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Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 Professors Hayriye Ayhan and Jim Dai Solutions to Homework 1 1. (a) Poisson (b) exponential (c) geometric (d) Bernoulli (e) binomial (f) normal 2. We are given that E [ X ] = 2 and V ar ( X ) = 16. Thus a) E [6 2 X ] = 6 2 E [ X ] = 2 and V ar (6 2 X ) = 4 V ar ( X ) = 64. b) E [( X 3) / 4] = 1 4 E [ X ] 3 4 = 1 / 4 and V ar (( X 3) / 4) = 1 16 V ar ( X ) = 1. 3. We know that a) 1 = 3 X k =1 P ( X = 2 k 1) = 3 X k =1 (2 k 1) c = 9 c so it follows that c = 1/9. b) E [ X ] = 3 i =1 (2 k 1) P ( X = 2 k 1) = (1 / 9)(1 + 9 + 25) = 35 / 9 . c) E [ X 2 ] = 3 i =1 (2 k 1) 2 P ( X = 2 k 1) = (1 / 9)(1 + 27 + 125) = 17 . d) V ar ( X ) = E [ X 2 ] ( E [ X ]) 2 = 17 (35 / 9) 2 = 1 . 8765 . e) Note that ( X 2) + = 0 with probability 1 / 9 + 3 / 9 2 with probability 5 / 9 Hence, E [( X 2) + ] = 2 * 5 / 9 = 10 / 9 ....
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This note was uploaded on 01/16/2012 for the course ISYE 3232 taught by Professor Billings during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Billings

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