ch03b - Section 3-10 3-117. a) E(X) = 300(0.4) = 120, V(X)...

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Section 3-10 3-117. a) E(X) = 300(0.4) = 120, V(X) = 300(0.4)(0.6) = 72 and 72 = σ X . Then, 0002 . 0 ) 54 . 3 ( 72 120 90 ) 90 ( = - = - 2245 Z P Z P X P b) ) 54 . 3 89 . 5 ( 72 120 90 72 120 70 ) 90 70 ( - < - = - < - 2245 < Z P Z P X P = 0.0002 – 0 = 0.0002 3-118. a) ( 29 ( 29 93 7 99 1 100 0 9 . 0 1 . 0 7 100 ... 9 . 0 1 . 0 1 100 9 . 0 1 . 0 0 100 ) 8 ( + + + = < X P = 0.2061 b) E(X) = 10, V(X) = 100(0.1)(0.9) = 9 and σ X = 3 . Then, 2524 . 0 ) 667 . 0 ( ) ( ) 8 ( 3 10 8 = - < = < 2245 < - Z P Z P X P c) 4971 . 0 ) 67 . 0 Z 67 . 0 ( P 3 10 12 Z 3 10 8 P ) 12 X 8 ( P = < < - = - < < - 2245 < < 3-119. 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. a) 1294 . 0 ) 13 . 1 Z ( P 1 ) 13 . 1 Z ( P 6 . 19 20 25 Z P ) 25 X ( P = - = = - 2245 b) P X P Z P Z ( ) ( ) ( . ) . 20 30 0 0 2 26 10 19 6 < < 2245 < < = < < 4881 . 0 5 . 0 98809 . 0 ) 0 Z ( P ) 26 . 2 Z ( P = - = < - = 3-120. Let X = number of defective inspected parts E(X) = 100(0.08) = 8 V(X) = 100(0.08)(0.92) = 7.36 a) P(X < 8) = P(X 7) = = - 7 0 100 ) 92 . 0 ( ) 08 . 0 ( 100 i i i i = 0.4471 b) 5 . 0 ) 0 ( 36 . 7 8 8 ) 8 ( = < = - < 2245 < Z P Z P X P 3-121. 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 = 0.005 and n = 2000. Also, E(X) = 2000(0.005) = 10 and V(X) = 2000(0.005)(0.995) = 9.95. 9441 . 0 055917 . 0 1 ) 59 . 1 ( 1 ) 59 . 1 ( 95 . 9 10 5 ) 5 ( = - = - < - = - = - 2245 Z P Z P Z P X P . 3-122. Let X denote the number of particles in 10 cm 2 of dust. Then, X is a Poisson random variable with λ = 10(1000) = 10,000. Also, E(X) = λ = 10,000 and V(X) = λ 2 = 10 8 . P X P Z P Z ( , ) , , ( ) . 2245 - = = 10 000 10 000 10 000 10 0 05 8 3-123. E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5
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a) P X P X P Z P Z ( ) ( . ) . . ( . ) = 2245 -  = ≤ - 2 2 5 2 5 5 4 5 118 = - = - = 1 118 1 0881 0119 P Z ( . ) . . b) 0793 . 0 ) 41 . 1 Z ( P 5 . 4 5 2 Z P ) 2 X ( P = - = - 2245 c) P X ( ) . . . . . . . = + + = 2 50 0 01 0 9 50 1 01 0 9 50 2 01 0 9 0118 0 50 1 49 2 48 The probability computed using the continuity correction is closer. d) P X P X P Z P Z ( ) ( . ) . . ( . ) . = 2245 -  = = 10 105 105 5 4 5 2 59 0 995 e) P X P X P Z P Z ( ) ( . ) . . ( . ) . < = 2245
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This note was uploaded on 03/15/2009 for the course IE 315 taught by Professor Kapur during the Spring '09 term at University of Washington.

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ch03b - Section 3-10 3-117. a) E(X) = 300(0.4) = 120, V(X)...

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