S10 Sol

# S10 Sol - Exam#1 Solutions IE 336 Spring 2010 1 The pdf in...

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Unformatted text preview: Exam #1 Solutions IE 336 Spring 2010 1. The pdf in this problem is discrete (not continuous). Therefore, to compute c and E ( X ), we must sum (not integrate) over the possible values the pdf can take. (a) (10 points) Determine c . Recall that ∑ N x =1 x = N ( N +1) 2 . 1 = X x p ( x ) = N/ 3 X x =1 cx + N X x = N/ 3+1 2 cx = c N/ 3 X x =1 x + 2 c N X x =1 x- N/ 3 X x =1 x = 2 c N X x =1 x- c N/ 3 X x =1 x = 2 c · N ( N + 1) 2- c · ( N/ 3)( N/ 3 + 1) 2 = c · 17 N 2 + 15 N 18 Therefore, c = 18 17 N 2 +15 N . (b) (10 points) Determine E ( X ). Recall that ∑ N x =1 x 2 = N ( N +1)(2 N +1) 6 . E ( X ) = X x xp ( x ) = N/ 3 X x =1 cx 2 + N X x = N/ 3+1 2 cx 2 = c N/ 3 X x =1 x 2 + 2 c · N X x =1 x 2- N/ 3 X x =1 x 2 = 2 c N X x =1 x 2- c N/ 3 X x =1 x 2 = 2 c · N ( N + 1)(2 N + 1) 6- c · ( N/ 3)( N/ 3 + 1)(2 N/ 3 + 1) 6 = c · 106 N 3 + 153 N 2 + 45 N 162 where c is from part (a)....
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## This note was uploaded on 05/05/2011 for the course IE 336 taught by Professor Bruce,s during the Spring '08 term at Purdue.

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S10 Sol - Exam#1 Solutions IE 336 Spring 2010 1 The pdf in...

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