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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ( = = Z P Z P X P b) ) 54 . 3 89 . 5 ( 72 120 90 72 120 70 ) 90 70 ( < = < < Z P Z P X P = 0.0002 – 0 = 0.0002 3-118. a) () () 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 = < = < < Z P Z P X P c) 4971 . 0 ) 67 . 0 Z 67 . 0 ( P 3 10 12 Z 3 10 8 P ) 12 8 ( = < < = < < X P < < 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 = = > = > > b) PX PZ () ( ) ( . 20 30 0 0 2 26 10 19 6 << ≅ << = . ) 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) = = 0.4471 = 7 0 100 ) 92 . 0 ( ) 08 . 0 ( 100 i i i i b) 5 . 0 ) 0 ( 36 . 7 8 8 ) 8 ( = < = < < 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 ( = = < = = Z P Z P Z P X P . 3-122. Let X denote the number of particles in 10 of dust. Then, X is a Poisson random variable with cm 2 λ = 10(1000) = 10,000. Also, E(X) = λ = 10,000 and V(X) = λ 2 = 10 8 . (, ) ,, >≅ > => = 10 000 10 000 10 000 10 00 5 8 . 3-123. E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
a) PX PZ () (. ) . . ≤= ≤ ≅ =≤ 22 5 25 5 45 118 ) =− = 1 118 1 0881 0119 ) . . b) 0793 . 0 ) 41 . 1 Z ( P 5 . 4 5 2 Z P ) 2 X ( P = = c) . . . . . . . + + = 2 50 0 01 09 50 1 0109 50 2 0118 05 0 14 9 24 8 The probability computed using the continuity correction is closer. d) ) . . ) . ≤= ≤ ≅ ≤ =≤ = 10 105 5 2 59 0 995 e) ( . ) . . ) . <= ≤ ≅ ≤ 10 9 5 95 5 212 0 983 3-124. E(X) = 50(0.1) = 5 and V(X) = 50(0.1)(0.9) = 4.5 a) ) . . ) . ≥= ≥ ≅ =≥ −= 21 5 15 5 165 0 951 b) . ) . ≥≅ − = 2 1414 0 921 c) () () . . . <= = 50 0 50 1 01 0 9 0 966 0 9 The probability computed using the continuity correction is closer. d ) ) ) . = 6 55 0 24 0 4052 e ) ) ) . >= = 6 7 65 0 707 0 24 Section 3-11 3-125. a) P(X < 9, Y < 2.5) = P(X < 9) P(Y < 2.5) 2457 . 0 ) 97725 . 0 )( 25143 . 0 ( ) 2 Z ( P ) 67 . 0 Z ( P 25 . 2 5 . 2 Z P 5 . 1 10 9 Z P = = < < = < < = b) P(X > 8, Y < 2.25) = P(X > 8)P(Y < 2.25) 7641 . 0 ) 84134 . 0 )( 90824 . 0 ( ) 1 Z ( P )) 33 . 1 Z ( P 1 ( ) 1 Z ( P ) 33 . 1 Z ( P 25 . 2 25 . 2 Z P 5 . 1 10 8 Z P = = < < < > = < > = =
Background image of page 2
c) P(8.5 X 11.5, Y > 1.75) = P(8.5 X 11.5)P(Y > 1.75) 5743 . 0 ) 84134 . 0 )( 68268 . 0 ( )) 1 Z ( P 1 )( 1 Z 1 ( P ) 1 Z ( P ) 1 Z 1 ( P 25 . 2 75 . 1 Z P 5 . 1 10 5 . 11 Z 5 . 1 10 5 . 8 P = = < = > = > = d) P(X < 13, 1.5 Y 1.8) = P(X < 13)P(1.5 Y 1.8) 1848 . 0
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue University-West Lafayette.

Page1 / 21

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online