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Unformatted text preview: Homework 4 Solutions Problem Solutions : Yates and Goodman, 2.7.4 2.8.3 2.8.5 2.9.2 2.10.2 2.10.3 3.1.2 3.2.4 and 3.3.5 Problem 2.7.4 Solution Given the distributions of D , the waiting time in days and the resulting cost, C , we can answer the following questions. (a) The expected waiting time is simply the expected value of D . E [ D ] = 4 X d =1 d P D ( d ) = 1(0 . 2) + 2(0 . 4) + 3(0 . 3) + 4(0 . 1) = 2 . 3 (1) (b) The expected deviation from the waiting time is E [ D D ] = E [ D ] E [ d ] = D D = 0 (2) (c) C can be expressed as a function of D in the following manner. C ( D ) = 90 D = 1 70 D = 2 40 D = 3 40 D = 4 (3) (d) The expected service charge is E [ C ] = 90(0 . 2) + 70(0 . 4) + 40(0 . 3) + 40(0 . 1) = 62 dollars (4) Problem 2.8.3 Solution From the solution to Problem 2.4.2, the PMF of X is P X ( x ) = . 2 x = 1 . 5 x = 0 . 3 x = 1 otherwise (1) The expected value of X is E [ X ] = X x xP X ( x ) = ( 1)(0 . 2) + 0(0 . 5) + 1(0 . 3) = 0 . 1 (2) The expected value of X 2 is E X 2 = X x x 2 P X ( x ) = ( 1) 2 (0 . 2) + 0 2 (0 . 5) + 1 2 (0 . 3) = 0 . 5 (3) The variance of X is Var[ X ] = E X 2 ( E [ X ]) 2 = 0 . 5 (0 . 1) 2 = 0 . 49 (4) 1 Problem 2.8.5 SolutionProblem 2....
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This note was uploaded on 04/03/2008 for the course EEE 350 taught by Professor Duman during the Fall '08 term at ASU.
 Fall '08
 Duman

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