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Unformatted text preview: CHAPTER 2 Section 21
21. Let "a", "b" denote a part above, below the specification
S = {aaa, aab, aba, abb, baa, bab, bba, bbb} 22. 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⎪
⎩
⎭
23. 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} 24. S = {0,1,2,...} = set of nonnegative integers 25. If only the number of tracks with errors is of interest, then S = {0,1,2, ...,24} 26. 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} 27.
28. 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} 29. S = {0,1,2,...,} in ppb. 210. S = {0,1,2,...,} in milliseconds 211.
212. s = small, m = medium, l = large; S = {s, m, l, ss, sm, sl, ….} 213
214. 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 21 without
air red blue black white 215.
PRESS 1 2 CAVI TY 1 2 3 5 4 6 7 8 1 2 3 4 5 6 7 216. memory 4 8 12 disk storage
200 300 400 200 300 217.
218. 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 } 219 a.) b.) 22 400 200 300 400 8 c.) d.) e.) 2.20 a.) 23 b.) c.) d.) e.) 24 221. 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, …} 222. a) A B C
b) A B C
c) 25 d.) A B C e.) 223. 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} 26 224. 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 = 95 225.
226. a.) A′ ∩ B = 10, B ′ =10, A ∪ B = 92
b.) 227. 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 228.
229. 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} 27 232.a.) The sample space contains all points in the positive XY plane.
b) A 10 c) 20
B d)
B 20 10 A 10 A e)
B 20 28 233 a) b) c) d) 29 Section 22
234. 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(∅)= 0 235. a) P(A) = 0.4
b) P(B) = 0.8
c) P(A') = 0.6
d) P(A∪B) = 1
e) P(A∩B) = 0.2 236. a) S = {1, 2, 3, 4, 5, 6}
b) 1/6
c) 2/6
d) 5/6 237. a) S = {1,2,3,4,5,6,7,8}
b) 2/8
c) 6/8 238. x
= 0.3, x = 6
20 239. a) 0.5 + 0.2 = 0.7
b) 0.3 + 0.5 = 0.8 240. a) 1/10
b) 5/10 241. a) 0.25
b) 0.75 242. Total possible: 1016, Only 108 valid, P(valid) = 108/1016 = 1/108 243. 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 108 244. a) 5*5*4 = 100
b) (5*5)/100 = 25/100=1/4 245. 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 = 0.84 246. 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) = 0.98 210 247. 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 = 0.92 248. 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 23
249. 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
250. a) P( 251. 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. 252. a) 70/100 = 0.70
b) (79+8670)/100 = 0.95
c) No, P( A ∩ B ) ≠ 0 253. a) 350/370
345 + 5 + 12 362
b)
=
370
370
345 + 5 + 8 358
c)
=
370
370
d) 345/370 254. a) 170/190 = 17/19
b) 7/190 255. a) P(unsatisfactory) = (5+102)/130 = 13/130
b) P(both criteria satisfactory) = 117/130 = 0.90, No 256. a) (207+350+357201204345+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 24 211 257. 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 86 258.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
259. a) 345/357 b) 5/13 260. a) 12/100 b) 12/28 c) 34/122 261. Need data from Table 22 on page 34
a) P(A) = 0.05 + 0.10 = 0.15
P( A ∩ B ) 0.04 + 0.07
b) P(AB) =
=
= 0.153
0.72
P( B)
c) P(B) = 0.72
P( A ∩ B ) 0.04 + 0.07
=
= 0.733
d) P(BA) =
0.15
P( B)
e) P(A ∩ B) = 0.04 +0.07 = 0.11
f) P(A ∪ B) = 0.15 + 0.72 – 0.11 = 0.76 262. 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) = 0.2 263. 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 ⎠ 264. 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) = 0.085 265. a) 4/499 = 0.0080
b) (5/500)(4/499) = 0.000080
c) (495/500)(494/499) = 0.98 266. a) 3/498 = 0.0060
b) 4/498 = 0.0080
c) ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 3 ⎞ = 4.82x10 −7
⎜
⎟⎜
⎟⎜
⎟
⎝ 500 ⎠ ⎝ 499 ⎠ ⎝ 498 ⎠ 267. a) P(gas leak) = (55 + 32)/107 = 0.813
b) P(electric failuregas leak) = (55/107)/(87/102) = 0.632
c) P(gas leak electric failure) = (55/107)/(72/107) = 0.764 212 268. No, if B ⊂ A , then P(A/B) = P( A ∩ B) P(B)
=
=1
P(B)
P(B) A
B 269. A B C Section 25
270. 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) = 0.30 271. 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
272. 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) = 0.014 273. 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
274. a) P(A) = 0.03
b) P(A') = 0.97
c) P(BA) = 0.40
d) P(BA') = 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 213 275. 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(RN)P(N) + P(RA)P(A) + P(RW)P(W)
= (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15)
= 0.028 276. Let B denote the event that a glass breaks.
Let L denote the event that large packaging is used.
P(B)= P(BL)P(L) + P(BL')P(L')
= 0.01(0.60) + 0.02(0.40) = 0.014 277. 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(UC)P(C) = (0.75)(0.03) = 0.0225
b) P(PR)P(R) = (0.50)(0.25) = 0.125 278. a) (0.88)(0.27) = 0.2376
b) (0.12)(0.13+0.52) = 0.0.078 279. 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
= 280. Let A and B denote the events that the first and second chips selected are defective, respectively.
a) P(B) = P(BA)P(A) + P(BA')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 26 ≠ 281. Because P( A B ) 282. 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.
283. 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. 214 284. 285. 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(BA) = P(A ∩ B)/P(A) = (22/100)/(30/100) = 0.733 286. 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. 287. 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.
288. 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⎠ 289. 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
8 215 290. 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 291. [1(0.1)(0.05)][1(0.1)(0.05)][1(0.2)(0.1)] = 0.9702 292. 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. 293. 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 27
294. Because, P( A B ) P(B) = P( A ∩ B ) = P( B A ) P(A), P( B A) =
295. 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) 296.
mainstorage 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 ' ) 216 = 0.05(0.25)
= 0.769
0.05(0.25) + 0.005(0.75) 297. 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 ' )
298. a) P(D)=P(DG)P(G)+P(DG’)P(G’)=(.005)(.991)+(.99)(.009)=0.013865
b) P(GD’)=P(G∩D’)/P(D’)=P(D’G)P(G)/P(D’)=(.995)(.991)/(1.013865)=0.9999 299. a) P(S) = 0.997(0.60) + 0.9995(0.27) + 0.897(0.13) = 0.9847
b) P(ChS) =(0.13)( 0.897)/0.9847 = 0.1184 Section 28
2100. Continuous: a, c, d, f, h, i; Discrete: b, e, and g Supplemental Exercises
2101. 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 } . 2102. a) 20/200
d) b) 135/200 c) 65/200 A 25 B
20 90 217 2103. 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(AB) = P(A ∩ B)/P(B) = 0.158 2104. 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
2105. 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 2106. (0.20)(0.30) +(0.7)(0.9) = 0.69 218 2107. 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 ⎠ 2108. 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 2109. 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 2110. P(Possess) = 0.95(0.99) +(0.05)(0.90) = 0.9855 2111. Let D denote the event that a container is incorrectly filled and let H denote the event that a container is
filled under highspeed 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 2112. 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 2113. 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) = ⎜ 2114. 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 2115. 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 219 2116. 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
2117. 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 2118. 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. 220 2119. a)
b)
c)
d)
. e)
f) 2120. 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, offcenter, and proper alignments, respectively.
Then,
a) P(E) = P(ES) P(S) + P(EO) P (O) + P(EP) P(P)
= 0.01(0.10) + 0.02(0.05) + 0.001(0.85)
= 0.00285
b) P(SE) = P ( E S) P (S)
P( E) 2121. = 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 2122. a) (0.4)(0.1) + (0.3)(0.1) +(0.2)(0.2) + (0.4)(0.1) = 0.15
b) P(4 or moreprovided) = (0.4)(0.1)/0.15 = 0.267 MindExpanding Exercises
2123. Let E denote a read error and let S, O, B, P denote skewed, offcenter, both, and proper alignments,
respectively.
P(E) = P(ES)P(S) + P(EO)P(O) + P(EB)P(B) + P(EP)P(P)
= 0.01(0.10) + 0.02(0.05) + 0.06(0.01) + 0.001(0.84) = 0.00344 2124. 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 = 3 221 2125. 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(x50)
100x
100x
50 ≤ x ≤ 100
Mean profit = [100(50)5(x50)](0.4) + 100x(0.55)5x(0.05)20x
5x
100(50)5(x50)
100(100)5(x100)
100x
100 ≤ x ≤ 200
Mean profit = [100(50)5(x50)](0.4) + [100(100)5(x100)](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 2126. 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(EX=0)P(X=0) + P(EX=1) P(X=1) + P(EX=2) P(X=2) + P(EX=3) P(X=3) + P(EX=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 B1. 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(EX=0) = 1, P(EX=1) = 0.05, P(EX=2) =
0.05 2 = 0.0025 , P(EX=3) = 0.05 3 = 1.25 × 10 −4 , P(EX=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
2127.
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' ) 222 2128. 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 21.4 on CD
S21. From the multiplication rule, the answer is 5 × 3 × 4 × 2 = 120 S22. From the multiplication rule, 3 ×4 ×3 = 36 S23. From the multiplication rule, 3×4×3×4 =1 4
4 S24. From equation S21, the answer is 10! = 3628800 S25. From the multiplication rule and equation S21, the answer is 5!5! = 14400 S26. From equation S23, S27. a) From equation S24, 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 S28. 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! () 223 S29. 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. S210. 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 S211. S212. 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 S213. 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 224 ( ) = 418!! = 3060. Therefore, the
!14
18
4 ( )( )( ) = 165!! × 143!! × 214!! = 2184 .
!
!
!12
6
1 4
1 14
2 S214. () 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 − S215. 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 () () 225 CHAPTER 3
Section 31 {0,1,2,...,1000} 31. The range of X is 32 The range of X is {0,1,2,. ..,50} 33. ,
The range of X is {0,1 2,. ..,99999} 34 ,
The range of X is {0,1 2,3,4,5} 35. 36 { } 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%. 37. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is
{0,1,2,.. .} 38 The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is
{0,1,2,.. .} 39. The range of X is 310 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 ⎭ 311 The range of X is {0,1,2, K,10000} 312 The range of X is {0,1,2,...,5000} {0,1,2,...,15} Section 32
313. 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
314 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 = 1/2 315. All probabilities are greater than or equal to zero and sum to one. 31 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
316 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)= 1P(X=1)=10.5714=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)=1 317. 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) = 1 318 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) = 1/4 319. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 320 P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 321. P(X = 0) = 0.023 = 8 x 106
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 322 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 = 0.512 323 P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = 0.5 million) = 0.1 324 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) = 0.931 325. 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 32 Section 33 326 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 / 6 327. 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 = 3/8 328 ⎧ 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
329. 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 33 330 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
331. 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, 332 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
333. 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) = 0.5 334 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) = 0.7 335. 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 34 336 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 34
337 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
339 Determine E(X) and V(X) for random variable in exercise 315
. 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
340 Determine E(X) and V(X) for random variable in exercise 315 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
341. Mean and variance for exercise 319 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 35 342 Mean and variance for exercise 320 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
343. Mean and variance for random variable in exercise 322 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
344 Mean and variance for exercise 323 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
345. 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 35
346 E(X) = (0+100)/2 = 50, V(X) = [(1000+1)21]/12 = 850 347. E(X) = (3+1)/2 = 2, V(X) = [(31+1)2 1]/12 = 0.667 348 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⎠ 36 349. 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 ⎠ ⎣
⎦ 350 ⎛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
351. 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) = 352 353 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 354 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). 37 Section 36
355. 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 356 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 357. ⎛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 38 358 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. 359. ⎛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⎟
⎝⎠
⎝⎠
360 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⎠ 39 361. 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 ⎪
⎭
⎩ 362 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) = 363. 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
364 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 365. ( )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 310 2 998 366 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.
367. 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
368 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
1 311 1 101 + ( )0.02 0.98
102
2 2 100 = 0.666 369. 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⎟
⎝⎠
⎝⎠ 370 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 37
371. 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
372 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 312 373. 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
374 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)
375. ⎡⎛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
376 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) = (10.2)30.2= 0.1024
b) By independence, (0.8)10 = 0.1074. (Also, P(X > 10) = 0.1074) 377 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=3.91x1019 µ = E (Y ) =
378 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 366. 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 366.
P(Y = 10) = (1 − 0.1897)9(0.1897) = 0.0286
c) E(Y) = 1/0.1897 = 5.27 313 379. 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. 380 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)x1p, 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.
381. 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.
382 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. 383. 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 = 1731.18 384 Let X denote the number of transactions until all computers have failed. Then, X is negative binomial
random variable with p = 108 and r = 3. a) E(X) = 3 x 108
b) V(X) = [3(1−1080]/(1016) = 3.0 x 1016 314 385 Let X denote a geometric random variable with parameter p. Let q = 1p.
∞ 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 38
386 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 315 1
p2 x −1 387. ( )( ) = (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) 388 0.2 0.1
0.0
0 1 2 3 x 316 389. ⎫
⎪
⎪
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⎟
⎟
⎝
⎠ 390 ⎛ 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.) 391. 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 317 392 . 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
393. 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)(10.2931) = 0.2811.
394 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! 318 395. a) For Exercise 386, the finite population correction is 96/99.
For Exercise 387, the finite population correction is 16/19.
Because the finite population correction for Exercise 386 is closer to one, the binomial approximation to
the distribution of X should be better in Exercise 386.
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
386.
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.
396 a.) From Exercise 392, 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 392, 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 39
397. 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 ) = 398 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 + 319 = 1 − 0.4832 = 0.5168 399. P( X = 0) = e − λ = 0.05 . Therefore, λ = −ln(0.05) = 2.996.
Consequently, E(X) = V(X) = 2.996. 3100 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! 3101. 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
3102 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
3103. 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. 320 3104 3105. = 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
3106 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
3107. 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⎟
⎝⎠
3108 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) = 15 321 3109. 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 3110 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 3111. 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 onehalf hour. Then, W is a Poisson random variable with
λ = 2.5. P (W < 2) = P(W = 0) + P(W = 1) = 0.2873
3112 X is a negative binomial with r=4 and p=0.0001
E ( X ) = r / p = 4 / 0.0001 = 40000 requests 3113. 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 3114 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. 3115 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 3116 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 3117. 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. 322 3118 Require f(1) + f(2) + f(3) + f(4) = 1. Therefore, c(1+2+3+4) = 1. Therefore, c = 0.1. 3119. 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
3120 3121. 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 3122
x
f(x) 3123. 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)]= 3124 ⎛ 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 − e0.32768 − e0.32768(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 − e0.32768 = 0.2794
E(Y) = 1/p = 3.58 323 3125. 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 − [e2 + e2(2) + (e222)/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) = e4 + (e441)/1!+ (e442)/2! =1  [0.01832+0.07326+0.1465] = 0.7619. 3126 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 324 3127. 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.
3128 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 e10t = 0.9, resulting in t = 0.0105 hours = 0.63 seconds 3129. 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) = e1 = 0.3679.
b) Let Y denote the number of flaws in one panel, then
P(Y ≥ 1) = 1 − P(Y=0) = 1 − e0.02 = 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
3130. 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): 325 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) = 3131. ⎛ 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 326 3132 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 3133 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≤
3134 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. 327 3135 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 3136 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(x1000)
mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x1000)](0.2) + 0.3x(0.5)  0.1x
0.05x
0.3(1000) +
0.3(2000) +
0.3x
2000 ≤ x ≤ 3000
0.05(x1000)
0.05(x2000)
mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x1000)](0.2) + [0.3(2000) + 0.05(x2000)](0.3) + 0.3x(0.2)  0.1x
0.05x
0.3(1000) +
0.3(2000) +
0.3(3000)+
3000 ≤ x
0.05(x1000)
0.05(x2000)
0.05(x3000)
mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x1000)](0.2) + [0.3(2000)+0.05(x2000)]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 3137 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. 328 CHAPTER 4 Section 42
∞ 41. ∞ ∫ 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 ∞ 42. ∞ ∫ 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 43 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 41 ∞ 44 ∫ 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 45 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
∞ 46. −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. 42 50.25 47 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 48. 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 49 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 410. 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 43
411. 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) = 0 412. 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 43 x 413. ∫ 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 414. 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 415. ∫ 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 416. 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) = e3000/1000 = 0.5
x 417. 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. 44 418 f ( x) = 2e −2 x , x > 0 419. ⎧
⎪0.2, 0 < x < 4
f ( x) = ⎨
⎪0.04, 4 ≤ x < 9
⎩ 420. ⎧
⎪0.25, − 2 < x < 1
f X ( x) = ⎨
⎪0.5, 1 ≤ x < 1.5
⎩ 421. 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 44
4 4 422. 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 4 423. 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 45 64
9 x)dx 1 424. 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
5 425. 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
⎝ 50.25 426. 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
120 427. 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 ∞ 428. E ( X ) = ∫ x 2 x − 3dx = − 2 x −1 ∞
1 =2 1 46 + (109.39 ) 2
x2 = 33.19 dx ∞ 429. ∫ 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 5.1 430. 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 45
431. 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 47 1205
1200 = 0.5 432. 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. 433. 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 = 0.7 434. 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 48 435 (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
⎩
436 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 ) = 437. 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 49 40 438. ∫ 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(40x) = 0.90 and x = 31
c) E(X) = (30 + 40)/2 = 35 and V(X) = (40 − 30) 2
= 8.33
12 Section 46
439. 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 440. 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 441 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 = 1.33 442. 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. 410 443. 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. 444. 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 = 5.16 411 445. ⎛
⎝ 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
446. ⎛
⎝ 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. 412 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. 447. ⎛
⎝ 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 > 448. ⎛
⎝ 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 413 449. ⎛
⎝ 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.
. 450. 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). 451. 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 = 72 414 452. 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, 453. 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. 454 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 415 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% 456. 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. 457. 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 416 458. ⎛
⎝ 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
459. ⎛
⎝ a) P(X > 0.0026) = P⎜ Z > 0.0026 − 0.002 ⎞
⎟
0.0004
⎠ = P(Z > 1.5)
= 1P(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⎜ 460. ⎛
⎝ 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 = 11.455 417 Section 47 461. 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
462. 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 <
463. 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
464. 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. 418 465. 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 ⎠
⎝
466. 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 >
467 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 ⎠
⎝ 468. 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, 469 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 419 469 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
470 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, 471 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 420 Section 49
0 472. ∫ 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 ∫ 473. = 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 474. = 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 1 475. 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 421 476. 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 477. 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 10minute interval equals one minus the
probability of zero calls in
a 10minute 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. 478. 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. 422 479. 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 0 480. = 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. 481. 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 482. = 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.
0 483. 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. 423 15 484. 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. 485. 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. 486. 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 485a). 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) =
487. 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! ⎥
⎣
⎦ 488. 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! 424 2! 3! ⎦ 489. a) From Exercise 488, P(Y > 3) = 0.01899. Let W denote the number of onehour
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. 490. 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 490, P(Y > 120) = ∞ 120 491. 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 490 part b., the probability of no calls in onehalf 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. 492. 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 . 425 493. 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 491 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.
∞ 494. 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.
∞ 495. 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 410
496 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 426 . 497. 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 () 498. 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) = 499 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 ) = 4100 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 427 4101. 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 4102 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 4103. 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 4104. 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 4105. 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
∞ 4106 Γ(r ) = ∫ x r −1e − x dx . Use integration by parts with u = x r −1 and dv =ex. Then,
0 r −1 − x Γ(r ) = − x e ∞
0 ∞ + (r − 1) ∫ x r − 2e − x dx = (r − 1)Γ(r − 1) .
0 428 ∞ 4107 ∫
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.
4108. If X is a chisquare 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 411
4109. 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
4110. 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 4111. 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
4112 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 ⎠ 3 429 = 0.1576 4113. 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) = 4114 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 4115 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 4116 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. 430 Section 411
4117 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 4118 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 2 431 4119 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. 4120 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 2 432 4121 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 4122 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 433 ) + 8 . 4056 = 881 . 65 hours 4123 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 4124 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 4125 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 4126 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 434 x4
8 − 13 x 3 + 55 x 2 − 100 x
9
9
9 4
2 2 . 4127. 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. 4128 a) This answer is the same as part a. of Exercise 4127.
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 4129. 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.
4130 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 4131. 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 4130). Then,
P (W ≥ 1) = 1 − P (W = 0) = 1 − 10 0.3935 0 (1 − 0.3935)10 = 0.9933 .
0 () 435 4132 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 4133 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 436 22 2 4134 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 4135. 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. 4136 a) P(X > 3.5) from part a. of Exercise 4135 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. 4137. 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 > 4138 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 . 437 0.0006 σ ) = 0.9973 . 4139. If P(0.002x < 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.
4140 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 4141. If P(X > 10,000) = 0.99, then P(Z > µ = 11,398 . 4142 = 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. 4143 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 438 = 2.72 and 4144 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
4145. 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 439 4146 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 MindExpanding Exercises
4147. 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! 4148. 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. 440 4149. 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 . 4150. 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(EA)P(A) + P(EA')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 4151. 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 1 441 4152. 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 48 on CD
S41. 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) S42. 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 ) = 0.24 442 S43. 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
S44. 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
S45 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)=0.54380.4562=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. 443 CHAPTER 5
Section 51
51. 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 52 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 53. 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 54 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(YX=1.5) = 2(1/3)+3(2/3) =2 1/3 e) Since fY1.5(y)≠fY(y), X and Y are not independent
55 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 56. 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 51 57. 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
58 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(YX=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.
59. f ( x, y ) ≥ 0 and f ( x, y ) = 1
R 510. 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) = 52 1
4 5
8 7
8 = 3
8 12
36 = 1/ 3 511. 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
512 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(YX=1) = 0.5
e) no, X and Y are not independent
513. Because X and Y denote the number of printers in each category, X ≥ 0, Y ≥ 0 and X + Y = 4
514. 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, 53 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 fY1(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 fY1(x)
0.920824
0.077543
0.001632 e) E(YX=1)=0(.920824)+1(0.077543)+2(0.001632)=0.080807
f) No, X and Y are not independent since, for example, fY(0)≠fY1(0). 54 515 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 fY3(y)
0.857
0.143
0
0
0 e) E(YX=3) = 0(0.857)+1(0.143) = 0.143
f) V(YX=3) = 02(0.857)+12(0.143) 0.1432= 0.123
g) no, X and Y are not independent 55 516 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
2.37x109
1.35x106
2.86x104
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 109 1.36 x 106 x=3
for 2.899 x 104 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 fY1(y)=f(y
)
8.1 x 1011
1.08 x 107
5.37 x 105
0.0119
0.988 e) E(YX=2) = E(Y)= n(p)= 4(0.997)=3.988
f) V(YX=2) = V(Y)=n(p)(1p)=4(0.997)(0.003)=0.0120
g) yes, X and Y are independent. 56 x=4 x = 1,2,3,4 0.0274 0.972 Section 52 517. 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
518 a) b) c) 519. 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
2 520 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 (1p1)= 20(0.05)(0.95) = 0.95 57 521. 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) = 1 522 ( )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
523. 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. 58 524 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
4 525. 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. 59 526 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 . 527. 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. )
528 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 = 4xy 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
529 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. 510 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(XY=2)=0(0.0204)+1(0.2449)+2(0.7347) = 1.7143
530 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!
= 531 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 511 532 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
533 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 −8 512 Section 53
33 534 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.
32 535. 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 10 535 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(YX=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. 513 2 (2.88) ( 2.52 −1) =0.3733 xydxdy = 0 ydy = 0 4
81
00 a) f X ( x) = 4
81 1 3 00 536 y2
2 4
4
−0 =
9
9 537.
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.
538 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
1 514 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
24 515 xy 2
2 3 =
0 x+2 dx
x 15
8 1 0 539. 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 Y1 (y) defined over this line segment 1
0 x 12
0
3 c) E(YX=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 y2 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 X2 (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
6 516 3x 540 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 541. 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
5 517 3 1 x 542 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
543. 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
544 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
3 518 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 545. 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(YX=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) = 5e5y for 0 < y
∞∞ 546 ∞ e − 2 x e −3 y dydx = c
0x 547. ∞ 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
2 519 −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
∞ 548 ∞ −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
2 520 15 − 6
e
2 ) 549. 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
0.50.5 550 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
45 521 12
15 (5) +
6 2
7.5 (7.5) = 19
9 551. 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
0 552 = 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. 522 553 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 5minute 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) fYX=x(y), for x = 2, 4, 6, 8 5
4 f(y2) 554 3
2
1
0
0 1 2 3 y 523 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 54
0.5 1 1 555. 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 ) =
000 556 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 .
0 524 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 1 557. 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 4 558 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 ) =
559. 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) =
−3 525 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, 560 f XYZ ( x, y, z )
and from part 559 a., f XY ( x, y ) =
f XY ( x, y ) f Z xy ( z ) = Therefore, f Z
561 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 562 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 0 526 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
0 563 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) =
564 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 1 527 + 0.4215 = 0.8190 565. 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 . () 566 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 55
567. 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) 528 568 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 569.
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
570 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) = 0.990.992=0.0099
V(Y) = 0.980.982=0.0196 σ XY = 0.9702 − (0.99)(0.98) = 0
ρ XY = σσ σ = 0 XY X 571 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(1p1)=20(1/3)(2/3)=4.44
V(X2) = np2(1p2)=20(1/3)(2/3)=4.44
E(X1X2) =n(n1)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. 529 − 2.267
(4.44)(4.44) = − 0 .5 1 572 From Exercise 540, c=8/81.
From Exercise 541, 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
573. Similarly to 549, 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 2.062 530 574
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= 575. 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 .
576. 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 531 Then, 577 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 578 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 56
579 a.) 0.04
0.03
0.02 z(0)
0.01
10 0.00
2 0
1 0 y 1 2 3 10
4 532 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
4 533 580 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 581. 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
582 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 2 534 x −µ x
σx 2 2(1− ρ 2 ) 2 + (1−ρ 2 ) x −µ x
σx 2 x−µ x σx 2 Now fYX=x is in the form of a normal distribution.
b) E(YX=x) = µ y + ρ σy
σx ( x − µ x ) . This answer can be seen from part 582a. Since the PDF is in the form of a normal distribution, then the mean can be obtained from the exponent.
c) V(YX=x) = σ 2 (1 − ρ2 ) . This answer can be seen from part 582a. Since the PDF is in the form
y
of a normal distribution, then the variance can be obtained from the exponent. 583
∞∞ ∞∞ 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. 584 − 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: 535 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 =1 585 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. 536 586 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 57
587. 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 <
588 ) = 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 537 589 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 > 590 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 591. )=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 < 592 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=ABC
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 < 593. 5 .9 − 6
) = P(Z < 0.82) = 0.206.
0.1225 a) Let X denote the average fillvolume 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 . 538 12 − 12.1 σ / 100 = 0.005. e.) P( X < 12) = 0.01 implies that P Z <
Then
594 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
595. 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
596 ( )( )( )( )( ) The sum of
x and
597. 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 =39/64 539 598 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)= 1.250.8752=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 = 599 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=20z 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) = 5100 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. 540 2
3 20 − x − z 5101. 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.
5102 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. () 5103 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 5104
00 0 y2
2 2
0 3 dx = 2c x3 3 = 18c . Therefore, c = 1/18.
0 541 11 1
1
18 5105. 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 5106 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 5107. 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/ 4 542 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 5108 a) f Z 0, 0 ( z ) = b)
4 z
4 dz = 2 . 0
1 5109. 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. 5110 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 . 543 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,
1P(Xa<25,000, X2<25,000, …, X6<25,000)=1P(X1<25,000)…P(X6<25,000)
=1(1e25/8)(1e2.5)(1e2.5)(1e1.25)(1e1.25)(1e1)=1.2592=0.7408
5111. 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.
5112 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 5113. 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 +
17.75 544 x2
2 18.25 ) = 0 .5
17.75 5114 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 .
5115. 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.
5116 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 .
. 5117 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
4 545 x 5118 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 5119 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 5120 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
0.246 546 5121 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. MindExpanding Exercises
5122. 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 )
5123 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 crossterm 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 crossterm 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 547 a b a b 0 0 f XY ( x, y )dydx = 5124
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 5125 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 58 on CD S51. S52. f Y ( y) = 1
4 at y = 3, 5, 7, 9 from Theorem S51. Because X ≥ 0 , the transformation is onetoone; that is ( )p (1 − p)
( y ) = ( )(0.25) (0.75) f Y ( y) = f X ( y ) =
If p = 0.25, S53. 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 S52, for y = 0, 1, 4, 9. for 10 ≤ y ≤ 22 − 102y 2 ) 22 = 18
10 −y y S54. 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 chisquare distribution with k = 2 degrees of
freedom). 548 S55. 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= S56. 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 ∞ S57. ∞ 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. 549 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 S58. 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 . ∞ S59. 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 S51 holds in this case also.
S510. 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
S511. 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. 550 0 ≤ y ≤ 1 and 0 ≤ a ≤ 4 . That is, for
y S512. 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 59 on CD e tx 1
=
m
x =1 m
m S513 . 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
2 551 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 ) = ∞ S514. 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 552 ∞ S515 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
= S516. 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 chisquare distribution with k 1 + k 2 degrees of freedom.
∞ S517. 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 =0 553 β S518. 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
12 554 ∞ S519. 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 ) = S520. − 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 = (λt)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 555 n S521. a) t
E (e ) = ∏ E (e tX i ) = (1 − λ )
tY −n i =1 b) From Exercise S520, Y has a gamma distribution with parameter λ and n. 2 S522. 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 S523. Because a chisquare distribution is a special case of the gamma distribution with λ = 1
k
and r = , from
2
2 Exercise S520. 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 S524. a) = 1 and M ( r ) (0) = µ r' b) From Exercise S520,
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 S520. Section 510 on CD S525. Use Chebychev's inequality with c = 4. Then,
S526. 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 . 556 P( X − 20 > 2σ ) ≤ S527. 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 S528. 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 S529. 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 S530. 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 ) . 557 CHAPTER 6 Section 61
61. 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. 61 62. 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 62 (272.82)2
19 63. 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 63 64. 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 65. Sample average: 64 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 65 66. 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 67. 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. 66 68 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 69. 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. 67 a. Dot Diagram of CRT data in exercise 65 (Data were rounded for the plot)
Dotplot for Exp 1Exp 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
611. 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
612 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 68 613. 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 63
614 Stemandleaf display of octane rating
Leaf Unit = 0.10 N = 83 834 represents 83.4 1
834
3
8433
4
853
7
86777
13
87456789
24
8823334556679
34
890233678899
(13) 900111344456789
36
9100011122256688
22
9222236777
14
93023347
8
942247
4
95
4
9615
2
97
2
988
1
99
1 1003 615 a.) Stemandleaf display for cycles to failure: unit = 100
1
1
5
10
22
33
(15)
22
11
5
2 12 represents 1200 0T3
0F
0S7777
0o88899
1*000000011111
1T22222223333
1F444445555555555
1S66667777777
1o888899
2*011
2T22 b) No, only 5 out of 70 coupons survived beyond 2000 cycles. 69 616 Stemandleaf display of percentage of cotton N = 64
Leaf Unit = 0.10
321 represents 32.1%
1
6
9
17
24
(14)
26
17
12
9
5
3 617. Stemandleaf display for Problem 24.yield: unit = 1
1
1
7
21
38
(11)
41
27
19
7
1 618 321
3256789
33114
3356666688
340111223
3455666667777779
35001112344
3556789
36234
366888
3713
37689 7o8
8*
8T223333
8F44444444555555
8S66666666667777777
8o88888999999
9*00000000001111
9T22233333
9F444444445555
9S666677
9o8 Descriptive Statistics
Variable
Octane Rating 619. 621. N
83 Median
Q1
90.400 88.600 Q3
92.200 Descriptive Statistics
Variable
cycles 620 12 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 610 622 a.) sample mean: x = 65.811 inches standard deviation s = 2.106 inches
b.) Stemandleaf display of female engineering student heights
Leaf Unit = 0.10 610 represents 61.0 inches
1
3
5
9
17
(4)
16
8
3
1 N = 37 610
6200
6300
640000
6500000000
660000
6700000000
6800000
6900
700 x
c.) median: ~ = 66.000 inches
623 Stemandleaf display for Problem 623. Strength: unit = 1.0 12 represents 12
1
1
2
4
6
11
22
28
39
47
(12)
41
32
22
12
8
5 5329
533
5342
53547
5366
5375678
53812345778888
539016999
54011166677889
541123666688
5420011222357899
543011112556
54400012455678
5452334457899
54623569
547357
54811257 611 624 Stemandleaf of concentration N = 60 Leaf Unit = 1.0 22 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 29
31
39
422223
45689
501223444
55666777899999
611244
6556677789
7022333
76777
801
89 The data have a symmetrical bellshaped 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
59.45 612 = 9215 .93
59 625 Stemandleaf display for Problem 625. 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
260.85 613 626 Stemandleaf of speed (in megahertz) N = 120
Leaf Unit = 1.0 634 represents 634 megahertz
2
7
16
35
48
(17)
55
36
24
17
5
3
1
1 6347
6424899
65223566899
660000001233455788899
670022455567899
6800001111233333458
690000112345555677889
70011223444556
710057889
72000012234447
7359
7468
75
763 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
683.00 614 120 = 78402 .925
119 627 a.)Stemandleaf display of Problem 627. Rating: unit = 0.10 12 represents 1.2
1
2
5
7
9
12
18
(7)
15
8
4
3
2 830
840
85000
8600
8700
88000
89000000
900000000
910000000
920000
930
940
9500 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. 615 628 a.) Stemandleaf diagram of NbOCl3 N = 27
Leaf Unit = 100
04 represents 40 grammole/liter x 103
6
7
(9)
11
7
7
3 0444444
05
1001122233
15679
2
25677
3124
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 629 87792869 − = 917030 .85 gram − mole/liter x10 −3
N
40 Median
1256 a.)Stemandleaf display for Problem 629. Height: unit = 0.10 12 represents 1.2 Female Students Male Students
061
1
0062
3
0063
5
000064
9
0000000065 17
2
6500
000066 (4)
3
660
0000000067 16
7
670000
0000068
8
17
680000000000
0069
3
(15) 69000000000000000
070
1
18
700000000
11
7100000
6
7200
4
7300
2 740
1
750
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. 616 Section 64
Frequency Tabulation for Exercise 614.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 630 20 10 0
83.0 85.5 88.0 90.5 93.0 octane data 617 95.5 98.0 100.5 631.
Frequency Tabulation for Exercise 615.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 618 2000 2250 632
Frequency Tabulation for Exercise 616.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 619 37.5 38.5 633.
Frequency Tabulation for Exercise 617.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 620 94 96 98 Frequency Tabulation for Exercise 614.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 634 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. 621 635
Frequency Tabulation for Problem 622. 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 stemleafdiagram). 20 Frequency 636 10 0
5320 5340 5360 5380 5400 5420 5440 5460 5480 5500 shear strength 622 637
Frequency Tabulation for exercise 624. 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 stemandleaf display.
Yes, the histogram of the distance data shows the same shape as the stemandleaf display in
exercise 625. 20 Frequency 638 10 0
220 228 236 244 252 260 268 276 284 292 distance 623 639 Histogram for the speed data in exercise 626. Yes, the histogram of the speed data shows the
same shape as the stemandleaf display in exercise 626 Frequency 20 10 0
620 720 770 speed (megahertz) Yes, the histogram of the wine rating data shows the same shape as the stemandleaf display in
exercise 627. 7
6
5 Frequency 640 670 4
3
2
1
0
85 90 rating 624 95 641
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 65
Descriptive Statistics of Oring 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 642 60
50
40
30 625 643. 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 644 2.5 2.0 626 TrMean
2.415
Q1
1.912 StDev
0.534
Q3
2.973 SE Mean
0.189 645. 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 646 90 80 627 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. 647.
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. 448 The golf course yardage data appear to be skewed. Also, there is an outlying data point above 7500 yards. 628 7500
7400 yardage 7300
7200
7100
7000
6900
6800 649 Boxplot for problem 649 100 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. 650 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 stemleafdiagram. 629 shear strength 5490 5440 5390 5340 651
Boxplot for problem 651
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. 652 The box plot and the stemleafdiagram show that the data are very symmetrical about the mean. It also
shows that there are no outliers in the data. 630 90 concentration 80
70
60
50
40
30 653 Boxplot for problem 653
285 Distance in yards 275
265
255
245
235
225 The plot indicates that most balls will fall somewhere in the 250275 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 625. 654 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 stemleafdiagram. 631 rating 95 90 85 655 Boxplot for problem 655
75 70
Height
65 60
Female
students Male
students We can see that the two distributions seem to be centered at different values. 632 656 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 66
Time Series Plot 13
12
11
10
Res_Time 657. 9
8
7
6
5
4 Index 5 10 15 Computer response time appears random. No trends or patterns are obvious. 633 20 658 a.) Stemleafplot 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 stemleafplot 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 659. 10 20 30 a) 634 40 260
250
240
230
Force 220
210
200
190
180
170
In d e x 10 20 30 b) Stemandleaf display for Problem 223.Force: unit = 1
3
6
14
18
(5)
17
14
10
3 40 12 represents 12 17558
18357
1900445589
201399
2100238
22005
235678
241555899
25158 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. 660
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 40 635 50 661 a) Sunspots 150 100 50 0
Index 10 20 30 40 50 60 70 80 90 100 b) Stemandleaf display for Problem 225.Sunspots: unit = 1
17
29
39
50
50
38
33
23
20
16
10
8
7
4 12 represents 12 001224445677777888
1001234456667
20113344488
300145567789
4011234567788
504579
60223466778
7147
82356
9024668
1013
118
12245
13128
HI154 The data appears to decrease between 1790 and 1835, the stem and leaf plot indicates skewed data. a.) Time Series Plot 16 miles flown 662 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.) Stemandleaf of miles fl N = 84 636 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 stemleafdiagram appear to be nearly normally distributed. Section 67
663 Normal Probability Plot for 61
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. 637 664 Normal Probability Plot for time
Exercise 62 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. 665 There is no evidence to doubt that data are normally distributed Normal Probability Plot for 65
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 638 50 60 70 80 43.975 StDev 95 11.5000 666 Normal Probability Plot for temperature
Data from exercise 613
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.
667 Normal Probability Plot for 614
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 639 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. 668 Normal Probability Plot for cycles to failure
Data from exercise 615
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. 669
Normal Probability Plot for 627
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. 640 670
Normal Probability Plot for concentration
Data from exercise 624
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.
671
Normal Probability Plot for 622...629
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. 672 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. 641 Supplemental Exercises
673. 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. 6 674 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). 642 675 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. 676 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. 677 643 StDev
0.948
1.023 SE Mean
0.150
0.162 Comparative boxplots for viscosity data
17 16 Viscosity
15 14 13 676 First half 676 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. 678 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. 679 a)Stemandleaf display for Problem 235: unit = 1 644 12 represents 12 1
8
18
(7)
15
12
7
5
3 0T3
0F4444555
0S6666777777
0o8888999
1*111
1T22233
1F45
1S77
1o899 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. 680 a.) Stemandleaf 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.) 645 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. 681
Normal Probability Plot for 676 First h...676 Second
ML Estimates  95% CI 99 676 First
half 95 676 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. 646 682 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. 683
Normal Probability Plot for 644...656
ML Estimates  95% CI 644 99 656 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. 647 884 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. 685 648 Boxplot for Exercise 685
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 “startup” 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 686 Trial 2 AD* 70
60
50
40
30
20
10
5
1
4.5 5.5 Data 649 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 687 = 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 = 688 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 689 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. 650 The value of µ may be quite yi = a + bxi 690 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 691 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 692 Using the results found in Exercise 690 with a = − x
and b = 1/s, the mean and standard deviation
s of the zi are z = 0 and sZ = 1. 693. 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
2 651 n +1 n xi + x n +1 xi
694 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 [ (
( + ) ) 652 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
695. 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. 696. 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. 653 CHAPTER 7 Section 72
2n 71. () 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.
72. () 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 73. 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 74. Since both estimators are unbiased: 71 Relative Efficiency = ˆ
ˆ
MSE (Θ1 ) V (Θ1 ) σ 2 / 7
2
=
=
=
2
ˆ
ˆ
MSE (Θ 2 ) V (Θ 2 ) 3σ / 2 21 75. ˆ
ˆ
MSE (Θ1 ) V (Θ1 ) 10
=
=
= 2 .5
ˆ ) V (Θ ) 4
ˆ
MSE (Θ 2
2 76. 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 . 77. 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. 72 78. 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 79. (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. 73 σ2
n 710 () 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. 711 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 74 712. 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. 75 716
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 [ [ = 717 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 1 718 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 76 X1
for
n1 p1 and X2
n2 Section 73
n 719. 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 720. 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. 77 721. 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 =1 722. 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 78 723. 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. 724 725 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. 79 726. 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 = 727 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 728 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
729 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) 710 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
1 730 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) 711 731 a)Using the results from Example 712 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. 732 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 75 733. 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
734. 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 712 735. 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 736.
n=6 σX =
σX 737. 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 738. 739. 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
= 713 740. 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.
2 741. 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
742.
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 714 743. 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
744. 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 .
745. 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 1 746. f ( x1 , x 2 , x3 , x 4 , x5 ) = 747. 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 =1 748. f ( x1 , x 2 , x 3 , x 4 ) = 1 for 0 ≤ x1 ≤ 1,0 ≤ x 2 ≤ 1,0 ≤ x 3 ≤ 1,0 ≤ x 4 ≤ 1 749. 1.5 2 2 2
X 1 − X 2 ~ N (100 − 105,
+)
25 30
~ N (−5,0.2233) 715 750. σ X1 − X 2 = 0.2233 = 0.4726 751. 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
752. No, because Central Limit Theorem states that with large samples (n ≥ 30), X is approximately normally
distributed. 753. 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 = 1 754. 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.
755. P ( X ≤ 37) = P ( Z ≤ −5.36) = 0 756. Binomial with p equal to the proportion of defective chips and n = 100. 757. 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 716 758 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. 759 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 =1 717 760 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 0 718 u = ln x and dv = x θ dx MindExpanding Exercises
P ( X 1 = 0, X 2 = 0) = M ( M − 1)
N ( N − 1) P ( X 1 = 0, X 2 = 1) = 761. 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) = 762 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. 719 763 (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 loglikelihood 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. 720 764. 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 765 (
)(
)
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 ) = 766 767. (
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 768. = 1 − (1 − p ) [ ]. From Exercise 765,
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 765,
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 721 769. ))( (( ) P F X ( n ) ≤ t = P X ( n ) ≤ F −1 (t ) = t n from Exercise 765 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 765 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 −1 770. 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 2 771 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.
772 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 (n1) 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. 722 Section 7.3.3 on CD S71 From Example S72 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 S72 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− − 22 723 and S73. 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+ S74 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 S72, ˆ
b.) µ
S76. m + x +1
m + λ0 + 1 = λ0 a) From Example S72, the Bayes estimate is ˆ
b.) µ S75. 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)=
0 724 CHAPTER 8 Section 82
81 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%.
82 a.) A zα = 2.33 would give result in a 98% twosided confidence interval.
b.) A zα = 1.29 would give result in a 80% twosided confidence interval.
c.) A zα = 1.15 would give result in a 75% twosided confidence interval. 83 a.) A zα = 1.29 would give result in a 90% onesided confidence interval.
b.) A zα = 1.65 would give result in a 95% onesided confidence interval.
c.) A zα = 2.33 would give result in a 99% onesided confidence interval. 84 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 81 85 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.
86 Interval (1): 3124.9 ≤ µ ≤ 3215.7 and Interval (2):: 3110.5 ≤ µ ≤ 3230.1
Interval (1): halflength =90.8/2=45.4 Interval (2): halflength =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
halflength, and therefore a smaller confidence interval. The 99% confidence level will make the interval
larger.
87 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. 88 95% Twosided 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
89 95% Twosided 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
810 9 5 99% Twosided 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 82 811 a.) 99% Twosided 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% Onesided 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
812 a.) 95% Twosided CI on the true mean life of a 75watt 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% Onesided 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 83 813 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% Twosided CI on the true mean compressive strength
zα/2 = z0.005 = 2.58 x − z0.005
3250 − 2.58 814 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 75watt 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
815 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.
816 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 84 817 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. 818 If n is doubled in Eq 87: 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 87: 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 83 t 0.025,15 = 2.131 t 0.05,10 = 1.812 t 0.005, 25 = 2.787 819 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.) 821 a.)
b.) 820 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 85 t 0.10, 20 = 1.325 822 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 − 2.131 823 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 − 2.539 824 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 86 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 825
ML Estimates  95% CI 99
95
90
80 Percent 825 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,n1 = 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. 87 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 826 70
60
50
40
30
20
10
5
1
2150 2250 2 350 Data b.) 95% twosided 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% lowerconfidence bound on mean strength x − t 0.05,11
2259.9 − 1.796 s s
n
35.6 ≤µ ≤µ
12
2241.4 ≤ µ 88 35.6
12 827 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% twosided confidence interval on mean rod diameter
For α = 0.05 and n = 15, tα/2,n1 = 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 ≤ 8.244 828 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 twosided
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). 89 829 95% lower bound confidence for the mean wall thickness
given x = 4.05 s = 0.08 n = 25 tα,n1 = 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 830 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% twosided confidence interval on mean percentage enrichment
For α = 0.01 and n = 12, tα/2,n1 = 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 ≤ 2.991 810 s = 0.0993 0.0993
12 831 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,n1 = 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 811 2 3 1 .6 7 S tD e v 95 Percent 832 ≤ µ ≤ x + t 0.025, 24 1 .3 6 9 4 4 Section 84 834 χ 02.05,10 = 18.31 χ 02.025,15 = 27.49 χ 02.01,12 = 26.22 χ 02.005, 25 = 46.93 833 χ 02.95, 20 = 10.85 χ 02.99,18 = 7.01 a.) χ 0.05, 24
2 = 36.42 b.)
c.)
835 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
836 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
837 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
838 99% twosided 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 812 χ 02.995,16 = 5.14 839 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
840 99% twosided confidence interval for σ for the hole diameter data. (Exercise 835)
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
841 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 85
842 95% Confidence Interval on the death rate from lung cancer. ˆ
p= 823
= 0.823
1000 ˆ
p − zα / 2
0.823 − 1.96 843 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. 813 844 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.) 845 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
846 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
847 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
0.017 814 848 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 86
849 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,n1 = 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.
850 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. 851 Given x = 317.2 s = 15.7 n = 10 for α=0.05 tα/2,n1 = t0.005,9 = 3.250 815 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. 852 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 . 853 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,n1 = 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 + 854 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% twosided confidence interval on mean rod diameter is 8.216 855 90% prediction interval on wall thickness on the next bottle tested. 816 ≤ µ ≤ 8.244 given x = 4.05 s = 0.08 n = 25 for tα/2,n1 = 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 + 856 To obtain a one sided prediction interval, use tα,n1 instead of tα/2,n1
Since we want a 95% one sided prediction interval, tα/2,n1 = 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
857 99% prediction interval for enrichment data given x = 2.9 s = 0.099 n = 12 for α = 0.01 and n = 12, tα/2,n1
= 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). 858 95% Prediction Interval on the volume of syrup of the next beverage dispensed
x = 1.10 s = 0.015 n = 25 tα/2,n1 = 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 ≤ 817 µ ≤ 1.106 859 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,n1 = 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 87
860 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.
861 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).
862 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 . 818 863 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).
864 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 . 865 95% tolerance interval on the diameter of the rods in exercise 827 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).
866 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 ≤ ∞)
.
867 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. 868 99% tolerance interval on rod enrichment data that have a 95% CL 819 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).
869 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 870 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 871 µ = 50
a) n = 16 σ unknown
x = 52 s = 1.5 52 − 50 to = =1 8 / 16 The Pvalue 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 Pvalue 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 Pvalue 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.
872 µ = 50, σ 2 = 5 820 ≥ 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. 821 873 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. 874 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% twosided 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% twosided CI (this is because of the value of α)
d.) 99% onesided 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% onesided upper bound on the confidence interval on σ2 comprehensive strength 822 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% twosided confidenceinterval 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% onesided 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% onesided 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% onesided 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 823 value can greatly affect the sample mean, standard deviation and confidence intervals. The farther from
the mean, the greater the effect. 875 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. 876 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) 877 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,n1 = 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 + 824 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 878 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. 825 879 a) Normal probability plot for the coefficient of restitution
Norm al Probability Plot for 879
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, n1 = 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. 880 95% Confidence Interval on the death rate from lung cancer. 826 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 881
ML Estimates  95% CI 99
95
90
80 Percent 881 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, n1 = 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 + 827 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 882 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, n1 = 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 + 828 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. 883 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. 884 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. 829 885 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 886 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 887 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 830 − 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)/(1p)
p=0.6004 between 10.19 and 10.41. 889 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) 890 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. 831 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 832 CHAPTER 9 Section 91
91 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. 92 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. 93 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.
94 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 91 95. 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
96 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.
97 Use n = 5, everything else held constant (from the values in exercise 96):
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 92 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.
98 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. 99 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.
910 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 93 b) β = P ( X ≤ 185 when µ = 195)
X − 195 185 − 195
=P
≤
20 / 16
20 / 16
= P(Z ≤ −2)
= 0.02275. 911 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. 912 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. 94 913 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
914 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).
915 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) = 95 = 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
916 Power Function Curve
1
0.8 1−β 0.6
0.4
0.2
0
175 180 185 190 195 200 µ
917. 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)=1P(Z≤ 9.13) ≈ 0
ˆ
b) β = P( p ≤ 0.8 when p=0.75) = P(Z ≤ 2.58)=0.99506. 96 918 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 919 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 97 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 92
920 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) Pvalue = 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)
≅0 921. 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) Pvalue = 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. 98 = (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. 922 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 99 = (1.96 + 1.29) 2 (0.9) 2
= 15.21,
(0.75) 2 923. 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) Pvalue = 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. 924 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) Pvalue = 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 ≅1 910 = (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. 925. 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 = Pvalue = 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. 926 n= (z + zβ ) σ 2
2 α /2 δ2 = (z 0.05 + z 0.20 )2 σ 2
(3250 − 3500) 2 n ≅1 911 = (1.65 + .84) 2 (60) 2
= 0.357,
(250) 2 927 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 = 1β = 10.97062 = 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. 928 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. 912 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=1β =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 twosided 99% confidence interval is
equivalent to a twosided hypothesis test at α = 0.01, the conclusions necessarily must be consistent. 929 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 = 1β = 10 = 1 + zβ ) σ 2
2 α δ2 = (z 0.05 + z 0.1 )2 σ 2
( 4 .5 − 4 ) 2 n≅2 913 = (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 93
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,n1 where tα/2,n1 = 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 <Pvalue < 2* 0.5 ; 0..8 < Pvalue <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 930 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. 914 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. 931 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,n1 where tα/2,n1 = 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.
Pvalue = 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 . 915 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.
932 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,n1 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 acrefeet at α = 0.01. The 0.10 < Pvalue < 0.25. 916 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 acrefeet at α=0.01. 917 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,n1 where tα/2,n1 = 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<Pvalue = 2* 0.1 Therefore, 0.1< Pvalue<0.2.
b)
Normal Probability Plot for 933
ML Estimates  95% CI 99
95
90
80 Percent 933 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. 918 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.
934 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,n1 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 Pvalue > 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.
935. 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,n1 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 ftlb/in at α = 0.05. The Pvalue < 0.0005
.
936.
In order to use t statistic in hypothesis testing, we need to assume that the underlying distribution is normal. 919 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,n1 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 <Pvalue < 0.005
937. 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,n1 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 Pvalue > 0.4, based on Table IV. Minitab gives Pvalue = 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 . 920 938 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, n1 = 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 PValue: 2*0.05<Pvalue<2*0.10 therefore 0.10< Pvalue<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 .
939 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,n1 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< Pvalue<0.005.
Minitab gives Pvalue = 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. 921 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.
940 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,n1 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.)Pvalue: 0.02<Pvalue<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.) 941 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,n1 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. 922 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<Pvalue = 2* 0.0025 Therefore, 0.002< Pvalue<0.005.
Minitab gives a pvalue 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 = 10.2 = 0.8. d) From the same OC chart, and for the specified power, we would need approximately 38 observations. d=
n = 75 .
942 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,n1 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 Pvalue 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 . 923 Section 94
943 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) Pvalue = P(χ2 > 8.96) for 14 degrees of freedom: c) λ= σ
0.0125
=
= 1 .2 5
σ0
0.01 0.5 < Pvalue < 0.9 power = 0.8, β=0.2 using chart VIk the required sample size is 50 944 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.) Pvalue: The χ2 is between 0.50 and 0.10. Therefore, 0.2<Pvalue<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 ≤ 76.80 924 945 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) Pvalue = P(χ2 > 4984.83) for 15 degrees of freedom Pvalue < 0.005 946 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.) Pvalue: The Pvalue<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. 947. a) In order to use χ2 statistic in hypothesis testing and confidence interval construction, we need to assume
that the underlying distribution is normal. 925 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. 948 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.
949 Using the chart in the Appendix, with
n = 30. 926 Section 95
950 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. 951 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
. 927 2 0.15(1− 0.15) / 85 952 a) Using the information from Exercise 848, 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 Pvalue: 2(1Φ(0.53))=2(10.70194)=0.59612 953. a) Using the information from Exercise 848, 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) Pvalue = 1 − Φ(0.53) = 0.29806 928 954 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.) Pvalue =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. 955. 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) Pvalue = Φ(0.452) = 0.67364 929 956 . 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) Pvalue = 1Φ(0.94) =1.8264= 0.67364
957. 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. 958 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. 930 Section 97
959. 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 34
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 Pvalue is between 0.05 and 0.1 using Table III. Pvalue = 0.0649 (found using Minitab) 931 960. 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 12
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) Pvalue = 0.2237 (found using Minitab) 932 961. 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) 58
7 9
4 10
6 11
4 1215
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 Pvalue is between 0.9 and 0.5 using Table III. Pvalue = 0.6325 (found using Minitab) 933 962 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 34
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) Pvalue = 0.0155 (found using Minitab) 934 963 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 23
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 Pvalue is between 0.5 and 0.1 using Table III Pvalue = 0.3048 (found using Minitab) 964 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) Pvalue = 0 (found using Minitab) 935 Section 98
965. 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.
Pvalue = 0.070 (using Minitab) 966 1. The variable of interest is calls by surgicalmedical patients.
2. H0:Calls by surgicalmedical patients are independent of Medicare status.
3. H1:Calls by surgicalmedical 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. Pvalue = 0.85 967. 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. Pvalue = 0.002 968 1. The variable of interest is characteristic among deflections and ranges. 936 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. Pvalue = 0.652 969. 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. Pvalue = 0.013 970 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. Pvalue ≈ 0 937 Supplemental Exercises a. Sample Size, n
50 p(1 − p)
n
Sampling Distribution
Normal b. 80 Normal p c. 971 100 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. 972 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 Pvalue
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 Pvalue decreases as the sample size increases.
973. 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. 938 974 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. 975. 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.
976 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 Pvalue: 1 − Φ (0.31) = 1 − 0.6217 = 0.3783 939 which n=100 n=400 z0 =
z0 = 86 − 85
16 / 100
86 − 85 = 0.63 Pvalue: 1 − Φ (0.63) = 1 − 0.7357 = 0.2643 = 1.25 Pvalue: 1 − Φ (1.25) = 1 − 0.8944 = 0.1056
16 / 400
86 − 85
z0 =
= 3.13 Pvalue: 1 − Φ (3.13) = 1 − 0.9991 = 0.0009
16 / 2500 n=2500 The data would be statistically significant when n=2500 at α=0.01 977. 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 Pvalue
7) x = 153.7 , s= 11.3, n=20 t0 = 153.7 − 150 = 1 .4 6 11.3 20 Pvalue = 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.
978 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=0.01 940 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. 979 a) 1) the parameter of interest is the standard deviation, σ 941 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 pvalue
7) n = 10, s = 15.7
χ2 =
0 ( ) 9(15.7) 2
= 5.546
400 Pvalue = P χ 2 < 5546 ;
. 0.1 < P − value < 0.5 8) The Pvalue 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 ) Pvalue = P χ 2 < 30.81 ; 0.01 < P − value < 0.025 8) The Pvalue 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 Pvalue, providing
0
more evidence against the null hypothesis. 980 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 = Pvalue = ( ) 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 Pvalue 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 = 1.0 942 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 = Pvalue = ( ) 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 Pvalue 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.
981. 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.
982 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 Pvalue
7) s = 0.004 and n = 10 χ 02 = ( ) 9(0.004) 2
= 0.000009
( 4) 2 P − value ≅ 0.
Pvalue = P χ < 0.00009 ;
The Pvalue 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.
2 943 983. 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
05
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 944 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) Pvalue = 0.0123 (found using Minitab)
χ2 =
0 984 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) Pvalue = 0.6409 (found using Minitab) 945 985 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 986 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 Pvalue = 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 946 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.
987 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 Pvalue ≅ 0 947 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.
988 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 Pvalue 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. 948 989 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution
is normal. Use the ttest 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,n1
7) find the sample statistic x = 3.265 s =2.127 n = 20 x−µ and calculate the tstatistic t0 = s/ n 8) Draw your conclusion and find the Pvalue. 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,n1 where tα/2,n1= 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 Pvalue is
between 0.1 and 0.2. Minitab gives a value of 0.124
d.) The confidence interval found in exercise 881 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 ≤ 4.260 949 2.127
20 990 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
2 950 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 991 The parameter of interest is the true,µ.
H0 : µ = µ0
H1 µ ≠ µ0 992 a.) Reject H0 if z0 < zαε 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 951 993 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 = 0.01 994 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 = 0.05 995 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 = 952 996 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 = 0.05 997 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 953
CHAPTER 10 Section 102
1) The parameter of interest is the difference in fill volume, µ1 − µ 2 ( note that ∆0=0) 101. 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) Pvalue = 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 ∆ = 0.04 101 (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 = 9 102. 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. 103 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 104. 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 2 102 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) Pvalue = 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. 105. x1 = 30.87 x2 = 30.68 103 σ 1 = 0.10 σ2 = 0.15
n1 = 12 n2 = 10
a) 90% twosided 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% twosided 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% uppersided 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. 104 106. 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) Pvalue = 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 107. + 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 = 0 105 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) Pvalue ≅ P ( z ≤ −7.25) = 1 − P ( z ≤ 7.25) = 1 − 1 ≅ 0
108. 99% level of confidence, E = 4, and z0.005 = 2.575 z
n ≅ 0.005
E
109. 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
1010. 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. 1011. Catalyst 1 Catalyst 2 106 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. 1012. 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) Pvalue = 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. 1013. 1) The parameter of interest is the difference in mean active concentration, µ1 − µ 2 107 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.
Pvalue = 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 twosided 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. 1014. (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. 108 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 1015 70
60
50
40
30
20
10
5
1
700 750 800 109 1016 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 103
1017. 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) Pvalue = 2P ( t > 0.230) > 2(0.40), Pvalue > 0.80 1010 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
1018. 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 ttest and confidence intervals involving the tdistribution
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. 1011 1019. 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 1012 1020. 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
AndersonDarling Normality Test
ASquared: 0.463
PValue: 0.220 195 Average: 192.067
StDev: 9.43751
N: 15 205 AndersonDarling Normality Test
ASquared: 0.295
PValue: 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 1013 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) Pvalue = 2P (t < 1.19) 0.75 < pvalue < 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*=2n1,
the sample sizes of 15 are inadequate. 1021. 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) Pvalue = 2P (t < −2.83) 2(0.005) < Pvalue < 2(0.010) = 0.010 < Pvalue < 0.020 1014 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 1022. 10.3 10.4 10.5 10.6 10.7 solution solution
Average: 10.4
StDev: 0.230940
N: 10 AndersonDarling Normality Test
ASquared: 0.269
PValue: 0.595 AndersonDarling Normality Test
ASquared: 0.211
PValue: 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) 1015 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) Pvalue = P(t < −4.64): Pvalue < 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 ftlb higher that supplier 1 using α = 0.05.
1023. Using the information provided in Exercise 920, 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. 1016 1024 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) Pvalue = P ( t > 1.79) 0.025 < Pvalue < 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 . 1017 1025. 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
Pvalue = 2P 1026. (t < −5.498) Pvalue < 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 1018 1027.
. 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) Pvalue = 2P(t > 3.03), 2(0.0025) < Pvalue < 2(0.005)
0.005 < Pvalue < 0.010 1019 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.
1028 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.
Pvalue = P(t > 0597), 0.25 < Pvalue < 0.40 1020 1029. If α = 0.01, construct a 99% twosided confidence interval on the difference to answer question 1028.
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. 1030. 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 onesided 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. 1021 a.)
Normal Probability Plot for Brand 1...Brand 2
ML Estimates Brand 1 99 Brand 2 95
90
80 Percent 1031 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.)Pvalue = 2P (t > 2.558) Pvalue ≈ 2(0.01)=0.02 1022 d.) d = 5
0 . 275
2 ( 9 . 09 ) e.) 1β =0.25 β=0.95 Power =10.95=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 1032 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 1023 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.)Pvalue = 2P d.) d= (t < −1.36) Pvalue ≈ 2(0.1)=0.2 0 .2
= 5 .0 7
2(0.01971)
β = 0.2 d = e.) 1β = 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 104
1033. 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. 1024 1034. It is only necessary for the differences to be normally distributed for the paired ttest to be appropriate and
reliable. Therefore, the ttest 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 1035. AndersonDarling Normality Test
ASquared: 0.318
PValue: 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. 1025 1036. 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 1037 d = 868.375 sd = 1290, n = 8
99% confidence interval: d − t α / 2, n −1 sd AndersonDarling Normality Test
ASquared: 0.439
PValue: 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.
1038. 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. 1026 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 1039. AndersonDarling Normality Test
ASquared: 0.315
PValue: 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. 1027 1040. 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.
1041. 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. 1028 1042. 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.
1043. 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.
1044. 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. 1029 2 = 3.53 , n=4 Section 105
1045 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 1046 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
1047. 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. 1030 1048. 2
1) The parameters of interest are the etchrate 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.
1049. 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. 1050. 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. 1031 c) 90% lowersided 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
1051 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% lowersided 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 1032 1052 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. 1053 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.
1054 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 1033 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.
1055 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.
1056. 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. 1034 1057. 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.
1058 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. 1059 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. 1060 1) The parameters of interest are the time to assemble standard deviations, σ1 , σ 2
2
2) H0 : σ1 = σ 2
2 1035 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 106
1061. 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 1036 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. Pvalue = 2(1−P(z < 1.49)) = 0.13622 1062. 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 polishinginduced defects produced by the two
polishing solutions the 0.01 level of significance. Pvalue = 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. 1063. a) Power = 1 − β 1037 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
1064. 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 1038 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
1065. 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.
Pvalue = 2(1−P(z < 3.42)) = 0.00062
1066. 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. 1039 1067 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
1068 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 AndersonDarling Normality Test
ASquared: 0.315
PValue: 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 AndersonDarling Normality Test
ASquared: 0.376
PValue: 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. 1040 1069 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 9hour
Average: 16.3556
StDev: 2.06949
N: 9 11 12 13 14 15 1hour
AndersonDarling Normality Test
ASquared: 0.171
PValue: 0.899 Average: 11.4833
StDev: 2.37016
N: 6 AndersonDarling Normality Test
ASquared: 0.158
PValue: 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).
1070. 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. 1041 1071 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,n1 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,n1 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,n1 where t0.05,7 =1.895.
7) d = 4.125
sd = 1.246
n=8
4.125 − 5
t0 =
= −1986
.
1.246 / 8 1042 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,n1 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. 1072. (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 zvalue used is different..
Which one is used depends upon the level of confidence considered acceptable. 1043 1073 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) 1074 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. 1044 1075 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.
1076 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 1045 = 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. 1077. 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. 1078
p2 a) 1) The parameters of interest are the proportion of lenses that are unsatisfactory after tumblepolishing, 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) 1046 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. 1079. 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
= 1080 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 Pvalue would be P(Z ≥ z0 ). 1047 2 H0 : µ1 = µ 2 1081. 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 1019. 1082 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 1018 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.
1083 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
AndersonDarling Normality Test
ASquared: 0.934
PValue: 0.011 Average: 40.25
StDev: 3.89280
N: 10 1048 AndersonDarling Normality Test
ASquared: 1.582
PValue: 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
AndersonDarling Normality Test
ASquared: 0.381
PValue: 0.329 Average: 41.25
StDev: 1.21952
N: 10 AndersonDarling Normality Test
ASquared: 0.440
PValue: 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.
1084 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 1049 = 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 1083d. 1085 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
AndersonDarling Normality Test
ASquared: 0.384
PValue: 0.323 Average: 755.6
StDev: 0.843274
N: 10 AndersonDarling Normality Test
ASquared: 0.682
PValue: 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. 1050 1086. 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. 1087. 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 AndersonDarling Normality Test
ASquared: 0.526
PValue: 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 Pvalue
5) The test statistic is
d
t0 =
sd / n
6) Reject H0 if the Pvalue is significantly small.
7) d = −0.222
sd = 1.30
n=9
−0.222
t0 =
= −0.512
130 / 9
.
8) Pvalue = 2P(T < 0.512) = 2P(T > 0.512) 2(0.25) < Pvalue < 2(0.40)
0.50 < Pvalue < 0.80
Since the Pvalue 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 1051 1088. 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 AndersonDarling Normality Test
ASquared: 0.518
PValue: 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 pvalue
5) The test statistic is
d
t0 =
sd / n
6) Reject H0 if the Pvalue is significantly small.
7) d = 0.133
sd = 2.065
n = 15
0.133
t0 =
= 0.25
2.065 / 15
8) Pvalue = 2P(T > 0.25)
2(0.40) < Pvalue
0.80 < Pvalue
Since the Pvalue 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= 1β , 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. 1052 1089 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) 1053 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) Pvalue = 2P(t < 2.74), 2(0.005) < Pvalue < 2(0.01)
0.001 < Pvalue < 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 MindExpanding Exercises
1090 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(1α)% 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. 1054 1091 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 . 1055 with respect to n1, n2 and 1092 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). 1093 . 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 1056 1094. 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 Twosample ttest : t 0 = −0.43 Pvalue = 0.68 xd = −10 1095 s d = 0.707 Paired ttest : t 0 = −3162
. a.) θ= p1
p2 and θˆ = ˆ
p1
ˆ
p2 Pvalue ≈ 0 and ln(θˆ) ~ N [ln(θ ), (n1 − x1 ) / n1 x1 + (n2 − x 2 ) / n2 x 2 ] The (1α) 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 (1α) 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 1057 1/ 4 1096 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. 1058 CHAPTER 11 Section 112
111. 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) 112. 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) 1 111 113. 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
7.0251E3
1.25965E3
5.57703
.00001
Analysis of Variance
Source
Sum of Squares
Df Mean Square
FRatio 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
Rsquared = 54.47 percent
Stnd. Error of Est. = 2.39287 ˆ
σ2 = 5.7258
If the calculations were to be done by hand use Equations (117) and (118). Regression Plot
y = 21.7883  0.0070251 x
S = 2.39287 RSq = 54.5 % RSq(adj) = 52.7 % 10 y 114. 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 = 1.679 112 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
FRatio 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
Rsquared = 76.73 percent
Stnd. Error of Est. = 2.96104 ˆ
σ2 = 8.76775
If the calculations were to be done by hand use Equations (117) and (118). 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 115. 40 35 30 25
25 30 35 40 45 Observed 113 50 116. 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
FRatio 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
Rsquared = 99.99 percent
Stnd. Error of Est. = 1.94284 ˆ
σ2 = 3.7746
If the calculations were to be done by hand use Equations (117) and (118). 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
117. a)
Predictor
Constant
x
S = 3.660 Coef
33.535
0.03540 StDev
2.614
0.01663 RSq = 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 RSq(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 114 P
0.047 118. 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
RSq = 80.0%
RSq(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 119. 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
RSq = 74.8%
RSq(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 115 F
53.50 P
0.000 1110. 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
RSq = 87.9%
RSq(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 1111. 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
RSq = 85.2%
RSq(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
92.22 116 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 (straightline) model seems plausible for this situation.
Predictor
Constant
x
S = 99.05 Coef
2625.39
36.962 StDev
45.35
2.967 RSq = 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 RSq(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 1112. ˆ
y = 0.470467 + 20.5673(1) = 21.038 2200
2100
2000
1900
1800
1700
1600 2100 2600 y 117 1113.
1114. 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
1115. 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 117 and 118 can be applied to the new variables using the facts that
i =1 ˆ
ˆ
β 1∗ = β 1
1116. ( 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 1117. and n 0.4 0.6 0.8 1 1.2 watershed 118 1.4 1.6 1.8 2 Section 115
1118. 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 111 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
1119. 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 112 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. 119 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
1120. = .00796 1
73.9 2
+
= 0.04091
20 33991.6 a) Refer to ANOVA table of Exercise 114.
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 104 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,n2 where −t0.005,26 = −2.78 or t0 > t0.005,26 = 2.78
7) Using the results from Exercise 104 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. 1110 1121. Refer to ANOVA of Exercise 115
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 ttest
5) The test statistic is t 0 = β1
se(β1) 6) Reject H0 if t0 < −tα/2,n2 where −t0.025,22 = −2.074 or t0 > t0.025,22 = 2.074
7) Using the results from Exercise 115
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 105
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 Fstatistic is the square of the tstatistic. The Ftest is a restricted to a twosided test, whereas the
ttest could be used for onesided 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 ttest
5) The test statistic is t0 = ˆ
β0
ˆ
se( β0 ) 6) Reject H0 if t0 < −tα/2,n2 where −t0.025,22 = −2.074 or t0 > t0.025,22 = 2.074
7) Using the results from Exercise 115 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. 1111 1122. Refer to ANOVA for Exercise 106
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 106 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. Pvalue < 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,n2 where −t0.005,10 = −3.17 or t0 > t0.005,10 = 3.17
7) Using the results from Exercise 106 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. Pvalue = 0.
d) H0: β0 = 0 H1: β 0 ≠ 0 t0 = − 6.3355 − 0
= − 3 .8
1.66765 Pvalue < 0.005; Reject H0 and conclude that the intercept should be included in the model. 1123. Refer to ANOVA table of Exercise 117
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. Pvalue = 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 1112 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. Pvalue = 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. Pvalue ≅ 0
1124. Refer to ANOVA of Exercise 118
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. Pvalue = 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. Pvalue = 0.12166. 1113 1125. Refer to ANOVA of Exercise 119
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. Pvalue = 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. Pvalue = 0.000078. 1126. Refer to ANOVA of Exercise 1111
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) Pvalue < 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. 1114 1127. Refer to ANOVA of Exercise 1112
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. Pvalue < 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. Pvalue = 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. Pvalue < 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 . Pvalue = 0.0064. 1115 1128. β1 t0 = After the transformation σ 2 / S xx ˆ
ˆ
σ ∗ = bσ . Therefore, t0∗ = 1129. 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 n1 degree of freedom.
2 xi2
ˆ
ˆ
β = 21.031461,σ = 3.611768, and xi2 = 14.7073 .
The tstatistic in part a. is 22.3314 and H 0 : β 0 = 0 is rejected at usual α values. b) From Exercise 1117, 1130. 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 116 and 117
1131. tα/2,n2 = 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. 1116 1132. tα/2,n2 = 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 1133 through 1135: These computer printouts were obtained from Statgraphics. For Minitab users, the
standard errors are obtained from the Regression subroutine.
1133. 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 1117 ) 1134. 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
1135. 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 Yx 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. 1136. 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 1118 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
1137. 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
1138. 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
1139. 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
1140. 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
1141. 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 1119 ) 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 117
Use the results of Exercise 114 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 1142. 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 1120 8 10 12 1143. Use the Results of exercise 115 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)
1144. 4.8 5.8 6.8
Taxes R 2 ≡ 76.73% ; Use the results of Exercise 116 to answer the following questions 1121 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 1145. 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 1122 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 1146. 0 10
2 1 0 Normal Score 1123 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 1147. 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 1148. 4 Fitted Value 0 0 10 10 20 20 30 30
0 10 20 30 40 80 x 130 180 Fitted Value 1124 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 1149. 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 1125 1 2 1150. 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. 1151. 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 n2 degrees of freedom is an F random variable with 1 and n2 degrees
of freedom, the usually ttest that compares  t0  to tα / 2, n − 2 is equivalent to comparing fα ,1, n − 2 = tα / 2, n − 2 . 1126 2
f 0 = t0 to 1152. 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
1153. 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 1154. 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 119
1155. 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 1127 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 .
1156. 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 . 1128 1157. 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 . Pvalue = (3.69E6)(2) = 7.38E6
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 . Pvalue = (0.3468)(2) = 0.6936
1/ 2 1158 Therefore, 1159 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 . Pvalue ≅ 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.
1160. 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 . Pvalue = 2(0.02274) = 0.04548 1129 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 1161. 1
0
1
2
2 1 0 Normal Score 1130 1 2 25 1162. 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 . Pvalue = 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 . Pvalue = (0.3373)(2) = 0.6746. Do not reject
Supplemental Exercises
n 1163. 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 1131 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 1164. 1.3 1 0.7
1.1 1.3 1.5 1.7
x Yes, a straight line relationship seems plausible. 1132 1.9 2.1 b) Model fitting results for: y
Independent variable
coefficient std. error
tvalue
sig.level
CONSTANT
0.966824
0.004845
199.5413
0.0000
x
1.543758
0.003074
502.2588
0.0000
RSQ. (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
FRatio
Pvalue
Model
1.96613
1
1.96613
252264.
.0000
Error
0.0000623515
8 0.00000779394
Total (Corr.)
1.96619
9
Rsquared = 0.999968
Stnd. error of est. = 2.79176E3
Rsquared (Adj. for d.f.) = 0.999964
DurbinWatson 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.
Pvalue < 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,n2 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 = 1133 1165. 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 1166 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 1134 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 1135 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 1167. 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 1136 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) 1168. 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.
1169. 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. 1170. 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 1171 70
60
50
40
30
16 17 index 1137 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 1172 0.4
0.3
0.2
0.3 0.4 0.5 0.6 0.7 x 1138 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 1173 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. 1174 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 1139 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 yxo ≤ 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 1175 a) y 940 930 920 920 930 940 x b) 3 4 5 Fitted Value Residual ˆ
y = 33.3 + 0.9636 x 1140 6 7 8 c)Predictor
Constant
Therm Coef
66.0
0.9299 S = 5.435 RSq = 71.2% SE Coef
194.2
0.2090 T
0.34
4.45 P
0.743
0.002 RSq(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 1141 Residuals Versus the Fitted Values
(response is IR) Residual 5 0 5 920 930 940 Fitted Value MindExpanding Exercises 1176. 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. 1177. 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 1142 = ( 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 1178. 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 1179. . 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 1180. 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. 1143 in which a Yi with small variance and ˆ
β 1= (
( wi xi y i )( ( wi ) 1181. 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
1182. 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 n1 degrees of freedom is
slightly smaller than the corresponding t value based on n2 degrees of freedom.
This confidence interval is shorter because 1144 ...
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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.
 Spring '11
 Dr.IL

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