HW4%20solution

HW4%20solution - IE 300 / GE 331 Spring 2009 Homework #4...

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IE 300 / GE 331 Spring 2009 Homework #4 Solutions Question 1 Associate a success with a paper that receives a grade that has not been received before. Let X i be the number of papers between the i th success and the ( i + 1)st success. Then we have X = 1 + 5 i = 1 X i and hence: E [ X ] = 1 + 5 X i = 1 E [ X i ] . After receiving i - 1 di erent grades so far ( i - 1 successes), each subsequent paper has probability (6- i ) / 6 of receiving a grade that has not been received before. Therefore, the random variable X i is geometric with parameter p i = (6 - i ) / 6, so E [ X i ] = 6 / (6 - i ). It follows that E [ X ] = 1 + 5 X i = 1 6 6 - i = 1 + 6 5 X i = 1 1 i = 14 . 7 . Question 2 Let X be the waiting time and Y be the number of customers found. For x < 0, we have F X ( x ) = 0, while for x 0, F X ( x ) = P ( X x ) = 1 2 P ( X x | Y = 0) + 1 2 P ( X x | Y = 1) . Since, P ( X x | Y = 0) = 1 , P ( X x | Y = 1) = 1 - e - λ x , we obtain F X ( x ) = 1 2 (2 - e - λ x ) , if x 0 , 0 , otherwise. CDF has a discontinuity at x = 0. X is neither discrete nor continuous. 1
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Question 3 (a) We first calculate the CDF of X . For x [0 , r ], we have F X ( x ) = P ( X x ) = π x 2 π r 2 = ± x r ² 2 . For x < 0, we have F X ( x ) = 0, and for x > r , we have F X ( x ) = 1. By di erentiating, we obtain the PDF f X ( x ) = 2 x r 2 , if 0 x r , 0 , otherwise. We have E [ X ] = Z r 0 2 x 2 r 2 dx = 2 r 3 . Also, E [ X 2 ] = Z r 0 2 x 3 r 2 dx = r 2 2 .
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HW4%20solution - IE 300 / GE 331 Spring 2009 Homework #4...

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