mont4e_sm_ch03_sec02

# mont4e_sm_ch03_sec02 - CHAPTER 3 Section 3-1 3-1. 3-2. 3-3....

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Unformatted text preview: CHAPTER 3 Section 3-1 3-1. 3-2. 3-3. 3-4. 3-5. The range of X is {0,1,2,...,1000} The range of X is {0,12,...,50} , The range of X is {0,12,...,99999} , The range of X is {0,12,3,4,5} , 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 {0,12,...,100} . Although the range actually obtained from lots typically might not , exceed 10%. { } 3-6. 3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is , {0,12,...} The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is , {0,12,...} The range of X is 3-8. 3-9. 3-10. {0,1,2,...,15} 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 ⎭ The range of X is 3-11. {0,1,2, … ,10000} 3-12.The range of X is {100, 101, …, 150} 3-13.The range of X is {0,1,2,…, 40000) Section 3-2 3-14. 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 a) P(X = 1.5) = 1/3 b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) P ( 0 ≤ X < 2) = P ( X = 0) + P ( X = 1.5) = 1/ 3 + 1/ 3 = e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2 2/3 3-15. All probabilities are greater than or equal to zero and sum to one. 3-1 a) P(X ≤ 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(-1 ≤ X ≤ 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X ≤ -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 3-16. All probabilities are greater than or equal to zero and sum to one. a) P(X≤ 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.5714=0.4286 c) P(2<X<6)=P(X=3)=0.1429 d) P(X≤1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1 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 Probabilities are nonnegative and sum to one. a) P(X = 2) = 3/4(1/4)2 = 3/64 b) P(X ≤ 2) = 3/4[1+1/4+(1/4)2] = 63/64 c) P(X > 2) = 1 − P(X ≤ 2) = 1/64 d) P(X ≥ 1) = 1 − P(X ≤ 0) = 1 − (3/4) = 1/4 3-17. 3-18. 3-19. X = number of successful surgeries. P(X=0)=0.1(0.33)=0.033 P(X=1)=0.9(0.33)+0.1(0.67)=0.364 P(X=2)=0.9(0.67)=0.603 P(X = 0) = 0.023 = 8 x 10-6 P(X = 1) = 3[0.98(0.02)(0.02)]=0.0012 P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576 P(X = 3) = 0.983 = 0.9412 X = number of wafers that pass P(X=0) = (0.2)3 = 0.008 P(X=1) = 3(0.2)2(0.8) = 0.096 P(X=2) = 3(0.2)(0.8)2 = 0.384 P(X=3) = (0.8)3 = 0.512 3-20. 3-21. 3-22. X: the number of computers that vote for a left roll when a right roll is appropriate. p=0.0001. P(X=0)=(1-p)4=0.99994=0.9996 P(X=1)=4*(1-p)3p=4*0.999930.0001=0.0003999 P(X=2)=C42(1-p)2p2=5.999*10-8 P(X=3)=C43(1-p)1p3=3.9996*10-12 P(X=4)=C40(1-p)0p4=1*10-16 P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 X = number of components that meet specifications P(X=0) = (0.05)(0.02) = 0.001 P(X=1) = (0.05)(0.98) + (0.95)(0.02) = 0.068 P(X=2) = (0.95)(0.98) = 0.931 3-23. 3-24. 3-25. 3-26. 3-2 3-27. 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 Section 3-3 3-28. x<0 ⎫ ⎧ 0, ⎪1 / 3 0 ≤ x < 1.5⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨2 / 3 1.5 ≤ x < 2⎬ ⎪5 / 6 2 ≤ x < 3 ⎪ ⎪ ⎪ ⎪ 3≤ x ⎪ ⎩1 ⎭ f X (0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3 where f X (1.5) = P ( X = 1.5) = 1 / 3 f X (2) = 1 / 6 f X (3) = 1 / 6 3-29. 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 ⎪ ⎪ ⎪1 2≤ x ⎪ ⎩ ⎭ f X (−2) = 1 / 8 f X (−1) = 2 / 8 where f X ( 0) = 2 / 8 f X (1) = 2 / 8 f X ( 2) = 1 / 8 a) P(X ≤ 1.25) = 7/8 b) P(X ≤ 2.2) = 1 c) P(-1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4 d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8 3-30. ⎧ 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 where f X (2) = 5 / 25 f X (3) = 7 / 25 f X (4) = 9 / 25 a) P(X < 1.5) = 4/25 b) P(X ≤ 3) = 16/25 c) P(X > 2) = 1 − P(X ≤ 2) = 1 − 9/25 = 16/25 d) P(1 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 1) = 9/25 − 4/25 = 5/25 = 1/5 3-31. 3-3 ...
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## This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.

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