 18 = 0 .3 6 50 ˆ p − zα / 2 0.36 − 1.96 b.) z n = α/2 E 2 n = 50 z α / 2 = 1 .9 6 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + zα / 2 n ˆ ˆ p (1 − p ) n 0.36(0.64) 0.36(0.64) ≤ p ≤ 0.36 + 1.96 50 50 0.227 ≤ p ≤ 0.493 2 1.96 p (1 − p ) = 0.02 0.36(1 − 0.36) = 2212.76 n ≅ 2213 c.) 8-45 z n = α/2 E 2 1.96 p (1 − p ) = 0.02 0.5(1 − 0.5) = 2401 The worst case would be for p = 0.5, thus with E = 0.05 and α = 0.01, zα/2 = z0.005 = 2.58 we obtain a sample size of: z n = α/2 E 8-46 2 2 2 2.58 p (1 − p ) = 0.5(1 − 0.5) = 665.64 , n ≅ 666 0.05 99% Confidence Interval on the fraction defective. 10 = 0.0125 800 ˆ ˆ p (1 − p ) ˆ ∞ ≤ p ≤ p + zα / 2 n ˆ p= ∞ ≤ p ≤ 0.0125 + 2.33 n = 800 z α = 2 .3 3 0.0125(0.9875) 800 ∞ ≤ p ≤ 0.0217 8-47 E = 0.017, α = 0.01, zα/2 = z0.005 = 2.58 z n = α/2 E 2 2 2.58 p (1 − p ) = 0.5(1 − 0.5) = 5758.13 , n ≅ 5759 0.017 8-14 8-48 95% Confidence Interval on the fraction defective produced with this tool. ˆ p= 13 z α / 2 = 1 .9 6 = 0.04333 n = 300 300 ˆ ˆ ˆ ˆ p (1 − p ) p (1 − p ) ˆ ˆ p − zα / 2 ≤ p ≤ p + zα / 2 n n 0.04333 − 1.96 0.04333(0.95667) 0.04333(0.95667) ≤ p ≤ 0.04333 + 1.96 300 300 0.02029 ≤ p ≤ 0.06637 Section 8-6 8-49 95% prediction interval on the life of the next tire given x = 60139.7 s = 3645.94 n = 16 for α=0.05 tα/2,n-1 = t0.025,15 = 2.131 x − t 0.025,15 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.025,15 s 1 + n n 1 1 ≤ x n +1 ≤ 60139.7 + 2.131(3645.94) 1 + 16 16 52131.1 ≤ x n +1 ≤ 68148.3 60139.7 − 2.131(3645.94) 1 + The prediction interval is considerably wider than the 95% confidence interval (58,197.3 ≤ µ ≤ 62,082.07), which is to be expected since we are going beyond our data. 8-50 99% prediction interval on the Izod impact data n = 20 x = 1.25 s = 0.25 t 0.005,19 = 2.861 x − t 0.005,19 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.005,19 s 1 + n n 1 1 ≤ x n +1 ≤ 1.25 + 2.861(0.25) 1 + 20 20 0.517 ≤ x n +1 ≤ 1.983 1.25 − 2.861(0.25) 1 + The lower bound of the 99% prediction interval is considerably lower than the 99% confidence interval (1.108 ≤ µ ≤ ∞), which is to be expected since we are going beyond our data. 8-51 Given x = 317.2 s = 15.7 n = 10 for α=0.05 tα/2,n-1 = t0.005,9 = 3.250 8-15 x − t 0.005,9 s 1 + 1 1 ≤ x n+1 ≤ x + t 0.005,9 s 1 + n n 1 1 ≤ x n+1 ≤ 317.2 − 3.250(15.7) 1 + 10 10 263.7 ≤ x n+1 ≤ 370.7 317.2 − 3.250(15.7) 1 + The length of the prediction interval is longer. 8-52 99% prediction interval on the polyunsaturated fat n = 6 x = 16.98 s = 0.319 t 0.005,5 = 4.032 x − t 0.005,5 s 1 + 1 1 ≤ x n+1 ≤ x + t 0.005,5 s 1 + n n 1 1 ≤ x n+1 ≤ 16.98 + 4.032(0.319) 1 + 6 6 15.59 ≤ x n+1 ≤ 18.37 16.98 − 4.032(0.319) 1 + The length of the prediction interval is a lot longer than the width of the confidence interval 16.455 ≤ µ ≤ 17.505 . 8-53 90% prediction interval on the next specimen of concrete tested given x = 2260 s = 35.57 n = 12 for α = 0.05 and n = 12, tα/2,n-1 = t0.05,11 = 1.796 x − t 0.05,11 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.05,11 s 1 + n n 1 1 ≤ x n +1 ≤ 2260 + 1.796(35.57) 1 + 12 12 2193.5 ≤ x n +1 ≤ 2326.5 2260 − 1.796(35.57) 1 + 8-54 95% prediction interval on the next rod diameter tested n = 15 x = 8.23 s = 0.025 t 0.025,14 = 2.145 x − t 0.025,14 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.025,14 s 1 + n n 1 1 ≤ x n +1 ≤ 8.23 − 2.145(0.025) 1 + 15 15 8.17 ≤ x n +1 ≤ 8.29 8.23 − 2.145(0.025) 1 + 95% two-sided confidence interval on mean rod diameter is 8.216 8-55 90% prediction interval on wall thickness on the next bottle tested. 8-16 ≤ µ ≤ 8.244 given x = 4.05 s = 0.08 n = 25 for tα/2,n-1 = t0.05,24 = 1.711 1 1 ≤ x n +1 ≤ x + t 0.05, 24 s 1 + n n x − t 0.05, 24 s 1 + 1 1 ≤ x n +1 ≤ 4.05 − 1.711(0.08) 1 + 25 25 3.91 ≤ x n +1 ≤ 4.19 4.05 − 1.711(0.08) 1 + 8-56 To obtain a one sided prediction interval, use tα,n-1 instead of tα/2,n-1 Since we want a 95% one sided prediction interval, tα/2,n-1 = t0.05,24 = 1.711 and x = 4.05 s = 0.08 n = 25 x − t 0.05, 24 s 1 + 1 ≤ x n +1 n 1 ≤ x n +1 25 3.91 ≤ x n +1 4.05 − 1.711(0.08) 1 + The prediction interval bound is a lot lower than the confidence interval bound of 4.023 mm 8-57 99% prediction interval for enrichment data given x = 2.9 s = 0.099 n = 12 for α = 0.01 and n = 12, tα/2,n-1 = t0.005,11 = 3.106 x − t 0.005,12 s 1 + 1 1 ≤ x n+1 ≤ x + t 0.005,12 s 1 + n n 1 1 ≤ x n+1 ≤ 2.9 − 3.106(0.099) 1 + 12 12 2.58 ≤ x n+1 ≤ 3.22 2.9 − 3.106(0.099) 1 + The prediction interval is much wider than the 99% CI on the population mean (2.813 ≤ µ ≤ 2.991). 8-58 95% Prediction Interval on the volume of syrup of the next beverage dispensed x = 1.10 s = 0.015 n = 25 tα/2,n-1 = t0.025,24 = 2.064 x − t 0.025, 24 s 1 + 1 1 ≤ x n+1 ≤ x + t 0.025, 24 s 1 + n n 1 1 ≤ x n+1 ≤ 1.10 − 2.064(0.015) 1 + 25 25 1.068 ≤ x n+1 ≤ 1.13 1.10 − 2.064(0.015) 1 + The prediction interval is wider than the confidence interval: 1.093 ≤ 8-17 µ ≤ 1.106 8-59 90% prediction interval the value of the natural frequency of the next beam of this type that will be tested. given x = 231.67, s=1.53 For α = 0.10 and n = 5, tα/2,n-1 = t0.05,4 = 2.132 x − t 0.05, 4 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.05, 4 s 1 + n n 1 1 ≤ x n +1 ≤ 231.67 − 2.132(1.53) 1 + 5 5 228.1 ≤ x n +1 ≤ 235.2 231.67 − 2.132(1.53) 1 + The 90% prediction in interval is greater than the 90% CI. Section 8-7 8-60 95% tolerance interval on the life of the tires that has a 95% CL given x = 60139.7 s = 3645.94 n = 16 we find k=2.903 x − ks, x + ks 60139.7 − 2.903(3645.94 ), 60139.7 + 2.903(3645.94 ) (49555.54, 70723.86) 95% confidence interval (58,197.3 ≤ µ ≤ 62,082.07) is shorter than the 95%tolerance interval. 8-61 99% tolerance interval on the Izod impact strength PVC pipe that has a 90% CL given x=1.25, s=0.25 and n=20 we find k=3.368 x − ks, x + ks 1.25 − 3.368(0.25), 1.25 + 3.368(0.25) (0.408, 2.092) The 99% tolerance interval is much wider than the 99% confidence interval on the population mean (1.090 ≤ µ ≤ 1.410). 8-62 99% tolerance interval on the brightness of television tubes that has a 95% CL given x = 317.2 s = 15.7 n = 10 we find k=4.433 x − ks, x + ks 317.2 − 4.433(15.7 ), 317.2 + 4.433(15.7 ) (247.60, 386.80) The 99% tolerance interval is much wider than the 95% confidence interval on the population mean 301.06 ≤ µ ≤ 333.34 . 8-18 8-63 99% tolerance interval on the polyunsaturated fatty acid in this type of margarine that has a confidence level of 95% x = 16.98 s = 0.319 n=6 and k = 5.775 x − ks, x + ks 16.98 − 5.775(0.319 ), 16.98 + 5.775(0.319 ) (15.14, 18.82) The 99% tolerance interval is much wider than the 99% confidence interval on the population mean (16.46 ≤ µ ≤ 17.51). 8-64 90% tolerance interval on the comprehensive strength of concrete that has a 90% CL given x = 2260 s = 35.57 n = 12 we find k=2.404 x − ks, x + ks 2260 − 2.404(35.57 ), 2260 + 2.404(35.57 ) (2174.5, 2345.5) The 90% tolerance interval is much wider than the 95% confidence interval on the population mean 2237.3 ≤ µ ≤ 2282.5 . 8-65 95% tolerance interval on the diameter of the rods in exercise 8-27 that has a 90% confidence level. x = 8.23 s = 0.0.25 n=15 and k=2.713 x − ks, x + ks 8.23 − 2.713(0.025), 8.23 + 2.713(0.025) (8.16, 8.30) The 95% tolerance interval is wider than the 95% confidence interval on the population mean (8.216 ≤ µ ≤ 8.244). 8-66 90% tolerance interval on wall thickness measurements that have a 90% CL given x = 4.05 s = 0.08 n = 25 we find k=2.077 x − ks, x + ks 4.05 − 2.077(0.08), 4.05 + 2.077(0.08) (3.88, 4.22) The lower bound of the 90% tolerance interval is much lower than the lower bound on the 95% confidence interval on the population mean (4.023 ≤ µ ≤ ∞) . 8-67 90% lower tolerance bound on bottle wall thickness that has confidence level 90%. given x = 4.05 s = 0.08 n = 25 and k = 1.702 x − ks 4.05 − 1.702(0.08) 3.91 The lower tolerance bound is of interest if we want to make sure the wall thickness is at least a certain value so that the bottle will not break. 8-68 99% tolerance interval on rod enrichment data that have a 95% CL 8-19 given x = 2.9 s = 0.099 n = 12 we find k=4.150 x − ks, x + ks 2.9 − 4.150(0.099 ), 2.9 + 4.150(0.099 ) (2.49, 3.31) The 99% tolerance interval is much wider than the 95% CI on the population mean (2.84 ≤ µ ≤ 2.96). 8-69 95% tolerance interval on the syrup volume that has 90% confidence level x = 1.10 s = 0.015 n = 25 and k=2.474 x − ks, x + ks 1.10 − 2.474(0.015), 1.10 + 2.474(0.015) (1.06, 1.14) Supplemental Exercises 8-70 α 1 + α 2 = α . Let α = 0.05 Interval for α 1 = α 2 = α / 2 = 0.025 The confidence level for x − 1.96σ / n ≤ µ ≤ x + 1.96σ / n is determined by the Where by the value of z0 which is 1.96. From Table II, we find Φ(1.96) = P(Z<1.96) = 0.975 and the confidence level is 95%. α 1 = 0.01, α 2 = 0.04 The confidence interval is x − 2.33σ / n ≤ µ ≤ x + 1.75σ / n , the confidence level is the same since α = 0.05 . The symmetric interval does not affect the level of significance. Interval for 8-71 µ = 50 a) n = 16 σ unknown x = 52 s = 1.5 52 − 50 to = =1 8 / 16 The P-value for t0 = 1, degrees of freedom = 15, is between 0.1 and 0.25. Thus we would conclude that the results are not very unusual. b) n = 30 to = 52 − 50 = 1 .3 7 8 / 30 The P-value for t0 = 1.37, degrees of freedom = 29, is between 0.05 and 0.1. Thus we would conclude that the results are somewhat unusual. c) n = 100 (with n > 30, the standard normal table can be used for this problem) zo = 52 − 50 = 2 .5 8 / 100 The P-value for z0 = 2.5, is 0.00621. Thus we would conclude that the results are very unusual. d) For constant values of x and s, increasing only the sample size, we see that the standard error of X decreases and consequently a sample mean value of 52 when the true mean is 50 is more unusual for the larger sample sizes. 8-72 µ = 50, σ 2 = 5 8-20 ≥ 7.44) or P ( s 2 ≤ 2.56) 15(7.44) 2 2 P ( s 2 ≥ 7.44) = P χ 15 ≥ = 0.05 ≤ P χ 15 ≥ 22.32 ≤ 0.10 52 2 Using Minitab P ( s ≥ 7.44) =0.0997 15(2.56) 2 2 P ( s 2 ≤ 2.56) = P χ 15 ≤ = 0.05 ≤ P χ 15 ≤ 7.68 ≤ 0.10 5 2 Using Minitab P ( s ≤ 2.56) =0.064 a.) For n = 16 find P ( s 2 ( ) ( ) n = 30 find P ( s 2 ≥ 7.44) or P ( s 2 ≤ 2.56) 29(7.44) 2 2 P ( s 2 ≥ 7.44) = P χ 29 ≥ = 0.025 ≤ P χ 29 ≥ 43.15 ≤ 0.05 5 2 Using Minitab P ( s ≥ 7.44) =0.044 29(2.56) 2 2 P ( s 2 ≤ 2.56) = P χ 29 ≤ = 0.01 ≤ P χ 29 ≤ 14.85 ≤ 0.025 5 Using Minitab P ( s ≤ 2.56) =0.014. b) For ( ) ( ) n = 71 P ( s 2 ≥ 7.44) or P ( s 2 ≤ 2.56) 70(7.44) 2 2 P ( s 2 ≥ 7.44) = P χ 15 ≥ = 0.005 ≤ P χ 70 ≥ 104.16 ≤ 0.01 5 2 Using Minitab P ( s ≥ 7.44) =0.0051 70(2.56) 2 2 P ( s 2 ≤ 2.56) = P χ 70 ≤ = P χ 70 ≤ 35.84 ≤ 0.005 5 2 Using Minitab P ( s ≤ 2.56) <0.001 c) For ( ( ) ) d) The probabilities get smaller as n increases. As n increases, the sample variance should approach the population variance; therefore, the likelihood of obtaining a sample variance much larger than the population variance will decrease. e) The probabilities get smaller as n increases. As n increases, the sample variance should approach the population variance; therefore, the likelihood of obtaining a sample variance much smaller than the population variance will decrease. 8-21 8-73 a) The data appear to follow a normal distribution based on the normal probability plot since the data fall along a straight line. b) It is important to check for normality of the distribution underlying the sample data since the confidence intervals to be constructed should have the assumption of normality for the results to be reliable (especially since the sample size is less than 30 and the central limit theorem does not apply). c) No, with 95% confidence, we can not infer that the true mean could be 14.05 since this value is not contained within the given 95% confidence interval. d) As with part b, to construct a confidence interval on the variance, the normality assumption must hold for the results to be reliable. e) Yes, it is reasonable to infer that the variance could be 0.35 since the 95% confidence interval on the variance contains this value. f) i) & ii) No, doctors and children would represent two completely different populations not represented by the population of Canadian Olympic hockey players. Since doctors nor children were the target of this study or part of the sample taken, the results should not be extended to these groups. 8-74 a.) The probability plot shows that the data appear to be normally distributed. Therefore, there is no evidence conclude that the comprehensive strength data are normally distributed. b.) 99% lower confidence bound on the mean x − t 0.01,8 25.12 − 2.896 s n x = 25.12, s = 8.42, n = 9 t 0.01,8 = 2.896 ≤µ 8.42 ≤µ 9 16.99 ≤ µ The lower bound on the 99% confidence interval shows that the mean comprehensive strength will most likely be greater than 16.99 Megapascals. c.) 98% lower confidence bound on the mean x − t 0.01,8 25.12 − 2.896 s n 8.42 x = 25.12, s = 8.42, n = 9 ≤ µ ≤ x − t 0.01,8 s n ≤ µ ≤ 25.12 − 2.896 9 16.99 ≤ µ ≤ 33.25 t 0.01,8 = 2.896 8.42 9 The bounds on the 98% two-sided confidence interval shows that the mean comprehensive strength will most likely be greater than 16.99 Megapascals and less than 33.25 Megapascals. The lower bound of the 99% one sided CI is the same as the lower bound of the 98% two-sided CI (this is because of the value of α) d.) 99% one-sided upper bound on the confidence interval on σ2 comprehensive strength s = 8.42, s 2 = 70.90 χ 02.99,8 = 1.65 8(8.42) 2 1.65 2 σ ≤ 343.74 σ2 ≤ The upper bound on the 99% confidence interval on the variance shows that the variance of the comprehensive strength will most likely be less than 343.74 Megapascals2. e.) 98% one-sided upper bound on the confidence interval on σ2 comprehensive strength 8-22 2 2 s = 8.42, s 2 = 70.90 χ 0.01,9 = 20.09 χ 0.99,8 = 1.65 8(8.42) 2 8(8.42) 2 2 ≤σ ≤ 20.09 1.65 2 28.23 ≤ σ ≤ 343.74 The bounds on the 98% two-sided confidence-interval on the variance shows that the variance of the comprehensive strength will most likely be less than 343.74 Megapascals2 and greater than 28.23 Megapascals2. The upper bound of the 99% one-sided CI is the same as the upper bound of the 98% twosided CI (this is because of the value of α) f.) 98% lower confidence bound on the mean x − t 0.01,8 23 − 2.896 s n 6.07 x = 23, s = 6.07, n = 9 t 0.01,8 = 2.896 ≤ µ ≤ x − t 0.01,8 s n 6.07 ≤ µ ≤ 23 − 2.896 9 17.14 ≤ µ ≤ 28.86 9 98% one-sided upper bound on the confidence interval on σ2 comprehensive strength 2 2 s = 6.07, s 2 = 36.9 χ 0.01,9 = 20.09 χ 0.99,8 = 1.65 8(6.07) 2 8(6.07) 2 ≤σ 2 ≤ 20.09 1.65 2 14.67 ≤ σ ≤ 178.64 Fixing the mistake decreased the values of the sample mean and the sample standard deviation. Since the sample standard deviation was decreased. The width of the confidence intervals were also decreased. g.) 98% lower confidence bound on the mean x − t 0.01,8 25 − 2.896 s n 8.41 x = 25, s = 8.41, n = 9 t 0.01,8 = 2.896 ≤ µ ≤ x − t 0.01,8 s ≤ µ ≤ 25 − 2.896 8.41 9 16.88 ≤ µ ≤ 33.12 n 9 98% one-sided upper bound on the confidence interval on σ2 comprehensive strength 2 2 s = 8.41, s 2 = 70.73 χ 0.01,9 = 20.09 χ 0.99,8 = 1.65 8(8.41) 2 8(8.41) 2 2 ≤σ ≤ 20.09 1.65 2 28.16 ≤ σ ≤ 342.94 Fixing the mistake did not have an affect on the sample mean or the sample standard deviation. They are very close to the original values. The width of the confidence intervals are also very similar. h.) When a mistaken value is near the sample mean, the mistake will not affect the sample mean, standard deviation or confidence intervals greatly. However, when the mistake is not near the sample mean, the 8-23 value can greatly affect the sample mean, standard deviation and confidence intervals. The farther from the mean, the greater the effect. 8-75 With σ = 8, the 95% confidence interval on the mean has length of at most 5; the error is then E = 2.5. a) n = z0.025 2 .5 b) n = z0.025 2 .5 2 2 82 = 1.96 64 = 39.34 2 .5 62 = 1.96 36 = 22.13 2 .5 2 = 40 2 = 23 As the standard deviation decreases, with all other values held constant, the sample size necessary to maintain the acceptable level of confidence and the length of the interval, decreases. 8-76 x = 15.33 s = 0.62 n = 20 k = 2.564 a.) 95% Tolerance Interval of hemoglobin values with 90% confidence x − ks, x + ks 15.33 − 2.564(0.62 ), 15.33 + 2.564(0.62 ) (13.74, 16.92) b.) 99% Tolerance Interval of hemoglobin values with 90% confidence k = 3.368 x − ks, x + ks 15.33 − 3.368(0.62 ), 15.33 + 3.368(0.62 ) (13.24, 17.42) 8-77 95% prediction interval for the next sample of concrete that will be tested. given x = 25.12 s = 8.42 n = 9 for α = 0.05 and n = 9, tα/2,n-1 = t0.025,8 = 2.306 x − t 0.025,8 s 1 + 1 1 ≤ x n+1 ≤ x + t 0.025,8 s 1 + n n 1 1 ≤ x n+1 ≤ 25.12 + 2.306(8.42) 1 + 9 9 4.65 ≤ x n+1 ≤ 45.59 25.12 − 2.306(8.42) 1 + 8-24 a.) There is no evidence to reject the assumption that the data are normally distributed. Normal Probability Plot for foam height ML Estimates - 95% CI 99 ML Estimates Mean 203.2 StDev 95 7.11056 90 80 Percent 8-78 70 60 50 40 30 20 10 5 1 178 188 198 208 218 228 Data b.) 95% confidence interval on the mean x − t 0.025,9 203.2 − 2.262 s n 7.50 x = 203.20, s = 7.5, n = 10 t 0.025,9 = 2.262 ≤ µ ≤ x − t 0.025,9 s n ≤ µ ≤ 203.2 + 2.262 10 197.84 ≤ µ ≤ 208.56 7.50 10 c.) 95% prediction interval on a future sample x − t 0.025,9 s 1 + 1 1 ≤ µ ≤ x − t 0.025,9 s 1 + n n 1 1 ≤ µ ≤ 203.2 + 2.262(7.50) 1 + 10 10 185.41 ≤ µ ≤ 220.99 d.) 95% tolerance interval on foam height with 99% confidence k = 4.265 x − ks, x + ks 203.2 − 4.265(7.5), 203.2 + 4.265(7.5) (171.21, 235.19) 203.2 − 2.262(7.50) 1 + e.) The 95% CI on the population mean has the smallest interval. This type of interval tells us that 95% of such intervals would contain the population mean. The 95% prediction interval, tell us where, most likely, the next data point will fall. This interval is quite a bit larger than the CI on the mean. The tolerance interval is the largest interval of all. It tells us the limits that will include 95% of the data with 99% confidence. 8-25 8-79 a) Normal probability plot for the coefficient of restitution Norm al Probability Plot for 8-79 ML Estimates - 95% CI 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 0.59 0.60 0.61 0.62 0.63 0.64 0.65 0 .66 Data b.) 99% CI on the true mean coefficient of restitution x = 0.624, s = 0.013, n = 40 ta/2, n-1 = t0.005, 39 = 2.7079 x − t 0.005,39 0.624 − 2.7079 s n 0.013 ≤ µ ≤ x + t 0.005,39 s n ≤ µ ≤ 0.624 + 2.7079 40 0.618 ≤ µ ≤ 0.630 0.013 40 c.) 99% prediction interval on the coefficient of restitution for the next baseball that will be tested. x − t 0.005,39 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.005,39 s 1 + n n 1 1 ≤ x n +1 ≤ 0.624 + 2.7079(0.013) 1 + 40 40 0.588 ≤ x n +1 ≤ 0.660 0.624 − 2.7079(0.013) 1 + d.) 99% tolerance interval on the coefficient of restitution with a 95% level of confidence ( x − ks, x + ks ) (0.624 − 3.213(0.013), 0.624 + 3.213(0.013)) (0.582, 0.666) e.)The confidence interval in part (b) describes the confidence interval on the population mean and we may interpret this to mean that 99% of such intervals will cover the population mean. The prediction interval tells us that within that within a 99% probability that the next baseball will have a coefficient of restitution between 0.588 and 0.660. And the tolerance interval captures 99% of the values of the normal distribution with a 95% level of confidence. 8-80 95% Confidence Interval on the death rate from lung cancer. 8-26 8 = 0 .2 n = 4 0 40 ˆ ˆ p (1 − p ) ˆ p − zα ≤p n ˆ p= 0.2 − 1.65 0.2(0.8) ≤p 40 0.0956 ≤ p a.)The normal probability shows that the data are mostly follow the straight line, however, there are some points that deviate from the line near the middle. It is probably safe to assume that the data are normal. Normal Probability Plot for 8-81 ML Estimates - 95% CI 99 95 90 80 Percent 8-81 z α = 1 .6 5 70 60 50 40 30 20 10 5 1 0 5 10 Data b.) 95% CI on the mean dissolved oxygen concentration x = 3.265, s = 2.127, n = 20 ta/2, n-1 = t0.025, 19 = 2.093 x − t 0.025,19 3.265 − 2.093 s n 2.127 ≤ µ ≤ x + t 0.025,19 s n ≤ µ ≤ 3.265 + 2.093 20 2.270 ≤ µ ≤ 4.260 2.127 20 c.) 95% prediction interval on the oxygen concentration for the next stream in the system that will be tested.. x − t 0.025,19 s 1 + 1 1 ≤ x n +1 ≤ x + t 0.025,19 s 1 + n n 1 1 ≤ x n +1 ≤ 3.265 + 2.093(2.127) 1 + 20 20 − 1.297 ≤ x n +1 ≤ 7.827 3.265 − 2.093(2.127) 1 + 8-27 d.) 95% tolerance interval on the values of the dissolved oxygen concentration with a 99% level of confidence ( x − ks, x + ks ) (3.265 − 3.168(2.127), 3.265 + 3.168(2.127)) (−3.473, 10.003) e.)The confidence interval in part (b) describes the confidence interval on the population mean and we may interpret this to mean that there is a 95% probability that the interval may cover the population mean. The prediction interval tells us that within that within a 95% probability that the next stream will have an oxygen concentration between –1.297 and 7.827mg/L. And the tolerance interval captures 95% of the values of the normal distribution with a 99% confidence level. a.) There is no evidence to support that the data are not normally distributed. The data points appear to fall along the normal probability line. Normal Probability Plot for tar content ML Estimates - 95% CI 99 ML Estimates Mean 1.529 StDev 95 0.0556117 90 80 Percent 8-82 70 60 50 40 30 20 10 5 1 1.4 1.5 1.6 1.7 Data b.) 99% CI on the mean tar content x = 1.529, s = 0.0566, n = 30 ta/2, n-1 = t0.005, 29 = 2.756 x − t 0.005, 29 1.529 − 2.756 s n 0.0566 ≤ µ ≤ x + t 0.005, 29 s n ≤ µ ≤ 1.529 + 2.756 30 1.501 ≤ µ ≤ 1.557 0.0566 30 e.) 99% prediction interval on the tar content for the next sample that will be tested.. x − t 0.005,19 s 1 + 1 1 ≤ x n+1 ≤ x + t 0.005,19 s 1 + n n 1 1 ≤ x n+1 ≤ 1.529 + 2.756(0.0566) 1 + 30 30 1.370 ≤ x n+1 ≤ 1.688 1.529 − 2.756(0.0566) 1 + 8-28 f.) 99% tolerance interval on the values of the tar content with a 95% level of confidence ( x − ks, x + ks ) (1.529 − 3.350(0.0566), 1.529 + 3.350(0.0566)) (1.339, 1.719) e.)The confidence interval in part (b) describes the confidence interval on the population mean and we may interpret this to mean that 95% of such intervals will cover the population mean. The prediction interval tells us that within that within a 95% probability that the sample will have a tar content between 1.370 and 1.688. And the tolerance interval captures 95% of the values of the normal distribution with a 99% confidence level. 8-83 a.) 95% Confidence Interval on the population proportion ˆ n=1200 x=8 p = 0.0067 zα/2=z0.025=1.96 ˆ p − za / 2 0.0067 − 1.96 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + za / 2 n ˆ ˆ p (1 − p ) n 0.0067(1 − 0.0067) 0.0067(1 − 0.0067) ≤ p ≤ 0.0067 + 1.96 1200 1200 0.0021 ≤ p ≤ 0.0113 b) No, there is not evidence to support the claim that the fraction of defective units produced is one percent or less. This is because the upper limit of the control limit is greater than 0.01. 8-84 a.) 99% Confidence Interval on the population proportion n=1600 ˆ p = 0.005 zα/2=z0.005=2.58 x=8 ˆ p − za / 2 0.005 − 2.58 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + za / 2 n ˆ ˆ p (1 − p ) n 0.005(1 − 0.005) 0.005(1 − 0.005) ≤ p ≤ 0.005 + 2.58 1600 1600 0.0004505 ≤ p ≤ 0.009549 b.) E = 0.008, α = 0.01, zα/2 = z0.005 = 2.58 z n = α /2 E 2 2.58 p (1 − p ) = 0.008 2 0.005(1 − 0.005) = 517.43 , n ≅ 518 c.) E = 0.008, α = 0.01, zα/2 = z0.005 = 2.58 z n = α/2 E 2 2.58 p (1 − p ) = 0.008 2 0.5(1 − 0.5) = 26001.56 , n ≅ 26002 d.)Knowing an estimate of the population proportion reduces the required sample size by a significant amount. A sample size of 518 is much more reasonable than a sample size of over 26,000. 8-29 8-85 ˆ p= 117 = 0.242 484 = 1.645 a) 90% confidence interval; zα / 2 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + zα / 2 n 0.210 ≤ p ≤ 0.274 ˆ p − zα / 2 ˆ ˆ p (1 − p ) n With 90% confidence, we believe the true proportion of new engineering graduates who were planning to continue studying for an advanced degree lies between 0.210 and 0.274. b) 95% confidence interval; zα / 2 = 1.96 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + zα / 2 n 0.204 ≤ p ≤ 0.280 ˆ p − zα / 2 ˆ ˆ p (1 − p ) n With 95% confidence, we believe the true proportion of new engineering graduates who were planning to continue studying for an advanced degree lies between 0.204 and 0.280. c) Comparison of parts a and b: The 95% confidence interval is larger than the 90% confidence interval. Higher confidence always yields larger intervals, all other values held constant. d) Yes, since both intervals contain the value 0.25, thus there in not enough evidence to determine that the true proportion is not actually 0.25. Mind Expanding Exercises 8-86 a.) P ( χ 12− α , 2 r < 2λTr < χ α2 , 2 r ) = 1 − α 2 2 =P χ 2 χα 2 1− α , 2 r 2 2Tr <λ< 2 ,2r 2Tr Then a confidence interval for µ= 1 λ 2Tr is χα 2 2 ,2r , 2Tr χ 2 1− α , 2 r 2 b) n = 20 , r = 10 , and the observed value of Tr is 199 + 10(29) = 489. A 95% confidence interval for 8-87 α1 = ∞ z α1 1 2π 1 λ 2 e − x2 dx = 1 − is z α1 2(489) 2(489) , = (28.62,101.98) 34.17 9.59 1 − ∞ 2π 2 e 8-30 − x2 dx Therefore, 1 − α 1 = Φ ( z α1 ) . To minimize L we need to minimize we need to minimize Φ −1 (1 − α 1 ) + Φ(1 − α 2 ) subject to α 1 + α 2 = α . Therefore, Φ −1 (1 − α 1 ) + Φ(1 − α + α 1 ) . ∂ Φ −1 (1 − α 1 ) = − 2π e ∂α 1 2 zα 1 2 ∂ Φ −1 (1 − α + α 1 ) = 2π e ∂α 1 2 zα −α 1 2 2 zα −α 1 2 zα 1 e =e 2 z α1 = z α − α1 . Consequently, α 1 = α − α 1 , 2α 1 = α and α 1 = α 2 = α . 2 2 Upon setting the sum of the two derivatives equal to zero, we obtain 8.88 . This is solved by a.) n=1/2+(1.9/.1)(9.4877/4) n=46 b.) (10-.5)/(9.4877/4)=(1+p)/(1-p) p=0.6004 between 10.19 and 10.41. 8-89 a) ~ P( X i ≤ µ ) = 1 / 2 ~ P(allX ≤ µ ) = (1 / 2) n i ~ P(allX i ≥ µ ) = (1 / 2) n P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) = 1 2 n + 1 2 n =2 1 2 n = 1 2 n −1 1 ~ 1 − P ( A ∪ B ) = P (min( X i ) < µ < max( X i )) = 1 − 2 ~ < max( X )) = 1 − α b.) P (min( X i ) < µ i n The confidence interval is min(Xi), max(Xi) 8-90 We would expect that 950 of the confidence intervals would include the value of µ. This is due to the definition of a confidence interval. 8-31 Let X bet the number of intervals that contain the true mean (µ). We can use the large sample approximation to determine the probability that P(930<X<970). Let p= 950 930 970 = 0.950 p1 = = 0.930 and p 2 = = 0.970 1000 1000 1000 The variance is estimated by p (1 − p ) 0.950(0.050) = n 1000 P (0.930 < p < 0.970) = P Z < =P Z< (0.970 − 0.950) 0.950(0.050) 1000 −P Z < (0.930 − 0.950) 0.950(0.050) 1000 0.02 − 0.02 −P Z < = P ( Z < 2.90) − P ( Z < −2.90) = 0.9963 0.006892 0.006892 8-32 CHAPTER 9 Section 9-1 9-1 a) H 0 : µ = 25, H1 : µ ≠ 25 Yes, because the hypothesis is stated in terms of the parameter of interest, inequality is in the alternative hypothesis, and the value in the null and alternative hypotheses matches. b) H 0 : σ > 10, H1 : σ = 10 No, because the inequality is in the null hypothesis. c) H 0 : x = 50, H1 : x ≠ 50 No, because the hypothesis is stated in terms of the statistic rather than the parameter. d) H 0 : p = 0.1, H1 : p = 0.3 No, the values in the null and alternative hypotheses do not match and both of the hypotheses are equality statements. e) H 0 : s = 30, H1 : s > 30 No, because the hypothesis is stated in terms of the statistic rather than the parameter. 9-2 a) α = P(reject H0 when H0 is true) = P( X ≤ 11.5 when µ = 12) = P X − µ 11.5 − 12 ≤ σ / n 0 .5 / 4 = P(Z ≤ −2) = 0.02275. The probability of rejecting the null hypothesis when it is true is 0.02275. ( ) b) β = P(accept H0 when µ = 11.25) = P X > 115 when µ = 11.25 = . P X − µ 11.5 − 11.25 > σ/ n 0 .5 / 4 P(Z > 1.0) = 1 − P(Z ≤ 1.0) = 1 − 0.84134 = 0.15866 The probability of accepting the null hypothesis when it is false is 0.15866. 9-3 a) α = P( X ≤ 11.5 | µ = 12) = P X − µ 11.5 − 12 ≤ σ / n 0.5 / 16 = P(Z ≤ −4) = 0. The probability of rejecting the null, when the null is true, is approximately 0 with a sample size of 16. b) β = P( X > 11.5 | µ =11.25) = P X − µ 11.5 − 11.25 > σ/ n 0.5 / 16 = P(Z > 2) = 1 − P(Z ≤ 2) = 1− 0.97725 = 0.02275. The probability of accepting the null hypothesis when it is false is 0.02275. 9-4 Find the boundary of the critical region if α = 0.01: 0.01 = P Z≤ c − 12 0 .5 / 4 What Z value will give a probability of 0.01? Using Table 2 in the appendix, Z value is −2.33. Thus, c − 12 0 .5 / 4 = −2.33, c = 11.4175 9-1 9-5. P Z≤ c − 12 0.5 / 4 = 0.05 What Z value will give a probability of 0.05? Using Table 2 in the appendix, Z value is −1.65. c − 12 Thus, = −1.65, c = 11.5875 0.5 / 4 9-6 a) α = P( = X ≤ 98.5) + P( X > 101.5) X − 100 98.5 − 100 P ≤ 2/ 9 2/ 9 +P X − 100 101.5 − 100 > 2/ 9 2/ 9 = P(Z ≤ −2.25) + P(Z > 2.25) = (P(Z ≤- 2.25)) + (1 − P(Z ≤ 2.25)) = 0.01222 + 1 − 0.98778 = 0.01222 + 0.01222 = 0.02444 b) β = P(98.5 ≤ X ≤ 101.5 when µ = 103) 98.5 − 103 X − 103 101.5 − 103 ≤ ≤ =P 2/ 9 2/ 9 2/ 9 = P(−6.75 ≤ Z ≤ −2.25) = P(Z ≤ −2.25) − P(Z ≤ −6.75) = 0.01222 − 0 = 0.01222 c) β = P(98.5 ≤ X ≤ 101.5 when µ = 105) 98.5 − 105 X − 105 101.5 − 105 ≤ ≤ =P 2/ 9 2/ 9 2/ 9 = P(−9.75≤ Z ≤ −5.25) = P(Z ≤ −5.25) − P(Z ≤ −9.75) =0−0 = 0. The probability of accepting the null hypothesis when it is actually false is smaller in part c since the true mean, µ = 105, is further from the acceptance region. A larger difference exists. 9-7 Use n = 5, everything else held constant (from the values in exercise 9-6): a) P( X ≤ 98.5) + P( X >101.5) =P X − 100 98.5 − 100 ≤ 2/ 5 2/ 5 +P X − 100 101.5 − 100 > 2/ 5 2/ 5 = P(Z ≤ −1.68) + P(Z > 1.68) = P(Z ≤ −1.68) + (1 − P(Z ≤ 1.68)) = 0.04648 + (1 − 0.95352) = 0.09296 b) β = P(98.5 ≤ X ≤ 101.5 when µ = 103) 98.5 − 103 X − 103 101.5 − 103 =P ≤ ≤ 2/ 5 2/ 5 2/ 5 = P(−5.03 ≤ Z ≤ −1.68) = P(Z ≤ −1.68) − P(Z ≤ −5.03) = 0.04648 − 0 = 0.04648 9-2 c) β = P(98.5 ≤ x ≤ 101.5 when µ = 105) =P 98.5 − 105 X − 105 101.5 − 105 ≤ ≤ 2/ 5 2/ 5 2/ 5 = P(−7.27≤ Z ≤ −3.91) = P(Z ≤ −3.91) − P(Z ≤ −7.27) = 0.00005 − 0 = 0.00005 It is smaller, because it is not likely to accept the product when the true mean is as high as 105. 9-8 a) α = P( X > 185 when µ = 175) X − 175 185 − 175 =P > 20 / 10 20 / 10 = P(Z > 1.58) = 1 − P(Z ≤ 1.58) = 1 − 0.94295 = 0.05705 b) β = P( X ≤ 185 when µ = 195) X − 195 185 − 195 =P ≤ 20 / 10 20 / 10 = P(Z ≤ −1.58) = 0.05705. 9-9 a) z= 190 − 175 20 / 10 = 2 .3 7 , Note that z is large, therefore reject the null hypothesis and conclude that the mean foam height is greater than 175 mm. b) P( X > 190 when µ = 175) X − 175 190 − 175 =P > 20 / 10 20 / 10 = P(Z > 2.37) = 1 − P(Z ≤ 2.37) = 1 − 0.99111 = 0.00889. The probability that a value of at least 190 mm would be observed (if the true mean height is 175 mm) is only 0.00889. Thus, the sample value of x = 190 mm would be an unusual result. 9-10 Using n = 16: a) α = P( X > 185 when µ = 175) X − 175 185 − 175 =P > 20 / 16 20 / 16 = P(Z > 2) = 1 − P(Z ≤ 2) = 1 − 0.97725 = 0.02275 9-3 b) β = P ( X ≤ 185 when µ = 195) X − 195 185 − 195 =P ≤ 20 / 16 20 / 16 = P(Z ≤ −2) = 0.02275. 9-11 a) P (X > c | µ = 175) = 0.0571 c − 175 = P(Z ≥ 1.58) 20 / 16 c − 175 , and c = 182.9 Thus, 1.58 = 20 / 16 P Z> b) If the true mean foam height is 195 mm, then β = P( X ≤ 182.9 when µ = 195) 182.9 − 195 =P Z≤ 20 / 16 = P(Z ≤ −2.42) = 0.00776 c) For the same level of α, with the increased sample size, β is reduced. 9-12 a) α = P( =P X ≤ 4.85 when µ = 5) + P( X X −5 4.85 − 5 ≤ 0.25 / 8 0.25 / 8 > 5.15 when µ = 5) +P X −5 5.15 − 5 > 0.25 / 8 0.25 / 8 = P(Z ≤ −1.7) + P(Z > 1.7) = P(Z ≤ −1.7) + (1 − P(Z ≤ 1.7) = 0.04457 + (1 − 0.95543) = 0.08914. b) Power = 1 − β X ≤ 5.15 when µ = 5.1) 4.85 − 5.1 X − 5.1 5.15 − 5.1 ≤ ≤ =P 0.25 / 8 0.25 / 8 0.25 / 8 β = P(4.85 ≤ = P(−2.83 ≤ Z ≤ 0.566) = P(Z ≤ 0.566) − P(Z ≤ −2.83) = 0.71566 − 0.00233 = 0.71333 1 − β = 0.2867. 9-4 9-13 Using n = 16: a) α = P( X ≤ 4.85 | µ = 5) + P( X > 5.15 | µ = 5) X −5 4.85 − 5 ≤ 0.25 / 16 0.25 / 16 =P +P X −5 5.15 − 5 > 0.25 / 16 0.25 / 16 = P(Z ≤ −2.4) + P(Z > 2.4) = P(Z ≤ −2.4) +(1 − P(Z ≤ 2.4)) = 2(1 − P(Z ≤ 2.4)) = 2(1 − 0.99180) = 2(0.0082) = 0.0164. b) β = P(4.85 ≤ X ≤ 5.15 | µ = 5.1) =P 4.85 − 5.1 X − 5 .1 5.15 − 5.1 ≤ ≤ 0.25 / 16 0.25 / 16 0.25 / 16 = P(−4 ≤ Z ≤ 0.8) = P(Z ≤ 0.8) − P(Z ≤ −4) = 0.78814 − 0 = 0.78814 1 − β = 0.21186 9-14 Find the boundary of the critical region if α = 0.05: 0.025 = P Z≤ c−5 0.25 / 8 What Z value will give a probability of 0.025? Using Table 2 in the appendix, Z value is −1.96. c−5 Thus, 0.25 / 8 c−5 0.25 / 8 = −1.96, c = 4.83 and = 1 .9 6 , c=5.17 The acceptance region should be (4.83 ≤ X ≤ 5.17). 9-15 Operating characteristic curve: x = 185 β=P Z≤ x−µ 20 / 10 =P Z≤ µ 178 181 184 187 190 193 196 199 185 − µ 20 / 10 P Z≤ 185 − µ 20 / 10 P(Z ≤ 1.11) = P(Z ≤ 0.63) = P(Z ≤ 0.16) = P(Z ≤ −0.32) = P(Z ≤ −0.79) = P(Z ≤ −1.26) = P(Z ≤ −1.74) = P(Z ≤ −2.21) = 9-5 = β 1−β 0.8665 0.7357 0.5636 0.3745 0.2148 0.1038 0.0409 0.0136 0.1335 0.2643 0.4364 0.6255 0.7852 0.8962 0.9591 0.9864 Operating Characteristic Curve 1 0.8 β 0.6 0.4 0.2 0 175 180 185 190 195 200 µ 9-16 Power Function Curve 1 0.8 1−β 0.6 0.4 0.2 0 175 180 185 190 195 200 µ 9-17. The problem statement implies H0: p = 0.6, H1: p > 0.6 and defines an acceptance region as ˆ p≤ 400 = 0 .8 0 500 and rejection region as ˆ a) α=P( p >0.80 | p=0.60) = P Z > ˆ p > 0 .8 0 0.80 − 0.60 0 .6 ( 0 .4 ) 500 = P(Z>9.13)=1-P(Z≤ 9.13) ≈ 0 ˆ b) β = P( p ≤ 0.8 when p=0.75) = P(Z ≤ 2.58)=0.99506. 9-6 9-18 X ~ bin(10, 0.3) Implicitly, H0: p = 0.3 and H1: p < 0.3 n = 10 ˆ Accept region: p > 0.1 ˆ Reject region: p ≤ 0.1 Use the normal approximation for parts a), b) and c): a) When p =0.3 α = ˆ P ( p < 0.1) = P Z ≤ 0 .1 − 0 .3 0.3(0.7) 10 = P ( Z ≤ −1.38) = 0.08379 b) When p = 0.2 β = P ˆ ( p > 0.1) = P Z> 0 .1 − 0 .2 0.2(0.8) 10 = P ( Z > −0.79) = 1 − P ( Z < −0.79) = 0.78524 c) Power = 1 − β = 1 − 078524 = 0.21476 9-19 X ~ bin(15, 0.4) H0: p = 0.4 and H1: p ≠ 0.4 p1= 4/15 = 0.267 p2 = 8/15 = 0.533 ˆ 0.267 ≤ p ≤ 0.533 ˆ ˆ p < 0.267 or p > 0.533 Accept Region: Reject Region: Use the normal approximation for parts a) and b) a) When p = 0.4, α =P Z< ˆ ˆ = P ( p < 0.267) + P ( p > 0.533) 0 .267 − 0 .4 +P Z > 0 .533 − 0 .4 0 .4 ( 0 .6 ) 0 .4 ( 0 .6 ) 15 15 = P ( Z < − 1 .05 ) + P ( Z > 1 .05 ) = P ( Z < − 1 .05 ) + (1 − P ( Z < 1 .05 )) = 0 .14686 + 0 .14686 = 0 .29372 9-7 b) When p = 0.2, ˆ β = P(0.267 ≤ p ≤ 0.533) = P 0.267 − 0.2 ≤Z≤ 0.533 − 0.2 0.2(0.8) 0.2(0.8) 15 15 = P (0.65 ≤ Z ≤ 3.22) = P ( Z ≤ 3.22) − P ( Z ≤ 0.65) = 0.99936 − 0.74215 = 0.25721 Section 9-2 9-20 a.) 1) The parameter of interest is the true mean water temperature, µ. 2) H0 : µ = 100 3) H1 : µ > 100 4) α = 0.05 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 > zα where z0.05 = 1.65 7) x = 98 , σ = 2 z0 = 98 − 100 2/ 9 = −3.0 8) Since -3.0 < 1.65 do not reject H0 and conclude the water temperature is not significantly different greater than 100 at α = 0.05. b) P-value = 1 − Φ ( −3.0) = 1 − 0.00135 = 0.99865 c) β = Φ z 0.05 + 100 − 104 2/ 9 = Φ(1.65 + −6) = Φ(-4.35) ≅0 9-21. a) 1) The parameter of interest is the true mean yield, µ. 2) H0 : µ = 90 3) H1 : µ ≠ 90 4) α = 0.05 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 < −z α/2 where −z0.025 = −1.96 or z0 > zα/2 where z0.025 = 1.96 7) x = 90.48 , σ = 3 z0 = 90.48 − 90 = 0 .3 6 3/ 5 8) Since −1.96 < 0.36 < 1.96 do not reject H0 and conclude the yield is not significantly different from 90% at α = 0.05. b) P-value = 2[1 − Φ(0.36)] = 2[1 − 0.64058] = 0.71884 c) n = (z α /2 + zβ δ 2 )σ 2 2 = ( z 0 .025 + z 0 .05 )2 3 2 (85 − 90 )2 n ≅ 5. 9-8 = (1 .96 + 1 . 65 )2 9 (− 5 )2 = 4 . 69 d) β = Φ z0.025 + 90 − 92 3/ 5 − Φ − z0.025 + 90 − 92 3/ 5 = Φ(1.96 + −1.491) − Φ(−1.96 + −1.491) = Φ(0.47) − Φ(−3.45) = Φ(0.47) − (1 − Φ(3.45)) = 0.68082 − ( 1 − 0.99972) = 0.68054. e) For α = 0.05, zα/2 = z0.025 = 1.96 σ σ x − z0.025 ≤ µ ≤ x + z0.025 n n 90.48 − 1.96 3 5 ≤ µ ≤ 90.48 + 196 . 3 5 87.85 ≤ µ ≤ 93.11 With 95% confidence, we believe the true mean yield of the chemical process is between 87.85% and 93.11%. Since 90% is contained in the confidence interval, our decision in (a) agrees with the confidence interval. 9-22 a) 1) The parameter of interest is the true mean crankshaft wear, µ. 2) H0 : µ = 3 3 ) H1 : µ ≠ 3 4) α = 0.05 x−µ 5) z0 = σ/ n 6) ) Reject H0 if z0 < −z α/2 where −z0.025 = −1.96 or z0 > zα/2 where z0.025 = 1.96 7) x = 2.78, σ = 0.9 z0 = 2.78 − 3 0.9 / 15 = − 0 .9 5 8) Since –0.95 > -1.96, do not reject the null hypothesis and conclude there is not sufficient evidence to support the claim the mean crankshaft wear is not equal to 3 at α = 0.05. b) β = Φ z 0.025 + 3 − 3.25 0.9 / 15 − Φ − z 0.025 + 3 − 3.25 0.9 / 15 = Φ(1.96 + −1.08) − Φ(−1.96 + −1.08) = Φ(0.88) − Φ(-3.04) = 0.81057 − (0.00118) = 0.80939 c.) n= (z + zβ ) σ 2 2 α /2 δ2 = (z 0.025 + z 0.10 )2 σ 2 (3.75 − 3) 2 n ≅ 16 9-9 = (1.96 + 1.29) 2 (0.9) 2 = 15.21, (0.75) 2 9-23. a) 1) The parameter of interest is the true mean melting point, µ. 2) H0 : µ = 155 3) H1 : µ ≠ 155 4) α = 0.01 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 < −z α/2 where −z0.005 = −2.58 or z0 > zα/2 where z0.005 = 2.58 7) x = 154.2, σ = 1.5 z0 = 154.2 − 155 = −1.69 1.5 / 10 8) Since –1.69 > -2.58, do not reject the null hypothesis and conclude there is not sufficient evidence to support the claim the mean melting point is not equal to 155 °F at α = 0.01. b) P-value = 2*P(Z <- 1.69) =2* 0.045514 = 0.091028 β = Φ z0.005 − c) = Φ 2.58 − δn δn − Φ − z0.005 − σ σ (155 − 150) 10 (155 − 150) 10 − Φ − 2.58 − 1.5 1.5 = Φ(-7.96)- Φ(-13.12) = 0 – 0 = 0 d) n= (z + zβ ) σ 2 2 α/2 δ2 = (z 0.005 + z 0.10 )2 σ 2 (150 − 155) 2 (2.58 + 1.29) 2 (1.5) 2 = = 1.35, (5) 2 n ≅ 2. 9-24 a.) 1) The parameter of interest is the true mean battery life in hours, µ. 2) H0 : µ = 40 3) H1 : µ > 40 4) α = 0.05 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 > zα where z0.05 = 1.65 7) x = 40.5 , σ = 1.25 z0 = 40.5 − 40 1.25 / 10 = 1 .2 6 8) Since 1.26 < 1.65 do not reject H0 and conclude the battery life is not significantly different greater than 40 at α = 0.05. b) P-value = 1 − Φ (1.26) = 1 − 0.8962 = 0.1038 c) β = Φ z 0.05 + 40 − 42 1.25 / 10 = Φ(1.65 + −5.06) = Φ(-3.41) ≅0.000325 d.) n= (z + zβ ) σ 2 2 α δ2 = (z 0.05 + z 0.10 )2 σ 2 (40 − 44) 2 n ≅1 9-10 = (1.65 + 1.29) 2 (1.25) 2 = 0.844, ( 4) 2 e.)95% Confidence Interval x + z 0 .05 σ / n ≤ µ 40 . 5 + 1 . 65 (1 . 25 ) / 10 ≤ µ 39 . 85 ≤ µ The lower bound of the 90 % confidence interval must be greater than 40 to verify that the true mean exceeds 40 hours. 9-25. a) 1) The parameter of interest is the true mean tensile strength, µ. 2) H0 : µ = 3500 3) H1 : µ ≠ 3500 4) α = 0.01 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 < −zα/2 where −z0.005 = −2.58 or z0 > zα/2 where z0.005 = 2.58 7) x = 3250 , σ = 60 z0 = 3250 − 3500 60 / 12 = −14.43 8) Since −14.43 < −2.58, reject the null hypothesis and conclude the true mean tensile strength is significantly different from 3500 at α = 0.01. b) Smallest level of significance = P-value = 2[1 − Φ (14.43) ]= 2[1 − 1] = 0 The smallest level of significance at which we are willing to reject the null hypothesis is 0. c) zα/2 = z0.005 = 2.58 x − z 0.005 σ 3250 − 2.58 ≤ µ ≤ x + z 0.005 n 31.62 12 σ n ≤ µ ≤ 3250 + 2.58 31.62 12 3205.31 ≤ µ ≤ 3294.69 With 95% confidence, we believe the true mean tensile strength is between 3205.31 psi and 3294.69 psi. We can test the hypotheses that the true mean tensile strength is not equal to 3500 by noting that the value is not within the confidence interval. 9-26 n= (z + zβ ) σ 2 2 α /2 δ2 = (z 0.05 + z 0.20 )2 σ 2 (3250 − 3500) 2 n ≅1 9-11 = (1.65 + .84) 2 (60) 2 = 0.357, (250) 2 9-27 a) 1) The parameter of interest is the true mean speed, µ. 2) H0 : µ = 100 3) H1 : µ < 100 4) α = 0.05 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 < −zα where −z0.05 = −1.65 7) x = 102.2 , σ = 4 z0 = 102.2 − 100 4/ 8 = 1 .5 6 8) Since 1.56> −1.65, do not reject the null hypothesis and conclude the there is insufficient evidence to conclude that the true speed strength is less than 100 at α = 0.05. b) β = Φ − z0.05 − c) n = (z (95 − 100) 8 4 = Φ(-1.65 - −3.54) = Φ(1.89) = 0.97062 Power = 1-β = 1-0.97062 = 0.02938 + zβ ) σ 2 2 α δ2 = (z 0.05 + z 0.15 )2 σ 2 (95 − 100) 2 = (1.65 + 1.03) 2 (4) 2 = 4.597, (5) 2 n≅5 d) x − z0.05 σ n 102.2 − 1.65 ≤µ 4 ≤µ 8 99.866 ≤ µ Since the lower limit of the CI is just slightly below 100, we are relatively confident that the mean speed is not less than 100 m/s. Also the sample mean is greater than 100. 9-28 a) 1) The parameter of interest is the true mean hole diameter, µ. 2) H0 : µ = 1.50 3) H1 : µ ≠ 1.50 4) α = 0.01 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 < −zα/2 where −z0.005 = −2.58 or z0 > zα/2 where z0.005 = 2.58 7) x = 1.4975 , σ = 0.01 z0 = 1.4975 − 1.50 0.01 / 25 = − 1 .2 5 8) Since −2.58 < -1.25 < 2.58, do not reject the null hypothesis and conclude the true mean hole diameter is not significantly different from 1.5 in. at α = 0.01. 9-12 b) β = Φ z0.005 − = Φ 2.58 − δn δn Φ − z0.005 − σ σ (1.495 − 1.5) 25 (1.495 − 1.5) 25 − Φ − 2.58 − 0.01 0.01 = Φ(5.08)- Φ(-0.08) = 1 – .46812 = 0.53188 power=1-β =0.46812. c) Set β = 1 − 0.90 = 0.10 n= ( zα / 2 + z β ) 2 σ 2 δ2 = ( z 0.005 + z 0.10 ) 2 σ 2 (1.495 − 1.50) 2 ≅ ( 2.58 + 1.29) 2 (0.01) 2 (−0.005) 2 = 59.908, n ≅ 60. d) For α = 0.01, zα/2 = z0.005 = 2.58 x − z0.005 σ n 1.4975 − 2.58 σ ≤ µ ≤ x + z0.005 0.01 25 n ≤ µ ≤ 1.4975 + 2.58 0.01 25 1.4923 ≤ µ ≤ 1.5027 The confidence interval constructed contains the value 1.5, thus the true mean hole diameter could possibly be 1.5 in. using a 99% level of confidence. Since a two-sided 99% confidence interval is equivalent to a two-sided hypothesis test at α = 0.01, the conclusions necessarily must be consistent. 9-29 a) 1) The parameter of interest is the true average battery life, µ. 2) H0 : µ = 4 3 ) H1 : µ > 4 4) α = 0.05 x−µ 5) z0 = σ/ n 6) Reject H0 if z0 > zα where z0.05 = 1.65 7) x = 4.05 , σ = 0.2 z0 = 4.05 − 4 = 1 .7 7 0.2 / 50 8) Since 1.77>1.65, reject the null hypothesis and conclude that there is sufficient evidence to conclude that the true average battery life exceeds 4 hours at α = 0.05. b) β = Φ z0.05 − c) n = (z (4.5 − 4) 50 0 .2 = Φ(1.65 – 17.68) = Φ(-16.03) = 0 Power = 1-β = 1-0 = 1 + zβ ) σ 2 2 α δ2 = (z 0.05 + z 0.1 )2 σ 2 ( 4 .5 − 4 ) 2 n≅2 9-13 = (1.65 + 1.29) 2 (0.2) 2 = 1.38, (0.5) 2 x − z0.05 d) σ ≤µ n 0 .2 ≤µ 50 4.05 − 1.65 4.003 ≤ µ Since the lower limit of the CI is just slightly above 4, we conclude that average life is greater than 4 hours at α=0.05. Section 9-3 a. 1) The parameter of interest is the true mean interior temperature life, µ. 2) H0 : µ = 22.5 3) H1 : µ ≠ 22.5 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if |t0| > tα/2,n-1 where tα/2,n-1 = 2.776 7) x = 22.496 , s = 0.378 n=5 t0 = 22.496 − 22.5 0.378 / 5 = −0.00237 8) Since –0.00237 >- 2.776, we cannot reject the null hypothesis. There is not sufficient evidence to conclude that the true mean interior temperature is not equal to 22.5 °C at α = 0.05. 2*0.4 <P-value < 2* 0.5 ; 0..8 < P-value <1.0 b.) The points on the normal probability plot fall along the line. Therefore, there is no evidence to conclude that the interior temperature data is not normally distributed. Normal Probability Plot for temp ML Estimates - 95% CI 99 ML Estimates Mean 22.496 StDev 95 0.338384 90 80 Percent 9-30 70 60 50 40 30 20 10 5 1 21.5 22.5 23.5 Data c.) d = δ | µ − µ 0 | | 22.75 − 22.5 | = = = 0 .6 6 σ σ 0.378 Using the OC curve, Chart VI e) for α = 0.05, d = 0.66, and n = 5, we get β ≅ 0.8 and power of 1−0.8 = 0.2. 9-14 d) d = δ | µ − µ 0 | | 22.75 − 22.5 | = = = 0 .6 6 σ σ 0.378 Using the OC curve, Chart VI e) for α = 0.05, d = 0.66, and β ≅ 0.1 (Power=0.9), n = 40 . e) 95% two sided confidence interval s x − t 0.025, 4 22.496 − 2.776 n 0.378 ≤ µ ≤ x + t 0.025, 4 s n ≤ µ ≤ 22.496 + 2.776 0.378 5 22.027 ≤ µ ≤ 22.965 5 We cannot conclude that the mean interior temperature is not equal to 22.5 since the value is included inside the confidence interval. 9-31 a. 1) The parameter of interest is the true mean female body temperature, µ. 2) H0 : µ = 98.6 3) H1 : µ ≠ 98.6 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if |t0| > tα/2,n-1 where tα/2,n-1 = 2.064 7) x = 98.264 , s = 0.4821 n=25 t0 = 98.264 − 98.6 = −3.48 0.4821 / 25 8) Since 3.48 > 2.064, reject the null hypothesis and conclude that the there is sufficient evidence to conclude that the true mean female body temperature is not equal to 98.6 °F at α = 0.05. P-value = 2* 0.001 = 0.002 b) d = δ | µ − µ0 | | 98 − 98.6 | = = = 1 .2 4 σ σ 0.4821 Using the OC curve, Chart VI e) for α = 0.05, d = 1.24, and n = 25, we get β ≅ 0 and power of 1−0 ≅ 1. c) d = δ | µ − µ0 | | 98.2 − 98.6 | = = = 0 .8 3 σ σ 0.4821 Using the OC curve, Chart VI e) for α = 0.05, d = 0.83, and β ≅ 0.1 (Power=0.9), n = 20 . 9-15 d) 95% two sided confidence interval x − t0.025, 24 s ≤ µ ≤ x + t0.025, 24 n s n 0.4821 0.4821 ≤ µ ≤ 98.264 + 2.064 25 25 98.065 ≤ µ ≤ 98.463 98.264 − 2.064 We can conclude that the mean female body temperature is not equal to 98.6 since the value is not included inside the confidence interval. e) N orm al P rob ab ility P lot for 9 -3 1 M L E stima te s - 9 5 % C I 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 97 98 99 D ata Data appear to be normally distributed. 9-32 a.) 1) The parameter of interest is the true mean rainfall, µ. 2) H0 : µ = 25 3) H1 : µ > 25 4) α = 0.01 5) t0 = x−µ s/ n 6) Reject H0 if t0 > tα,n-1 where t0.01,19 = 2.539 7) x = 26.04 s = 4.78 n = 20 t0 = 26.04 − 25 4.78 / 20 = 0 .9 7 8) Since 0.97 < 2.539, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean rainfall is greater than 25 acre-feet at α = 0.01. The 0.10 < P-value < 0.25. 9-16 b.) the data on the normal probability plot fall along the straight line. Therefore there is evidence that the data are normally distributed. Normal Probability Plot for rainfall ML Estimates - 95% CI 99 ML Estimates Mean 26.035 StDev 95 4.66361 90 Percent 80 70 60 50 40 30 20 10 5 1 20 30 40 Data c.) d = δ | µ − µ 0 | | 27 − 25 | = = = 0 .4 2 σ σ 4.78 Using the OC curve, Chart VI h) for α = 0.01, d = 0.42, and n = 20, we get β ≅ 0.7 and power of 1−0.7 = 0.3. d) d = δ | µ − µ 0 | | 27.5 − 25 | = = = 0 .5 2 σ σ 4.78 Using the OC curve, Chart VI h) for α = 0.05, d = 0.42, and β ≅ 0.1 (Power=0.9), n = 75 . e) 95% two sided confidence interval x − t 0.025, 24 26.03 − 2.776 s n ≤µ 4.78 ≤µ 20 23.06 ≤ µ Since the lower limit of the CI is less than 25, we conclude that there is insufficient evidence to indicate that the true mean rainfall is not greater than 25 acre-feet at α=0.01. 9-17 a. 1) The parameter of interest is the true mean sodium content, µ. 2) H0 : µ = 130 3) H1 : µ ≠ 130 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if |t0| > tα/2,n-1 where tα/2,n-1 = 2.045 7) x = 129.753 , s = 0.929 n=30 t0 = 129.753 − 130 = −1.456 0.929 30 8) Since 1.456 < 2.064, do not reject the null hypothesis and conclude that the there is not sufficient evidence that the true mean sodium content is different from 130mg at α = 0.05. From table IV the t0 value is found between the values of 0.05 and 0.1 with 29 degrees of freedom, so 2*0.05<P-value = 2* 0.1 Therefore, 0.1< P-value<0.2. b) Normal Probability Plot for 9-33 ML Estimates - 95% CI 99 95 90 80 Percent 9-33 70 60 50 40 30 20 10 5 1 127 128 129 130 131 132 Data The assumption of normality appears to be reasonable. c) d = δ | µ − µ 0 | | 130.5 − 130 | = = = 0.538 σ σ 0.929 Using the OC curve, Chart VI e) for α = 0.05, d = 0.53, and n = 30, we get β ≅ 0.2 and power of 1−0.20 = 0.80. 9-18 d) d = δ | µ − µ 0 | | 130.1 − 130 | = = = 0 .1 1 σ σ 0.929 Using the OC curve, Chart VI e) for α = 0.05, d = 0.11, and β ≅ 0.25 (Power=0.75), n = 100 . d) 95% two sided confidence interval x − t 0.025, 29 129.753 − 2.045 s n 0.929 ≤ µ ≤ x + t 0.025, 29 s n ≤ µ ≤ 129.753 + 2.045 30 129.406 ≤ µ ≤ 130.100 0.929 30 We can conclude that the mean sodium content is equal to 130 because that value is inside the confidence interval. 9-34 a.)1) The parameter of interest is the true mean tire life, µ. 2) H0 : µ = 60000 3) H1 : µ > 60000 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if t0 > tα,n-1 where 7) t 0.05 ,15 = 1 .753 n = 16 x = 60,139.7 s = 3645.94 60139.7 − 60000 t0 = 3645.94 / 16 = 0 .1 5 8) Since 0.15 < 1.753., do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean tire life is greater than 60,000 kilometers at α = 0.05. The P-value > 0.40. b.) d = δ | µ − µ 0 | | 61000 − 60000 | = = = 0 .2 7 σ σ 3645.94 Using the OC curve, Chart VI g) for α = 0.05, d = 0.27, and β ≅ 0.1 (Power=0.9), n = 4. Yes, the sample size of 16 was sufficient. 9-35. In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean impact strength, µ. 2) H0 : µ = 1.0 3) H1 : µ > 1.0 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if t0 > tα,n-1 where t0.05,19 = 1.729 7) x = 1.25 s = 0.25 n = 20 t0 = 1.25 − 1.0 = 4 .4 7 0.25 / 20 8) Since 4.47 > 1.729, reject the null hypothesis and conclude there is sufficient evidence to indicate that the true mean impact strength is greater than 1.0 ft-lb/in at α = 0.05. The P-value < 0.0005 . 9-36. In order to use t statistic in hypothesis testing, we need to assume that the underlying distribution is normal. 9-19 1) The parameter of interest is the true mean current, µ. 2) H0 : µ = 300 3) H1 : µ > 300 4) α = 0.05 5) t0 = x−µ s/ n t 0.05,9 = 1.833 n = 10 x = 317.2 s = 15.7 317.2 − 300 6) Reject H0 if t0 > tα,n-1 where 7) t0 = 15.7 / 10 = 3 .4 6 8) Since 3.46 > 1.833, reject the null hypothesis and conclude there is sufficient evidence to indicate that the true mean current is greater than 300 microamps at α = 0.05. The 0.0025 <P-value < 0.005 9-37. a.) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean coefficient of restitution, µ. 2) H0 : µ = 0.635 3) H1 : µ > 0.635 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if t0 > tα,n-1 where t0.05,39 = 1.685 7) x = 0.624 s = 0.013 n = 40 t0 = 0.624 − 0.635 0.013 / 40 = − 5 .3 5 8) Since –5.25 < 1.685, do not reject the null hypothesis and conclude that there is not sufficient evidence to indicate that the true mean coefficient of restitution is greater than 0.635 at α = 0.05. b.)The P-value > 0.4, based on Table IV. Minitab gives P-value = 1. c) d = δ | µ − µ 0 | | 0.64 − 0.635 | = = = 0 .3 8 σ σ 0.013 Using the OC curve, Chart VI g) for α = 0.05, d = 0.38, and n = 40, we get β ≅ 0.25 and power of 1−0.25 = 0.75. d) d = δ | µ − µ 0 | | 0.638 − 0.635 | = = = 0 .2 3 σ σ 0.013 Using the OC curve, Chart VI g) for α = 0.05, d = 0.23, and β ≅ 0.25 (Power=0.75), n = 40 . 9-20 9-38 a)In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean oxygen concentration, µ. 2) H0 : µ = 4 3 ) H1 : µ ≠ 4 4) α = 0.01 5) t0 = x−µ s/ n 6) Reject H0 if |t0 |>ta/2, n-1 = t0.005, 19 = 2.861 7) x = 3.265, s = 2.127, n = 20 t0 = 3.265 − 4 2.127 / 20 = − 1 .5 5 8) Since -2.861<-1.55 <1.48, do not reject the null hypothesis and conclude that there is insufficient evidence to indicate that the true mean oxygen not equal 4 at α = 0.01. b.) The P-Value: 2*0.05<P-value<2*0.10 therefore 0.10< P-value<0.20 c.) d = δ | µ − µ0 | | 3 − 4 | = = = 0 .4 7 σ σ 2.127 Using the OC curve, Chart VI f) for α = 0.01, d = 0.47, and n = 20, we get β ≅ 0.70 and power of 1−0.70 = 0.30. d.) d = δ | µ − µ 0 | | 2 .5 − 4 | = = = 0 .7 1 σ σ 2.127 Using the OC curve, Chart VI f) for α = 0.01, d = 0.71, and β ≅ 0.10 (Power=0.90), n = 40 . 9-39 a)In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean cigar tar content, µ. 2) H0 : µ = 1.5 3) H1 : µ > 1.5 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if t0 > tα,n-1 where t0.05,29 =1.699 7) x = 1.529 s = 0.0566 n = 30 t0 = 1.529 − 1.5 0.0566 / 30 = 2.806 8) Since 2.806 > 1.699, reject the null hypothesis and conclude that there is sufficient evidence to indicate that the true mean tar content is greater than 1.5 at α = 0.05. b.) From table IV the t0 value is found between the values of 0.0025 and 0.005 with 29 degrees of freedom. Therefore, 0.0025< P-value<0.005. Minitab gives P-value = 0.004. c) d = δ | µ − µ 0 | | 1 .6 − 1 .5 | = = = 1 .7 7 σ σ 0.0566 Using the OC curve, Chart VI g) for α = 0.05, d = 1.77, and n = 30, we get β ≅ 0 and power of 1−0 = 1. 9-21 e.) d= δ | µ − µ 0 | | 1 .6 − 1 .5 | = = = 1 .7 7 σ σ 0.0566 Using the OC curve, Chart VI g) for α = 0.05, d = 1.77, and β ≅ 0.20 (Power=0.80), n = 4. 9-40 a) 1) The parameter of interest is the true mean height of female engineering students, µ. 2) H0 : µ = 65 3) H1 : µ ≠ 65 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if |t0 | > tα/2,n-1 where t0.025,59 =2.0281 7) x = 65.811 inches s = 2.106 inches n = 37 t0 = 65.811 − 65 2.11 / 37 = 2 .3 4 8) Since 2.34 > 2.0281, reject the null hypothesis and conclude that there is sufficient evidence to indicate that the true mean height of female engineering students is not equal to 65 at α = 0.05. b.)P-value: 0.02<P-value<0.05. c) d= 62 − 65 2.11 = 1.42 , n=37 so, from the OC Chart VI e) for α = 0.05, we find that β≅0. Therefore, the power ≅ 1. d.) 9-41 d= 64 − 65 2.11 * n = 40 . = 0 .4 7 for α = 0.05, and β≅0.2 (Power=0.8). a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean concentration of suspended solids, µ. 2) H0 : µ = 55 3) H1 : µ ≠ 55 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if |t0 | > tα/2,n-1 where t0.025,59 =2.000 7) x = 59.87 s = 12.50 n = 60 t0 = 59.87 − 55 12.50 / 60 = 3.018 8) Since 3.018 > 2.000, reject the null hypothesis and conclude that there is sufficient evidence to indicate that the true mean concentration of suspended solids is not equal to 55 at α = 0.05. 9-22 b) From table IV the t0 value is found between the values of 0.001 and 0.0025 with 59 degrees of freedom, so 2*0.001<P-value = 2* 0.0025 Therefore, 0.002< P-value<0.005. Minitab gives a p-value of 0.0038 c) 50 − 55 d= 12.50 = 0.4 , n=60 so, from the OC Chart VI e) for α = 0.05, d= 0.4 and n=60 we find that β≅0.2. Therefore, the power = 1-0.2 = 0.8. d) From the same OC chart, and for the specified power, we would need approximately 38 observations. d= n = 75 . 9-42 50 − 55 12.50 = 0 .4 Using the OC Chart VI e) for α = 0.05, d = 0.4, and β ≅ 0.10 (Power=0.90), a)In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean distance, µ. 2) H0 : µ = 280 3) H1 : µ > 280 4) α = 0.05 5) t0 = x−µ s/ n 6) Reject H0 if t0 > tα,n-1 where t0.05,99 =1.6604 7) x = 260.3 s = 13.41 n = 100 t0 = 260.3 − 280 13.41 / 100 = −14.69 8) Since –14.69 < 1.6604, do not reject the null hypothesis and conclude that there is insufficient evidence to indicate that the true mean distance is greater than 280 at α = 0.05. b.) From table IV the t0 value is found above the value 0.005, therefore, the P-value is greater than 0.995. c) d = δ | µ − µ 0 | | 290 − 280 | = = = 0 .7 5 σ σ 13.41 Using the OC curve, Chart VI g) for α = 0.05, d = 0.75, and n = 100, we get β ≅ 0 and power of 1−0 = 1. f.) d= δ | µ − µ 0 | | 290 − 280 | = = = 0 .7 5 σ σ 13.41 Using the OC curve, Chart VI g) for α = 0.05, d = 0.75, and β ≅ 0.20 (Power=0.80), n = 15 . 9-23 Section 9-4 9-43 a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true standard deviation of the diameter, σ. However, the answer can be found by performing a hypothesis test on σ2. 2) H0 : σ2 = 0.0001 3) H1 : σ2 > 0.0001 4) α = 0.01 ( n − 1)s2 5) χ2 = 0 σ2 2 6) Reject H0 if χ2 > χα ,n −1 where χ2.01,14 = 29.14 0 0 7) n = 15, s2 = 0.008 14(0.008)2 = 8.96 0.0001 σ 8) Since 8.96 < 29.14 do not reject H0 and conclude there is insufficient evidence to indicate the true standard deviation of the diameter exceeds 0.01 at α = 0.01. χ2 = 0 ( n − 1)s2 2 = b) P-value = P(χ2 > 8.96) for 14 degrees of freedom: c) λ= σ 0.0125 = = 1 .2 5 σ0 0.01 0.5 < P-value < 0.9 power = 0.8, β=0.2 using chart VIk the required sample size is 50 9-44 a.) In order to use χ2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true variance of sugar content, σ2. However, the answer can be found by performing a hypothesis test on σ2. 2) H0 : σ2 = 18 3) H1 : σ2 ≠ 18 4) α = 0.05 5) χ2 = 0 ( n − 1)s2 σ2 2 6) Reject H0 if χ2 < χ1− α / 2 ,n −1 where 0 χ 02.975,9 = 2.70 or 2 χ2 > χα ,2,n −1 where 0 χ 02.025,9 = 19.02 7) n = 10, s = 4.8 χ2 0 = (n − 1) s 2 σ2 9(4.8) 2 = = 11.52 18 8) Since 11.52 < 19.02 do not reject H0 and conclude there is insufficient evidence to indicate the true variance of sugar content is significantly different from 18 at α = 0.01. b.) P-value: The χ2 is between 0.50 and 0.10. Therefore, 0.2<P-value<1 0 c.) The 95% confidence interval includes the value 18, therefore, we could not be able to conclude that the variance was not equal to 18. 9(4.8) 2 9(4.8) 2 ≤σ2 ≤ 19.02 2.70 2 10.90 ≤ σ ≤ 76.80 9-24 9-45 a) In order to use χ2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the standard deviation of tire life, σ. However, the answer can be found by performing a hypothesis test on σ2. 2) H0 : σ2 = 40,000 3) H1 : σ2 > 40,000 4) α = 0.05 5) χ2 = 0 (n − 1) s 2 σ2 2 6) Reject H0 if χ2 > χα ,n −1 where χ2.05,15 = 25.00 0 0 7) n = 16, s2 = (3645.94)2 χ2 0 = (n − 1) s 2 15(3645.94) 2 = = 4984.83 40000 σ2 8) Since 4984.83 > 25.00 reject H0 and conclude there is strong evidence to indicate the true standard deviation of tire life exceeds 200 km at α = 0.05. b) P-value = P(χ2 > 4984.83) for 15 degrees of freedom P-value < 0.005 9-46 a) In order to use χ2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true standard deviation of Izod impact strength, σ. However, the answer can be found by performing a hypothesis test on σ2. 2) H0 : σ2 = (0.10)2 3) H1 : σ2 ≠ (0.10)2 4) α = 0.01 5) χ2 = 0 ( n − 1)s2 σ2 2 6) Reject H0 if χ2 < χ1− α / 2 ,n −1 where 0 χ 02.995,19 = 6.84 27 or 2 χ2 > χα ,2,n −1 where χ 0.005,19 0 2 = 38.58 7) n = 20, s = 0.25 χ2 = 0 (n − 1) s 2 σ2 = 19(0.25) 2 = 118.75 (0.10) 2 8) Since 118.75 > 38.58 reject H0 and conclude there is sufficient evidence to indicate the true standard deviation of Izod impact strength is significantly different from 0.10 at α = 0.01. b.) P-value: The P-value<0.005 c.) 99% confidence interval for σ: First find the confidence interval for σ2 : 2 For α = 0.01 and n = 20, χα / 2 , n −1 = χ 02.995,19 = 6.84 2 and χ1− α / 2 ,n −1 = χ 02.005,19 = 38.58 19(0.25) 2 19(0.25) 2 ≤σ 2 ≤ 38.58 6.84 2 0.03078 ≤ σ ≤ 0.1736 Since 0.01 falls below the lower confidence bound we would conclude that the population standard deviation is not equal to 0.01. 9-47. a) In order to use χ2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal. 9-25 1) The parameter of interest is the true standard deviation of titanium percentage, σ. However, the answer can be found by performing a hypothesis test on σ2. 2) H0 : σ2 = (0.25)2 3) H1 : σ2 ≠ (0.25)2 4) α = 0.01 5) χ2 = 0 ( n − 1)s2 σ2 2 2 6) Reject H0 if χ2 < χ1− α / 2 ,n −1 where χ2.995,50 = 27.99 or χ2 > χα ,2,n −1 where χ2.005,50 = 79.49 0 0 0 0 7) n = 51, s = 0.37 χ2 = 0 ( n − 1)s2 = 50(0.37) 2 = 109.52 σ (0.25) 2 8) Since 109.52 > 79.49 we would reject H0 and conclude there is sufficient evidence to indicate the true standard deviation of titanium percentage is significantly different from 0.25 at α = 0.01. 2 b) 95% confidence interval for σ: First find the confidence interval for σ2 : 2 2 For α = 0.05 and n = 51, χα / 2 , n −1 = χ2.025,50 = 71.42 and χ1− α / 2 ,n −1 = χ2.975,50 = 32.36 0 0 50(0.37) 2 50(0.37) 2 ≤σ 2 ≤ 71.42 32.36 0.096 ≤ σ2 ≤ 0.2115 Taking the square root of the endpoints of this interval we obtain, 0.31 < σ < 0.46 Since 0.25 falls below the lower confidence bound we would conclude that the population standard deviation is not equal to 0.25. 9-48 Using the chart in the Appendix, with λ= 0.012 = 1.22 and n = 15 we find 0.008 λ= 40 = 1.49 and β = 0.10, we find 18 β = 0.80. 9-49 Using the chart in the Appendix, with n = 30. 9-26 Section 9-5 9-50 a) The parameter of interest is the true proportion of engine crankshaft bearings exhibiting surface roughness. 2) H0 : p = 0.10 3) H1 : p >0.10 4) α = 0.05 5) z0 = x − np 0 np 0 (1 − p 0 ) or z0 = ˆ p − p0 ; Either approach will yield the same conclusion p 0 (1 − p 0 ) n 6) Reject H0 if z0 > zα where zα = z0.05 = 1.65 13 7) x = 10 n = 85 p = = 0.043 300 z0 = x − np 0 np 0 (1 − p 0 ) = 10 − 85(0.10) 85(0.10)(0.90) = 0.54 8) Since 0.53 < 1.65, do not reject the null hypothesis and conclude the true proportion of crankshaft bearings exhibiting surface roughness is not significantly greater than 0.10, at α = 0.05. 9-51 p= 0.15, p0=0.10, n=85, and zα/2=1.96 β =Φ =Φ p0 − p + zα / 2 p0 (1− p0 ) / n p(1− p) / n −Φ p0 − p − zα / 2 p0 (1− p0 ) / n 0.10− 0.15+1.96 0.10(1− 0.10) / 85 0.15(1− 0.15) / 85 p(1− p) / n −Φ 0.10− 0.15−1.96 0.10(1− 0.10) / 85 = Φ(0.36) − Φ(−2.94) = 0.6406− 0.0016= 0.639 n= = zα / 2 p0 (1− p0 ) − z β p(1− p) 2 p − p0 1.96 0.10(1− 0.10) −1.28 0.15(1− 0.15) 0.15− 0.10 = (10.85) 2 = 11763 ≅ 118 . 9-27 2 0.15(1− 0.15) / 85 9-52 a) Using the information from Exercise 8-48, test 1) The parameter of interest is the true fraction defective integrated circuits 2) H0 : p = 0.05 3) H1 : p ≠ 0.05 4) α = 0.05 5) z0 = x − np 0 np 0 (1 − p 0 ) or z0 = ˆ p − p0 p 0 (1 − p 0 ) n ; Either approach will yield the same conclusion 6) Reject H0 if z0 < − zα/2 where −zα/2 = −z0.025 = −1.96 or z0 > zα/2 where zα/2 = z0.025 = 1.96 13 7) x = 13 n = 300 p = = 0.043 300 z0 = x − np 0 np 0 (1 − p 0 ) = 13 − 300(0.05) 300(0.05)(0.95) = −0.53 8) Since −0.53 > −1.65, do not reject null hypothesis and conclude the true fraction of defective integrated circuits is not significantly less than 0.05, at α = 0.05. b.) The P-value: 2(1-Φ(0.53))=2(1-0.70194)=0.59612 9-53. a) Using the information from Exercise 8-48, test 1) The parameter of interest is the true fraction defective integrated circuits 2) H0 : p = 0.05 3) H1 : p < 0.05 4) α = 0.05 5) z0 = x − np0 np0 (1 − p0 ) or z0 = ˆ p − p0 p0 (1 − p0 ) n ; Either approach will yield the same conclusion 6) Reject H0 if z0 < − zα where −zα = −z0.05 = −1.65 13 7) x = 13 n = 300 p = = 0.043 300 z0 = x − np0 13 − 300(0.05) = = −0.53 np0 (1 − p0 ) 300(0.05)(0.95) 8) Since −0.53 > −1.65, do not null hypothesis and conclude the true fraction of defective integrated circuits is not significantly less than 0.05, at α = 0.05. b) P-value = 1 − Φ(0.53) = 0.29806 9-28 9-54 a) 1) The parameter of interest is the true proportion of engineering students planning graduate studies 2) H0 : p = 0.50 3) H1 : p ≠ 0.50 4) α = 0.05 x − np 0 z0 = 5) np 0 (1 − p 0 ) or z0 = ˆ p − p0 p 0 (1 − p 0 ) n ; Either approach will yield the same conclusion 6) Reject H0 if z0 < − zα/2 where −zα/2 = −z0.025 = −1.96 or z0 > zα/2 where zα/2 = z0.025 = 1.96 13 7) x = 117 n = 484 p = = 0.043 300 x − np 0 z0 = np 0 (1 − p 0 ) = 117 − 484(0.5) 484(0.5)(0.5) = −11.36 8) Since −11.36 > −1.65, reject the null hypothesis and conclude the true proportion of engineering students planning graduate studies is significantly different from 0.5, at α = 0.05. b.) P-value =2[1 − Φ(11.36)] ≅ 0 c.) ˆ p= 117 = 0.242 484 ˆ p − zα / 2 0.242 − 1.96 ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + zα / 2 n ˆ ˆ p (1 − p ) n 0.242(0.758) 0.242(0.758) ≤ p ≤ 0.242 − 1.96 484 484 0.204 ≤ p ≤ 0.280 Since the 95% confidence interval does not contain the value 0.5, then conclude that the true proportion of engineering students planning graduate studies is significantly different from 0.5. 9-55. a) 1) The parameter of interest is the true percentage of polished lenses that contain surface defects, p. 2) H0 : p = 0.02 3) H1 : p < 0.02 4) α = 0.05 5) z0 = x − np0 np0 (1 − p0 ) or z0 = ˆ p − p0 p0 (1 − p0 ) n ; Either approach will yield the same conclusion 6) Reject H0 if z0 < − zα where −zα = −z0.05 = −1.65 7) x = 6 n = 250 ˆ p= 6 = 0.024 250 ˆ p − p0 z0 = = p0 (1 − p0 ) n 0.024 − 0.02 = 0.452 0.02(1 − 0.02) 250 8) Since 0.452 > −1.65 do not reject the null hypothesis and conclude the machine cannot be qualified at the 0.05 level of significance. b) P-value = Φ(0.452) = 0.67364 9-29 9-56 . a) 1) The parameter of interest is the true percentage of football helmets that contain flaws, p. 2) H0 : p = 0.1 3) H1 : p > 0.1 4) α = 0.01 5) z0 = x − np0 np0 (1 − p0 ) or ˆ p − p0 p0 (1 − p0 ) n z0 = ; Either approach will yield the same conclusion 6) Reject H0 if z0 > zα where zα = z0.01 = 2.33 7) x = 16 n = 200 ˆ p= 16 = 0 .0 8 200 ˆ p − p0 z0 = = p 0 (1 − p 0 ) n 0.08 − 0.10 0.10(1 − 0.10) 200 = −0.94 8) Since -0.452 < 2.33 do not reject the null hypothesis and conclude the proportion of football helmets with flaws does not exceed 10%. b) P-value = 1-Φ(0.94) =1-.8264= 0.67364 9-57. The problem statement implies that H0: p = 0.6, H1: p > 0.6 and defines an acceptance region as 315 p≤ = 0.63 and rejection region as p > 0.63 500 a) The probability of a type 1 error is ˆ α = P( p ≥ 0.63 | p = 0.6 ) = P Z ≥ . 0.63 − 0.6 = P(Z ≥ 1.37 ) = 1 − P( Z < 1.37) = 0.08535 0.6(0.4) 500 b) β = P( P ≤ 0.63 | p = 0.75) = P(Z ≤ −6.196) = 0. 9-58 1) The parameter of interest is the true proportion of batteries that fail before 48 hours, p. 2) H0 : p = 0.002 3) H1 : p < 0.002 4) α = 0.01 5) z0 = x − np0 np0 (1 − p0 ) or z0 = ˆ p − p0 p0 (1 − p0 ) n ; Either approach will yield the same conclusion 6) Reject H0 if z0 < -zα where -zα = -z0.01 = -2.33 7) x = 15 n = 5000 ˆ p= 15 = 0.003 5000 ˆ p − p0 z0 = = p 0 (1 − p 0 ) n 0.003 − 0.002 0.002(1 − 0.998) 5000 = 1.58 8) Since 1.58 > -2.33 do not reject the null hypothesis and conclude the proportion of proportion of cell phone batteries that fail is not less than 0.2% at α=0.01. 9-30 Section 9-7 9-59. Expected Frequency is found by using the Poisson distribution e −λ λ x P( X = x) = where λ = [0( 24) + 1(30) + 2(31) + 3(11) + 4( 4)] / 100 = 1.41 x! Value Observed Frequency Expected Frequency 0 24 30.12 1 30 36.14 2 31 21.69 3 11 8.67 4 4 2.60 Since value 4 has an expected frequency less than 3, combine this category with the previous category: Value Observed Frequency Expected Frequency 0 24 30.12 1 30 36.14 2 31 21.69 3-4 15 11.67 The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3 a) 1) The variable of interest is the form of the distribution for X. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is k (Oi − Ei )2 i =1 2 χ0 = Ei 6) Reject H0 if χ2 > χ 2.05,3 = 7.81 o 0 7) (24− 30.12)2 + (30− 36.14)2 + (31− 21.69)2 + (15−11.67)2 = 7.23 χ= 2 0 30.12 36.14 21.69 11.67 8) Since 7.23 < 7.81 do not reject H0. We are unable to reject the null hypothesis that the distribution of X is Poisson. b) The P-value is between 0.05 and 0.1 using Table III. P-value = 0.0649 (found using Minitab) 9-31 9-60. Expected Frequency is found by using the Poisson distribution e −λ λx P( X = x) = where λ = [1(1) + 2(11) + x! + 7(10) + 8(9)] / 75 = 4.907 Estimated mean = 4.907 Value Observed Frequency Expected Frequency 1 1 2.7214 2 11 6.6770 3 8 10.9213 4 13 13.3977 5 11 13.1485 6 12 10.7533 7 10 7.5381 8 9 4.6237 Since the first category has an expected frequency less than 3, combine it with the next category: Value Observed Frequency Expected Frequency 1-2 12 9.3984 3 8 10.9213 4 13 13.3977 5 11 13.1485 6 12 10.7533 7 10 7.5381 8 9 4.6237 The degrees of freedom are k − p − 1 = 7 − 1 − 1 = 5 a) 1) The variable of interest is the form of the distribution for the number of flaws. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.01 5) The test statistic is χ2 = 0 6) Reject H0 if χ2 o > χ 2.01,5 0 k ( O i − E i )2 i =1 Ei = 15.09 7) χ2 = 0 (12 − 9.3984) 2 + + ( 9 − 4.6237) 2 = 6.955 9.3984 4.6237 8) Since 6.955 < 15.09 do not reject H0. We are unable to reject the null hypothesis that the distribution of the number of flaws is Poisson. b) P-value = 0.2237 (found using Minitab) 9-32 9-61. Estimated mean = 10.131 Value Rel. Freq Observed (Days) Expected (Days) 5 0.067 2 6 0.067 2 8 0.100 3 9 0.133 4 10 0.200 6 11 0.133 4 12 0.133 4 13 0.067 2 14 0.033 1 15 0.067 2 1.0626 1.7942 3.2884 3.7016 3.7501 3.4538 2.9159 2.2724 1.6444 1.1106 Since there are several cells with expected frequencies less than 3, the revised table would be: Value Observed (Days) Expected (Days) 5-8 7 9 4 10 6 11 4 12-15 9 6.1452 3.7016 3.7501 3.4538 7.9433 The degrees of freedom are k − p − 1 = 5 − 1 − 1 = 3 a) 1) The variable of interest is the form of the distribution for the number of calls arriving to a switchboard from noon to 1pm during business days. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is χ2 = 0 k ( O i − E i )2 i =1 Ei 6) Reject H0 if χ2 > χ2.05,3 = 7.81 o 0 χ 02 7) (7 − 6 . 1452 )2 + (4 − 3 . 7016 )2 + (6 − 3 . 7501 )2 + (4 − 3 . 4538 )2 + (9 − 7 . 9433 )2 = 1.72 = 6 . 1452 3 . 7016 3 . 7501 3 . 4538 7 . 9433 8) Since 1.72 < 7.81 do not reject H0. We are unable to reject the null hypothesis that the distribution for the number of calls is Poisson. b) The P-value is between 0.9 and 0.5 using Table III. P-value = 0.6325 (found using Minitab) 9-33 9-62 Use the binomial distribution to get the expected frequencies with the mean = np = 6(0.25) = 1.5 Value Observed Expected 0 4 8.8989 1 21 17.7979 2 10 14.8315 3 13 6.5918 4 2 1.6479 The expected frequency for value 4 is less than 3. Combine this cell with value 3: Value Observed Expected 0 4 8.8989 1 21 17.7979 2 10 14.8315 3-4 15 8.2397 The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3 a) 1) The variable of interest is the form of the distribution for the random variable X. 2) H0: The form of the distribution is binomial with n = 6 and p = 0.25 3) H1: The form of the distribution is not binomial with n = 6 and p = 0.25 4) α = 0.05 5) The test statistic is χ2 = 0 6) Reject H0 if χ2 o > χ2.05,3 0 k ( O i − E i )2 i =1 Ei = 7.81 7) χ2 = 0 ( 4 − 8.8989) 2 + 8.8989 + (15 − 8.2397) 2 8.2397 = 10.39 8) Since 10.39 > 7.81 reject H0. We can conclude that the distribution is not binomial with n = 6 and p = 0.25 at α = 0.05. b) P-value = 0.0155 (found using Minitab) 9-34 9-63 The value of p must be estimated. Let the estimate be denoted by psample 0( 39) + 1(23) + 2(12) + 3(1) = 0.6667 75 sample mean = ˆ p sample = sample mean 0.6667 = = 0.02778 n 24 Value Observed Expected 0 39 38.1426 1 23 26.1571 2 12 8.5952 3 1 1.8010 Since value 3 has an expected frequency less than 3, combine this category with that of value 2: Value Observed Expected 0 39 38.1426 1 23 26.1571 2-3 13 10.3962 The degrees of freedom are k − p − 1 = 3 − 1 − 1 = 1 a) 1) The variable of interest is the form of the distribution for the number of underfilled cartons, X. 2) H0: The form of the distribution is binomial 3) H1: The form of the distribution is not binomial 4) α = 0.05 5) The test statistic is χ2 = 0 6) Reject H0 if χ2 o > χ 2.05,1 0 χ2 = 0 k ( O i − E i )2 i =1 Ei = 384 . ( 39 − 38.1426) 2 + (23 − 26.1571)2 + (13 − 10.3962) 2 = 1.053 381426 . 26.1571 10.39 8) Since 1.053 < 3.84 do not reject H0. We are unable to reject the null hypothesis that the distribution of the number of underfilled cartons is binomial at α = 0.05. 7) b) The P-value is between 0.5 and 0.1 using Table III P-value = 0.3048 (found using Minitab) 9-64 Estimated mean = 49.6741 use Poisson distribution with λ=49.674 All expected frequencies are greater than 3. The degrees of freedom are k − p − 1 = 26 − 1 − 1 = 24 a) 1) The variable of interest is the form of the distribution for the number of cars passing through the intersection. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is χ2 = 0 k ( O i − E i )2 i =1 Ei 6) Reject H0 if χ2 > χ2.05,24 = 36.42 o 0 7) Estimated mean = 49.6741 χ2 = 769.57 0 8) Since 769.57 >>> 36.42, reject H0. We can conclude that the distribution is not Poisson at α = 0.05. b) P-value = 0 (found using Minitab) 9-35 Section 9-8 9-65. 1. The variable of interest is breakdowns among shift. 2. H0: Breakdowns are independent of shift. 3. H1: Breakdowns are not independent of shift. 4. α = 0.05 5. The test statistic is: χ= r c 2 0 (O E ij i =1 j =1 6. The critical value is χ .05 , 6 2 − E ij ) 2 ij = 12.592 7. The calculated test statistic is χ 0 = 11.65 2 8. χ 2 > χ 2.05,6 , do not reject H0 and conclude that the data provide insufficient evidence to claim that 0/ 0 machine breakdown and shift are dependent at α = 0.05. P-value = 0.070 (using Minitab) 9-66 1. The variable of interest is calls by surgical-medical patients. 2. H0:Calls by surgical-medical patients are independent of Medicare status. 3. H1:Calls by surgical-medical patients are not independent of Medicare status. 4. α = 0.01 5. The test statistic is: χ= r c 2 0 (O E ij i =1 j =1 6. The critical value is χ .01,1 2 = 6.637 7. The calculated test statistic is χ 0 2 8. χ 02 > χ 02.01,1 / − E ij ) 2 ij = 0.033 , do not reject H0 and conclude that the evidence is not sufficient to claim that surgical- medical patients and Medicare status are dependent. P-value = 0.85 9-67. 1. The variable of interest is statistics grades and OR grades. 2. H0: Statistics grades are independent of OR grades. 3. H1: Statistics and OR grades are not independent. 4. α = 0.01 5. The test statistic is: χ= r (O c 2 0 i =1 j =1 6. The critical value is χ 2 .01, 9 2 χ 02 > χ 0.01,9 E ij = 21.665 7. The calculated critical value is 8. − E ij ) 2 ij χ = 25.55 2 0 Therefore, reject H0 and conclude that the grades are not independent at α = 0.01. P-value = 0.002 9-68 1. The variable of interest is characteristic among deflections and ranges. 9-36 2. H0: Deflection and range are independent. 3. H1: Deflection and range are not independent. 4. α = 0.05 5. The test statistic is: r χ= (O c 2 0 E ij i =1 j =1 6. The critical value is χ 2 0.05 , 4 = 9.488 7. The calculated test statistic is 8. χ 02 > χ 02.05, 4 / − E ij ) 2 ij χ = 2 .4 6 2 0 , do not reject H0 and conclude that the evidence is not sufficient to claim that the data are not independent at α = 0.05. P-value = 0.652 9-69. 1. The variable of interest is failures of an electronic component. 2. H0: Type of failure is independent of mounting position. 3. H1: Type of failure is not independent of mounting position. 4. α = 0.01 5. The test statistic is: 2 χ0 = r c (O E ij i =1 j =1 6. The critical value is χ 0.01, 3 2 = 11.344 7. The calculated test statistic is 8. 2 χ 02 > χ 0.01,3 / − E ij ) 2 ij χ = 10.71 2 0 , do not reject H0 and conclude that the evidence is not sufficient to claim that the type of failure is not independent of the mounting position at α = 0.01. P-value = 0.013 9-70 1. The variable of interest is opinion on core curriculum change. 2. H0: Opinion of the change is independent of the class standing. 3. H1: Opinion of the change is not independent of the class standing. 4. α = 0.05 5. The test statistic is: χ= r c 2 0 i =1 j =1 6. The critical value is χ 0.05 , 3 2 χ 02 >>> χ 02.05,3 − E ij ) 2 ij E ij = 7.815 7. The calculated test statistic is 8. (O χ = 26.97 . 2 .0 , reject H0 and conclude that the opinions on the change are not independent of class standing. P-value ≈ 0 9-37 Supplemental Exercises a. Sample Size, n 50 p(1 − p) n Sampling Distribution Normal b. 80 Normal p c. 9-71 100 Normal p Sample Mean = p Sample Variance = Sample Mean p Sample Variance p(1 − p) 50 p(1 − p) 80 p(1 − p) 100 d) As the sample size increases, the variance of the sampling distribution decreases. 9-72 a. n 50 b. 100 c. 500 1000 d. Test statistic 0.095 − 0.10 z0 = = − 0 .1 2 0.10(1 − 0.10) / 50 0.095 − 0.10 z0 = = −0.15 0.10(1 − 0.10) / 100 0.095 − 0.10 z0 = = −0.37 0.10(1 − 0.10) / 500 0.095 − 0.10 z0 = = −0.53 0.10(1 − 0.10) / 1000 P-value 0.4522 conclusion Do not reject H0 0.4404 Do not reject H0 0.3557 Do not reject H0 0.2981 Do not reject H0 e. The P-value decreases as the sample size increases. 9-73. σ = 12, δ = 205 − 200 = 5, α = 0.025, z0.025 = 1.96, 2 a) n = 20: β = Φ 1.96 − 5 20 = Φ (0.163) = 0.564 12 b) n = 50: β = Φ 1.96 − 5 50 = Φ (−0.986) = 1 − Φ (0.986) = 1 − 0.839 = 0.161 12 c) n = 100: β = Φ 1.96 − 5 100 = Φ (−2.207) = 1 − Φ (2.207) = 1 − 0.9884 = 0.0116 12 d) β, which is the probability of a Type II error, decreases as the sample size increases because the variance of the sample mean decreases. Consequently, the probability of observing a sample mean in the acceptance region centered about the incorrect value of 200 ml/h decreases with larger n. 9-38 9-74 σ = 14, δ = 205 − 200 = 5, α = 0.025, z0.025 = 1.96, 2 a) n = 20: β = Φ 1.96 − 5 20 = Φ (0.362) = 0.6406 14 b) n = 50: β = Φ 1.96 − 5 50 = Φ (−0.565) = 1 − Φ (0.565) = 1 − 0.7123 = 0.2877 14 c) n = 100: β = Φ 1.96 − 5 100 = Φ (−1.611) = 1 − Φ (1.611) = 1 − 0.9463 = 0.0537 14 d) The probability of a Type II error increases with an increase in the standard deviation. 9-75. σ = 8, δ = 204 − 200 = −4, a) n = 20: β = Φ 1.96 − α = 0.025, z0.025 = 1.96. 2 4 20 = Φ( −0.28) = 1 − Φ(0.28) = 1 − 0.61026 = 0.38974 8 Therefore, power = 1 − β = 0.61026 b) n = 50: β = Φ 1.96 − 4 50 = Φ( −2.58) = 1 − Φ(2.58) = 1 − 0.99506 = 0.00494 8 Therefore, power = 1 − β = 0.995 c) n = 100: β = Φ 1.96 − 4 100 = Φ( −3.04) = 1 − Φ(3.04) = 1 − 0.99882 = 0.00118 8 Therefore, power = 1 − β = 0.9988 d) As sample size increases, and all other values are held constant, the power increases because the variance of the sample mean decreases. Consequently, the probability of a Type II error decreases, implies the power increases. 9-76 α=0.01 β = Φ z 0.01 + 85 − 86 = Φ (2.33 − 0.31) = Φ (2.02) = 0.9783 16 / 25 85 − 86 n=100 β = Φ z 0.01 + = Φ (2.33 − 0.63) = Φ (1.70) = 0.9554 16 / 100 85 − 86 n=400 β = Φ z 0.01 + = Φ (2.33 − 1.25) = Φ (1.08) = 0.8599 16 / 400 85 − 86 n=2500 β = Φ z 0.01 + = Φ (2.33 − 3.13) = Φ (−0.80) = 0.2119 16 / 2500 a.) n=25 b.) n=25 z0 = 86 − 85 16 / 25 = 0 .3 1 P-value: 1 − Φ (0.31) = 1 − 0.6217 = 0.3783 9-39 which n=100 n=400 z0 = z0 = 86 − 85 16 / 100 86 − 85 = 0.63 P-value: 1 − Φ (0.63) = 1 − 0.7357 = 0.2643 = 1.25 P-value: 1 − Φ (1.25) = 1 − 0.8944 = 0.1056 16 / 400 86 − 85 z0 = = 3.13 P-value: 1 − Φ (3.13) = 1 − 0.9991 = 0.0009 16 / 2500 n=2500 The data would be statistically significant when n=2500 at α=0.01 9-77. a) Rejecting a null hypothesis provides a stronger conclusion than failing to reject a null hypothesis. Therefore, place what we are trying to demonstrate in the alternative hypothesis. Assume that the data follow a normal distribution. b) 1) the parameter of interest is the mean weld strength, µ. 2) H0 : µ = 150 3) H1 : µ > 150 4) Not given 5) The test statistic is: x − µ0 t0 = s/ n 6) Since no critical value is given, we will calculate the P-value 7) x = 153.7 , s= 11.3, n=20 t0 = 153.7 − 150 = 1 .4 6 11.3 20 P-value = P( t ≥ 1.46) = 0.05 < p − value < 0.10 8) There is some modest evidence to support the claim that the weld strength exceeds 150 psi. If we used α = 0.01 or 0.05, we would not reject the null hypothesis, thus the claim would not be supported. If we used α = 0.10, we would reject the null in favor of the alternative and conclude the weld strength exceeds 150 psi. 9-78 a.) α=0.05 β = Φ z 0.05 + n=100 0 .5 − 0 .6 0.5(0.5) / 100 = Φ (1.65 − 2.0) = Φ (−0.35) = 0.3632 Power = 1 − β = 1 − 0.3632 = 0.6368 n=150 β = Φ z 0.05 + 0 .5 − 0 .6 0.5(0.5) / 100 = Φ (1.65 − 2.45) = Φ (−0.8) = 0.2119 Power = 1 − β = 1 − 0.2119 = 0.7881 n=300 β = Φ z 0.05 + 0 .5 − 0 .6 0.5(0.5) / 300 = Φ (1.65 − 3.46) = Φ (−1.81) = 0.03515 Power = 1 − β = 1 − 0.03515 = 0.96485 b.) α=0.01 9-40 n=100 β = Φ z 0.01 + 0 .5 − 0 .6 0.5(0.5) / 100 = Φ (2.33 − 2.0) = Φ (0.33) = 0.6293 Power = 1 − β = 1 − 0.6293 = 0.3707 n=150 β = Φ z 0.01 + 0 .5 − 0 .6 0.5(0.5) / 100 = Φ (2.33 − 2.45) = Φ (−0.12) = 0.4522 Power = 1 − β = 1 − 0.4522 = 0.5478 n=300 β = Φ z 0.01 + 0 .5 − 0 .6 0.5(0.5) / 300 = Φ (2.33 − 3.46) = Φ (−1.13) = 0.1292 Power = 1 − β = 1 − 0.1292 = 0.8702 Decreasing the value of α decreases the power of the test for the different sample sizes. c.) α=0.05 n=100 β = Φ z 0.05 + 0 .5 − 0 .8 0.5(0.5) / 100 = Φ (1.65 − 6.0) = Φ (−4.35) ≅ 0.0 Power = 1 − β = 1 − 0 ≅ 1 The true value of p has a large effect on the power. The further p is away from p0 the larger the power of the test. d.) n= = n= = zα / 2 p0 (1− p0 ) − z β p(1− p) 2 p − p0 2.58 0.5(1− 0.50) −1.65 0.6(1− 0.6) 2 = (4.82) 2 = 23.2 ≅ 24 0.6 − 0.5 zα / 2 p0 (1− p0 ) − zβ p(1− p) 2 p − p0 2.58 0.5(1− 0.50) −1.65 0.8(1− 0.8) 0.8 − 0.5 2 = (2.1) 2 = 4.41≅ 5 The true value of p has a large effect on the power. The further p is away from p0 the smaller the sample size that is required. 9-79 a) 1) the parameter of interest is the standard deviation, σ 9-41 2) H0 : σ2 = 400 3) H1 : σ2 < 400 4) Not given 5) The test statistic is: χ2 = 0 ( n − 1) s2 σ2 6) Since no critical value is given, we will calculate the p-value 7) n = 10, s = 15.7 χ2 = 0 ( ) 9(15.7) 2 = 5.546 400 P-value = P χ 2 < 5546 ; . 0.1 < P − value < 0.5 8) The P-value is greater than any acceptable significance level, α, therefore we do not reject the null hypothesis. There is insufficient evidence to support the claim that the standard deviation is less than 20 microamps. b) 7) n = 51, s = 20 χ2 = 0 ( 50(15.7)2 = 30.81 400 ) P-value = P χ 2 < 30.81 ; 0.01 < P − value < 0.025 8) The P-value is less than 0.05, therefore we reject the null hypothesis and conclude that the standard deviation is significantly less than 20 microamps. c) Increasing the sample size increases the test statistic χ2 and therefore decreases the P-value, providing 0 more evidence against the null hypothesis. 9-80 a) 1) the parameter of interest is the variance of fatty acid measurements, σ2 2) H0 : σ2 = 1.0 3) H1 : σ2 ≠ 1.0 4) α=0.01 5) The test statistic is: 6) χ2 = 0 ( n − 1) s2 σ2 χ 02.995,5 = 0.41 reject H0 if χ 02 < 0.41 or χ 02.005,5 = 16.75 reject H0 if χ 02 > 16.75 7) n = 6, s = 0.319 5(0.319) 2 = 0.509 12 P χ 2 < 0.509 ; 0.01 < P − value < 0.02 χ 02 = P-value = ( ) 8) Since 0.509>0.41, do not reject the null hypothesis and conclude that there is insufficient evidence to conclude that the variance is not equal to 1.0. The P-value is greater than any acceptable significance level, α, therefore we do not reject the null hypothesis. b) 1) the parameter of interest is the variance of fatty acid measurements, σ2 (now n=51) 2) H0 : σ2 = 1.0 9-42 3) H1 : σ2 ≠ 1.0 4) α=0.01 5) The test statistic is: 6) χ 2 0.995 , 50 χ2 = 0 ( n − 1) s2 σ2 2 = 27.99 reject H0 if χ 0 < 27.99 or χ 02.005,5 = 79.49 reject H0 if χ 02 > 79.49 7) n = 51, s = 0.319 50(0.319) 2 = 5 .0 9 12 P χ 2 < 5.09 ; P − value < 0.01 χ 02 = P-value = ( ) 8) Since 5.09<27.99, reject the null hypothesis and conclude that there is sufficient evidence to conclude that the variance is not equal to 1.0. The P-value is smaller than any acceptable significance level, α, therefore we do reject the null hypothesis. c.) The sample size changes the conclusion tha t is drawn. With a small sample size, the results are inconclusive. A larger sample size helps to make sure that the correct conclusion is drawn. 9-81. Assume the data follow a normal distribution. a) 1) The parameter of interest is the standard deviation, σ. 2) H0 : σ2 = (0.00002)2 3) H1 : σ2 < (0.00002)2 4) α = 0.01 5) The test statistic is: χ2 = 0 ( n − 1) s2 σ2 6) χ2.99,7 = 1.24 reject H0 if χ2 < 124 . 0 0 7) s = 0.00001 and α = 0.01 7(0.00001) 2 χ= = 1 .7 5 (0.00002) 2 2 0 1.75 > 1.24, do not reject the null hypothesis; that is, there is insufficient evidence to conclude the standard deviation is at most 0.00002 mm. b) Although the sample standard deviation is less than the hypothesized value of 0.00002, it is not significantly less (when α = 0.01) than 0.00002 to conclude the standard deviation is at most 0.00002 mm. The value of 0.00001 could have occurred as a result of sampling variation. 9-82 Assume the data follow a normal distribution. 1) The parameter of interest is the standard deviation of the concentration, σ. 2) H0 : σ2 =42 3) H1 : σ2 < 42 4) not given 5) The test statistic is: χ2 = 0 ( n − 1) s2 σ2 6) will be determined based on the P-value 7) s = 0.004 and n = 10 χ 02 = ( ) 9(0.004) 2 = 0.000009 ( 4) 2 P − value ≅ 0. P-value = P χ < 0.00009 ; The P-value is approximately 0, therefore we reject the null hypothesis and conclude that the standard deviation of the concentration is less than 4 grams per liter. 2 9-43 9-83. Create a table for the number of nonconforming coil springs (value) and the observed number of times the number appeared. One possible table is: Value Obs 0 0 1 0 2 0 3 1 4 4 5 3 6 4 7 6 8 4 9 3 10 0 11 3 12 3 13 2 14 1 15 1 16 0 17 2 18 1 The value of p must be estimated. Let the estimate be denoted by psample 0(0) + 1(0) + 2(0) + + 19(2) = 9.325 40 sample mean 9.325 = = = 0.1865 n 50 sample mean = ˆ p sample Value Observed Expected 0 0 0.00165 1 0 0.01889 2 0 0.10608 3 1 0.38911 4 4 1.04816 5 3 2.21073 6 4 3.80118 7 6 5.47765 8 4 6.74985 9 3 7.22141 10 0 6.78777 11 3 5.65869 12 3 4.21619 13 2 2.82541 14 1 1.71190 15 1 0.94191 16 0 0.47237 17 2 0.21659 18 1 0.09103 19 2 0.03515 Since several of the expected values are less than 3, some cells must be combined resulting in the following table: Value 0-5 6 7 8 9 10 11 12 ≥13 Observed 8 4 6 4 3 0 3 3 9 Expected 3.77462 3.80118 5.47765 6.74985 7.22141 6.78777 5.65869 4.21619 6.29436 The degrees of freedom are k − p − 1 = 9 − 1 − 1 = 7 a) 1) The variable of interest is the form of the distribution for the number of nonconforming coil springs. 2) H0: The form of the distribution is binomial 3) H1: The form of the distribution is not binomial 9-44 19 2 4) α = 0.05 5) The test statistic is χ2 = 0 6) Reject H0 if χ2 0 > χ 2.05,7 0 k ( O i − E i )2 i =1 Ei = 14.07 7) (8 - 3.77462)2 (4 − 3.8.011) 2 (9 − 6.29436) 2 + ++ = 17.929 3.77462 38011 . 6.29436 8) Since 17.929 > 14.07 reject H0. We are able to conclude the distribution of nonconforming springs is not binomial at α = 0.05. b) P-value = 0.0123 (found using Minitab) χ2 = 0 9-84 Create a table for the number of errors in a string of 1000 bits (value) and the observed number of times the number appeared. One possible table is: Value 0 1 2 3 4 5 Obs 3 7 4 5 1 0 The value of p must be estimated. Let the estimate be denoted by psample 0(3) + 1(7) + 2(4) + 3(5) + 4(1) + 5(0) = 1 .7 20 sample mean 1 .7 = = = 0.0017 n 1000 sample mean = ˆ p sample Value Observed Expected 0 3 3.64839 1 7 6.21282 2 4 5.28460 3 5 2.99371 4 1 1.27067 5 0 0.43103 Since several of the expected values are less than 3, some cells must be combined resulting in the following table: Value 0 1 2 ≥3 Observed 3 7 4 6 Expected 3.64839 6.21282 5.28460 4.69541 The degrees of freedom are k − p − 1 = 4 − 1 − 1 = 2 a) 1) The variable of interest is the form of the distribution for the number of errors in a string of 1000 bits. 2) H0: The form of the distribution is binomial 3) H1: The form of the distribution is not binomial 4) α = 0.05 5) The test statistic is 6) Reject H0 if 7) χ 2 0 χ2 0 > χ2.05,2 0 k (Oi − Ei )2 i =1 2 χ0 = Ei = 5.99 (3 − 3.64839)2 + = 3.64839 (6 − 4.69541)2 + 4.69541 = 0.88971 8) Since 0.88971 < 9.49 do not reject H0. We are unable to reject the null hypothesis that the distribution of the number of errors is binomial at α = 0.05. b) P-value = 0.6409 (found using Minitab) 9-45 9-85 We can divide the real line under a standard normal distribution into eight intervals with equal probability. These intervals are [0,.32), [0.32, 0.675), [0.675, 1.15), [1.15, ∞) and their negative counterparts. The probability for each interval is p = 1/8 = .125 so the expected cell frequencies are E = np = (100) (0.125) = 12.5. The table of ranges and their corresponding frequencies is completed as follows. Interval Obs. Frequency. Exp. Frequency. x ≤ 5332.5 1 12.5 5332.5< x ≤ 5357.5 4 12.5 5357.5< x ≤ 5382.5 7 12.5 5382.5< x ≤ 5407.5 24 12.5 5407.5< x ≤ 5432.5 30 12.5 20 12.5 5432.5< x ≤ 5457.5 5457.5< x ≤ 5482.5 15 12.5 x ≥ 5482.5 5 12.5 The test statistic is: χ 02 = (1 - 12.5) 12 . 5 2 + ( 4 − 12 . 5 ) 2 + 12 . 5 + and we would reject if this value exceeds (15 - 12.5) 12 . 5 2 + ( 5 − 12 . 5 ) 2 = 63 . 36 12 . 5 2 2 χ 20.05,5 = 11.07 . Since χ o > χ 0.05,5 , reject the hypothesis that the data are normally distributed 9-86 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean concentration of suspended solids, µ. 2) H0 : µ = 50 3) H1 : µ < 50 4) α = 0.05 5) Since n>>30 we can use the normal distribution z0 = x−µ s/ n 6) Reject H0 if z0 <- zα where z0.05 =1.65 7) x = 59.87 s = 12.50 n = 60 z0 = 59.87 − 50 12.50 / 60 = 6 .1 2 8) Since 6.12>-1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean concentration of suspended solids is less than 50 ppm at α = 0.05. b) The P-value = Φ(6.12) ≅ 1 . c.) We can divide the real line under a standard normal distribution into eight intervals with equal probability. These intervals are [0,.32), [0.32, 0.675), [0.675, 1.15), [1.15, ∞) and their negative 9-46 counterparts. The probability for each interval is p = 1/8 = .125 so the expected cell frequencies are E = np = (60) (0.125) = 7.5. The table of ranges and their corresponding frequencies is completed as follows. Interval Obs. Frequency. Exp. Frequency. x ≤ 45.50 9 7.5 45.50< x ≤ 51.43 5 7.5 7.5 51.43< x ≤ 55.87 7 55.87< x ≤ 59.87 11 7.5 59.87< x ≤ 63.87 4 7.5 7.5 63.87< x ≤ 68.31 9 68.31< x ≤ 74.24 8 7.5 6 7.5 x ≥ 74.24 The test statistic is: χ 2o = (9 − 7.5) 2 (5 − 7.5) 2 + + 7 .5 7 .5 + (8 − 7.5) 2 (6 − 7.5) 2 + = 5 .0 6 7 .5 7 .5 and we would reject if this value exceeds χ 20.05,5 = 11.07 . Since it does not, we cannot reject the hypothesis that the data are normally distributed. 9-87 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean overall distance for this brand of golf ball, µ. 2) H0 : µ = 270 3) H1 : µ < 270 4) α = 0.05 5) Since n>>30 we can use the normal distribution z0 = x−µ s/ n 6) Reject H0 if z0 <- zα where z0.05 =1.65 7) x = 1.25 s = 0.25 n = 100 z0 = 260.30 − 270.0 13.41 / 100 = −7.23 8) Since –7.23<-1.65, reject the null hypothesis and conclude there is sufficient evidence to indicate that the true mean distance is less than 270 yds at α = 0.05. b) The P-value ≅ 0 9-47 c) We can divide the real line under a standard normal distribution into eight intervals with equal probability. These intervals are [0,.32), [0.32, 0.675), [0.675, 1.15), [1.15, ∞) and their negative counterparts. The probability for each interval is p = 1/8 = .125 so the expected cell frequencies are E = np = (100) (0.125) = 12.5. The table of ranges and their corresponding frequencies is completed as follows. Interval Obs. Frequency. Exp. Frequency. x ≤ 244.88 16 12.5 244.88< x ≤ 251.25 6 12.5 17 12.5 251.25< x ≤ 256.01 256.01< x ≤ 260.30 9 12.5 260.30< x ≤ 264.59 13 12.5 8 12.5 264.59< x ≤ 269.35 269.35< x ≤ 275.72 19 12.5 x ≥ 275.72 12 12.5 The test statistic is: χ 2o = (16 − 12.5) 2 (6 − 12.5) 2 + + 12.5 12.5 + (19 − 12.5) 2 (12 − 12.5) 2 + = 12 12.5 12.5 and we would reject if this value exceeds χ 20.05,5 = 11.07 . Since it does, we can reject the hypothesis that the data are normally distributed. 9-88 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean coefficient of restitution, µ. 2) H0 : µ = 0.635 3) H1 : µ > 0.635 4) α = 0.01 5) Since n>30 we can use the normal distribution z0 = x−µ s/ n 6) Reject H0 if z0 > zα where z0.05 =2.33 7) x = 0.324 s = 0.0131 n = 40 z0 = 0.624 − 0.635 0.0131 / 40 = − 5 .3 1 8) Since –5.31< 2.33, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean coefficient of restitution is greater than 0.635 at α = 0.01. b) The P-value Φ(5.31) ≅ 1.. c.) If the lower bound of the CI was above the value 0.635 then we could conclude that the mean coefficient of restitution was greater than 0.635. 9-48 9-89 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. Use the t-test to test the hypothesis that the true mean is 2.5 mg/L. 1) State the parameter of interest: The parameter of interest is the true mean dissolved oxygen level, µ. 2) State the null hypothesis H0 : µ = 2.5 3) State the alternative hypothesis H1 : µ ≠ 2.5 4) Give the significance level α = 0.05 5) Give the statistic t0 = x−µ s/ n 6) Reject H0 if |t0 | <tα/2,n-1 7) find the sample statistic x = 3.265 s =2.127 n = 20 x−µ and calculate the t-statistic t0 = s/ n 8) Draw your conclusion and find the P-value. b) Assume the data are normally distributed. 1) The parameter of interest is the true mean dissolved oxygen level, µ. 2) H0 : µ = 2.5 3) H1 : µ ≠ 2.5 4) α = 0.05 5)Test statistic t0 = x−µ s/ n 6) Reject H0 if |t0 | >tα/2,n-1 where tα/2,n-1= t0.025,19b =2.093 7) x = 3.265 s =2.127 n = 20 t0 = 3.265 − 2.5 2.127 / 20 = 1.608 8) Since 1.608 < 2.093, do not reject the null hypotheses and conclude that the true mean is not significantly different from 2.5 mg/L c.) The value of 1.608 is found between the columns of 0.05 and 0.1 of table IV. Therefore the P-value is between 0.1 and 0.2. Minitab gives a value of 0.124 d.) The confidence interval found in exercise 8-81 b. agrees with the hypothesis test above. The value of 2.5 is within the 95% confidence limits. The confidence interval shows that the interval is quite wide due to the large sample standard deviation value. x − t 0.025,19 3.265 − 2.093 s n 2.127 ≤ µ ≤ x + t 0.025,19 s n ≤ µ ≤ 3.265 + 2.093 20 2.270 ≤ µ ≤ 4.260 9-49 2.127 20 9-90 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. d= δ | µ − µ 0 | | 73 − 75 | = = =2 σ σ 1 Using the OC curve for α = 0.05, d = 2, and n = 10, we get β ≅ 0.0 and power of 1−0.0 ≅ 1. d= δ | µ − µ 0 | | 72 − 75 | = = =3 σ σ 1 Using the OC curve for α = 0.05, d = 3, and n = 10, we get β ≅ 0.0 and power of 1−0.0 ≅ 1. b) d = δ | µ − µ 0 | | 73 − 75 | = = =2 σ σ 1 Using the OC curve, Chart VI e) for α = 0.05, d = 2, and β ≅ 0.1 (Power=0.9), n* = 5 . d= Therefore, n= n* + 1 5 + 1 = =3 2 2 δ | µ − µ 0 | | 72 − 75 | = = =3 σ σ 1 Using the OC curve, Chart VI e) for α = 0.05, d = 3, and β ≅ 0.1 (Power=0.9), n* = 3 . Therefore, n= n* + 1 3 + 1 = =2 2 2 9-50 c) σ = 2 . δ | µ − µ 0 | | 73 − 75 | = = =1 σ σ 2 d= Using the OC curve for α = 0.05, d = 1, and n = 10, we get β ≅ 0.10 and power of 1−0.10 ≅ 0.90. d= δ | µ − µ0 | | 72 − 75 | = = = 1 .5 σ σ 2 Using the OC curve for α = 0.05, d = 1.5, and n = 10, we get β ≅ 0.04 and power of 1−0.04 ≅ 0.96. d= δ | µ − µ 0 | | 73 − 75 | = = =1 σ σ 2 Using the OC curve, Chart VI e) for α = 0.05, d = 1, and β ≅ 0.1 (Power=0.9), n = 10 . * d= Therefore, n * + 1 10 + 1 = = 5 .5 n= 2 2 n≅6 δ | µ − µ 0 | | 72 − 75 | = = = 1 .5 σ σ 2 Using the OC curve, Chart VI e) for α = 0.05, d = 3, and β ≅ 0.1 (Power=0.9), n* = 7 . Therefore, n= n* + 1 7 + 1 = =4 2 2 Increasing the standard deviation lowers the power of the test and increases the sample size required to obtain a certain power. Mind Expanding Exercises 9-91 The parameter of interest is the true,µ. H0 : µ = µ0 H1 µ ≠ µ0 9-92 a.) Reject H0 if z0 < -zα-ε or z0 > zε X − µ0 X − µ0 X − µ0 X − µ0 <− = | µ = µ0 ) + P( > | µ = µ0 ) P σ/ n σ/ n σ/ n σ/ n P ( z0 < − zα − ε ) + P ( z0 > zε ) = Φ(− zα −ε ) + 1 − Φ( zε ) P( = ((α − ε )) + (1 − (1 − ε )) = α b.) β = P(zε ≤ X ≤ zε when µ1 = µ 0 + d ) or β = P( − zα − ε < Z 0 < z ε | µ1 = µ 0 + δ) x−µ0 β = P( − z α − ε < < z ε | µ 1 = µ 0 + δ) σ2 /n = P( − z α − ε − = Φ( z ε − δ σ2 /n δ σ2 /n < Z < zε − ) − Φ( − z α − ε − δ ) σ2 /n δ ) σ2 /n 9-51 9-93 1) The parameter of interest is the true mean number of open circuits, λ. 2) H0 : λ = 2 3 ) H1 : λ > 2 4) α = 0.05 5) Since n>30 we can use the normal distribution z0 = X −λ λ/n 6) Reject H0 if z0 > zα where z0.05 =1.65 7) x = 1038/500=2.076 n = 500 z0 = 2.076 − 2 2 / 500 = 0 .8 5 8) Since 0.85< 1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean number of open circuits is greater than 2 at α = 0.01 9-94 1) The parameter of interest is the true standard deviation of the golf ball distance, λ. 2) H0 : σ = 10 3) H1 : σ < 10 4) α=0.05 5) Since n>30 we can use the normal distribution S −σ0 z0 = σ 02 /(2n) 6) Reject H0 if z0 < zα where z0.05 =-1.65 7) s = 13.41 n = 100 z0 = 13.41 − 10 10 2 /(200) = 4 .8 2 8) Since 4.82 > -1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true standard deviation is less than 10 at α = 0.05 9-95 θ = µ + 1.645σ S −σ0 95% percentile using z0 = σ 02 /(2n) 95% percentile: X + 1.645 s 2 /(2n) S .E.(θ ) = σ / n = s 2n n = s / 3n = 9-52 9-96 1) The parameter of interest is the true standard deviation of the golf ball distance, λ. 2) H0 : θ = 285 3) H1 : σ > 285 4) α=0.05 5) Since n>30 we can use the normal distribution ˆ Θ − ϑ0 z0 = σ 2 /(3n) 6) Reject H0 if z0 > zα where z0.05 =1.65 7) ˆ Θ= 282.36 n = 100 z0 = 282.36 − 285 10 2 /(300) = − 4 .5 7 8) Since -4.82 > 1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true 95% is greater than 285 at α = 0.05 9-97 1) The parameter of interest is the true mean number of open circuits, λ. 2) H0 : λ = λ0 3 ) H1 : λ ≠ λ 0 4) α = 0.05 5) test statistic 2λ χ= n X i − λ0 i =1 2 0 2λ n Xi i =1 6) Reject H0 if 7) compute 2λ 2 χ 02 > χ a / 2, 2 n or χ 02 < χ 12−a / 2, 2 n n Xi and plug into i =1 2λ χ= 2 0 n X i − λ0 i =1 2λ n Xi i =1 8) make conclusions alternative hypotheses 1 ) H0 : λ = λ 0 H1 : λ > λ 0 Reject H0 if 2 χ 02 > χ a , 2 n 2 ) H0 : λ = λ 0 H1 : λ < λ 0 Reject H0 if 2 χ 02 < χ a , 2 n 9-53 CHAPTER 10 Section 10-2 1) The parameter of interest is the difference in fill volume, µ1 − µ 2 ( note that ∆0=0) 10-1. a) 2) H0 : 3) H1 : µ1 − µ 2 = 0 µ1 − µ 2 ≠ 0 or or µ1 = µ 2 µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is z0 = ( x1 − x2 ) − ∆ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 7) x1 = 16.015 x2 = 16.005 σ1 = 0.02 n1 = 10 σ 2 = 0.025 n2 = 10 z0 = (16.015 − 16.005) (0.02) 2 (0.025) 2 + 10 10 = 0 .9 9 8) since -1.96 < 0.99 < 1.96, do not reject the null hypothesis and conclude there is no evidence that the two machine fill volumes differ at α = 0.05. b) P-value = 2(1 − Φ(0.99)) = 2(1 − 0.8389) = 0.3222 c) Power = 1 − β , where β = Φ zα / 2 − ∆ − ∆0 2 σ1 n1 =Φ + σ2 2 ∆ − ∆0 − Φ − zα / 2 − 2 σ1 σ 2 +2 n1 n2 n2 0.04 1.96 − 2 2 − Φ − 1.96 − 0.04 2 (0.02) (0.025) (0.02) (0.025) 2 + + 10 10 10 10 = Φ (1.96 − 3.95) − Φ (− 1.96 − 3.95) = Φ (− 1.99 ) − Φ (− 5.91) = 0.0233 − 0 = 0.0233 Power = 1 −0.0233 = 0.9967 d) ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ 2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n 2 n1 n 2 (0.02) 2 (0.025) 2 (0.02) 2 (0.025) 2 + ≤ µ1 − µ 2 ≤ (16.015 − 16.005) + 1.96 + 10 10 10 10 −0.0098 ≤ µ1 − µ 2 ≤ 0.0298 (16.015 − 16.005) − 1.96 With 95% confidence, we believe the true difference in the mean fill volumes is between −0.0098 and 0.0298. Since 0 is contained in this interval, we can conclude there is no significant difference between the means. e) Assume the sample sizes are to be equal, use α = 0.05, β = 0.05, and ∆ = 0.04 10-1 (z n≅ ( 2 + z β ) σ 12 + σ 2 2 α /2 δ 2 ) = (1.96 + 1.645) ((0.02) 2 (0.04) 2 + (0.025) 2 2 ) = 8.35, n = 9, use n1 = n2 = 9 10-2. 1) The parameter of interest is the difference in breaking strengths, µ1 − µ 2 and ∆0 = 10 2) H0 : µ1 − µ 2 = 10 3) H1 : µ1 − µ 2 > 10 4) α = 0.05 5) The test statistic is z0 = ( x1 − x2 ) − ∆ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 > zα = 1.645 7) x1 = 162.5 x2 = 155.0 δ = 10 σ1 = 1.0 n1 = 10 σ 2 = 1.0 n2 = 12 z0 = (162.5 − 155.0) − 10 = −5.84 (1.0) 2 (10) 2 . + 10 12 8) Since -5.84 < 1.645 do not reject the null hypothesis and conclude there is insufficient evidence to support the use of plastic 1 at α = 0.05. 10-3 β= Φ 1.645 − n= (12 − 10) = Φ (− 3.03) = 0.0012 , Power = 1 – 0.0012 = 0.9988 11 + 10 12 2 ( zα 2 + z β ) 2 (σ 12 + σ 2 ) (∆ − ∆ 0 ) 2 = (1.645 + 1.645)2 (1 + 1) = 5.42 ≅ 6 (12 − 10) 2 Yes, the sample size is adequate 10-4. a) 1) The parameter of interest is the difference in mean burning rate, µ1 − µ 2 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 10-2 1 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is z0 = ( x1 − x2 ) − ∆ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 7) x1 = 18 x2 = 24 σ1 = 3 n1 = 20 σ2 = 3 n2 = 20 z0 = (18 − 24) (3) 2 (3) 2 + 20 20 = −6.32 8) Since −6.32 < −1.96 reject the null hypothesis and conclude the mean burning rates differ significantly at α = 0.05. b) P-value = 2(1 − Φ (6.32)) ∆ − ∆0 c) β = Φ zα / 2 − 2 σ1 n1 + − Φ − zα / 2 − σ2 2 n2 2.5 = Φ 1.96 − = 2(1 − 1) = 0 2 2 (3) (3) + 20 20 − Φ −1.96 − ∆ − ∆0 2 σ1 σ 2 +2 n1 n2 2.5 2 (3) (3) 2 + 20 20 = Φ(1.96 − 2.64) − Φ( −1.96 − 2.64) = Φ( −0.68) − Φ( −4.6) = 0.24825 − 0 = 0.24825 d) ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ 2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n 2 n1 n 2 (3) 2 (3)2 (3) 2 (3) 2 + ≤ µ1 − µ 2 ≤ (18 − 24) + 1.96 + 20 20 20 20 −7.86 ≤ µ1 − µ 2 ≤ −4.14 (18 − 24) − 1.96 We are 95% confident that the mean burning rate for solid fuel propellant 2 exceeds that of propellant 1 by between 4.14 and 7.86 cm/s. 10-5. x1 = 30.87 x2 = 30.68 10-3 σ 1 = 0.10 σ2 = 0.15 n1 = 12 n2 = 10 a) 90% two-sided confidence interval: 2 2 (x1 − x2 ) − zα / 2 σ 1 + σ 2 n1 (30.87 − 30.68) − 1.645 n2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 σ 12 n1 + 2 σ2 n2 (0.10) 2 (0.15) 2 (0.10) 2 (0.15) 2 + ≤ µ1 − µ 2 ≤ (30.87 − 30.68) + 1.645 + 12 10 12 10 0.0987 ≤ µ1 − µ 2 ≤ 0.2813 We are 90% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.0987 and 0.2813 fl. oz. b) 95% two-sided confidence interval: ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ 2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n 2 n1 n 2 (30.87 − 30.68) − 1.96 (0.10) 2 (0.15) 2 (0.10) 2 (0.15)2 + ≤ µ1 − µ 2 ≤ ( 30.87 − 30.68) + 1.96 + 12 10 12 10 0.0812 ≤ µ1 − µ 2 ≤ 0.299 We are 95% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.0812 and 0.299 fl. oz. Comparison of parts a and b: As the level of confidence increases, the interval width also increases (with all other values held constant). c) 95% upper-sided confidence interval: µ1 − µ 2 ≤ ( x1 − x2 ) + zα 2 σ1 σ 2 +2 n1 n2 µ1 − µ 2 ≤ ( 30.87 − 30.68) + 1.645 (0.10) 2 (0.15) 2 + 12 10 µ1 − µ 2 ≤ 0.2813 With 95% confidence, we believe the fill volume for machine 1 exceeds the fill volume of machine 2 by no more than 0.2813 fl. oz. 10-4 10-6. a) 1) The parameter of interest is the difference in mean fill volume, µ1 − µ 2 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is z0 = ( x1 − x2 ) − ∆ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 7) x1 = 30.87 x2 = 30.68 σ1 = 0.10 n1 = 12 σ 2 = 0.15 n2 = 10 z0 = (30.87 − 30.68) (0.10) 2 (0.15) 2 + 12 10 = 3 .4 2 8) Since 3.42 > 1.96 reject the null hypothesis and conclude the mean fill volumes of machine 1 and machine 2 differ significantly at α = 0.05. b) P-value = 2(1 − Φ(3.42)) = 2(1 − 0.99969) = 0.00062 use α = 0.05, β = 0.10, and ∆ = 0.20 c) Assume the sample sizes are to be equal, ( zα / 2 + zβ ) ( n≅ 2 2 σ1 (∆ − ∆0 ) x1 = 89.6 2 ) = (1.96 + 1.28) ( (0.10) 2 ( −0.20) 2 2 + (0.15) 2 ) = 8.53, n = 9, use n1 = n2 = 9 x2 = 92.5 2 σ1 10-7. + σ2 2 σ 2 = 1.2 2 n2 = 20 = 1.5 n1 = 15 a) 95% confidence interval: ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ 2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n 2 n1 n 2 15 1.2 . 15 1.2 . + ≤ µ1 − µ 2 ≤ ( 89.6 − 92.5) + 1.96 + 15 20 15 20 −3.684 ≤ µ1 − µ 2 ≤ −2.116 (89.6 − 92.5) − 1.96 With 95% confidence, we believe the mean road octane number for formulation 2 exceeds that of formulation 1 by between 2.116 and 3.684. b) 1) The parameter of interest is the difference in mean road octane number, µ1 − µ 2 and ∆0 = 0 10-5 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 < 0 or µ1 < µ 2 4) α = 0.05 5) The test statistic is z0 = ( x1 − x 2 ) − ∆ 0 σ 12 n1 + 2 σ2 n2 6) Reject H0 if z0 < −zα = −1.645 7) x1 = 89.6 x2 = 92.5 2 σ1 = 1.5 n1 = 15 σ 2 = 1.2 2 n2 = 20 z0 = (89.6 − 92.5) = − 7 .2 5 1 .5 1 .2 + 15 20 8) Since −7.25 < -1.645 reject the null hypothesis and conclude the mean road octane number for formulation 2 exceeds that of formulation 1 using α = 0.05. c) P-value ≅ P ( z ≤ −7.25) = 1 − P ( z ≤ 7.25) = 1 − 1 ≅ 0 10-8. 99% level of confidence, E = 4, and z0.005 = 2.575 z n ≅ 0.005 E 10-9. 2 (σ + σ ) = 2.575 (9 + 9) = 7.46, n = 8, use n1 = n2 = 8 4 2 1 2 2 95% level of confidence, E = 1, and z0.025 =1.96 z n ≅ 0.025 E 10-10. 2 2 (σ 2 1 +σ 2 2 ) 2 1.96 = (1.5 + 1.2) = 10.37, n = 11, use n1 = n2 = 11 1 Case 1: Before Process Change µ1 = mean batch viscosity before change x1 = 750.2 Case 2: After Process Change µ2 = mean batch viscosity after change x2 = 756.88 σ1 = 20 σ 2 = 20 n1 = 15 n2 = 8 90% confidence on µ1 − µ 2 , the difference in mean batch viscosity before and after process change: ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ 2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n 2 n1 n 2 (750.2 − 756.88) − 1.645 (20) 2 (20) 2 ( 20)2 (20) 2 + ≤ µ1 − µ 2 ≤ ( 750.2 − 756 / 88) + 1.645 + 15 8 15 8 −2108 ≤ µ1 − µ 2 ≤ 7.72 . We are 90% confident that the difference in mean batch viscosity before and after the process change lies within −21.08 and 7.72. Since 0 is contained in this interval we can conclude with 90% confidence that the mean batch viscosity was unaffected by the process change. 10-11. Catalyst 1 Catalyst 2 10-6 x1 = 65.22 x2 = 68.42 σ1 = 3 σ2 = 3 n1 = 10 n2 = 10 a) 95% confidence interval on µ1 − µ 2 , the difference in mean active concentration ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ 2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n 2 n1 n 2 (3) 2 (3) 2 (3) 2 (3) 2 + ≤ µ1 − µ 2 ≤ ( 65.22 − 68.42) + 1.96 + 10 10 10 10 −5.83 ≤ µ1 − µ 2 ≤ −0.57 (65.22 − 68.42) − 196 . We are 95% confident that the mean active concentration of catalyst 2 exceeds that of catalyst 1 by between 0.57 and 5.83 g/l. b) Yes, since the 95% confidence interval did not contain the value 0, we would conclude that the mean active concentration depends on the choice of catalyst. 10-12. a) 1) The parameter of interest is the difference in mean batch viscosity before and after the process change, µ1 − µ 2 2) H0 : µ1 − µ 2 = 10 3) H1 : µ1 − µ 2 < 10 4) α = 0.10 5) The test statistic is z0 = ( x1 − x2 ) − ∆ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 < −zα where z0.1 = −1.28 7) x1 = 750.2 ∆0 = 1 0 x2 = 756.88 σ1 = 20 σ 2 = 20 n1 = 15 n2 = 8 z0 = (750.2 − 756.88) − 10 = −1.90 (20)2 (20)2 + 15 8 8) Since −1.90 < −1.28 reject the null hypothesis and conclude the process change has increased the mean by less than 10. b) P-value = P( z ≤ −190) = 1 − P( z ≤ 190) = 1 − 0.97128 = 0.02872 . . c) Parts a and b above give evidence that the mean batch viscosity change is less than 10. This conclusion is also seen by the confidence interval given in a previous problem since the interval did not contain the value 10. Since the upper endpoint is 7.72, then this also gives evidence that the difference is less than 10. 10-13. 1) The parameter of interest is the difference in mean active concentration, µ1 − µ 2 10-7 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is z0 = ( x1 − x2 ) − ∆ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 7) 7) x1 = 65.22 x2 = 68.42 δ = 0 σ1 = 3 n1 = 10 σ2 = 3 n2 = 10 z0 = ( 65 . 22 − 68 . 42 ) − 0 9 9 + 10 10 = − 2 . 385 8) Since −2.385 < −1.96 reject the null hypothesis and conclude the mean active concentrations do differ significantly at α = 0.05. P-value = 2 (1 − Φ(2.385)) = 2(1 − 0.99146) = 0.0171 The conclusions reached by the confidence interval of the previous problem and the test of hypothesis conducted here are the same. A two-sided confidence interval can be thought of as representing the “acceptance region” of a hypothesis test, given that the level of significance is the same for both procedures. Thus if the value of the parameter under test that is specified in the null hypothesis falls outside the confidence interval, this is equivalent to rejecting the null hypothesis. 10-14. (5) β = Φ 1.96 − (5) − Φ − 1.96 − 32 32 32 32 + + 10 10 10 10 = Φ (− 1.77 ) − Φ(− 5.69 ) = 0.038364 − 0 = 0.038364 Power = 1 − β = 1− 0.038364 = 09616. it would appear that the sample sizes are adequate to detect the difference of 5, based on the power. Calculate the value of n using α and β. n≅ (z ( 2 + z β ) σ 12 + σ 2 2 α/2 (∆ − ∆ 0 )2 ) = (1.96 + 1.77) (9 + 9) = 10.02, Therefore, 10 is just slightly 2 (5) 2 too few samples. 10-8 The data from the first sample n=15 appear to be normally distributed. 99 95 90 80 Percent 70 60 50 40 30 20 10 5 1 700 750 800 The data from the second sample n=8 appear to be normally distributed 99 95 90 80 Percent 10-15 70 60 50 40 30 20 10 5 1 700 750 800 10-9 10-16 The data all appear to be normally distributed based on the normal probability plot below. 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 55 65 75 Section 10-3 10-17. a) 1) The parameter of interest is the difference in mean rod diameter, µ1 − µ 2 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α / 2, n1 + n 2 − 2 where − t 0.025,30 = −2.042 or t0 > t α / 2, n1 + n 2 − 2 where t 0.025,30 = 2.042 7) ) x1 = 8.73 x2 = 8.68 sp = 2 s1 = 0.35 s2 = 0.40 2 = n1 = 15 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 n2 = 17 t0 = (8.73 − 8.68) 1 1 0.614 + 15 17 14(0.35) + 16(0.40) = 0.614 30 = 0.230 8) Since −2.042 < 0.230 < 2.042, do not reject the null hypothesis and conclude the two machines do not produce rods with significantly different mean diameters at α = 0.05. b) P-value = 2P ( t > 0.230) > 2(0.40), P-value > 0.80 10-10 c) 95% confidence interval: t0.025,30 = 2.042 ( x1 − x2 ) − t α / 2,n + n 1 2 −2 (sp ) (8.73 − 8.68) − 2.042(0.614) 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + t α / 2, n1 + n 2 − 2 (sp ) + n1 n 2 n1 n 2 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( 8.73 − 8.68) + 2.042(0.643) + 15 17 15 17 − 0.394 ≤ µ1 − µ 2 ≤ 0.494 Since zero is contained in this interval, we are 95% confident that machine 1 and machine 2 do not produce rods whose diameters are significantly different. 2 10-18. Assume the populations follow normal distributions and σ1 = σ 2 . The assumption of equal variances may be 2 permitted in this case since it is known that the t-test and confidence intervals involving the t-distribution are robust to this assumption of equal variances when sample sizes are equal. Case 2: ATC µ2 = mean foam expansion for ATC x2 = 6.9 Case 1: AFCC µ1 = mean foam expansion for AFCC x1 = 4.7 s1 = 0.6 n1 = 5 s2 = 0.8 n2 = 5 95% confidence interval: t0.025,8 = 2.306 ( x1 − x2 ) − t α / 2,n + n 1 2 −2 (sp ) sp = 4(0.60) + 4(0.80) = 0.7071 8 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + t α / 2, n1 + n 2 − 2 (sp ) + n1 n 2 n1 n 2 ( 4 . 7 − 6 . 9 ) − 2 . 306 ( 0 .7071 ) 11 11 + ≤ µ 1 − µ 2 ≤ (4 . 7 − 6 . 9 ) + 2 .306 ( 0 . 7071 ) + 55 55 −3.23 ≤ µ1 − µ 2 ≤ −117 . Yes, with 95% confidence, we believe the mean foam expansion for ATC exceeds that of AFCC by between 1.17 and 3.23. 10-11 10-19. a) 1) The parameter of interest is the difference in mean catalyst yield, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 < 0 or µ1 < µ 2 4) α = 0.01 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α , n1 + n 2 − 2 where − t 0.01, 25 = −2.485 7) x1 = 86 s1 = 3 sp = x2 = 89 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 11(3) 2 + 14(2) 2 = = 2.4899 25 s2 = 2 n1 = 12 n2 = 15 t0 = (86 − 89) 1 1 2.4899 + 12 15 = − 3 .1 1 8) Since −3.11 < −2.787, reject the null hypothesis and conclude that the mean yield of catalyst 2 significantly exceeds that of catalyst 1 at α = 0.01. b) 99% confidence interval: t0.005,19 = 2.861 ( x1 − x2 ) − t α / 2,n + n 1 2 −2 ( sp ) 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + t α / 2, n1 + n 2 − 2 (sp ) + n1 n 2 n1 n 2 (86 − 89 ) − 2 . 787 ( 2 . 4899 ) 1 1 1 1 + ≤ µ 1 − µ 2 ≤ (86 − 89 ) + 2 . 787 ( 2 . 4899 ) + 12 15 12 15 − 5.688 ≤ µ1 − µ 2 ≤ −0.3122 We are 95% confident that the mean yield of catalyst 2 exceeds that of catalyst 1 by between 0.3122 and 5.688 10-12 10-20. a) According to the normal probability plots, the assumption of normality appears to be met since the data fall approximately along a straight line. The equality of variances does not appear to be severely violated either since the slopes are approximately the same for both samples. Normal Probability Plot Normal Probability Plot .999 .999 .99 .99 .95 .80 Probability Probability .95 .50 .20 .05 .80 .50 .20 .05 .01 .01 .001 .001 180 190 200 210 175 185 type1 Average: 196.4 StDev: 10.4799 N: 15 175 185 195 205 type2 Anderson-Darling Normality Test A-Squared: 0.463 P-Value: 0.220 195 Average: 192.067 StDev: 9.43751 N: 15 205 Anderson-Darling Normality Test A-Squared: 0.295 P-Value: 0.549 180 190 type2 200 210 type1 b) 1) The parameter of interest is the difference in deflection temperature under load, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 < 0 or µ1 < µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α , n1 + n 2 − 2 where − t 0.05,28 = −1.701 10-13 7) Type 1 Type 2 x1 = 196.4 sp = x2 = 192.067 s1 = 10.48 s2 = 9.44 n1 = 15 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 14(10.48) 2 + 14( 9.44) 2 = 9.97 28 n2 = 15 = t0 = (196.4 − 192.067) 11 9.97 + 15 15 = 1 .1 9 8) Since 1.19 > −1.701 do not reject the null hypothesis and conclude the mean deflection temperature under load for type 2 does not significantly exceed the mean deflection temperature under load for type 1 at the 0.05 level of significance. c) P-value = 2P (t < 1.19) 0.75 < p-value < 0.90 d) ∆ = 5 Use sp as an estimate of σ: µ − µ1 5 d= 2 = = 0.251 2 sp 2(9.97) Using Chart VI g) with β = 0.10, d = 0.251 we get n ≅ 100. So, since n*=2n-1, the sample sizes of 15 are inadequate. 10-21. n1 = n2 = 51 ; Therefore, a) 1) The parameter of interest is the difference in mean etch rate, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α / 2, n1 + n 2 − 2 where − t 0.025,18 = −2.101 or t0 > t α / 2, n1 + n 2 − 2 where t 0.025,18 = 2.101 7) x1 = 9.97 s1 = 0.422 n1 = 10 sp = x2 = 10.4 = s2 = 0.231 2 2 ( n1 − 1)s1 + ( n2 − 1)s2 n1 + n 2 − 2 9(0.422) 2 + 9(0.231) 2 = 0.340 18 n2 = 10 t0 = (9.97 − 10.4) 1 1 0.340 + 10 10 = − 2 .8 3 8) Since −2.83 < −2.101 reject the null hypothesis and conclude the two machines mean etch rates do significantly differ at α = 0.05. b) P-value = 2P (t < −2.83) 2(0.005) < P-value < 2(0.010) = 0.010 < P-value < 0.020 10-14 c) 95% confidence interval: t0.025,18 = 2.101 ( x1 − x2 ) − t α / 2,n + n 1 2 −2 (sp ) (9 .97 − 10 .4) − 2.101(. 340 ) 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + t α / 2, n1 + n 2 − 2 (sp ) + n1 n 2 n1 n 2 1 1 1 1 + ≤ µ1 − µ 2 ≤ (9.97 − 10 .4 ) + 2 .101(. 340 ) + 10 10 10 10 − 0.7495 ≤ µ1 − µ 2 ≤ −0.1105 We are 95% confident that the mean etch rate for solution 2 exceeds the mean etch rate for solution 1 by between 0.1105 and 0.7495. d) According to the normal probability plots, the assumption of normality appears to be met since the data from both the samples fall approximately along a straight line. The equality of variances does not appear to be severely violated either since the slopes are approximately the same for both samples. Normal Probability Plot Normal Probability Plot .99 .95 .95 Probability .999 .99 Probability .999 .80 .50 .20 .80 .50 .20 .05 .05 .01 .01 .001 .001 9.5 10.0 10.0 10.5 10.1 10.2 Average: 9.97 StDev: 0.421769 N: 10 10-22. 10.3 10.4 10.5 10.6 10.7 solution solution Average: 10.4 StDev: 0.230940 N: 10 Anderson-Darling Normality Test A-Squared: 0.269 P-Value: 0.595 Anderson-Darling Normality Test A-Squared: 0.211 P-Value: 0.804 a) 1) The parameter of interest is the difference in mean impact strength, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 < 0 or µ1 < µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x 2 ) − ∆ 0 2 s12 s 2 + n1 n 2 6) Reject the null hypothesis if t0 < − t α , ν where 2 s12 s 2 + n1 n 2 ν= 2 1 s n1 2 n1 − 1 t 0.05, 23 = 1.714 since 2 = 23.72 2 2 + s n2 n2 − 1 ν ≅ 23 (truncated) 10-15 7) x1 = 290 x2 = 321 s1 = 12 n1 = 10 s2 = 22 n2 = 16 (290 − 321) t0 = = − 4 .6 4 (12) 2 (22) 2 + 10 16 8) Since −4.64 < −1.714 reject the null hypothesis and conclude that supplier 2 provides gears with higher mean impact strength at the 0.05 level of significance. b) P-value = P(t < −4.64): P-value < 0.0005 c) 1) The parameter of interest is the difference in mean impact strength, µ 2 − µ1 2) H0 : µ 2 − µ1 = 25 3) H1 : µ 2 − µ1 > 25 or µ 2 > µ1 + 25 4) α = 0.05 5) The test statistic is t0 = ( x2 − x1 ) − δ 2 s1 s2 +2 n1 n 2 6) Reject the null hypothesis if t0 > t α , ν = 1.708 where 2 s 12 s2 +2 n1 n2 ν= s 12 n1 2 n1 − 1 + = 23 . 72 2 s2 n2 n2 − 1 ν ≅ 23 7) x1 = 290 x2 = 321 ∆ 0 =25 s1 = 12 t0 = s2 = 22 n1 = 10 n2 = 16 (321 − 290) − 25 = 0.898 (12)2 (22)2 + 10 16 8) Since 0.898 < 1.714, do not reject the null hypothesis and conclude that the mean impact strength from supplier 2 is not at least 25 ft-lb higher that supplier 1 using α = 0.05. 10-23. Using the information provided in Exercise 9-20, and t0.025,25 = 2.06, we find a 95% confidence interval on the difference, µ 2 − µ1 : ( x2 − x1 ) − t 0.025, 25 2 s12 s 2 s2 s2 + ≤ µ 2 − µ1 ≤ ( x2 − x1 ) + t 0.025, 25 1 + 2 n1 n2 n1 n2 31 − 2.069(6.682) ≤ µ 2 − µ1 ≤ 31 + 2.069(6.682) 17.175 ≤ µ 2 − µ1 ≤ 44.825 Since the 95% confidence interval represents the differences that µ 2 − µ1 could take on with 95% confidence, we can conclude that Supplier 2 does provide gears with a higher mean impact strength than Supplier 1. This is visible from the interval (17.175, 44.825) since zero is not contained in the interval and the differences are all positive, meaning that µ 2 − µ1 > 0. 10-16 10-24 a) 1) The parameter of interest is the difference in mean speed, µ1 − µ 2 , ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 > 0 or µ1 > µ 2 4) α = 0.10 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 1 1 + n1 n 2 sp 6) Reject the null hypothesis if t0 > t α , n1 + n 2 − 2 where t 0.10,14 =1.345 7) Case 1: 25 mil Case 2: 20 mil x1 = 1.15 sp = x2 = 1.06 s1 = 0.11 s2 = 0.09 = n1 = 8 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 7( 011)2 + 7(0.09) 2 . = 0.1005 14 n2 = 8 t0 = (1.15 − 1.06) 11 0.1005 + 88 = 1 .7 9 8) Since 1.79 > 1.345 reject the null hypothesis and conclude reducing the film thickness from 25 mils to 20 mils significantly increases the mean speed of the film at the 0.10 level of significance (Note: since increase in film speed will result in lower values of observations). b) P-value = P ( t > 1.79) 0.025 < P-value < 0.05 c) 90% confidence interval: t0.025,14 = 2.145 ( x1 − x2 ) − t α / 2,n + n 1 2 −2 (1.15 − 1.06) − 2.145(.1005) (sp ) 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + t α / 2, n1 + n 2 − 2 (sp ) + n1 n 2 n1 n 2 11 11 + ≤ µ1 − µ 2 ≤ (1.15 − 1.06) + 2.145(.1005) + 88 88 − 0.0178 ≤ µ1 − µ 2 ≤ 0.1978 We are 90% confident the mean speed of the film at 20 mil exceeds the mean speed for the film at 25 mil by between -0.0178 and 0.1978 µJ/in2 . 10-17 10-25. 1) The parameter of interest is the difference in mean melting point, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.02 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α / 2, n1 + n 2 − 2 where − t 0.0025, 40 = −2.021 or t0 > t α / 2, n1 + n 2 − 2 where t 0.025, 40 = 2.021 7) x1 = 420 sp = x2 = 426 s1 = 4 n1 = 21 = s2 = 3 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 20(4) 2 + 20(3) 2 = 3.536 40 n2 = 21 t0 = (420 − 426) 1 1 3.536 + 21 21 = −5.498 8) Since −5.498< −2.021 reject the null hypothesis and conclude that the data do not support the claim that both alloys have the same melting point at α = 0.02 P-value = 2P 10-26. (t < −5.498) P-value < 0.0010 | µ1 − µ 2 | 3 = = 0.375 2σ 2 ( 4) Using the appropriate chart in the Appendix, with β = 0.10 and α = 0.05 we have: n* = 75, so d= n= n* + 1 = 38 , n1 = n2 =38 2 10-18 10-27. . a) 1) The parameter of interest is the difference in mean wear amount, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 2 s1 s2 +2 n1 n2 6) Reject the null hypothesis if t0 < − t 0.025, 26 where − t 0.025, 26 = −2.056 or t0 > t 0.025, 26 where t 0.025, 26 = 2.056 since ν= 2 2 s12 s 2 + n1 n 2 s12 n1 2 n1 − 1 + = 26.98 2 s2 n2 n2 − 1 ν ≅ 26 (truncated) 7) x1 = 20 s1 = 2 n1 = 25 x2 = 15 s2 = 8 n2 = 25 t0 = (20 − 15) (2) 2 (8) 2 + 25 25 = 3 .0 3 8) Since 3.03 > 2.056 reject the null hypothesis and conclude that the data support the claim that the two companies produce material with significantly different wear at the 0.05 level of significance. b) P-value = 2P(t > 3.03), 2(0.0025) < P-value < 2(0.005) 0.005 < P-value < 0.010 10-19 c) 1) The parameter of interest is the difference in mean wear amount, µ1 − µ 2 2 ) H0 : µ 1 − µ 2 = 0 3 ) H1 : µ 1 − µ 2 > 0 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 2 s1 s2 +2 n1 n2 6) Reject the null hypothesis if t0 > t 0.05,27 where 7) x1 = 20 t 0.05, 26 = 1.706 since x2 = 15 s1 = 2 s2 = 8 n1 = 25 n2 = 25 (20 − 15) t0 = (2) 2 (8) 2 + 25 25 = 3 .0 3 8) Since 3.03 > 1.706 reject the null hypothesis and conclude that the data support the claim that the material from company 1 has a higher mean wear than the material from company 2 using a 0.05 level of significance. 10-28 1) The parameter of interest is the difference in mean coating thickness, µ1 − µ 2 , with ∆0 = 0. 2) H0 : µ1 − µ 2 = 0 3 ) H1 : µ 1 − µ 2 > 0 4) α = 0.01 5) The test statistic is t0 = 6) Reject the null hypothesis if t0 > ( x1 − x2 ) − δ 2 s1 s2 +2 n1 n 2 t 0.01,18 where t 0.01,18 s 12 s2 +2 n1 n2 ν= s 12 n1 2 n1 − 1 + = 2.552 since 2 18 . 37 2 s2 n2 n2 − 1 ν ≅ 18 (truncated) 7) x1 = 103.5 x2 = 99.7 s1 = 10.2 n1 = 11 s2 = 20.1 n2 = 13 t0 = (103.5 − 99.7) (10.2) 2 (20.1) 2 + 11 13 = 0.597 8) Since 0.597 < 2.552, do not reject the null hypothesis and conclude that increasing the temperature does not significantly reduce the mean coating thickness at α = 0.01. P-value = P(t > 0597), 0.25 < P-value < 0.40 10-20 10-29. If α = 0.01, construct a 99% two-sided confidence interval on the difference to answer question 10-28. t0.005,19 = 2.878 (x1 − x2 ) − tα / 2 ,ν 2 2 s1 s 2 + ≤ µ 1 − µ 2 ≤ ( x1 − x 2 ) + tα / 2 ,ν n1 n 2 (10 . 2 ) 2 ( 20 . 1) 2 + ≤ µ 1 − µ 2 ≤ (103 . 5 − 99 . 7 ) − 2 . 878 11 13 (103 . 5 − 99 . 7 ) − 2 . 878 2 2 s1 s 2 + n1 n 2 (10 . 2 ) 2 ( 20 . 1) 2 + 11 13 − 14.52 ≤ µ1 − µ 2 ≤ 22.12 . Since the interval contains 0, we are 99% confident there is no difference in the mean coating thickness between the two temperatures; that is, raising the process temperature does not significantly reduce the mean coating thickness. 10-30. 95% confidence interval: t0.025,26 = 2.056 ( x1 − x2 ) − tα , ν 2 s1 s2 s2 s2 + 2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + t α , ν 1 + 2 n1 n 2 n1 n 2 (2) 2 (8) 2 (2) 2 (8) 2 (20 − 15) − 2.056 + ≤ µ1 − µ 2 ≤ (20 − 15) + 2.056 + 25 25 25 25 1.609 ≤ µ1 − µ 2 ≤ 8.391 95% lower one-sided confidence interval: t 0.05, 26 = 1.706 ( x1 − x 2 ) − t α, ν 2 s1 s 2 + 2 ≤ µ1 − µ 2 n1 n 2 (20 − 15) − 1.706 (2)2 + (8)2 25 25 ≤ µ1 − µ 2 2.186 ≤ µ1 − µ 2 For part a): We are 95% confident the mean abrasive wear from company 1 exceeds the mean abrasive wear from company 2 by between 1.609 and 8.391 mg/1000. For part c): We are 95% confident the mean abrasive wear from company 1 exceeds the mean abrasive wear from company 2 by at least 2.19mg/1000. 10-21 a.) Normal Probability Plot for Brand 1...Brand 2 ML Estimates Brand 1 99 Brand 2 95 90 80 Percent 10-31 70 60 50 40 30 20 10 5 1 244 254 264 274 284 294 Data b . 1) The parameter of interest is the difference in mean overall distance, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α / 2, n1 + n 2 − 2 where − t 0.025,18 = −2.101 or t0 > t α / 2, n1 + n 2 − 2 where t 0.025,18 = 2.101 7) x1 = 275.7 sp = x2 = 265.3 s1 = 8.03 9(8.03) 2 + 9(10.04) 2 = = 9. 09 20 s2 = 10.04 n1 = 10 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 n2 = 10 t0 = (275.7 − 265.3) 1 1 9.09 + 10 10 = 2.558 8) Since 2.558>2.101 reject the null hypothesis and conclude that the data do not support the claim that both brands have the same mean overall distance at α = 0.05. It appears that brand 1 has the higher mean differnce. c.)P-value = 2P (t > 2.558) P-value ≈ 2(0.01)=0.02 10-22 d.) d = 5 0 . 275 2 ( 9 . 09 ) e.) 1-β =0.25 β=0.95 Power =1-0.95=0.05 3 = 0.165 2(9.09) β=0..27 d = n*=100 n = 100 + 1 = 50 . 5 2 Therefore, n=51 f.) (x1 − x 2 ) − tα ,ν s p 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x 2 ) + tα ,ν s p + n1 n 2 n1 n 2 1 1 1 1 + ≤ µ1 − µ 2 ≤ (275.7 − 265.3) + 2.101(9.09) + 10 10 10 10 (275.7 − 265.3) − 2.101(9.09) 1.86 ≤ µ1 − µ 2 ≤ 18.94 Normal Probability Plot for Club1...Club2 ML Estimates - 95% CI C lub1 99 C lub2 95 90 Percent 80 70 60 50 40 30 20 10 5 1 0 .75 0.77 0.79 0.81 0.83 0.85 0.87 0.89 Data 10-32 a.) The data appear to be normally distributed and the variances appear to be approximately equal. The slopes of the lines on the normal probability plots are almost the same. b) 1) The parameter of interest is the difference in mean coefficient of restitution, µ1 − µ 2 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 sp 1 1 + n1 n 2 6) Reject the null hypothesis if t0 < − t α / 2, n1 + n 2 − 2 where − t 0.025, 22 = −2.074 or t0 > t α / 2, n1 + n 2 − 2 where t 0.025, 22 = 2.074 10-23 7) x1 = 0.8161 sp = x2 = 0.8271 s1 = 0.0217 s2 = 0.0175 n1 = 12 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n 2 − 2 11(0.0217 ) 2 + 11(0.0175 ) 2 = 0.01971 22 n2 = 12 = t0 = ( 0 . 8161 − 0 . 8271 ) 1 1 + 12 12 0 . 01971 = − 1 . 367 8) Since –1.367 > -2.074 do not reject the null hypothesis and conclude that the data do not support the claim that there is a difference in the mean coefficients of restitution for club1 and club2 at α = 0.05 c.)P-value = 2P d.) d= (t < −1.36) P-value ≈ 2(0.1)=0.2 0 .2 = 5 .0 7 2(0.01971) β = 0.2 d = e.) 1-β = 0.8 β ≅0 0 .1 = 2 .5 3 2(0.01971) Power ≅1 n*=4 , n = n * +1 = 2 .5 n ≅ 3 2 f.) 95% confidence interval (x1 − x 2 ) − tα ,ν s p 1 1 1 1 + ≤ µ1 − µ 2 ≤ ( x1 − x 2 ) + tα ,ν s p + n1 n 2 n1 n 2 (0.8161 − 0.8271) − 2.074(0.01971) 1 1 1 1 + ≤ µ1 − µ 2 ≤ (0.8161 − 0.8271) + 2.074(0.01971) + 12 12 12 12 − 0.0277 ≤ µ1 − µ 2 ≤ 0.0057 Zero is included in the confidence interval, so we would conclude that there is not a significant difference in the mean coefficient of restitution’s for each club at α=0.05. Section 10-4 10-33. d = 0.2736 sd = 0.1356, n = 9 95% confidence interval: d − t α / 2, n −1 0.2736 − 2.306 sd n 0.1356 9 ≤ µ d ≤ d + t α / 2, n −1 ≤ µ d ≤ 0.2736 + 2.306 sd n 0.1356 9 0.1694 ≤ µd ≤ 0.3778 With 95% confidence, we believe the mean shear strength of Karlsruhe method exceeds the mean shear strength of the Lehigh method by between 0.1694 and 0.3778. Since 0 is not included in this interval, the interval is consistent with rejecting the null hypothesis that the means are the same. The 95% confidence interval is directly related to a test of hypothesis with 0.05 level of significance, and the conclusions reached are identical. 10-24 10-34. It is only necessary for the differences to be normally distributed for the paired t-test to be appropriate and reliable. Therefore, the t-test is appropriate. Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 0.12 0.22 0.32 0.42 0.52 diff Average: 0.273889 StDev: 0.135099 N: 9 10-35. Anderson-Darling Normality Test A-Squared: 0.318 P-Value: 0.464 1) The parameter of interest is the difference between the mean parking times, µd. 2 ) H0 : µ d = 0 3) H1 : µ d ≠ 0 4) α = 0.10 5) The test statistic is t0 = d sd / n 6) Reject the null hypothesis if t0 < − t 0.05,13 where − t 0.05,13 = −1.771 or t0 > t 0.05,13 where t 0.05,13 = 1.771 7) d = 1.21 sd = 12.68 n = 14 t0 = 1.21 = 0.357 12.68 / 14 8) Since −1.771 < 0.357 < 1.771 do not reject the null and conclude the data do not support the claim that the two cars have different mean parking times at the 0.10 level of significance. The result is consistent with the confidence interval constructed since 0 is included in the 90% confidence interval. 10-25 10-36. According to the normal probability plots, the assumption of normality does not appear to be violated since the data fall approximately along a straight line. Normal Probability Plot .999 Pr ob abi lity .99 .95 .80 .50 .20 .05 .01 .001 -20 0 20 diff Average: 1.21429 StDev: 12.6849 N: 14 10-37 d = 868.375 sd = 1290, n = 8 99% confidence interval: d − t α / 2, n −1 sd Anderson-Darling Normality Test A-Squared: 0.439 P-Value: 0.250 where di = brand 1 - brand 2 ≤ µ d ≤ d + t α / 2, n −1 n 868.375 − 3.499 sd n 1290 1290 ≤ µ d ≤ 868.375 + 3.499 8 8 −727.46 ≤ µd ≤ 2464.21 Since this confidence interval contains zero, we are 99% confident there is no significant difference between the two brands of tire. 10-38. a) d = 0.667 sd = 2.964, n = 12 95% confidence interval: d − t α / 2, n −1 sd n 0.667 − 2.201 ≤ µ d ≤ d + t α / 2, n −1 sd n 2.964 2.964 ≤ µ d ≤ 0.667 + 2.201 12 12 −1.216 ≤ µd ≤ 2.55 Since zero is contained within this interval, we are 95% confident there is no significant indication that one design language is preferable. 10-26 b) According to the normal probability plots, the assumption of normality does not appear to be violated since the data fall approximately along a straight line. Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 -5 0 5 diff Average: 0.666667 StDev: 2.96444 N: 12 10-39. Anderson-Darling Normality Test A-Squared: 0.315 P-Value: 0.502 1) The parameter of interest is the difference in blood cholesterol level, µd where di = Before − After. 2) H0 : µ d = 0 3) H1 : µ d > 0 4) α = 0.05 5) The test statistic is t0 = d sd / n 6) Reject the null hypothesis if t0 > t 0.05,14 where t 0.05,14 = 1.761 7) d = 26.867 sd = 19.04 n = 15 t0 = 26.867 = 5.465 19.04 / 15 8) Since 5.465 > 1.761 reject the null and conclude the data support the claim that the mean difference in cholesterol levels is significantly less after fat diet and aerobic exercise program at the 0.05 level of significance. 10-27 10-40. a) 1) The parameter of interest is the mean difference in natural vibration frequencies, µd where di = finite element − Equivalent Plate. 2) H0 : µ d = 0 3) H1 : µ d ≠ 0 4) α = 0.05 5) The test statistic is t0 = d sd / n 6) Reject the null hypothesis if t0 < − t 0.025, 6 where − t 0.025, 6 = −2.447 or t0 > t 0.005,6 where t 0.005,6 = 2.447 7) d = −5.49 sd = 5.924 n =7 −5.49 = −2.45 5.924 / 7 8) Since −2.447< −2.45 < 2.447, do not reject the null and conclude the data suggest that the two methods do not produce significantly different mean values for natural vibration frequency at the 0.05 level of significance. t0 = b) 95% confidence interval: d − t α / 2, n −1 − 5.49 − 2.447 sd n ≤ µ d ≤ d + t α / 2, n −1 sd n 5.924 5.924 ≤ µ d ≤ −5.49 + 2.447 7 7 −10.969 ≤ µd ≤ -0.011 With 95% confidence, we believe that the mean difference between the natural vibration frequency from the equivalent plate method and the natural vibration frequency from the finite element method is between −10.969 and -0.011 cycles. 10-41. 1) The parameter of interest is the difference in mean weight, µd where di =Weight Before − Weight After. 2) H0 : µ d = 0 3) H1 : µ d > 0 4) α = 0.05 5) The test statistic is t0 = d sd / n 6) Reject the null hypothesis if t0 > t 0.05,9 where t 0.05,9 = 1.833 7) d = 17 s d = 6.41 n = 10 t0 = 17 = 8.387 6.41 / 10 8) Since 8.387 > 1.833 reject the null and conclude there is evidence to conclude that the mean weight loss is significantly greater than 0; that is, the data support the claim that this particular diet modification program is significantly effective in reducing weight at the 0.05 level of significance. 10-28 10-42. 1) The parameter of interest is the mean difference in impurity level, µd where di = Test 1 − Test 2. 2) H0 : µ d = 0 3) H1 : µ d ≠ 0 4) α = 0.01 5) The test statistic is t0 = 6) Reject the null hypothesis if t0 < d sd / n − t 0.005, 7 where − t 0.005, 7 = −3.499 or t0 > t 0.005, 7 where t 0.005, 7 = 3.499 7) d = −0.2125 sd = 0.1727 t0 = n =8 − 0.2125 = −3.48 0.1727 / 8 8) Since −3.48 > -3.499 cannot reject the null and conclude the tests give significantly different impurity levels at α=0.01. 10-43. 1) The parameter of interest is the difference in mean weight loss, µd where di = Before − After. 2) H0 : µ d = 10 3) H1 : µ d > 10 4) α = 0.05 5) The test statistic is t0 = d − ∆0 sd / n 6) Reject the null hypothesis if t0 > t 0.05,9 where t 0.05,9 = 1.833 7) d = 17 sd = 6.41 n = 10 t0 = 17 − 10 6.41 / 10 = 3 .4 5 8) Since 3.45 > 1.833 reject the null and conclude there is evidence to support the claim that this particular diet modification program is effective in producing a mean weight loss of at least 10 lbs at the 0.05 level of significance. 10-44. Use sd as an estimate for σ: n= (z α + z β )σ d 10 2 (1.645 + 1.29)6.41 = 10 Yes, the sample size of 10 is adequate for this test. 10-29 2 = 3.53 , n=4 Section 10-5 10-45 a) f0.25,5,10 = 1.59 b) f0.10,24,9 = 2.28 1 1 = = 0.529 f0.25,10,5 189 . d) f0.75,5,10 = e) f0.90,24,9 = 1 f 0 . 10 , 9 , 24 1 c) f0.05,8,15 = 2.64 a) f0.25,7,15 = 1.47 d) f0.75,7,15 = b) f0.10,10,12 = 2.19 10-46 f) f0.95,8,15 = e) f0.90,10,12 = f0.05,15,8 = 1 f 0 .25 ,15 , 7 1 f 0 . 10 ,12 ,10 c) f0.01,20,10 = 4.41 10-47. f) f0.99,20,10 = = 1 = 0 . 525 1 . 91 1 = 0.311 3.22 = 1 = 0 . 596 1 . 68 = 1 = 0 . 438 2 . 28 1 1 = = 0.297 f0.01,10,20 3.37 2 1) The parameters of interest are the variances of concentration, σ1 , σ 2 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is f0 = s12 2 s2 6) Reject the null hypothesis if f0 < f0.975,9 ,15 where f0.975,9 ,15 = 0.265 or f0 > f0.025,9 ,15 where f0.025,9,15 =3.12 7) n1 = 10 n2 = 16 s1 = 4.7 s2 = 5.8 f0 = ( 4 .7 ) 2 = 0.657 (5.8) 2 8) Since 0.265 < 0.657 < 3.12 do not reject the null hypothesis and conclude there is insufficient evidence to indicate the two population variances differ significantly at the 0.05 level of significance. 10-30 10-48. 2 1) The parameters of interest are the etch-rate variances, σ1 , σ 2 . 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is f0 = s12 2 s2 6) Reject the null hypothesis if f0 < f0.975,9 ,9 = 0.248 or f0 > f0.025,9 ,9 = 4.03 7) n1 = 10 n2 = 10 s1 = 0.422 s2 = 0.231 (0.422) 2 f0 = = 3.337 (0.231) 2 8) Since 0.248 < 3.337 < 4.03 do not reject the null hypothesis and conclude the etch rate variances do not differ at the 0.05 level of significance. 10-49. With λ = 2 = 1.4 β = 0.10, and α = 0.05 , we find from Chart VI o that n1* = n2* = 100. Therefore, the samples of size 10 would not be adequate. 10-50. a) 90% confidence interval for the ratio of variances: s12 2 s2 f 1−α / 2, n1 −1, n2 −1 σ 12 s12 ≤ 2 ≤ 2 f α / 2, n −1, n σ2 s2 1 2 −1 σ 12 (0.35) (0.35) 0.412 ≤ 2 ≤ 2 .3 3 (0.40) (0.40) σ2 σ 12 0.3605 ≤ 2 ≤ 2.039 σ2 0.6004 ≤ σ1 ≤ 1.428 σ2 b) 95% confidence interval: s12 2 s2 f 1−α / 2, n1 −1, n2 −1 σ 12 s12 ≤ 2 ≤ 2 f α / 2, n −1, n σ2 s2 1 2 −1 σ2 (0.35) (0.35) 0.342 ≤ 12 ≤ 2 .8 2 (0.40) (0.40) σ2 σ 12 ≤ 2.468 2 σ2 σ 0.5468 ≤ 1 ≤ 1.5710 σ2 0.299 ≤ The 95% confidence interval is wider than the 90% confidence interval. 10-31 c) 90% lower-sided confidence interval: 2 s1 s2 2 f1− α , n1 −1, n 2 −1 ≤ 2 σ1 σ2 2 σ 12 (0.35) 0.500 ≤ 2 (0.40) σ2 σ 12 0.438 ≤ 2 σ2 σ 0.661 ≤ 1 σ2 10-51 a) 90% confidence interval for the ratio of variances: s12 f 1−α / 2, n1 −1, n2 −1 2 s2 (0.6) 2 (0.8) 0.156 ≤ 2 0.08775 ≤ 2 σ1 σ2 2 2 σ1 σ2 2 ≤ σ 12 s12 ≤ 2 ≤ 2 f α / 2, n −1, n σ2 s2 1 (0.6)2 (0.8) 2 2 −1 6.39 ≤ 3.594 b) 95% confidence interval: 2 s1 s2 2 f1− α / 2, n1 −1, n 2 −1 ≤ (0.6) 2 0104 ≤ . ( 0.8) 2 0.0585 ≤ 2 σ1 σ2 2 2 σ1 σ2 2 ≤ ≤ ( 0.6) 2 ( 0.8) 2 2 s1 s2 2 fα / 2, n1 −1, n 2 −1 9.60 2 σ1 ≤ 5.4 σ2 2 The 95% confidence interval is wider than the 90% confidence interval. c) 90% lower-sided confidence interval: 2 s1 s2 2 f1− α , n1 −1, n 2 −1 ≤ (0.6) 2 ( 0.8) 2 0.243 ≤ 2 σ1 σ2 2 2 σ1 σ2 2 σ 12 0.137 ≤ 2 σ2 10-32 10-52 2 1) The parameters of interest are the thickness variances, σ1 , σ 2 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.02 5) The test statistic is s12 f0 = 2 s2 6) Reject the null hypothesis if f0 < f0.99,7 ,7 where f0.99,7 ,7 = 0.143 or f0 > f0.01,7,7 where f0.01,7,7 = 6.99 7) n1 = 8 s1 = 0.11 n2 = 8 s2 = 0.09 f0 = (0.11) 2 = 1 .4 9 (0.09) 2 8) Since 0.143 < 1.49 < 6.99 do not reject the null hypothesis and conclude the thickness variances do not significantly differ at the 0.02 level of significance. 10-53 2 1) The parameters of interest are the strength variances, σ1 , σ 2 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is f0 = s12 2 s2 6) Reject the null hypothesis if f0 < f0.975,9,15 where f0.975,9,15 = 0.265 or f0 > f0.025,9,15 where f0.025,9,15 =3.12 7) n1 = 10 n2 = 16 s1 = 12 s2 = 22 (12) 2 f0 = = 0.297 (22) 2 8) Since 0.265 < 0.297 < 3.12 do not reject the null hypothesis and conclude the population variances do not significantly differ at the 0.05 level of significance. 10-54 2 1) The parameters of interest are the melting variances, σ1 , σ 2 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is f0 = s12 2 s2 6) Reject the null hypothesis if f0 < f0.975,20,20 where f0.975,20,20 =0.4058 or f0 > f0.025,20,20 where f0.025,20,20 =2.46 10-33 7) n1 = 21 s1 = 4 n2 = 21 s2 = 3 ( 4) 2 = 1 .7 8 (3) 2 f0 = 8) Since 0.4058 < 1.78 < 2.46 do not reject the null hypothesis and conclude the population variances do not significantly differ at the 0.05 level of significance. 10-55 2 1) The parameters of interest are the thickness variances, σ1 , σ 2 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.01 5) The test statistic is f0 = 6) Reject the null hypothesis if f0 < s12 2 s2 f 0.995,10,12 where f 0.995,10,12 =0.1766 or f0 > f 0.005,10,12 where f 0.005,10,12 = 5.0855 7) n1 = 11 s1 = 10.2 n2 = 13 s2 = 20.1 f0 = (10.2) 2 = 0.2575 (20.1) 2 8) Since 0.1766 <0.2575 < 5.0855 do not reject the null hypothesis and conclude the thickness variances are not equal at the 0.01 level of significance. 10-56. 1) The parameters of interest are the time to assemble standard deviations, σ1 , σ 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.02 5) The test statistic is f0 = 6) Reject the null hypothesis if f0 < 7) n1 = 25 s12 2 s2 f 1−α / 2, n1 −1,n2 −1 =0..365 or f0 > f α / 2, n1 −1, n2 −1 = 2.86 n 2 = 21 s1 = 0.98 s2 = 1.02 2 f0 = (0.98) = 0.923 (1.02) 2 8) Since 0.365 < 0.923 < 2.86 do not reject the null hypothesis and conclude there is no evidence to support the claim that men and women differ significantly in repeatability for this assembly task at the 0.02 level of significance. 10-34 10-57. 98% confidence interval: s12 2 s2 f 1−α / 2, n1 −1, n2 −1 (0.923)0.365 ≤ 0.3369 ≤ σ 12 s12 ≤ 2 ≤ 2 f α / 2, n −1, n σ2 s2 1 2 −1 σ 12 ≤ (0.923)2.86 2 σ2 σ 12 ≤ 2.640 2 σ2 Since the value 1 is contained within this interval, we can conclude that there is no significant difference between the variance of the repeatability of men and women for the assembly task at a 98% confidence level. 10-58 For one population standard deviation being 50% larger than the other, then λ = 2. Using n =8, α = 0.01 and Chart VI p, we find that β ≅ 0.85. Therefore, we would say that n = n1 = n2 = 8 is not adequate to detect this difference with high probability. 10-59 1) The parameters of interest are the overall distance standard deviations, σ1 , σ 2 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is f0 = 6) Reject the null hypothesis if f0 < 7) n1 = 10 2 s1 s2 2 f .0.975,9,9 =0.248 or f0 > f 0.025,9,9 = 4.03 s1 = 8.03 n2 = 10 s2 = 10.04 2 f0 = (8.03) = 0.640 (10.04) 2 8) Since 0.248 < 0.640 < 4.04 do not reject the null hypothesis and conclude there is no evidence to support the claim that there is a difference in the standard deviation of the overall distance of the two brands at the 0.05 level of significance. 95% confidence interval: 2 s1 s2 2 f1− α / 2, n1 −1, n 2 −1 ≤ 2 σ1 σ2 2 ≤ 2 s1 s2 2 fα / 2, n1 −1, n 2 −1 σ 12 (0.640)0.248 ≤ 2 ≤ (0.640)4.03 σ2 2 σ 0.159 ≤ 12 ≤ 2.579 σ2 Since the value 1 is contained within this interval, we can conclude that there is no significant difference in the variance of the distances at a 95% significance level. 10-60 1) The parameters of interest are the time to assemble standard deviations, σ1 , σ 2 2 2) H0 : σ1 = σ 2 2 10-35 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is s12 2 s2 f0 = 6) Reject the null hypothesis if f0 < 7) n1 = 12 f .0.975,11,911 =0.288 or f0 > f 0.025,11,11 = 3.474 s1 = 0.0217 n2 = 12 s2 = 0.0175 2 f0 = (0.0217) = 1.538 (0.0175) 2 8) Since 0.288<1.538<3.474 do not reject the null hypothesis and conclude there is no evidence to support the claim that there is a difference in the standard deviation of the coefficient of restitution between the two clubs at the 0.05 level of significance. 95% confidence interval: 2 s1 s2 2 f1− α / 2, n1 −1, n 2 −1 ≤ (1.538)0.288 ≤ 0.443 ≤ 2 σ1 σ2 2 ≤ 2 s1 s2 2 fα / 2, n1 −1, n 2 −1 σ 12 ≤ (1.538)3.474 2 σ2 σ 12 ≤ 5.343 2 σ2 Since the value 1 is contained within this interval, we can conclude that there is no significant difference in the variances in the variances of the coefficient of restitution at a 95% significance level. Section 10-6 10-61. 1) the parameters of interest are the proportion of defective parts, p1 and p2 p1 = p2 3) H1 : p1 ≠ p2 2 ) H0 : 4) α = 0.05 5) Test statistic is ˆˆ p1 − p2 z0 = where 11 ˆ ˆ p (1 − p ) + n1 n2 ˆ p= x1 + x 2 n1 + n 2 6) Reject the null hypothesis if z0 < − z0.025 where − z0.025 = −1.96 or z0 > z0.025 where z0.025 = 1.96 7) n1 = 300 x1 = 15 n2 = 300 x2 = 8 10-36 p1 = 0.05 p= p2 = 0.0267 z0 = 15 + 8 = 0.0383 300 + 300 0.05 − 0.0267 = 1 .4 9 1 1 0.0383(1 − 0.0383) + 300 300 8) Since −1.96 < 1.49 < 1.96 do not reject the null hypothesis and conclude that yes the evidence indicates that there is not a significant difference in the fraction of defective parts produced by the two machines at the 0.05 level of significance. P-value = 2(1−P(z < 1.49)) = 0.13622 10-62. 1) the parameters of interest are the proportion of satisfactory lenses, p1 and p2 p1 = p2 3) H1 : p1 ≠ p2 2 ) H0 : 4) α = 0.05 5) Test statistic is z0 = ˆˆ p1 − p2 ˆ ˆ p (1 − p ) where 11 + n1 n2 6) Reject the null hypothesis if z0 < − 7) n1 = 300 ˆ p2 = 0.653 z 0.005 where − z 0.005 = −258 or z0 > z 0.005 where z 0.005 = 2.58 x2 = 196 ˆ p1 = 0.843 x1 + x 2 n1 + n 2 n2 = 300 x1 = 253 ˆ p= z0 = ˆ p= 253 + 196 = 0.748 300 + 300 0.843 − 0.653 = 5 .3 6 1 1 0.748(1 − 0.748) + 300 300 8) Since 5.36 > 2.58 reject the null hypothesis and conclude that yes the evidence indicates that there is significant difference in the fraction of polishing-induced defects produced by the two polishing solutions the 0.01 level of significance. P-value = 2(1−P(z < 5.36)) = 0 By constructing a 99% confidence interval on the difference in proportions, the same question can be answered by considering whether or not 0 is contained in the interval. 10-63. a) Power = 1 − β 10-37 zα β= /2 pq Φ 1 1 + n1 n 2 ˆˆ σp 1 − ( p1 − p 2 ) − zα /2 pq −Φ 1 1 + n1 n 2 ˆˆ σp ˆ − p2 1 − ( p1 − p 2 ) ˆ − p2 300(0.05) + 300(0.01) = 0 .0 3 q = 0.97 300 + 300 0.05(1 − 0.05) 0.01(1 − 0.01) σ p1 − p 2 = + = 0.014 300 300 p= 1.96 0.03(0.97) β= = Φ 1 1 1 1 + − (0.05− 0.01) −1.96 0.03(0.97) + − (0.05 − 0.01) 300 300 300 300 −Φ 0.014 0.014 Φ(− 0.91) − Φ (− 4.81) = 0.18141 − 0 = 0.18141 Power = 1 − 0.18141 = 0.81859 zα / 2 b) n ( p1 + p2 )(q1 + q2 ) + z 2 = 2 p1q1 + p2 q2 β ( p1 − p2 )2 1.96 (0.05 + 0.01)(0.95 + 0.99) + 1.29 2 = 2 0.05(0.95) + 0.01(0.99) = 382.11 (0.05 − 0.01)2 n = 383 zα / 2 pq 10-64. a) β = Φ 11 + − ( p1 − p2 ) n1 n2 ˆˆ ˆ σ p −p 1 p= −Φ 1 . 96 11 + − ( p1 − p2 ) n1 n2 ˆˆ ˆ σ p −p 2 1 300(0.05) + 300(0.02) = 0.035 300 + 300 σ p1 − p 2 = β =Φ − zα / 2 pq 2 q = 0.965 0.05(1 − 0.05) 0.02(1 − 0.02) + = 0.015 300 300 0 . 035 ( 0 . 965 ) 1 1 + 300 300 0 . 015 − (0 . 05 − 0 . 02 ) − 1 . 96 −Φ = Φ( −0.04) − Φ( −3.96) = 0.48405 − 0.00004 = 0.48401 Power = 1 − 0.48401 = 0.51599 10-38 0 . 035 ( 0 . 965 ) 1 1 + 300 300 0 . 015 − (0 . 05 − 0 . 02 ) zα /2 β 2 b) n = 196 . 2 ( p1 + p2 )(q1 + q 2 ) + z p1q1 + p 2q 2 ( p1 − p2 )2 ( 0.05 + 0.02)( 0.95 + 0.98) 2 = 2 + 129 0.05(0.95) + 0.02(0.98) . = 790.67 ( 0.05 − 0.02) 2 n = 791 10-65. 1) the parameters of interest are the proportion of residents in favor of an increase, p1 and p2 2) H0 : p1 = p 2 3) H1 : p1 ≠ p 2 4) α = 0.05 5) Test statistic is ˆˆ p1 − p2 z0 = ˆ ˆ p (1 − p ) where 11 + n1 n2 ˆ p= x1 + x 2 n1 + n 2 6) Reject the null hypothesis if z0 < − z0.025 where − z0.025 = −1.96 or z0 > z0.025 where z0.025 = 1.96 7) n1 = 500 n2 = 400 x1 = 385 x2 = 267 p1 = 0.77 p2 = 0.6675 z0 = ˆ p= 385 + 267 = 0.724 500 + 400 0.77 − 0.6675 = 3 .4 2 1 1 0.724(1 − 0.724) + 500 400 8) Since 3.42 > 1.96 reject the null hypothesis and conclude that yes the data do indicate a significant difference in the proportions of support for increasing the speed limit between residents of the two counties at the 0.05 level of significance. P-value = 2(1−P(z < 3.42)) = 0.00062 10-66. 95% confidence interval on the difference: ( p1 − p 2 ) − zα / 2 p1(1 − p1) p2 (1 − p2 ) p (1 − p1) p 2 (1 − p 2 ) + ≤ p1 − p 2 ≤ ( p1 − p 2 ) + zα / 2 1 + n1 n2 n1 n2 (005 − 00267) − 196 . . . 005(1 − 005) 00267(1 − 00267) . . . . 005(1− 005) 00267(1 − 00267) . . . . + ≤ p1 − p2 ≤ (005 − 00267) + 196 . . . + 300 300 300 300 −0.0074 ≤ p1 − p2 ≤ 0.054 Since this interval contains the value zero, we are 95% confident there is no significant difference in the fraction of defective parts produced by the two machines and that the difference in proportions is between − 0.0074 and 0.054. 10-39 10-67 95% confidence interval on the difference: p1(1 − p1) p 2 (1 − p 2 ) p (1 − p1) p2 (1 − p 2 ) + ≤ p1 − p2 ≤ ( p1 − p2 ) + zα / 2 1 + n1 n2 n1 n2 ( p1 − p 2 ) − zα / 2 ( 0.77 − 0.6675) − 196 . 0.77(1 − 0.77) 0.6675(1 − 0.6675) 0.77(1 − 0.77) 0.6675(1 − 0.6675) + ≤ p1 − p 2 ≤ ( 0.77 − 0.6675) + 196 . + 500 400 500 400 0.0434 ≤ p1 − p 2 ≤ 0.1616 Since this interval does not contain the value zero, we are 95% confident there is a significant difference in the proportions of support for increasing the speed limit between residents of the two counties and that the difference in proportions is between 0.0434 and 0.1616. Supplemental Exercises 10-68 a) Assumptions that must be met are normality, equality of variance, independence of the observations and of the populations. Normality and equality of variances appears to be reasonable, see normal probability plot. The data appear to fall along a straight line and the slopes appear to be the same. Independence of the observations for each sample is assumed. It is also reasonable to assume that the two populations are independent. Normal Probability Plot Normal Probability Plot .999 .999 .99 .99 .95 Probability Probability .95 .80 .50 .20 .80 .50 .20 .05 .05 .01 .01 .001 .001 100 97 98 99 100 101 102 103 102 103 104 Anderson-Darling Normality Test A-Squared: 0.315 P-Value: 0.522 Average: 105.069 StDev: 1.96256 N: 35 2 2) H0 : σ1 = σ 2 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is s12 2 s2 6) Reject H0 if f0 < f0.975,24 ,34 where f0.975,24 ,34 = 1 f 0.025,34, 24 = 1 = 0.459 2.18 or f0 > f0.025,24 ,34 where f0.025,24 ,34 =2.07 7) s1 = 1.53 n1 = 25 s2 =1.96 n2 = 35 f0 = 106 107 108 109 Anderson-Darling Normality Test A-Squared: 0.376 P-Value: 0.394 2 b) 1) the parameters of interest are the variances of resistance of products, σ1 , σ 2 2 f0 = 105 vendor 2 vendor 1 Average: 99.576 StDev: 1.52896 N: 25 101 104 (1.53) 2 = 0.609 (1.96) 2 8) Since 0.609 > 0.459, cannot reject H0 and conclude the variances are significantly different at α = 0.05. 10-40 10-69 a) Assumptions that must be met are normality, equality of variance, independence of the observations and of the populations. Normality and equality of variances appears to be reasonable, see normal probability plot. The data appear to fall along a straight line and the slopes appear to be the same. Independence of the observations for each sample is assumed. It is also reasonable to assume that the two populations are independent. . Normal Probability Plot Normal Probability Plot .99 .95 .95 Probability .999 .99 Probability .999 .80 .50 .20 .80 .50 .20 .05 .05 .01 .01 .001 .001 14 15 16 17 18 19 20 8 9 10 9-hour Average: 16.3556 StDev: 2.06949 N: 9 11 12 13 14 15 1-hour Anderson-Darling Normality Test A-Squared: 0.171 P-Value: 0.899 Average: 11.4833 StDev: 2.37016 N: 6 Anderson-Darling Normality Test A-Squared: 0.158 P-Value: 0.903 x2 = 11.483 b) x1 = 16.36 s1 = 2.07 s2 = 2.37 n1 = 9 n2 = 6 t α / 2, n1 + n 2 − 2 = t 0.005,13 where t 0.005,13 = 3.012 99% confidence interval: sp = (x1 − x 2 ) − t α / 2, n + n − 2 (s p ) 1 2 8(2.07) 2 + 5(2.37) 2 = 2 .1 9 13 1 1 1 1 + ≤ µ 1 − µ 2 ≤ (x1 − x 2 ) + tα / 2, n1 + n2 − 2 (s p ) + n1 n 2 n1 n 2 (16 .36 − 11 .483 ) − 3 .012 (2 .19 ) 11 11 + ≤ µ 1 − µ 2 ≤ (16 .36 − 11 . 483 ) + 3 . 012 (2 . 19 ) + 96 96 1.40 ≤ µ1 − µ 2 ≤ 8.36 c) Yes, we are 99% confident the results from the first test condition exceed the results of the second test condition by between 1.40 and 8.36 (×106 PA). 10-70. 2 a) 95% confidence interval for σ1 / σ 2 2 95% confidence interval on f 0.975,8,5 = 1 f 0.025,5,8 = 2 σ1 σ2 2 : 1 = 0.2075 , 4.82 f 0.025,8,5 = 6.76 σ 12 s12 s12 f 0.975,8,5 ≤ 2 ≤ 2 f 0.025,8,5 2 σ 2 s2 s2 σ2 4.285 4.285 (0.2075) ≤ 12 ≤ (6.76) 5.617 5.617 σ2 0.1583 ≤ σ 12 ≤ 5.157 2 σ2 b) Since the value 1 is contained within this interval, with 95% confidence, the population variances do not differ significantly and can be assumed to be equal. 10-41 10-71 a) 1) The parameter of interest is the mean weight loss, µd where di = Initial Weight − Final Weight. 2) H0 : µ d = 3 3) H1 : µ d > 3 4) α = 0.05 5) The test statistic is t0 = d − ∆0 sd / n 6) Reject H0 if t0 > tα,n-1 where t0.05,7 = 1.895. 7) d = 4.125 sd = 1.246 n=8 4.125 − 3 t0 = = 2.554 1.246 / 8 8) Since 2.554 > 1.895, reject the null hypothesis and conclude the average weight loss is significantly greater than 3 at α = 0.05. b) 2) H0 : µ d = 3 3) H1 : µ d > 3 4) α = 0.01 5) The test statistic is t0 = d − ∆0 sd / n 6) Reject H0 if t0 > tα,n-1 where t0.01,7 = 2.998. 7) d = 4.125 sd = 1.246 n=8 4.125 − 3 t0 = = 2.554 1.246 / 8 8) Since 2.554 <2.998, do not reject the null hypothesis and conclude the average weight loss is not significantly greater than 3 at α = 0.01. c) 2) H0 : µ d = 5 3) H1 : µ d > 5 4) α = 0.05 5) The test statistic is t0 = d − ∆0 sd / n 6) Reject H0 if t0 > tα,n-1 where t0.05,7 =1.895. 7) d = 4.125 sd = 1.246 n=8 4.125 − 5 t0 = = −1986 . 1.246 / 8 10-42 8) Since −1.986 < 1.895, do not reject the null hypothesis and conclude the average weight loss is not significantly greater than 5 at α = 0.05. Using α = 0.01 2) H0 : µ d = 5 3) H1 : µ d > 5 4) α = 0.01 5) The test statistic is t0 = d − ∆0 sd / n 6) Reject H0 if t0 > tα,n-1 where t0.01,7 = 2.998. 7) d = 4.125 sd = 1.246 n=8 4.125 − 5 t0 = = −1986 . 1.246 / 8 8) Since −1.986 < 2.998, do not reject the null hypothesis and conclude the average weight loss is not significantly greater than 5 at α = 0.01. 10-72. (x1 − x2 ) − zα / 2 σ 12 n1 + a) 90% confidence interval: 2 σ2 n2 ≤ µ1 − µ 2 ≤ ( x1 − x2 ) + zα / 2 σ 12 n1 + 2 σ2 n2 zα / 2 = 1.65 52 4 2 52 4 2 + ≤ µ1 − µ 2 ≤ (88 − 91) + 1.65 + 20 20 20 20 − 5.362 ≤ µ1 − µ 2 ≤ −0.638 (88 − 91) − 1.65 Yes, with 90% confidence, the data indicate that the mean breaking strength of the yarn of manufacturer 2 exceeds that of manufacturer 1 by between 5.362 and 0.638. b) 98% confidence interval: zα / 2 = 2.33 52 4 2 52 4 2 + ≤ µ1 − µ 2 ≤ (88 − 91) + 2.33 + 20 20 20 20 − 6.340 ≤ µ1 − µ 2 ≤ 0.340 (88 − 91) − 2.33 Yes, we are 98% confident manufacturer 2 produces yarn with higher breaking strength by between 0.340 and 6.340 psi. c) The results of parts a) and b) are different because the confidence level or z-value used is different.. Which one is used depends upon the level of confidence considered acceptable. 10-43 10-73 a) 1) The parameters of interest are the proportions of children who contract polio, p1 , p2 2 ) H0 : p 1 = p 2 3 ) H1 : p 1 ≠ p 2 4) α = 0.05 5) The test statistic is ˆˆ p1 − p2 z0 = ˆ ˆ p (1 − p ) 11 + n1 n2 6) Reject H0 if z0 < − zα /2 or z0 > zα /2 where zα /2 = 1.96 7) p1 = x1 110 = = 0.00055 n1 201299 (Placebo) x2 33 = = 0.00016 n2 200745 x1 + x 2 = 0.000356 n1 + n 2 (Vaccine) p2 = p= 0.00055 − 0.00016 z0 = = 6.55 1 1 + 201299 200745 8) Since 6.55 > 1.96 reject H0 and conclude the proportion of children who contracted polio is significantly different at α = 0.05. b) α = 0.01 Reject H0 if z0 < − zα /2 or z0 > zα /2 where zα /2 =2.58 z0 = 6.55 Since 6.55 > 2.58, reject H0 and conclude the proportion of children who contracted polio is different at α = 0.01. c) The conclusions are the same since z0 is so large it exceeds zα/2 in both cases. 0.000356(1 − 0.000356) 10-74 a) α = 0.10 n≅ ( zα / 2 ) zα / 2 = 1.65 2 ( 2 σ1 ( E) 2 b) α = 0.10 n≅ + σ2 2 ( zα / 2 ) ) ≅ (1.65) (25 + 16) = 49.61 , 2 (15) 2 . n = 50 zα / 2 = 2.33 2 ( 2 σ1 + σ2 2 ) ≅ ( 2.33) (25 + 16) = 98.93 , 2 n = 99 ( E) 2 (15) 2 . c) As the confidence level increases, sample size will also increase. d) α = 0.10 zα / 2 = 1.65 n≅ 2 ( zα / 2 )2 ( σ1 + σ 2 ) (1.65) 2 (25 + 16) 2 ≅ = 198.44 , ( E) 2 e) α = 0.10 n≅ ( zα / 2 ) (0.75) 2 n = 199 zα / 2 = 2.33 2 ( 2 σ1 + σ2 2 ) ≅ ( 2.33) (25 + 16) = 395.70 , 2 n =396 ( E)2 (0.75) 2 f) As the error decreases, the required sample size increases. 10-44 10-75 p1 = x1 387 = = 0.258 n1 1500 ( p1 − p2 ) ± zα / 2 p2 = x2 310 = = 0.2583 n2 1200 p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2 a) zα / 2 = z0.025 = 1.96 ( 0.258 − 0.2583) ± 1.96 0.258( 0.742) 0.2583(0.7417) + 1500 1200 − 0.0335 ≤ p1 − p2 ≤ 0.0329 Since zero is contained in this interval, we are 95% confident there is no significant difference between the proportion of unlisted numbers in the two cities. b) zα / 2 = z0.05 = 1.65 0.258(0.742) 0.2583(0.7417) + 1500 1200 −0.0282 ≤ p1 − p2 ≤ 0.0276 . ( 0.258 − 0.2583) ± 165 Again, the proportion of unlisted numbers in the two cities do not differ. x 774 = 0.258 c) p1 = 1 = n1 3000 x2 620 = = 0.2583 n2 2400 95% confidence interval: p2 = 0.258( 0.742) 0.2583(0.7417) + 3000 2400 −0.0238 ≤ p1 − p2 ≤ 0.0232 90% confidence interval: ( 0.258 − 0.2583) ± 1.96 ( 0.258 − 0.2583) ± 1.65 0.258(0.742) 0.2583(0.7417) + 3000 2400 − 0.0201 ≤ p1 − p2 ≤ 0.0195 Increasing the sample size decreased the error and width of the confidence intervals, but does not change the conclusions drawn. The conclusion remains that there is no significant difference. 10-76 a) 1) The parameters of interest are the proportions of those residents who wear a seat belt regularly, p1 , p2 2 ) H0 : p 1 = p 2 3 ) H1 : p 1 ≠ p 2 4) α = 0.05 5) The test statistic is p1 − p2 z0 = 1 1 p(1 − p) + n1 n2 6) Reject H0 if z0 < − zα /2 or z0 > zα /2 where z0.025 = 1.96 7) p1 = x1 165 = = 0.825 n1 200 p2 = p= x1 + x 2 = 0.807 n1 + n 2 x2 198 = = 0.792 n2 250 z0 = 0 . 825 − 0 . 792 1 1 0 . 807 (1 − 0 . 807 ) + 200 250 10-45 = 0 . 8814 8) Since −1.96 < 0.8814 < 1.96 do not reject H0 and conclude that evidence is insufficient to claim that there is a difference in seat belt usage α = 0.05. b) α = 0.10 Reject H0 if z0 < − zα /2 or z0 > zα /2 where z0.05 = 1.65 z0 = 0.8814 Since −1.65 < 0.8814 < 1.65, do not reject H0 and conclude that evidence is insufficient to claim that there is a difference in seat belt usage α = 0.10. c) The conclusions are the same, but with different levels of confidence. d) n1 =400, n2 =500 α = 0.05 Reject H0 if z0 < − zα /2 or z0 > zα /2 where z0.025 = 1.96 z0 = 0.825 − 0.792 = 1.246 1 1 0.807(1 − 0.807) + 400 500 Since −1.96 < 1.246 < 1.96 do not reject H0 and conclude that evidence is insufficient to claim that there is a difference in seat belt usage α = 0.05. α = 0.10 Reject H0 if z0 < − zα /2 or z0 > zα /2 where z0.05 = 1.65 z0 =1.012 Since −1.65 < 1.246 < 1.65, do not reject H0 and conclude that evidence is insufficient to claim that there is a difference in seat belt usage α = 0.10. As the sample size increased, the test statistic has also increased, since the denominator of z0 decreased. However, the decrease (or sample size increase) was not enough to change our conclusion. 10-77. a) Yes, there could be some bias in the results due to the telephone survey. b) If it could be shown that these populations are similar to the respondents, the results may be extended. 10-78 p2 a) 1) The parameters of interest are the proportion of lenses that are unsatisfactory after tumble-polishing, p1, 2 ) H0 : p 1 = p 2 3 ) H1 : p 1 ≠ p 2 4) α = 0.01 5) The test statistic is z0 = p1 − p2 p(1 − p) 1 1 + n1 n2 6) Reject H0 if z0 < − zα /2 or z0 > zα /2 where zα /2 = 2.58 7) x1 =number of defective lenses x 47 p1 = 1 = = 0.1567 n1 300 p2 = p= x1 + x 2 = 0.2517 n1 + n 2 x2 104 = = 0..3467 n2 300 z0 = 01567 − 0.3467 . 0.2517(1 − 0.2517) 10-46 1 1 + 300 300 = −5.36 8) Since −5.36 < −2.58 reject H0 and conclude there is strong evidence to support the claim that the two polishing fluids are different. b) The conclusions are the same whether we analyze the data using the proportion unsatisfactory or proportion satisfactory. The proportion of defectives are different for the two fluids. 10-79. n= (0.9 + 0.6)(0.1 + 0.4) 2.575 + 1.28 0.9(0.1) + 0.6(0.4) 2 ( 0 .9 − 0 .6 ) 2 5.346 = 59.4 0.09 n = 60 = 10-80 The parameter of interest is µ1 − 2µ 2 H 0: µ1 = 2µ 2 H1: µ1 > 2µ 2 → H 0: µ1 − 2µ 2 = 0 H1: µ1 − 2µ 2 > 0 Let n1 = size of sample 1 X1 estimate for µ1 Let n2 = size of sample 2 X2 estimate for µ2 X1 − 2 X2 is an estimate for µ1 − 2µ 2 The variance is V( X1 − 2 X2 ) = V( X1 ) + V(2 X2 ) = 2 σ1 4σ 2 +2 n1 n2 The test statistic for this hypothesis would then be: ( X − 2 X2 ) − 0 Z0 = 1 2 σ1 4σ1 +2 n1 n2 We would reject the null hypothesis if z0 > zα/2 for a given level of significance. The P-value would be P(Z ≥ z0 ). 10-47 2 H0 : µ1 = µ 2 10-81. H1 : µ1 ≠ µ 2 n1 = n2 =n β = 0.10 α = 0.05 2 Assume normal distribution and σ1 = σ 2 = σ 2 2 µ1 = µ 2 + σ d= 1 |µ1 − µ 2 | σ = = 2σ 2σ 2 From Chart VI, n∗ = 50 n= n ∗ + 1 50 + 1 = = 25.5 2 2 n1 = n2 =26 a) α = 0.05, β = 0.05 ∆= 1.5 Use sp = 0.7071 to approximate σ in equation 10-19. 10-82 d= ∆ 1 .5 = = 1.06 ≅ 1 2( s p ) 2(.7071) From Chart VI (e), n∗ = 20 n= n ∗ + 1 20 + 1 = = 10.5 2 2 n= 11 would be needed to reject the null hypothesis that the two agents differ by 0.5 with probability of at least 0.95. b) The original size of n = 5 in Exercise 10-18 was not appropriate to detect the difference since it is necessary for a sample size of 16 to reject the null hypothesis that the two agents differ by 1.5 with probability of at least 0.95. 10-83 a) No. Normal Probability Plot Normal Probability Plot .99 .95 .95 Probability .999 .99 Probability .999 .80 .50 .20 .80 .50 .20 .05 .05 .01 .01 .001 .001 23.9 24.4 30 24.9 Average: 24.67 StDev: 0.302030 N: 10 35 40 volkswag mercedes Anderson-Darling Normality Test A-Squared: 0.934 P-Value: 0.011 Average: 40.25 StDev: 3.89280 N: 10 10-48 Anderson-Darling Normality Test A-Squared: 1.582 P-Value: 0.000 b) The normal probability plots indicate that the data follow normal distributions since the data appear to fall along a straight line. The plots also indicate that the variances could be equal since the slopes appear to be the same. Normal Probability Plot Normal Probability Plot .999 .99 .80 .95 Probability .999 .95 Probability .99 .50 .20 .05 .80 .50 .20 .05 .01 .01 .001 .001 24.5 24.6 24.7 24.8 24.9 39.5 40.5 mercedes Average: 24.74 StDev: 0.142984 N: 10 41.5 42.5 volkswag Anderson-Darling Normality Test A-Squared: 0.381 P-Value: 0.329 Average: 41.25 StDev: 1.21952 N: 10 Anderson-Darling Normality Test A-Squared: 0.440 P-Value: 0.230 c) By correcting the data points, it is more apparent the data follow normal distributions. Note that one unusual observation can cause an analyst to reject the normality assumption. d) 95% confidence interval on the ratio of the variances, σ 2 / σ 2 V M s2 = 149 . V f9,9,0.025 = 4.03 s2 = 0.0204 M f9 ,9,0.975 = 1 f9,9 ,0.025 = 1 = 0.248 4.03 2 σ2 sV s2 f 9,9,0.975 < V < V f 9,9, 0.025 2 2 2 σM sM sM 2 σV 1.49 1.49 0.248 < 2 < 4.03 σM 0.0204 0.0204 18.114 < 2 σV < 294.35 2 σM Since the does not include the value of unity, we are 95% confident that there is evidence to reject the claim that the variability in mileage performance is the same for the two types of vehicles. There is evidence that the variability is greater for a Volkswagen than for a Mercedes. 10-84 2 1) the parameters of interest are the variances in mileage performance, σ1 , σ 2 2 2 2) H0 : σ1 = σ 2 Where Volkswagen is represented by variance 1, Mercedes by variance 2. 2 2 3) H1 : σ1 ≠ σ 2 2 4) α = 0.05 5) The test statistic is f0 = 2 s1 s2 2 6) Reject H0 if f0 < f0.975,9 ,9 where f0.975,9 ,9 = 1 f0.025,9 ,9 or f0 > f0.025,9 ,9 where f0.025,9 ,9 = 4.03 10-49 = 1 = 0.248 4.03 7) s1 = 1.22 s2 = 0.143 n1 = 10 n2 = 10 f0 = (122)2 . = 72.78 (0.143)2 8) Since 72.78 > 4.03, reject H0 and conclude that there is a significant difference between Volkswagen and Mercedes in terms of mileage variability. Same conclusions reached in 10-83d. 10-85 a) Underlying distributions appear to be normal since the data fall along a straight line on the normal probability plots. The slopes appear to be similar, so it is reasonable to assume that Normal Probability Plot Normal Probability Plot .999 .99 .99 .95 .95 Probability .999 Probability 2 σ 12 = σ 2 . .80 .50 .20 .05 .80 .50 .20 .05 .01 .01 .001 .001 751 752 753 754 755 754 755 ridgecre Average: 752.7 StDev: 1.25167 N: 10 756 757 valleyvi Anderson-Darling Normality Test A-Squared: 0.384 P-Value: 0.323 Average: 755.6 StDev: 0.843274 N: 10 Anderson-Darling Normality Test A-Squared: 0.682 P-Value: 0.051 b) 1) The parameter of interest is the difference in mean volumes, µ1 − µ 2 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3 ) H1 : µ 1 − µ 2 ≠ 0 o r µ 1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − δ sp 1 1 + n1 n2 6) Reject H0 if t0 < − t α / 2 , ν or z0 > t α / 2, ν where t α / 2 , ν = t 0.025,18 = 2.101 7) x1 = 752.7 x2 = 755.6 s1 = 1.252 9(1.252)2 + 9(0.843) 2 = 1.07 18 s2 = 0.843 n1 = 10 sp = n2 = 10 t0 = (752.7 − 755.6) 1 1 1.07 + 10 10 = − 6 .0 6 8) Since −6.06 < −2.101, reject H0 and conclude there is a significant difference between the two winery’s with respect to mean fill volumes. 10-50 10-86. d=2/2(1.07)=0.93, giving a power of just under 80%. Since the power is relatively low, an increase in the sample size would increase the power of the test. 10-87. a) The assumption of normality appears to be valid. This is evident by the fact that the data lie along a straight line in the normal probability plot. Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 -2 -1 0 1 2 diff Average: -0.222222 StDev: 1.30171 N: 9 Anderson-Darling Normality Test A-Squared: 0.526 P-Value: 0.128 b) 1) The parameter of interest is the mean difference in tip hardness, µd 2 ) H0 : µ d = 0 3) H1 : µ d ≠ 0 4) No significance level, calculate P-value 5) The test statistic is d t0 = sd / n 6) Reject H0 if the P-value is significantly small. 7) d = −0.222 sd = 1.30 n=9 −0.222 t0 = = −0.512 130 / 9 . 8) P-value = 2P(T < -0.512) = 2P(T > 0.512) 2(0.25) < P-value < 2(0.40) 0.50 < P-value < 0.80 Since the P-value is larger than any acceptable level of significance, do not reject H0 and conclude there is no difference in mean tip hardness. c) β = 0.10 µd = 1 1 1 d= = = 0.769 σ d 1.3 From Chart VI with α = 0.01, n = 30 10-51 10-88. a) According to the normal probability plot the data appear to follow a normal distribution. This is evident by the fact that the data fall along a straight line. Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 -3 -2 -1 0 1 2 3 diff Average: 0.133333 StDev: 2.06559 N: 15 Anderson-Darling Normality Test A-Squared: 0.518 P-Value: 0.158 b) 1) The parameter of interest is the mean difference in depth using the two gauges, µd 2 ) H0 : µ d = 0 3) H1 : µ d ≠ 0 4) No significance level, calculate p-value 5) The test statistic is d t0 = sd / n 6) Reject H0 if the P-value is significantly small. 7) d = 0.133 sd = 2.065 n = 15 0.133 t0 = = 0.25 2.065 / 15 8) P-value = 2P(T > 0.25) 2(0.40) < P-value 0.80 < P-value Since the P-value is larger than any acceptable level of significance, do not reject H0 and conclude there is no difference in mean depth measurements for the two gauges. c) Power = 0.8, Therefore, since Power= 1-β , β= 0.20 µ d = 1.65 d= 1.65 σd = 1.65 = 0.799 (2.065) From Chart VI (f) with α = 0.01 and β = 0.20, we find n =30. 10-52 10-89 a.) The data from both depths appear to be normally distributed, but the slopes are not equal. Therefore, it may not be assumed that 2 σ 12 = σ 2 . Normal Probability Plot for surface...bottom ML Estimates surface 99 bottom 95 90 Percent 80 70 60 50 40 30 20 10 5 1 4 5 6 7 8 Data b.) 1) The parameter of interest is the difference in mean HCB concentration, µ1 − µ 2 , with ∆0 = 0 2) H0 : µ1 − µ 2 = 0 or µ1 = µ 2 3) H1 : µ1 − µ 2 ≠ 0 or µ1 ≠ µ 2 4) α = 0.05 5) The test statistic is t0 = ( x1 − x2 ) − ∆ 0 2 s1 s2 +2 n1 n2 6) Reject the null hypothesis if t0 < − t 0.025,15 where − t 0.025,15 = −2.131 or t0 > t 0.025,15 where t 0.025,15 = 2.131 since 2 s12 s 2 + n1 n 2 ν= 2 1 s n1 2 n1 − 1 2 = 15.06 2 2 + s n2 n2 − 1 ν ≅ 15 (truncated) 10-53 7) x1 = 4.804 n1 = 10 s1 = 0.631 x2 = 5.839 n2 = 10 t0 = s2 = 1.014 (4.804 − 5.839) (0.631) 2 (1.014) 2 + 10 10 = −2.74 8) Since –2.74 < -2.131 reject the null hypothesis and conclude that the data support the claim that the mean HCB concentration is different at the two depths sampled at the 0.05 level of significance. b) P-value = 2P(t < -2.74), 2(0.005) < P-value < 2(0.01) 0.001 < P-value < 0.02 c) Assuming the sample sizes were equal: ∆ = 2 α = 0.05 a. n1 = n2 = 10 d= 2 =1 2(1) From Chart VI (e) we find β = 0.20, and then calculate Power = 1- β = 0.80 d.)Assuming the sample sizes were equal: ∆ = 2 α = 0.05 d= 2 = 0 .5 , 2(1) β = 0.0 From Chart VI (e) we find n*=50 and n= 50 + 1 = 25.5 , so 2 n=26 Mind-Expanding Exercises 10-90 The estimate of µ is given by X . Therefore, X = be: V (X ) = 2 1 σ 12 σ 2 σ2 + +3 4 n1 n2 n3 1 X1 + X 2 − X 3 . The variance of X can be shown to 2 ( ) . Using s1, s2, and s3 as estimates for σ1 σ2 and σ3 respectively. a) A 100(1-α)% confidence interval on µ is then: X − Zα / 2 2 2 2 2 s2 s2 1 s1 s 2 1 s1 s 2 + + 3 ≤ µ ≤ X + Zα / 2 + +3 4 n1 n 2 n3 4 n1 n 2 n3 b) A 95% confidence interval for µ is 2 2 2 2 2 2 1 (4.6 + 5.2) − 6.1 −1.96 1 0.7 + 0.6 + 0.8 ≤ µ ≤ 1 (4.6 + 5.2) − 6.1 +1.96 1 0.7 + 0.6 + 0.8 2 4 100 120 130 2 4 100 120 130 −1.2 − 0.163 ≤ µ ≤ −1.2 + 0.163 −1.363 ≤ µ ≤ −1.037 Since zero is not contained in this interval, and because the possible differences (-1.363, -1.037) are negative, we can conclude that there is sufficient evidence to indicate that pesticide three is more effective. 10-54 10-91 The V (X1 − X 2 ) = C1 n1 + C 2 n2 partial σ 12 n1 subject to derivatives of 2 σ2 + σ 12 n1 and suppose this is to equal a constant k. Then, we are to minimize n2 + 2 σ2 n2 = k . Using a Lagrange multiplier, we minimize by setting the f (n1 , n2 , λ ) = C1n1 + C2 n2 + λ σ 12 n1 + 2 σ2 n2 −k λ equal to zero. These equations are λσ 2 ∂ f (n1 , n2 , λ ) = C1 − 21 = 0 ∂n1 n1 λσ 2 ∂ f (n1 , n2 , λ ) = C2 − 2 2 = 0 ∂n2 n2 σ2 σ2 ∂ f (n1 , n2 , λ ) = 1 + 2 = k ∂λ n1 n2 Upon adding equations (1) and (2), we obtain (1) (2) (3) C1 + C 2 − λ σ 12 n1 Substituting from equation (3) enables us to solve for λ to obtain + 2 σ2 n2 =0 C1 + C 2 =λ k Then, equations (1) and (2) are solved for n1 and n2 to obtain n1 = σ 12 (C1 + C2 ) kC1 n2 = 2 σ 2 (C1 + C2 ) kC2 It can be verified that this is a minimum and that with these choices for n1 and n2. V (X1 − X 2 ) = σ 12 n1 + 2 σ2 n2 . 10-55 with respect to n1, n2 and 10-92 Maximizing the probability of rejecting H P − zα / 2 < x1 − x2 2 σ 12 σ 2 n1 + 0 is equivalent to minimizing < zα / 2 | µ1 − µ 2 = δ =P − zα / 2 − n2 δ 2 σ 12 σ 2 n1 + < Z < zα / 2 − n2 δ 2 2 σ1 σ 2 n1 where z is a standard normal random variable. This probability is minimized by maximizing + n2 δ 2 σ 12 σ 2 n1 Therefore, we are to minimize 2 2 σ1 σ 2 n1 + n2 f (n1 ) = σ 12 n1 + of f(n1) with respect to n1 and setting it equal to zero results in the equation n1 = σ1N σ1 + σ 2 and n2 = 2 σ2 N − n1 a) 2 − σ 12 σ2 + = 0. 2 ( N − n1 )2 n1 σ2N σ1 + σ 2 α = P( Z > z ε or Z < − z α −ε ) where Z has a standard normal distribution. Then, α = P( Z > zε ) + P( Z < − zα −ε ) = ε + α − ε = α b) β = P( − zα − ε < Z 0 < z ε | µ1 = µ 0 + δ) β = P( − z α − ε < n2 . Taking the derivative Also, it can be verified that the solution minimizes f(n1). 10-93 . subject to n1 + n2 = N. From the constraint, n2 = N − n1, and we are to minimize Upon solving for n1, we obtain + x−µ0 < z ε | µ 1 = µ 0 + δ) σ2 /n δ δ = P( − z α − ε − < Z < zε − ) σ2 /n σ2 /n δ δ = Φ( z ε − ) − Φ( − z α − ε − ) 2 /n σ σ2 /n 10-56 10-94. The requested result can be obtained from data in which the pairs are very different. Example: pair 1 2 3 4 5 sample 1 100 10 50 20 70 sample 2 110 20 59 31 80 x1 = 50 x2 = 60 s 1 = 36.74 s 2 = 36.54 s pooled = 36.64 Two-sample t-test : t 0 = −0.43 P-value = 0.68 xd = −10 10-95 s d = 0.707 Paired t-test : t 0 = −3162 . a.) θ= p1 p2 and θˆ = ˆ p1 ˆ p2 P-value ≈ 0 and ln(θˆ) ~ N [ln(θ ), (n1 − x1 ) / n1 x1 + (n2 − x 2 ) / n2 x 2 ] The (1-α) confidence Interval for ln(θ) will use the relationship ln(θˆ) − ln(θ ) Z= n1 − x1 n − x2 +2 n1 x1 n2 x2 ln(θˆ) − Z α 1/ 4 n1 − x1 n − x2 +2 n1 x1 n2 x 2 2 ≤ ln(θ ) ≤ ln(θˆ) + Z α 2 1/ 4 n1 − x1 n − x2 +2 n1 x1 n2 x2 b.) The (1-α) confidence Interval for θ use the CI developed in part (a.) where θ = e^( ln(θ)) θˆ − e Zα 2 n1 − x1 n −x +22 n1 x1 n2 x2 1/ 4 ≤ θ ≤ θˆ + e Zα 2 n1 − x1 n −x +22 n1 x1 n2 x2 1/ 4 c.) θˆ − e Zα 2 n1 − x1 n −x +22 n1 x1 n2 x 2 1.96 .25 100 − 27 100 −19 + 2700 1900 1.42 − e − 1.317 ≤ θ ≤ 4.157 ≤ θ ≤ θˆ + e Zα 2 n1 − x1 n −x +22 n1 x1 n2 x 2 1/ 4 1.96 ≤ θ ≤ 1.42 + e .25 100 − 27 100 −19 + 2700 1900 1/ 4 Since the confidence interval contains the value 1, we conclude that there is no difference in the proportions at the 95% level of significance 10-57 1/ 4 10-96 2 H 0 : σ 12 = σ 2 2 H1 : σ 12 ≠ σ 2 2 σ < S12 < f α2/ 2 ,n1 −1,n2 −1 | 12 = δ ≠ 1 1−α / 2 , n1 −1, n2 −1 S2 σ 2 β =P f 2 2 σ f σ 2 =P 2 1 where S12 / σ 12 S22 / σ 22 2 2 σ σ < S12 / σ12 < 22 f α / 2 ,n1 −1,n2 −1 | 12 = δ S2 / σ 2 σ σ2 1 2 2 1−α / 2 , n1 −1, n2 −1 2 has an F distribution with n1 − 1 and n2 − 1 degrees of freedom. 10-58 CHAPTER 11 Section 11-2 11-1. a) yi = β 0 + β 1 xi + ε i S xx = 157.42 − 43 = 25.348571 14 2 572 S xy = 1697.80 − 43(14 ) = −59.057143 β1 = S xy S xx = −59.057143 = −2.330 25.348571 β 0 = y − β1 x = 572 14 43 − ( −2.3298017)( 14 ) = 48.013 ˆ ˆˆ y = β 0 + β1 x ˆ y = 48.012962 − 2.3298017(4.3) = 37.99 ˆ c) y = 48.012962 − 2.3298017(3.7) = 39.39 ˆ d) e = y − y = 46.1 − 39.39 = 6.71 b) 11-2. a) yi = β 0 + β1 xi + ε i 2 S xx = 143215.8 − 1478 = 33991.6 20 S xy = 1083.67 − ˆ β1 = S xy S xx = (1478 )(12.75 ) 20 = 141.445 141.445 = 0.00416 33991.6 .75 ˆ β 0 = 1220 − (0.0041617512)( 1478 ) = 0.32999 20 ˆ y = 0.32999 + 0.00416 x SS E 0.143275 ˆ σ 2 = MS E = = = 0.00796 n−2 18 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 y -50 0 50 100 x ˆ y = 0.32999 + 0.00416(85) = 0.6836 ˆ c) y = 0.32999 + 0.00416(90) = 0.7044 ˆ d) β = 0.00416 b) 1 11-1 11-3. a) ˆ y = 0.3299892 + 0.0041612( 9 x + 32) 5 ˆ y = 0.3299892 + 0.0074902 x + 0.1331584 ˆ y = 0.4631476 + 0.0074902 x ˆ b) β = 0.00749 1 a) Regression Analysis - Linear model: Y = a+bX Dependent variable: Games Independent variable: Yards -------------------------------------------------------------------------------Standard T Prob. Parameter Estimate Error Value Level Intercept 21.7883 2.69623 8.081 .00000 Slope -7.0251E-3 1.25965E-3 -5.57703 .00001 -------------------------------------------------------------------------------Analysis of Variance Source Sum of Squares Df Mean Square F-Ratio Prob. Level Model 178.09231 1 178.09231 31.1032 .00001 Residual 148.87197 26 5.72585 -------------------------------------------------------------------------------Total (Corr.) 326.96429 27 Correlation Coefficient = -0.738027 R-squared = 54.47 percent Stnd. Error of Est. = 2.39287 ˆ σ2 = 5.7258 If the calculations were to be done by hand use Equations (11-7) and (11-8). Regression Plot y = 21.7883 - 0.0070251 x S = 2.39287 R-Sq = 54.5 % R-Sq(adj) = 52.7 % 10 y 11-4. 5 0 1500 2000 2500 3000 x b) ˆ y = 21.7883 − 0.0070251(1800) = 9.143 c) −0.0070251(-100) = 0.70251 games won. 1 = 142.35 yds decrease required. 0.0070251 ˆ e) y = 21.7883 − 0.0070251(1917) = 8.321 ˆ e= y−y d) = 10 − 8.321 = 1.679 11-2 a) Regression Analysis - Linear model: Y = a+bX Dependent variable: SalePrice Independent variable: Taxes -------------------------------------------------------------------------------Standard T Prob. Parameter Estimate Error Value Level Intercept 13.3202 2.57172 5.17948 .00003 Slope 3.32437 0.390276 8.518 .00000 -------------------------------------------------------------------------------Analysis of Variance Source Sum of Squares Df Mean Square F-Ratio Prob. Level Model 636.15569 1 636.15569 72.5563 .00000 Residual 192.89056 22 8.76775 -------------------------------------------------------------------------------Total (Corr.) 829.04625 23 Correlation Coefficient = 0.875976 R-squared = 76.73 percent Stnd. Error of Est. = 2.96104 ˆ σ2 = 8.76775 If the calculations were to be done by hand use Equations (11-7) and (11-8). ˆ y = 13.3202 + 3.32437 x ˆ b) y = 13.3202 + 3.32437(7.5) = 38.253 c) ˆ y = 13.3202 + 3.32437(5.8980) = 32.9273 ˆ y = 32.9273 ˆ e = y − y = 30.9 − 32.9273 = −2.0273 d) All the points would lie along the 45% axis line. That is, the regression model would estimate the values exactly. At this point, the graph of observed vs. predicted indicates that the simple linear regression model provides a reasonable fit to the data. Plot of Observed values versus predicted 50 45 Predicted 11-5. 40 35 30 25 25 30 35 40 45 Observed 11-3 50 11-6. a) Regression Analysis - Linear model: Y = a+bX Dependent variable: Usage Independent variable: Temperature -------------------------------------------------------------------------------Standard T Prob. Parameter Estimate Error Value Level Intercept -6.3355 1.66765 -3.79906 .00349 Slope 9.20836 0.0337744 272.643 .00000 -------------------------------------------------------------------------------Analysis of Variance Source Sum of Squares Df Mean Square F-Ratio Prob. Level Model 280583.12 1 280583.12 74334.4 .00000 Residual 37.746089 10 3.774609 -------------------------------------------------------------------------------Total (Corr.) 280620.87 11 Correlation Coefficient = 0.999933 R-squared = 99.99 percent Stnd. Error of Est. = 1.94284 ˆ σ2 = 3.7746 If the calculations were to be done by hand use Equations (11-7) and (11-8). ˆ y = −6.3355 + 9.20836 x ˆ b) y = −6.3355 + 9.20836(55) = 500.124 c) If monthly temperature increases by 1 F, y increases by 9.20836. d) ˆ y = −6.3355 + 9.20836(47) = 426.458 y = 426.458 ˆ e = y − y = 424.84 − 426.458 = −1.618 11-7. a) Predictor Constant x S = 3.660 Coef 33.535 -0.03540 StDev 2.614 0.01663 R-Sq = 20.1% Analysis of Variance Source DF SS Regression 1 60.69 Error 18 241.06 Total 19 301.75 T 12.83 -2.13 P 0.000 0.047 R-Sq(adj) = 15.7% MS 60.69 13.39 F 4.53 ˆ σ2 = 13.392 ˆ y = 33.5348 − 0.0353971x ˆ b) y = 33.5348 − 0.0353971(150) = 28.226 ˆ c) y = 29.4995 ˆ e = y − y = 31.0 − 29.4995 = 1.50048 11-4 P 0.047 11-8. a) y 60 50 40 850 950 1050 x Predictor Coef StDev T P Constant -16.509 9.843 -1.68 0.122 x 0.06936 0.01045 6.64 0.000 S = 2.706 R-Sq = 80.0% R-Sq(adj) = 78.2% Analysis of Variance Source DF SS MS F P Regression 1 322.50 322.50 44.03 0.000 Error 11 80.57 7.32 Total 12 403.08 ˆ σ2 = 7.3212 ˆ y = −16.5093 + 0.0693554 x ˆ ˆ b) y = 46.6041 e = y − y = 1.39592 ˆ c) y = −16.5093 + 0.0693554(950) = 49.38 a) 9 8 7 6 5 y 11-9. 4 3 2 1 0 60 70 80 90 100 x Yes, a linear regression would seem appropriate, but one or two points appear to be outliers. Predictor Constant x S = 1.318 Coef SE Coef T P -10.132 1.995 -5.08 0.000 0.17429 0.02383 7.31 0.000 R-Sq = 74.8% R-Sq(adj) = 73.4% Analysis of Variance Source DF Regression 1 Residual Error 18 Total 19 b) c) ˆ σ 2 = 1.737 SS 92.934 31.266 124.200 MS 92.934 1.737 ˆ y = −10.132 + 0.17429 x ˆ y = 4.68265 at x = 85 and 11-5 F 53.50 P 0.000 11-10. a) 250 y 200 150 100 0 10 20 30 40 x Yes, a linear regression model appears to be plausable. Predictor Coef StDev T P Constant 234.07 13.75 17.03 0.000 x -3.5086 0.4911 -7.14 0.000 S = 19.96 R-Sq = 87.9% R-Sq(adj) = 86.2% Analysis of Variance Source DF SS MS F P Regression 1 20329 20329 51.04 0.000 Error 7 2788 398 Total 8 23117 b) c) d) a) 40 30 y 11-11. ˆ ˆ σ2 = 398.25 and y = 234.071 − 3.50856 x ˆ y = 234.071 − 3.50856(30) = 128.814 ˆ y = 156.883 e = 15.1175 20 10 0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 x Yes, a simple linear regression model seems appropriate for these data. Predictor Constant x S = 3.716 Coef StDev T P 0.470 1.936 0.24 0.811 20.567 2.142 9.60 0.000 R-Sq = 85.2% R-Sq(adj) = 84.3% Analysis of Variance Source Regression Error Total DF 1 16 17 SS 1273.5 220.9 1494.5 MS 1273.5 13.8 F 92.22 11-6 P 0.000 b) ˆ σ2 = 13.81 ˆ y = 0.470467 + 20.5673 x c) d) ˆ y = 10.1371 e = 1.6629 a) y 2600 2100 1600 0 5 10 15 20 25 x Yes, a simple linear regression (straight-line) model seems plausible for this situation. Predictor Constant x S = 99.05 Coef 2625.39 -36.962 StDev 45.35 2.967 R-Sq = 89.6% Analysis of Variance Source DF SS Regression 1 1522819 Error 18 176602 Total 19 1699421 b) c) T 57.90 -12.46 P 0.000 0.000 R-Sq(adj) = 89.0% MS 1522819 9811 F 155.21 P 0.000 ˆ σ2 = 9811.2 ˆ y = 2625.39 − 36.962 x ˆ y = 2625.39 − 36.962(20) = 1886.15 d) If there were no error, the values would all lie along the 45 axis. The plot indicates age was reasonable regressor variable. 2600 2500 2400 2300 FITS1 11-12. ˆ y = 0.470467 + 20.5673(1) = 21.038 2200 2100 2000 1900 1800 1700 1600 2100 2600 y 11-7 11-13. 11-14. ˆ ˆ ˆ ˆ β 0 + β1 x = ( y − β1 x ) + β1 x = y a) The slopes of both regression models will be the same, but the intercept will be shifted. ˆ b) y = 2132.41 − 36.9618 x ˆ β 0 = 2625.39 vs. ˆ β 1 = −36.9618 11-15. Let ˆ β 0 ∗ = 2132.41 ˆ β 1 ∗ = −36.9618 ∗ ∗ x i = x i − x . Then, the model is Yi ∗ = β 0 + β1∗ xi∗ + ε i . n xi∗ = Equations 11-7 and 11-8 can be applied to the new variables using the facts that i =1 ˆ ˆ β 1∗ = β 1 11-16. ( yi − βxi ) (− xi ) = 2[ Therefore, ˆ β= yi xi xi ˆ y = 21.031461x . i =1 ( yi − βxi ) 2 . Upon setting the derivative equal to zero, we obtain 2 yi xi − β 2 xi ] = 0 . The model seems very appropriate - an even better fit. 45 40 35 chloride 30 25 20 15 10 5 0 0 0.2 yi∗ = 0 . Then, ˆ β 0∗ = 0 . The least squares estimate minimizes 2 11-17. and n 0.4 0.6 0.8 1 1.2 watershed 11-8 1.4 1.6 1.8 2 Section 11-5 11-18. a) 1) The parameter of interest is the regressor variable coefficient, β1 2) H 0 : β 1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.05 5) The test statistic is f0 = MS R SS R / 1 = MS E SS E /(n − 2) 6) Reject H0 if f0 > fα,1,12 where f0.05,1,12 = 4.75 7) Using results from Exercise 11-1 ˆ SS R = β1S xy = −2.3298017(−59.057143) = 137.59 SS E = S yy − SS R = 159.71429 − 137.59143 = 22.123 137.59 f0 = = 74.63 22.123 / 12 8) Since 74.63 > 4.75 reject H 0 and conclude that compressive strength is significant in predicting intrinsic permeability of concrete at α = 0.05. We can therefore conclude model specifies a useful linear relationship between these two variables. P − value ≅ 0.000002 b) ˆ σ2 = MS E = c) se ( β ) = ˆ 0 11-19. SS E 22.123 = = 1.8436 n−2 12 ˆ σ2 1 x2 + n S xx = 1 . 8436 ˆ and se ( β ) = 1 1 3 . 0714 2 + 14 25 . 3486 ˆ σ2 S xx = 1 . 8436 = 0 . 2696 25 . 3486 = 0 . 9043 a) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 : β1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.05 5) The test statistic is f0 = MS R SS R / 1 = MS E SS E /( n − 2 ) 6) Reject H0 if f0 > fα,1,18 where f0.05,1,18 = 4.414 7) Using the results from Exercise 11-2 SS R = β1S xy = ( 0.0041612)(141.445) = 0.5886 SS E = S yy − SS R 2 .75 = (8.86 − 1220 ) − 0.5886 = 0.143275 f0 = 0.5886 = 73.95 0.143275 / 18 8) Since 73.95 > 4.414, reject H 0 and conclude the model specifies a useful relationship at α = 0.05. 11-9 P − value ≅ 0.000001 b) ˆ se( β 1 ) = ˆ σ2 S xx = .00796 = 4.8391x10 − 4 33991.6 x 1 ˆ ˆ se( β 0 ) = σ 2 + n S xx 11-20. = .00796 1 73.9 2 + = 0.04091 20 33991.6 a) Refer to ANOVA table of Exercise 11-4. 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 : β1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.01 5) The test statistic is MS R SS R / 1 = MS E SS E /(n − 2) f0 = 6) Reject H0 if f0 > fα,1,26 where f0.01,1,26 = 7.721 7) Using the results of Exercise 10-4 MS R = 31.1032 MS E f0 = 8) Since 31.1032 > 7.721 reject H 0 and conclude the model is useful at α = 0.01. P − value = 0.000007 ˆ b) se ( β ) = 1 ˆ se ( β 0 ) = ˆ σ2 S xx ˆ σ2 = 5 . 7257 = . 001259 3608611 . 43 1 x + n S xx = 5 .7257 1 2110 .13 2 + 28 3608611 .43 = 2 .6962 c) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 : β1 = − 0 .01 3) H 1 : β1 ≠ −0 .01 4) α = 0.01 ˆ 5) The test statistic is t = β1 + .01 0 ˆ se( β1 ) 6) Reject H0 if t0 < −tα/2,n-2 where −t0.005,26 = −2.78 or t0 > t0.005,26 = 2.78 7) Using the results from Exercise 10-4 t0 = − 0.0070251 + .01 = 2.3618 0.00125965 8) Since 2.3618 < 2.78 do not reject H 0 and conclude the intercept is not zero at α = 0.01. 11-10 11-21. Refer to ANOVA of Exercise 11-5 a) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 : β1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.05, using t-test 5) The test statistic is t 0 = β1 se(β1) 6) Reject H0 if t0 < −tα/2,n-2 where −t0.025,22 = −2.074 or t0 > t0.025,22 = 2.074 7) Using the results from Exercise 11-5 3.32437 = 8.518 0.390276 8) Since 8.518 > 2.074 reject H 0 and conclude the model is useful α = 0.05. t0 = b) 1) The parameter of interest is the slope, β1 2) H 0 : β 1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.05 5) The test statistic is f0 = MS R SS R / 1 = MS E SS E / ( n − 2) 6) Reject H0 if f0 > fα,1,22 where f0.01,1,22 = 4.303 7) Using the results from Exercise 10-5 636.15569 / 1 = 72.5563 f0 = 192.89056 / 22 8) Since 72.5563 > 4.303, reject H 0 and conclude the model is useful at a significance α = 0.05. The F-statistic is the square of the t-statistic. The F-test is a restricted to a two-sided test, whereas the t-test could be used for one-sided alternative hypotheses. c) se ( β ) = ˆ 1 ˆ se ( β 0 ) = ˆ σ2 = S xx ˆ σ2 8 . 7675 = . 39027 57 . 5631 1 x + n S xx = 8 . 7675 1 6 . 4049 2 + 24 57 . 5631 = 2 . 5717 d) 1) The parameter of interest is the intercept, β0. 2) H 0 : β0 = 0 3) H 1 : β0 ≠ 0 4) α = 0.05, using t-test 5) The test statistic is t0 = ˆ β0 ˆ se( β0 ) 6) Reject H0 if t0 < −tα/2,n-2 where −t0.025,22 = −2.074 or t0 > t0.025,22 = 2.074 7) Using the results from Exercise 11-5 t0 = 13.3201 = 5.179 2.5717 8) Since 5.179 > 2.074 reject H 0 and conclude the intercept is not zero at α = 0.05. 11-11 11-22. Refer to ANOVA for Exercise 10-6 a) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 : β1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.01 f0 = 5) The test statistic is MS R SS R / 1 = MS E SS E /(n − 2) 6) Reject H0 if f0 > fα,1,22 where f0.01,1,10 = 10.049 7) Using the results from Exercise 10-6 f0 = 280583.12 / 1 = 74334.4 37.746089 / 10 8) Since 74334.4 > 10.049, reject H 0 and conclude the model is useful α = 0.01. P-value < 0.000001 ˆ ˆ b) se( β1 ) = 0.0337744, se( β 0 ) = 1.66765 c) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 : β1 = 10 3) H 1: β1 ≠ 10 4) α = 0.01 5) The test statistic is t0 = ˆ β 1 − β 1, 0 ˆ se( β 1 ) 6) Reject H0 if t0 < −tα/2,n-2 where −t0.005,10 = −3.17 or t0 > t0.005,10 = 3.17 7) Using the results from Exercise 10-6 t0 = 9.21 − 10 = −23.37 0.0338 8) Since −23.37 < −3.17 reject H 0 and conclude the slope is not 10 at α = 0.01. P-value = 0. d) H0: β0 = 0 H1: β 0 ≠ 0 t0 = − 6.3355 − 0 = − 3 .8 1.66765 P-value < 0.005; Reject H0 and conclude that the intercept should be included in the model. 11-23. Refer to ANOVA table of Exercise 11-7 a) H 0 : β1 = 0 H 1 : β1 ≠ 0 α = 0.01 f 0 = 4.53158 f 0.01,1,18 = 8.285 f 0 > f α ,1,18 / Therefore, do not reject H0. P-value = 0.04734. Insufficient evidence to conclude that the model is a useful relationship. b) ˆ se( β1 ) = 0.0166281 ˆ se( β ) = 2.61396 0 11-12 c) H 0 : β1 = −0.05 H 1 : β1 < −0.05 α = 0.01 − 0.0354 − ( −0.05 ) = 0.87803 0.0166281 t .01,18 = 2.552 t0 = t 0 < −t α ,18 / Therefore, do not rejectH0. P-value = 0.804251. Insufficient evidence to conclude that β1 is ≥ -0.05. d) H0 : β 0 = 0 H 1 : β0 ≠ 0 α = 0.01 t 0 = 12.8291 t .005,18 = 2.878 t 0 > tα / 2,18 Therefore, reject H0. P-value ≅ 0 11-24. Refer to ANOVA of Exercise 11-8 a) H0 : β1 = 0 H1 : β1 ≠ 0 α = 0.05 f 0 = 44.0279 f .05 ,1,11 = 4.84 f 0 > f α ,1,11 Therefore, rejectH0. P-value = 0.00004. b) ˆ se( β1 ) = 0.0104524 ˆ se( β ) = 9.84346 0 c) H 0 : β0 = 0 H1 : β0 ≠ 0 α = 0.05 t0 = −1.67718 t.025 ,11 = 2.201 | t0 |< −tα / 2 ,11 / Therefore, do not rejectH0. P-value = 0.12166. 11-13 11-25. Refer to ANOVA of Exercise 11-9 a) H 0 : β1 = 0 H 1 : β1 ≠ 0 α = 0.05 f 0 = 53.50 f .05,1,18 = 4.414 f 0 > f α ,1,18 Therefore, reject H0. P-value = 0.000009. b) ˆ se( β1 ) = 0.0256613 ˆ se( β ) = 2.13526 0 c) H 0 : β0 = 0 H1 : β0 ≠ 0 α = 0.05 t 0 = - 5.079 t.025,18 = 2.101 | t 0 |> tα / 2,18 Therefore, reject H0. P-value = 0.000078. 11-26. Refer to ANOVA of Exercise 11-11 a) H 0 : β1 = 0 H 1 : β1 ≠ 0 α = 0.01 f 0 = 92.224 f .01,1,16 = 8.531 f 0 > f α ,1,16 Therefore, reject H 0 . b) P-value < 0.00001 c) ˆ se( β1 ) = 2.14169 ˆ se( β ) = 1.93591 0 d) H 0 : β0 = 0 H1 : β0 ≠ 0 α = 0.01 t0 = 0.243 t.005 ,16 = 2.921 t0 > tα / 2 ,16 / Therefore, do not rejectH0. Conclude, Yes, the intercept should be removed. 11-14 11-27. Refer to ANOVA of Exercise 11-12 a) H 0 : β1 = 0 H 1 : β1 ≠ 0 α = 0.01 f 0 = 155.2 f .01,1,18 = 8.285 f 0 > f α ,1,18 Therefore, reject H0. P-value < 0.00001. b) ˆ se( β1 ) = 45.3468 ˆ se( β ) = 2.96681 0 c) H 0 : β1 = −30 H 1 : β1 ≠ −30 α = 0.01 − 36.9618 − ( −30 ) = −2.3466 2.96681 t .005 ,18 = 2.878 t0 = | t 0 |> −t α / 2 ,18 / Therefore, do not reject H0. P-value = 0.0153(2) = 0.0306. d) H 0 : β0 = 0 H1 : β0 ≠ 0 α = 0.01 t0 = 57.8957 t.005 ,18 = 2.878 t 0 > t α / 2,18 , therefore, reject H0. P-value < 0.00001. e) H0 : β 0 = 2500 H1 : β0 > 2500 α = 0.01 2625.39 − 2500 = 2.7651 45.3468 t.01,18 = 2.552 t0 = t0 > tα ,18 , therefore reject H 0 . P-value = 0.0064. 11-15 11-28. β1 t0 = After the transformation σ 2 / S xx ˆ ˆ σ ∗ = bσ . Therefore, t0∗ = 11-29. ˆ β ˆ σ a) ˆ β 1∗ = bˆ ∗ ˆ ˆ β 1 , S xx = a 2 S xx , x ∗ = ax , β 0∗ = bβ 0 , and a ˆ b β1 / a ˆ (bσ ) 2 / a 2 S xx = t0 . has a t distribution with n-1 degree of freedom. 2 xi2 ˆ ˆ β = 21.031461,σ = 3.611768, and xi2 = 14.7073 . The t-statistic in part a. is 22.3314 and H 0 : β 0 = 0 is rejected at usual α values. b) From Exercise 11-17, 11-30. d= | −0.01 − (−0.005) | 2.4 27 3608611.96 = 0.76 , S xx = 3608611.96 . Assume α = 0.05, from Chart VI and interpolating between the curves for n = 20 and n = 30, β ≅ 0.05 . Sections 11-6 and 11-7 11-31. tα/2,n-2 = t0.025,12 = 2.179 a) 95% confidence interval on β1 . ˆ ˆ se ( β ) β ±t α / 2,n− 2 1 1 − 2 .3298 ± t .025 ,12 ( 0 .2696 ) − 2 .3298 ± 2 .179 ( 0 .2696 ) − 2 .9173 . ≤ β 1 ≤ −1 .7423 . b) 95% confidence interval on β0 . ˆ ˆ β ±t se ( β ) 0 .025 ,12 0 48 . 0130 ± 2 . 179 ( 0 . 5959 ) 46 . 7145 ≤ β 0 ≤ 49 .3115 . c) 95% confidence interval on µ when x 0 = 2.5 . ˆ µ Y | x 0 = 48 .0130 − 2 .3298 ( 2 .5) = 42 .1885 ˆ ˆn µ Y | x ± t .025 ,12 σ 2 ( 1 + 0 ( x0 − x ) 2 S xx ) 1 42 .1885 ± ( 2 .179 ) 1 .844 ( 14 + ( 2 .5 − 3 .0714 ) 2 25 . 3486 ) 42 .1885 ± 2.179 ( 0 .3943 ) ˆ 41 .3293 ≤ µ Y | x 0 ≤ 43 .0477 d) 95% on prediction interval when x 0 = 2.5 . ˆ ˆ y 0 ± t .025,12 σ 2 (1 + 1 + n ( x0 − x ) 2 S xx ) 1 42.1885 ± 2.179 1.844 (1 + 14 + ( 2.5 − 3.0714 ) 2 25.348571 ) 42.1885 ± 2.179 (1.4056 ) 39.1257 ≤ y 0 ≤ 45.2513 It is wider because it depends on both the error associated with the fitted model as well as that with the future observation. 11-16 11-32. tα/2,n-2 = t0.005,18 = 2.878 ˆ a) β 1 ˆ ± (t 0.005,18 )se( β 1 ) . 0.0041612 ± (2.878)(0.000484) 0.0027682 ≤ β1 ≤ 0.0055542 ( ) ˆ ˆ b) β 0 ± t 0.005,18 se( β 0 ) . 0.3299892 ± (2.878)(0.04095) 0.212135 ≤ β 0 ≤ 0.447843 c) 99% confidence interval on µ when x 0 = 85 F . ˆ µ Y | x 0 = 0 .683689 ˆ ˆn µ Y | x ± t.005 ,18 σ 2 ( 1 + 0 ( x0 − x ) 2 S xx ) 1 0 .683689 ± ( 2 .878 ) 0 .00796 ( 20 + ( 85 − 73 . 9 ) 2 33991 .6 ) 0 .683689 ± 0 .0594607 ˆ 0 .6242283 ≤ µ Y | x 0 ≤ 0 .7431497 d) 99% prediction interval when x0 = 90 F . ˆ y0 = 0.7044949 2 (x −x) ˆ ˆ y0 ± t.005,18 σ 2 (1 + 1 + 0S xx ) n 2 1 0.7044949 ± 2.878 0.00796(1 + 20 + (90 − 73..96) ) 33991 0.7044949 ± 0.2640665 0.4404284 ≤ y0 ≤ 0.9685614 Note for Problems 11-33 through 11-35: These computer printouts were obtained from Statgraphics. For Minitab users, the standard errors are obtained from the Regression subroutine. 11-33. 95 percent confidence intervals for coefficient estimates -------------------------------------------------------------------------------Estimate Standard error Lower Limit Upper Limit CONSTANT 21.7883 2.69623 16.2448 27.3318 Yards -0.00703 0.00126 -0.00961 -0.00444 -------------------------------------------------------------------------------- a) −0.00961 ≤ β1 ≤ −0.00444. b) 16.2448 ≤ β0 ≤ 27.3318. c) d) 1 9.143 ± (2.056) 5.72585( 28 + 9.143 ± 1.2287 ˆ 7.9143 ≤ µ Y | x0 ≤ 10.3717 (1800 − 2110.14 ) 2 3608325 .5 1 9.143 ± (2.056) 5.72585(1 + 28 + 9.143 ± 5.0709 ) (1800 − 2110.14 ) 2 3608325.5 4.0721 ≤ y0 ≤ 14.2139 11-17 ) 11-34. 95 percent confidence intervals for coefficient estimates -------------------------------------------------------------------------------Estimate Standard error Lower Limit Upper Limit CONSTANT 13.3202 2.57172 7.98547 18.6549 Taxes 3.32437 0.39028 2.51479 4.13395 -------------------------------------------------------------------------------- a) 2.51479 ≤ β1 ≤ 4.13395. b) 7.98547 ≤ β0 ≤ 18.6549. c) 1 38.253 ± (2.074) 8.76775( 24 + 38.253 ± 1.5353 ˆ 36.7177 ≤ µY | x0 ≤ 39.7883 d) 38.253 ± (2.074) 8.76775(1 + 1 24 + ( 7.5 − 6.40492 ) 2 57.563139 ) ( 7.5− 6.40492 ) 2 ) 57.563139 38.253 ± 6.3302 31.9228 ≤ y0 ≤ 44.5832 11-35. 99 percent confidence intervals for coefficient estimates -------------------------------------------------------------------------------Estimate Standard error Lower Limit Upper Limit CONSTANT -6.33550 1.66765 -11.6219 -1.05011 Temperature 9.20836 0.03377 9.10130 9.93154 -------------------------------------------------------------------------------- a) 9.10130 ≤ β1 ≤ 9.31543 b) −11.6219 ≤ β0 ≤ −1.04911 1 c) 500124 ± (2.228) 3.774609( 12 + . (55−46.5)2 ) 3308.9994 500.124 ± 1.4037586 498.72024 ≤ µ Y|x 0 ≤ 50152776 . 1 d) 500.124 ± (2.228) 3.774609(1 + 12 + (55− 46.5)2 ) 3308.9994 500.124 ± 4.5505644 495.57344 ≤ y0 ≤ 504.67456 It is wider because the prediction interval includes error for both the fitted model and from that associated with the future observation. 11-36. a) − 0 . 07034 ≤ β 1 ≤ − 0 . 00045 b) 28 . 0417 ≤ β 0 ≤ 39 . 027 1 c) 28 . 225 ± ( 2 . 101 ) 13 . 39232 ( 20 + ( 150 − 149 . 3 ) 2 48436 . 256 ) 28 .225 ± 1 .7194236 26 .5406 ≤ µ y | x 0 ≤ 29 .9794 11-18 d) 28 . 225 ± ( 2 . 101 ) 13 . 39232 (1 + 1 20 + ( 150 − 149 . 3 ) 2 48436 . 256 28 . 225 ± 7 . 87863 20 . 3814 ≤ y 0 ≤ 36 . 1386 11-37. a) 0 . 03689 ≤ β 1 ≤ 0 . 10183 b) − 47 .0877 ≤ β 0 ≤ 14 .0691 1 c) 46 . 6041 ± ( 3 . 106 ) 7 . 324951 ( 13 + ( 910 − 939 ) 2 67045 . 97 ) 46 .6041 ± 2 .514401 44 .0897 ≤ µ y | x 0 ≤ 49 .1185 ) d) 46 . 6041 ± ( 3 . 106 ) 7 . 324951 (1 + 46 . 6041 ± 8 . 779266 1 13 + ( 910 − 939 ) 2 67045 . 97 ) 37 . 8298 ≤ y 0 ≤ 55 . 3784 11-38. a) 0 . 11756 ≤ β 1 ≤ 0 . 22541 b) − 14 . 3002 ≤ β 0 ≤ − 5 . 32598 1 c) 4.76301 ± ( 2.101) 1.982231( 20 + ( 85 − 82 .3 ) 2 3010 .2111 ) 4.76301 ± 0.6772655 4.0857 ≤ µ y| x0 ≤ 5.4403 d) 4 .76301 ± ( 2 .101 ) 1 . 982231 (1 + 1 20 + ( 85 − 82 . 3 ) 2 3010 . 2111 ) 4 .76301 ± 3 .0345765 1 .7284 ≤ y 0 ≤ 7 .7976 11-39. a) 201.552≤ β1 ≤ 266.590 b) − 4.67015≤ β0 ≤ −2.34696 ( 30 − 24 . 5 ) 2 1651 . 4214 ) (1− 0 .806111 ) 2 3 .01062 ) c) 128 . 814 ± ( 2 . 365 ) 398 . 2804 ( 1 + 9 128 .814 ± 16 .980124 111 .8339 ≤ µ y | x 0 ≤ 145 .7941 11-40. a) 14 . 3107 ≤ β 1 ≤ 26 . 8239 b) − 5 . 18501 ≤ β 0 ≤ 6 . 12594 1 c) 21 .038 ± ( 2 .921 ) 13 .8092 ( 18 + 21 .038 ± 2 .8314277 18 .2066 ≤ µ y | x 0 ≤ 23 .8694 1 d) 21 .038 ± ( 2 .921) 13 .8092 (1 + 18 + (1− 0 .806111 ) 2 3.01062 ) 21 .038 ± 11 .217861 9 .8201 ≤ y 0 ≤ 32 .2559 11-41. a) − 43 .1964 ≤ β 1 ≤ − 30 .7272 b) 2530 .09 ≤ β 0 ≤ 2720 . 68 1 c) 1886 .154 ± ( 2 . 101 ) 9811 . 21( 20 + ( 20 −13 .3375 ) 2 1114 .6618 ) 1886 . 154 ± 62 . 370688 1823 . 7833 ≤ µ y | x 0 ≤ 1948 . 5247 11-19 ) d) 1886 . 154 ± ( 2 . 101 ) 9811 . 21 (1 + 1886 . 154 ± 217 . 25275 1 20 ( 20 −13 . 3375 ) 2 1114 . 6618 + ) 1668 . 9013 ≤ y 0 ≤ 2103 . 4067 Section 11-7 Use the results of Exercise 11-4 to answer the following questions. = 0.544684 ; The proportion of variability explained by the model. 148.87197 / 26 2 RAdj = 1 − = 1 − 0.473 = 0.527 326.96429 / 27 a) R 2 b) Yes, normality seems to be satisfied since the data appear to fall along the straight line. N o r m a l P r o b a b ility P lo t 99.9 99 95 80 cu m u l. p ercen t 50 20 5 1 0.1 -3 . 9 -1 . 9 0.1 2.1 4.1 6.1 R esid u als c) Since the residuals plots appear to be random, the plots do not include any serious model inadequacies. Residuals vs. Predicted Values Regression of Games on Yards 6.1 4.1 4.1 Residuals 6.1 Residuals 11-42. 2.1 0.1 2.1 0.1 -1.9 -1.9 -3.9 -3.9 1400 1700 2000 2300 2600 0 2900 2 4 6 Predicted Values Yards 11-20 8 10 12 11-43. Use the Results of exercise 11-5 to answer the following questions. a) SalePrice Taxes Predicted Residuals 25.9 29.5 27.9 25.9 29.9 29.9 30.9 28.9 35.9 31.5 31.0 30.9 30.0 36.9 41.9 40.5 43.9 37.5 37.9 44.5 37.9 38.9 36.9 45.8 4.9176 5.0208 4.5429 4.5573 5.0597 3.8910 5.8980 5.6039 5.8282 5.3003 6.2712 5.9592 5.0500 8.2464 6.6969 7.7841 9.0384 5.9894 7.5422 8.7951 6.0831 8.3607 8.1400 9.1416 29.6681073 30.0111824 28.4224654 28.4703363 30.1405004 26.2553078 32.9273208 31.9496232 32.6952797 30.9403441 34.1679762 33.1307723 30.1082540 40.7342742 35.5831610 39.1974174 43.3671762 33.2311683 38.3932520 42.5583567 33.5426619 41.1142499 40.3805611 43.7102513 -3.76810726 -0.51118237 -0.52246536 -2.57033630 -0.24050041 3.64469225 -2.02732082 -3.04962324 3.20472030 0.55965587 -3.16797616 -2.23077234 -0.10825401 -3.83427422 6.31683901 1.30258260 0.53282376 4.26883165 -0.49325200 1.94164328 4.35733807 -2.21424985 -3.48056112 2.08974865 b) Assumption of normality does not seem to be violated since the data appear to fall along a straight line. Normal Probability Plot 99.9 cumulative percent 99 95 80 50 20 5 1 0.1 -4 -2 0 2 4 6 8 Residuals c) There are no serious departures from the assumption of constant variance. This is evident by the random pattern of the residuals. Plot of Residuals versus Predicted Plot of Residuals versus Taxes 6 4 4 Residuals 8 6 Residuals 8 2 2 0 0 -2 -2 -4 -4 26 29 32 35 38 41 44 3.8 Predicted Values d) 11-44. 4.8 5.8 6.8 Taxes R 2 ≡ 76.73% ; Use the results of Exercise 11-6 to answer the following questions 11-21 7.8 8.8 9.8 a) R 2 = 99.986% ; The proportion of variability explained by the model. b) Yes, normality seems to be satisfied since the data appear to fall along the straight line. Normal Probability Plot 99.9 99 cumulative percent 95 80 50 20 5 1 0.1 -2.6 -0.6 1.4 3.4 5.4 Residuals c) There might be lower variance at the middle settings of x. However, this data does not indicate a serious departure from the assumptions. Plot of Residuals versus Temperature Plot of Residuals versus Predicted 3.4 Residuals 5.4 3.4 Residuals 5.4 1.4 1.4 -0.6 -0.6 -2.6 -2.6 180 280 380 480 580 21 680 31 51 61 71 81 a) R = 20.1121% b) These plots indicate presence of outliers, but no real problem with assumptions. 2 Residuals Vers x us Resi dual sVersus the Fi tted Val ues (response is y) (resp nse is y) o 10 10 Residual 11-45. 41 Temperature Predicted Values l a u0 d i s e R 0 -10 -10 100 200 300 2 3 x 24 25 26 27 Fitte V lu dae 11-22 28 29 30 31 c) The normality assumption appears marginal. Normal Probability Plot of the Residuals (response is y) Residual 10 0 -10 -2 -1 0 1 2 Normal Score a) y 60 50 40 850 950 1050 x b) ˆ y = 0.677559 + 0.0521753 x H0 : β1 = 0 H 1 : β1 ≠ 0 α = 0.05 f 0 = 7.9384 f.05,1,12 = 4.75 f 0 > fα ,1,12 Reject Ho. ˆ σ 2 = 25.23842 ˆ2 d) σ orig = 7.324951 c) The new estimate is larger because the new point added additional variance not accounted for by the model. e) Yes, e14 is especially large compared to the other residuals. f) The one added point is an outlier and the normality assumption is not as valid with the point included. Normal Probability Plot of the Residuals (response is y) 10 Residual 11-46. 0 -10 -2 -1 0 Normal Score 11-23 1 2 g) Constant variance assumption appears valid except for the added point. Residuals Versus the Fitted Values Residuals Versus x (response is y) (response is y) 10 Residual Residual 10 0 0 -10 -10 850 950 45 1050 50 11-47. 55 Fitted Value x a) R 2 = 71.27% b) No major departure from normality assumptions. Normal Probability Plot of the Residuals (response is y) 3 Residual 2 1 0 -1 -2 -2 -1 0 1 2 Normal Score c) Assumption of constant variance appears reasonable. Residuals Versus x Residuals Versus the Fitted Values (response is y) (response is y) 3 3 2 1 Residual Residual 2 0 -1 1 0 -1 -2 -2 60 70 80 90 100 0 1 2 3 x 5 6 7 a) R = 0.879397 b) No departures from constant variance are noted. 2 Residuals Versus x Residuals Versus the Fitted Values (response is y) (response is y) 30 20 20 10 10 Residual 30 Residual 11-48. 4 Fitted Value 0 0 -10 -10 -20 -20 -30 -30 0 10 20 30 40 80 x 130 180 Fitted Value 11-24 230 8 c) Normality assumption appears reasonable. Normal Probability Plot of the Residuals (response is y) 30 20 Residual 10 0 -10 -20 -30 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Normal Score a) R 2 = 85 . 22 % b) Assumptions appear reasonable, but there is a suggestion that variability increases slightly with 11-49. Residuals Versus x Residuals Versus the Fitted Values (response is y) (response is y) 5 Residual 5 0 0 -5 -5 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 0 10 20 x 30 40 Fitted Value c) Normality assumption may be questionable. There is some “ bending” away from a straight line in the tails of the normal probability plot. Normal Probability Plot of the Residuals (response is y) 5 Residual Residual y. 0 -5 -2 -1 0 Normal Score 11-25 1 2 11-50. a) R 2 = 0 . 896081 89% of the variability is explained by the model. b) Yes, the two points with residuals much larger in magnitude than the others. Normal Probability Plot of the Residuals (response is y) 2 Normal Score 1 0 -1 -2 -200 -100 0 100 Residual 2 c) Rnew model = 0.9573 Larger, because the model is better able to account for the variability in the data with these two outlying data points removed. d) σ ol d ˆ2 ˆ σ 2 new model model = 9811 . 21 = 4022 . 93 Yes, reduced more than 50%, because the two removed points accounted for a large amount of the error. 11-51. Using R 2 = 1 − SSyyE , S F0 = ( n − 2)(1 − SSyyE ) S = SS E S yy S yy − SS E SS E n− 2 = S yy − SS E ˆ σ2 Also, SS E = ˆ ˆ ( y i − β 0 − β 1 xi ) 2 = ˆ ( yi − y − β1 ( xi − x )) 2 = ˆ2 ( y i − y ) + β1 = ˆ ( y i − y ) − β1 ˆ2 S yy − SS E = β1 Therefore, F0 = 2 ( xi − x ) ˆ β 12 ˆ σ / S xx 2 ˆ ( xi − x ) 2 − 2 β 1 2 ( xi − x ) ( yi − y )(xi − x ) 2 2 2 = t0 Because the square of a t random variable with n-2 degrees of freedom is an F random variable with 1 and n-2 degrees of freedom, the usually t-test that compares | t0 | to tα / 2, n − 2 is equivalent to comparing fα ,1, n − 2 = tα / 2, n − 2 . 11-26 2 f 0 = t0 to 11-52. a) f = 0 . 9 ( 23 ) = 207 . Reject H 0 : β 1 = 0 . 0 1 − 0 .9 b) Because f . 05 ,1 , 23 = 4 . 28 , H0 2 is rejected if 23 R > 4 . 28 . 2 1− R That is, H0 is rejected if 23 R 2 > 4 .28 (1 − R 2 ) 27 .28 R 2 > 4 .28 R 2 > 0 .157 11-53. Yes, the larger residuals are easier to identify. 1.10269 -0.75866 -0.14376 0.66992 -2.49758 -2.25949 0.50867 0.46158 0.10242 0.61161 0.21046 -0.94548 0.87051 0.74766 -0.50425 0.97781 0.11467 0.38479 1.13530 -0.82398 11-54. For two random variables X1 and X2, V ( X 1 + X 2 ) = V ( X 1 ) + V ( X 2 ) + 2 Cov ( X 1 , X 2 ) Then, ˆ ˆ ˆ V (Yi − Yi ) = V (Yi ) + V (Yi ) − 2Cov (Yi , Yi ) ˆ ˆ = σ 2 + V ( β 0 + β 1 xi ) − 2σ 2 =σ 2 +σ 2 [ [+ 1 n = σ 2 1− (1 + n − 2σ )] ( xi − x ) S xx ( xi − x ) 2 S xx 2 [+ [+ 1 n 21 n ( xi − x ) 2 S xx ( xi − x ) S xx 2 a) Because ei is divided by an estimate of its standard error (when σ2 is estimated by σ2 ), ri has approximate unit variance. b) No, the term in brackets in the denominator is necessary. c) If xi is near x and n is reasonably large, ri is approximately equal to the standardized residual. d) If xi is far from x , the standard error of ei is small. Consequently, extreme points are better fit by least squares regression than points near the middle range of x. Because the studentized residual at any point has variance of approximately one, the studentized residuals can be used to compare the fit of points to the regression line over the range of x. Section 11-9 11-55. a) y = −0 .0280411 + 0 .990987 x ˆ b) H 0 : β 1 = 0 H 1 : β1 ≠ 0 f 0 = 79 . 838 α = 0.05 f .05 ,1 ,18 = 4 . 41 f 0 >> f α ,1 ,18 Reject H0 . c) r = 0 . 816 = 0 . 903 11-27 d) H 0 : ρ = 0 H1 : ρ ≠ 0 t0 = α = 0.05 R n−2 1− R t . 025 ,18 = 2 . 101 2 = 0 . 90334 18 = 8 . 9345 1 − 0 . 816 t 0 > t α / 2 ,18 Reject H0 . e) H 0 : ρ = 0 . 5 α = 0.05 H 1 : ρ ≠ 0 .5 z 0 = 3 . 879 z .025 = 1 . 96 z 0 > zα / 2 Reject H0 . f) tanh(arcta nh 0.90334 - z.025 17 ) ≤ ρ ≤ tanh(arcta nh 0.90334 + z.025 17 . ) where z.025 = 196 . 0 . 7677 ≤ ρ ≤ 0 . 9615 . 11-56. a) y = 69 .1044 + 0 . 419415 x ˆ b) H 0 : β 1 = 0 H 1 : β1 ≠ 0 α = 0.05 f 0 = 35 . 744 f .05 ,1, 24 = 4 .260 f 0 > f α ,1, 24 Reject H0 . c) r = 0.77349 d) H 0 : ρ = 0 H1 : ρ ≠ 0 t0 = 0 .77349 α = 0.05 24 1− 0 .5983 = 5 .9787 t .025 , 24 = 2 .064 t 0 > t α / 2 , 24 Reject H0 . e) H 0 : ρ = 0 . 6 α = 0.05 H 1 : ρ ≠ 0 .6 z 0 = (arctanh 0 .77349 − arctanh 0 .6 )( 23 ) 1 / 2 = 1 .6105 z .025 = 1 .96 z 0 > zα / 2 / Do not reject H0 . f) tanh(arcta nh 0.77349 - z ) ≤ ρ ≤ tanh(arcta nh 0.77349 + 23 .025 z .025 23 . ) where z.025 = 196 . 0 . 5513 ≤ ρ ≤ 0 . 8932 . 11-28 11-57. a) r = -0.738027 b) H 0 : ρ = 0 H1 : ρ ≠ 0 t0 = α = 0.05 −0.738027 26 1− 0.5447 = −5.577 t.025, 26 = 2.056 | t0 |> tα / 2, 26 Reject H0 . P-value = (3.69E-6)(2) = 7.38E-6 c) tanh(arcta nh - 0.738 - z.025 25 ) ≤ ρ ≤ tanh(arcta nh - 0.738 + where z.025 = 1 .96 . z.025 25 ) − 0 .871 ≤ ρ ≤ −0 .504 . d) H 0 : ρ = − 0 .7 H 1 : ρ ≠ − 0 .7 α = 0.05 z 0 = (arctanh − 0.738 − arctanh − 0.7 )( 25 )1 / 2 = −0.394 z.025 = 1.96 | z 0 |< zα / 2 Do not reject H0 . P-value = (0.3468)(2) = 0.6936 1/ 2 11-58 Therefore, 11-59 and 1 − R 2 = SS E . ˆS R = β 1 xx S yy S yy R n−2 T0 = = 1− R2 ˆ β1 1/ 2 n−2 S xx S yy = 1/ 2 SS E S yy ˆ β1 ˆ σ 2 S xx n = 50 r = 0.62 a) H 0 : ρ = 0 α = 0.01 H1 : ρ ≠ 0 t0 = r n−2 1− r 2 = 0 . 62 48 1 − ( 0 .62 ) 2 = 5 . 475 t .005 , 48 = 2 . 682 t 0 > t 0 .005 , 48 Reject H0 . P-value ≅ 0 b) tanh(arctanh 0.62 - z ) ≤ ρ ≤ tanh(arctanh 0.62 + z ) 47 47 .005 where .005 z.005 = 2.575. 0 .3358 ≤ ρ ≤ 0 . 8007 . c) Yes. 11-60. n = 10000, r = 0.02 a) H 0 : ρ = 0 α = 0.05 H1 : ρ ≠ 0 t0 = r n−2 1− r 2 = 0.02 10000 1−( 0.02) 2 = 2.0002 t.025,9998 = 1.96 t 0 > tα / 2,9998 Reject H0 . P-value = 2(0.02274) = 0.04548 11-29 where σ 2 = ˆ SS E . n−2 b) Since the sample size is so large, the standard error is very small. Therefore, very small differences are found to be "statistically" significant. However, the practical significance is minimal since r = 0.02 is essentially zero. a) r = 0.933203 b) H0 :ρ = 0 H1 : ρ ≠ 0 t0 = n−2 r 1− r 2 = α = 0.05 0 . 933203 15 1 − ( 0 . 8709 ) = 10 . 06 t.025 ,15 = 2 . 131 t 0 > t α / 2 ,1 5 Reject H0. c) y = 0.72538 + 0.498081x ˆ H 0 : β1 = 0 H 1 : β1 ≠ 0 α = 0.05 f 0 = 101 . 16 f .05 ,1 ,15 = 4 . 543 f 0 >> f α ,1 ,15 Reject H0. Conclude that the model is significant at α = 0.05. This test and the one in part b are identical. d) H0 : β0 = 0 H1 : β0 ≠ 0 α = 0.05 t 0 = 0 . 468345 t . 025 ,15 = 2 . 131 t 0 > tα / / 2 , 15 Do not reject H0. We cannot conclude β0 is different from zero. e) No problems with model assumptions are noted. Residuals Versus x Residuals Versus the Fitted Values (response is y) (response is y) 3 2 2 1 1 Residual 3 0 0 -1 -1 -2 -2 10 20 30 40 5 50 15 Fitted Value x Normal Probability Plot of the Residuals (response is y) 3 2 Residual Residual 11-61. 1 0 -1 -2 -2 -1 0 Normal Score 11-30 1 2 25 11-62. n = 25 r = 0.83 a) H 0 : ρ = 0 H 1 : ρ ≠ 0 α = 0.05 t0 = n−2 r 1− r 2 = 0 .83 = 7 . 137 23 1− ( 0 . 83 ) 2 t.025 , 23 = 2 . 069 t 0 > tα / 2 , 23 Reject H0 . P-value = 0. b) tanh(arctanh 0.83 - ) ≤ ρ ≤ tanh(arctanh 0.83 + z.025 22 z.025 22 ) where z.025 = 196 . 0 . 6471 ≤ ρ ≤ 0 . 9226 . . H 0 : ρ = 0. 8 c) H 1 : ρ ≠ 0 .8 α = 0.05 z 0 = (arctanh 0.83 − arctanh 0.8)( 22 )1 / 2 = 0.4199 z.025 = 1.96 z 0 > zα / 2 / H 0 . P-value = (0.3373)(2) = 0.6746. Do not reject Supplemental Exercises n 11-63. n n ˆ ( y i − yi ) = a) yi − i =1 ˆ ˆ y i = nβ 0 + β 1 ˆ yi and i =1 xi from normal equation i =1 Then, n ˆ ˆ ( nβ 0 + β1 xi ) − ˆ yi i =1 n n ˆ ˆ = nβ 0 + β 1 ˆ ˆ ( β 0 + β 1 xi ) xi − i =1 n ˆ ˆ = nβ 0 + β 1 i =1 n ˆ ˆ xi − nβ 0 − β1 i =1 n b) i =1 ˆ ( y i − yi )xi = n and i =1 ˆ β0 ˆ β0 n i =1 n i =1 n i =1 ˆ yi xi = β 0 ˆ xi + β1 ˆ xi + β1 n i =1 n i =1 xi = 0 i =1 yi xi − n i =1 2 n i =1 ˆ yi xi ˆ xi + β1 xi − n i =1 ˆ xi 2 − β 0 n i =1 xi 2 from normal equations. Then, ˆ ˆ (β 0 + β1 xi ) xi = n i =1 ˆ xi − β1 n i =1 xi 2 = 0 11-31 c) 1 n n ˆ yi = y i =1 ˆ y= ˆˆ ( β 0 +β1 x) 1n 1 ˆ ˆ ˆ yi = ( β 0 + β1 xi ) n i=1 n 1ˆ ˆ = (nβ 0 + β1 xi ) n 1 ˆ ˆ = (n( y − β1 x ) + β1 xi ) n 1 ˆ ˆ = (ny − nβ1 x + β1 xi ) n ˆ ˆ = y − βx + β x 1 =y a) Plot of y vs x 2.2 1.9 1.6 y 11-64. 1.3 1 0.7 1.1 1.3 1.5 1.7 x Yes, a straight line relationship seems plausible. 11-32 1.9 2.1 b) Model fitting results for: y Independent variable coefficient std. error t-value sig.level CONSTANT -0.966824 0.004845 -199.5413 0.0000 x 1.543758 0.003074 502.2588 0.0000 -------------------------------------------------------------------------------R-SQ. (ADJ.) = 1.0000 SE= 0.002792 MAE= 0.002063 DurbWat= 2.843 Previously: 0.0000 0.000000 0.000000 0.000 10 observations fitted, forecast(s) computed for 0 missing val. of dep. var. y = −0.966824 + 154376x . c) Analysis of Variance for the Full Regression Source Sum of Squares DF Mean Square F-Ratio P-value Model 1.96613 1 1.96613 252264. .0000 Error 0.0000623515 8 0.00000779394 -------------------------------------------------------------------------------Total (Corr.) 1.96619 9 R-squared = 0.999968 Stnd. error of est. = 2.79176E-3 R-squared (Adj. for d.f.) = 0.999964 Durbin-Watson statistic = 2.84309 2) H 0 : β 1 = 0 3) H 1 : β 1 ≠ 0 4) α = 0.05 5) The test statistic is f0 = SS R / k SS E /(n − p ) 6) Reject H0 if f0 > fα,1,8 where f0.05,1,8 = 5.32 7) Using the results from the ANOVA table f0 = 1.96613 / 1 = 255263.9 0.0000623515 / 8 8) Since 2552639 > 5.32 reject H0 and conclude that the regression model is significant at α = 0.05. P-value < 0.000001 d) 95 percent confidence intervals for coefficient estimates -------------------------------------------------------------------------------Estimate Standard error Lower Limit Upper Limit CONSTANT -0.96682 0.00485 -0.97800 -0.95565 x 1.54376 0.00307 1.53667 1.55085 -------------------------------------------------------------------------------- −0.97800 ≤ β 0 ≤ −0.95565 e) 2) H 0 : β 0 = 0 3) H 1 : β 0 ≠ 0 4) α = 0.05 5) The test statistic is t 0 = β0 se(β0 ) 6) Reject H0 if t0 < −tα/2,n-2 where −t0.025,8 = −2.306 or t0 > t0.025,8 = 2.306 7) Using the results from the table above −0.96682 = −199.34 0.00485 8) Since −199.34 < −2.306 reject H 0 and conclude the intercept is significant at α = 0.05. t0 = 11-33 11-65. ˆ a) y = 93.34 + 15.64 x b) H 0 : β1 = 0 H1 : β1 ≠ 0 f 0 = 12.872 α = 0.05 f .05,1,14 = 4.60 f 0 > f 0.05,1,14 Reject H0 . Conclude that β1 ≠ 0 at α = 0.05. c) (7.961 ≤ β1 ≤ 23.322) d) (74.758 ≤ β 0 ≤ 111.923) ˆ e) y = 93.34 + 15.64(2.5) = 132.44 [ − 1 132.44 ± 2.145 136.27 16 + ( 2.57.2.325) 017 2 132.44 ± 6.47 ˆ 125.97 ≤ µY |x0 = 2.5 ≤ 138.91 a) There is curvature in the data. 10 5 Vapor Pressure (mm Hg) 800 15 y 11-66 700 600 500 400 300 200 100 0 280 3 30 0 38 0 T em p erature (K ) 0 500 1000 1500 2000 2500 3000 3500 x b) y = - 1956.3 + 6.686 x c) Source Regression Residual Error Total DF 1 9 10 SS 491662 124403 616065 MS 491662 13823 11-34 F 35.57 P 0.000 d) Residuals Versus the Fitted Values (response is Vapor Pr) Residual 200 100 0 -100 -200 -100 0 100 200 300 400 500 600 Fitted Value There is a curve in the residuals. e) The data are linear after the transformation. 7 6 Ln(VP) 5 4 3 2 1 0 .0 0 2 7 0 .0 0 3 2 0 .0 0 3 7 1/T lny = 20.6 - 5201 (1/x) Analysis of Variance Source Regression Residual Error Total DF 1 9 10 SS 28.511 0.004 28.515 MS 28.511 0.000 11-35 F 66715.47 P 0.000 Residuals Versus the Fitted Values (response is y*) 0.02 Residual 0.01 0.00 -0.01 -0.02 -0.03 1 2 3 4 5 6 7 Fitted Value There is still curvature in the data, but now the plot is convex instead of concave. a) 15 10 y 11-67. 5 0 0 500 1000 1500 2000 2500 3000 3500 x b) c) ˆ y = −0.8819 + 0.00385 x H 0 : β1 = 0 H1 : β1 ≠ 0 α = 0.05 f 0 = 122.03 f 0 > f 0.05,1, 48 Reject H0 . Conclude that regression model is significant at α = 0.05 11-36 d) No, it seems the variance is not constant, there is a funnel shape. Residuals Versus the Fitted Values (response is y) 3 2 1 Residual 0 -1 -2 -3 -4 -5 0 5 10 Fitted Value e) 11-68. ˆ y ∗ = 0.5967 + 0.00097 x . Yes, the transformation stabilizes the variance. ˆ y ∗ = 1.2232 + 0.5075 x where y∗ = 1 . No, model does not seem reasonable. The residual plots indicate a y possible outlier. 11-69. ˆ y = 0.7916 x Even though y should be zero when x is zero, because the regressor variable does not normally assume values near zero, a model with an intercept fits this data better. Without an intercept, the MSE is larger because there are fewer terms and the residuals plots are not satisfactory. 11-70. 2 ˆ y = 4.5755 + 2.2047 x , r= 0.992, R = 98.40% 2 The model appears to be an excellent fit. Significance of regressor is strong and R is large. Both regression coefficients are significant. No, the existence of a strong correlation does not imply a cause and effect relationship. a) 110 100 90 80 days 11-71 70 60 50 40 30 16 17 index 11-37 18 b) The regression equation is ˆ y = −193 + 15.296 x Analysis of Variance Source Regression Residual Error Total DF 1 14 15 SS 1492.6 7926.8 9419.4 MS 1492.6 566.2 F 2.64 P 0.127 Cannot reject Ho; therefore we conclude that the model is not significant. Therefore the seasonal meteorological index (x) is not a reliable predictor of the number of days that the ozone level exceeds 0.20 ppm (y). 95% CI on β1 c) ˆ ˆ β 1 ± tα / 2 , n − 2 se ( β 1 ) 15 . 296 ± t .025 ,12 ( 9 . 421 ) 15 . 296 ± 2 . 145 ( 9 . 421 ) − 4 . 912 ≤ β 1 ≤ 35 . 504 d) ) The normality plot of the residuals is satisfactory. However, the plot of residuals versus run order exhibits a strong downward trend. This could indicate that there is another variable should be included in the model, one that changes with time. 40 2 30 20 Residual Normal Score 1 0 -1 10 0 -10 -20 -30 -2 -40 -40 -30 -20 -10 0 10 20 30 40 Residual 2 4 6 8 10 Observation Order a) 0.7 0.6 0.5 y 11-72 0.4 0.3 0.2 0.3 0.4 0.5 0.6 0.7 x 11-38 0.8 0.9 1.0 12 14 16 ˆ b) y = .6714 − 2964 x c) Analysis of Variance Source Regression Residual Error Total DF 1 6 7 SS 0.03691 0.13498 0.17189 MS 0.03691 0.02250 F 1.64 P 0.248 R2 = 21.47% d) There appears to be curvature in the data. There is a dip in the middle of the normal probability plot and the plot of the residuals versus the fitted values shows curvature. 1.5 0.2 0.1 0.5 Residual Normal Score 1.0 0.0 0.0 -0.5 -0.1 -1.0 -0.2 -1.5 -0.2 -0.1 0.0 0.1 0.4 0.2 0.5 0.6 Fitted Value Residual The correlation coefficient for the n pairs of data (xi, zi) will not be near unity. It will be near zero. The data for the 11-73 pairs (xi, zi) where z i = y i2 will not fall along the straight line correlation coefficient near unity. These data will fall on a line y i = xi which has a slope near unity and gives a y i = xi that has a slope near zero and gives a much smaller correlation coefficient. 11-74 a) 8 7 6 y 5 4 3 2 1 2 3 4 5 6 7 x b) ˆ y = −0.699 + 1.66 x c) Source Regression Residual Error Total DF 1 8 9 SS 28.044 9.860 37.904 MS 28.044 1.233 11-39 F 22.75 P 0.001 d) µ y| x = 4.257 x=4.25 0 4.257 ± 2.306 1.2324 1 (4.25 − 4.75) 2 + 10 20.625 4.257 ± 2.306(0.3717) 3.399 ≤ µ y|xo ≤ 5.114 e) The normal probability plot of the residuals appears straight, but there are some large residuals in the lower fitted values. There may be some problems with the model. 2 1 Residual Normal Score 1 0 0 -1 -1 -2 -2 -1 0 1 2 2 11-75 a) y 940 930 920 920 930 940 x b) 3 4 5 Fitted Value Residual ˆ y = 33.3 + 0.9636 x 11-40 6 7 8 c)Predictor Constant Therm Coef 66.0 0.9299 S = 5.435 R-Sq = 71.2% SE Coef 194.2 0.2090 T 0.34 4.45 P 0.743 0.002 R-Sq(adj) = 67.6% Analysis of Variance Source Regression Residual Error Total DF 1 8 9 SS 584.62 236.28 820.90 MS 584.62 29.53 F 19.79 P 0.002 Reject the hull hypothesis and conclude that the model is significant. 77.3% of the variability is explained by the model. d) H 0 : β1 = 1 H 1 : β1 ≠ 1 t0 = α=.05 ˆ β1 − 1 0.9299 − 1 = = −0.3354 ˆ 0.2090 se( β1 ) t a / 2,n − 2 = t .025,8 = 2.306 Since t 0 > −t a / 2,n − 2 , we cannot reject Ho and we conclude that there is not enough evidence to reject the claim that the devices produce different temperature measurements. Therefore, we assume the devices produce equivalent measurements. e) The residual plots to not reveal any major problems. Normal Probability Plot of the Residuals (response is IR) Normal Score 1 0 -1 -5 0 5 Residual 11-41 Residuals Versus the Fitted Values (response is IR) Residual 5 0 -5 920 930 940 Fitted Value Mind-Expanding Exercises 11-76. a) ˆ β1 = S xY S xx ˆ ˆ β 0 = Y − β1 x , ˆˆ ˆ ˆˆ Cov(β0 , β1 ) = Cov(Y , β1 ) − xCov(β1, β1 ) ˆ Cov(Y , β1 ) = Cov(Y , S xY ) Cov( = Sxx Yi , Yi ( xi − x )) nSxx σ ˆˆ ˆ Cov(β1, β1 ) = V (β1 ) = S xx 2 − xσ 2 ˆˆ Cov( β 0 , β1 ) = S xx b) The requested result is shown in part a. 11-77. ˆ ˆ (Yi − β 0 − β1 xi ) 2 = n−2 n−2 ˆ ) − E(β ) x = 0 ˆ E (ei ) = E (Yi ) − E ( β 0 1 i a) MS E = ei2 V (ei ) = σ 2 [1 − ( 1 + n E ( MS E ) = E (ei2 ) n−2 = ( xi − x ) 2 S xx )] Therefore, V (ei ) n−2 2 1 σ [1 − ( n + ( x S− x ) )] 2 = i xx n−2 σ [n − 1 − 1] = =σ2 n−2 2 11-42 = ( xi − x )σ 2 nSxx = 0. Therefore, b) Using the fact that SSR = MSR , we obtain { ˆ ˆ ˆ E ( MS R ) = E ( β12 S xx ) = S xx V ( β1 ) + [ E ( β1 )]2 = S xx 11-78. ˆ β1 = σ2 S xx } + β12 = σ 2 + β12 S xx S x1Y S x1 x1 n E i =1 ˆ E ( β1 ) = = S x1 x1 β1S x x + β 2 11 = n Yi ( x1i − x1 ) n ( β 0 + β1x1i + β 2 x2 i )( x1i − x1 ) i =1 S x1 x1 x2 i ( x1i − x1 ) i =1 S x1 x1 = β1 + β2Sx x 12 S x1 x1 No, β1 is no longer unbiased. σ ˆ V (β1 ) = S xx 2 11-79. . To minimize ˆ V ( β 1 ), S xx should be maximized. Because S xx = n ( xi − x ) 2 , Sxx is i =1 maximized by choosing approximately half of the observations at each end of the range of x. From a practical perspective, this allocation assumes the linear model between Y and x holds throughout the range of x and observing Y at only two x values prohibits verifying the linearity assumption. It is often preferable to obtain some observations at intermediate values of x. n 11-80. One might minimize a weighted some of squares wi ( yi − β 0 − β1 xi ) 2 i =1 ( w i large) receives greater weight in the sum of squares. ∂ β0 ∂ β1 n wi ( yi − β 0 − β1 xi )2 = −2 i =1 n n wi ( yi − β 0 − β1 xi ) i =1 wi ( yi − β 0 − β1xi )2 = −2 i =1 n wi ( yi − β 0 − β1 xi )xi i =1 Setting these derivatives to zero yields ˆ β0 ˆ wi + β1 ˆ β0 ˆ wi xi + β1 wi xi = 2 wi xi = wi yi wi xi yi as requested. 11-43 in which a Yi with small variance and ˆ β 1= ( ( wi xi y i )( ( wi ) 11-81. wi ˆ y = y+r = y+ sy sx ) wi xi2 − ( wi y i ˆ β0 = wi ) − wi xi ) 2 wi xi ˆ β1 wi − . (x − x) ( yi − y ) 2 ( x − x ) S xy ( xi − x ) 2 S xx S yy = y+ wi y i S xy (x − x) S xx ˆ ˆ ˆ ˆ = y + β1 x − β1 x = β 0 + β1 x 11-82. a) ∂ β1 n n ( yi − β 0 − β1 xi ) 2 = −2 i =1 ( yi − β 0 − β1 xi )xi i =1 Upon setting the derivative to zero, we obtain β0 xi + β1 2 xi = xi yi Therefore, ˆ β1 = ˆ b ) V ( β1 ) = V c) ˆ β 1 ± tα / 2, n−1 xi yi − β 0 xi xi (Yi − β 0) xi 2 xi 2 = xi ( yi − β 0 ) xi xi σ 2 2 2 = [ 2 xi ]2 = σ2 xi 2 ˆ σ2 xi 2 xi 2 ≥ ( xi − x) 2 . Also, the t value based on n-1 degrees of freedom is slightly smaller than the corresponding t value based on n-2 degrees of freedom. This confidence interval is shorter because 11-44 ...
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This note was uploaded on 09/26/2011 for the course ENGINEERIN 522 taught by Professor Dr.il during the Spring '11 term at Gannon.

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38148911-Solution-Manual-for-Applied-Statistics-and-Probability-for-Engineers-c02to11

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