hw5 - Solution of HW5 by Yan Wang and Wenhua Gao 17 The...

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Solution of HW5 by Yan Wang and Wenhua Gao 17. The mean of X EX = X x xP ( X = x ) = ( - 3)(0 . 2) + ( - 1)(0 . 3) + (1 . 5)(0 . 4) + (2)(0 . 1) = - 0 . 1 . To compute the variance of X , Frst we compute EX 2 . EX 2 = X x x 2 P ( X = x ) = ( - 3) 2 (0 . 2) + ( - 1) 2 (0 . 3) + (1 . 5) 2 (0 . 4) + (2) 2 (0 . 1) = 3 . 4 . Then var( X ) = EX 2 - ( EX ) 2 = 3 . 4 - ( - 0 . 1) 2 = 3 . 39 . s.d.( X ) = p var( X ) = 1 . 841195 . 19. The mean of X EX = 10 X k =1 kP ( X = k ) = 10 X k =1 k · 1 10 = 5 . 5 . To compute the variance of X , Frst we compute EX 2 . EX = 10 X k =1 k 2 P ( X = k ) = 10 X k =1 k 2 · 1 10 = 38 . 5 . var( X ) = EX 2 - ( EX ) 2 = 38 . 5 - (5 . 5) 2 = 8 . 25 . 20. The mean of X EX = n X k =1 kP ( X = k ) = n X k =1 k · 1 n = 1 n n X k =1 k = n + 1 2 . To compute the variance of X , Frst we compute EX 2 . EX = n X k =1 k 2 P ( X = k ) = n X k =1 k 2 · 1 n = 1 n n X k =1 k 2 = ( n + 1)(2 n + 1) 6 . var(
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This note was uploaded on 04/03/2008 for the course MATH 3C taught by Professor Schonmann during the Spring '07 term at UCLA.

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hw5 - Solution of HW5 by Yan Wang and Wenhua Gao 17 The...

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