Homework05Solutions

Homework05Solutions - CHEN 3010 Applied Data Analysis...

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CHEN 3010 Applied Data Analysis Fall 2004 Homework 05 Solutions Problem 1 3-132. Let X denote the production yield on a day. Then, P(X > 1400) = PZ > => = < = 1400 1500 10000 11 10 8413 () . 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. The answer is then P(Y = 5) = 5 5 0 8413 0 1587 0 4215 50 = (. )(. ) . b) P(Y = 0) = 5 0 0 8413 0 1587 0 0001 05 = ) . Problem 2 3-134. a) P(C 1 )P(C 2 )P(C 3 ) = (0.90)(0.99)(0.95) = 0.8465 b) 1 – 0.8465 = 0.1535 Problem 3 3-136. a) P(C 1 or C 2 or C 3 ) = 1 – P(C 1 ) P(C 2 ) P(C 3 ) = 1 – (0.10)(0.01)(0.05) = 0.99995 Problem 4 3-144. Let D denote the width of the casing minus the width of the door. Then, D is normally distributed. a) E(D) = 1/8 V(D) = 1 8 1 16 5 256 22 + = b) PD ) ( . ) . . >= > = <= = 1 4 14 18 5 256 08 91 89 1 0 8133 0 1867 c) ( . ) . < =< = 0 01 8 5 256 90 1867 Problem 5 4-16. Use n = 5, everything else held constant: a) P( X 98.5) + P( X 101.5) = P X 100 25 98 5 100 / . / + P X 100 101 5 100 / . / = P(Z 1.68) + P(Z 1.68) = 0.093 b) β = P(98.5 X 101.5 when μ = 103) = P X 98 5 103 103 101 5 103 . // . /
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This note was uploaded on 10/11/2011 for the course CHEN 3010 taught by Professor Kompala,dh during the Spring '08 term at Colorado.

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Homework05Solutions - CHEN 3010 Applied Data Analysis...

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