# 08_29 - STAT 410 Example 9 Examples for Fall 2008 Suppose a...

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STAT 410 Examples for 08/29/2008 Fall 2008 Example 9 : Suppose a discrete random variable X has the following probability distribution: P( X = 0 ) = p , P( X = k ) = ! 2 1 k k , k = 1, 2, 3, … a) Find the value of p that would make this a valid probability distribution. Must have = + 1 ! 2 1 k k k p = 1. Since a k k e k a 0 ! = = , 1 ! 2 1 2 1 1 - = = e k k k . Therefore, p + ( 1 2 1 - e ) = 1 and p = 2 1 2 e - . b) Find E ( X ). E ( X ) = x x p x all ) ( = 0 ( ) 2 1 2 e - + = 1 ! 2 1 k k k k = ( ) = - 1 ! 1 2 1 k k k = ( ) = - - 1 1 ! 1 2 1 2 1 k k k = = 0 ! 2 1 2 1 n n n = 2 2 1 e . c) Find the variance of X, Var ( X ). E ( X ( X – 1 ) ) = ( ) = - 2 ! 2 1 1 k k k k k = ( ) = - 2 ! 2 2 1 k k k = ( ) = - - 2 2 ! 2 2 1 4 1 k k k = = 0 ! 2 1 4 1 n n n = 4 2 1 e .

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E ( X 2 ) = E ( X ( X – 1 ) ) + E ( X ) = 2 1 4 3 e . Var
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08_29 - STAT 410 Example 9 Examples for Fall 2008 Suppose a...

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