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05 Discrete random variables examples - solutions

05 Discrete random variables examples - solutions -...

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University of California, Los Angeles Department of Statistics Statistics 100A Instructor: Nicolas Christou Examples - solutions Example 1: E ( N ) = E ( Cπr 2 ) = CπEr 2 = ( σ 2 + μ 2 ) . The mean and variance of the distribution of r are: E ( r ) = 23 . 5, var ( r ) = 1 . 54. Therefore, E ( N ) = 8 π (1 . 54 + 23 . 4 2 ) = 13800 . 39 13801 . Or simply compute E ( N ) as: E ( N ) = E ( Cπr 2 ) = X r r 2 P ( r ) = 8 π 21 2 (0 . 05) + 22 2 (0 . 20) + . . . + 26 2 (0 . 05) = 13800 . 39 13801 . Example 2: It is given that P ( X = i ) = cP ( X = i - 1) for i = 1 , 2. Let P(X=0)=p. P ( X = 1) = cP ( X = 0) = cp , and P ( X = 2) = cP ( X = 1) = c 2 p . Also, P ( X = 0) + P ( X = 1) + P ( X = 2) = 1. Or p + cp + c 2 p = 1 p = 1 1+ c + c 2 . The expected value of X is: E ( X ) = 0( p ) + 1( cp ) + 2( c 2 p ) = c 1 + c + c 2 + 2 c 2 1 + c + c 2 = c (1 + 2 c ) 1 + c + c 2 . Example 3: There are 8 white, 4 black and 2 orange balls. Two balls are selected without replacement. For each black we win $2, for each white we lose $1. We neither win nor we lose anything if we select an orange ball. a. Color X ($) P ( X ) W W - 2 8 14 7 13 = 56 182 W O or OW - 1 2 8 14 2 13 = 32 182 OO 0 2 14 1 13 = 2 182 BW or W B 1 2 4 14 8 13 = 64 182 BO or OB 2 2 4 14 2 13 = 16 182 BB 4 4 14 3 13 = 12 182 b.
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