# ch03 - CHAPTER 3 Section 3-1 3-1 3-2 3-3 3-4 3-5 The range...

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CHAPTER 3 Section 3-1 3-1. The range of X is {} 1000 ,..., 2 , 1 , 0 3-2 The range of X is 012 50 , , ,. .., 3-3. The range of X is 99999 3-4 The range of X is 012345 ,,,,, 3-5. The range of X is { . Because 490 parts are conforming, a nonconforming part must be selected in 491 selections. } } 491 ,..., 2 , 1 3-6 The range of X is { . Although the range actually obtained from lots typically might not exceed 10%. 100 3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is .. 3-8 The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is 3-9. The range of X is 15 ,..., 2 , 1 , 0 3-10 The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8. Therefore the range of X is 1 4 3 8 1 2 5 8 6 8 ,,,, 3-11 The range of X is } 10000 , , 2 , 1 , 0 { 3-12 The range of X is 5000 ,..., 2 , 1 , 0 Section 3-2 3-13. 6 / 1 ) 3 ( 6 / 1 ) 2 ( 3 / 1 ) 5 . 1 ( ) 5 . 1 ( 3 / 1 6 / 1 6 / 1 ) 0 ( ) 0 ( = = = = = = + = = = X X X X f f X P f X P f 3-14 a) P(X=1.5) = 1/3 b) P(0.5< X < 2.7) = P(X=1.5) + P(X=2) = 1/6 + 1/3 = 1/2 c) P(X > 3) = 0 d) P ( 0 X < 2) = P(X=0) + P(X=1.5) = 1/3 + 1/3 = 2/3 e) P(X=0 or X=2) = 1/3 + 1/6 = 1/2 3-15. All probabilities are greater than or equal to zero and sum to one. 3-1

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a) P(X 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(-1 X 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 3-16 All probabilities are greater than or equal to zero and sum to one. a) P(X 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.5714=0.4286 c) P(2<X<6)=P(X=3)=0.1429 d) P(X 1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1 3-17. Probabilities are nonnegative and sum to one. a) P(X = 4) = 9/25 b) P(X 1) = 1/25 + 3/25 = 4/25 c) P(2 X < 4) = 5/25 + 7/25 = 12/25 d) P(X > 10) = 1 3-18 Probabilities are nonnegative and sum to one. a) P(X = 2) = 3/4(1/4) 2 = 3/64 b) P(X 2) = 3/4[1+1/4+(1/4) 2 ] = 63/64 c) P(X > 2) = 1 P(X 2) = 1/64 d) P(X 1) = 1 P(X 0) = 1 (3/4) = 1/4 3-19. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 3-20 P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 3-21. P(X = 0) = 0.02 3 = 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) = 0.98 3 = 0.9412 3-22 X = number of wafers that pass P(X=0) = (0.2) 3 = 0.008 P(X=1) = 3(0.2) 2 (0.8) = 0.096 P(X=2) = 3(0.2)(0.8) 2 = 0.384 P(X=3) = (0.8) 3 = 0.512 3-23 P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 3-24 X = number of components that meet specifications P(X=0) = (0.05)(0.02) = 0.001 P(X=1) = (0.05)(0.98) + (0.95)(0.02) = 0.068 P(X=2) = (0.95)(0.98) = 0.931 3-25. X = number of components that meet specifications P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663 P(X=3) = (0.95)(0.98)(0.99) = 0.92169 3-2
Section 3-3 3-26 where < < < < = x x x x x x F 3 1 3 2 6 / 5 2 5 . 1 3 / 2 5 . 1 0 3 / 1 0 , 0 ) ( 6 / 1 ) 3 ( 6 / 1 ) 2 ( 3 / 1 ) 5 . 1 ( ) 5 . 1 ( 3 / 1 6 / 1 6 / 1 ) 0 ( ) 0 ( = = = = = = + = = = X X X X f f X P f X P f 3-27. where < < < < < = x x x x x x x F 2 1 2 1 8 / 7 1 0 8 / 5 0 1 8 / 3 1 2 8 / 1 2 , 0 ) ( 8 / 1 ) 2 ( 8 / 2 ) 1 ( 8 / 2 ) 0 ( 8 / 2 ) 1 ( 8 / 1 ) 2 ( = = = = = X X X X X f f f f f a) P(X 1.25) = 7/8 b) P(X 2.2) = 1 c) P(-1.1 < X 1) = 7/8 1/8 = 3/4 d) P(X > 0) = 1 P(X 0) = 1 5/8 = 3/8 3-28 where < < < < < = x x x x x x x F 4 1 4 3 25 / 16 3 2 25 / 9 2 1 25 / 4 1 0 25 / 1 0 , 0 ) ( 25 / 9 ) 4 ( 25 / 7 ) 3 ( 25 / 5 ) 2 ( 25 / 3 ) 1 ( 25 / 1 ) 0 ( = = = = = X X X X X f f f f f a) P(X < 1.5) = 4/25

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ch03 - CHAPTER 3 Section 3-1 3-1 3-2 3-3 3-4 3-5 The range...

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