Assignment2

# Assignment2 - ASSIGNMENT#2 SOLUTIONS(1(a Is not a...

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ASSIGNMENT #2 SOLUTIONS (1) (a) Is not a probability function, since p (1) = 1 - 2 5 = - 1 5 < 0. (b) It is a probability function, since 0 p ( y ) 1 for all y and 4 X y =0 y 2 30 = 1 30 4 × 5 × 9 6 = 1 (c) It is not a probability function, since 5 X y =0 1 5 = 6 5 > 1. (2) (a) We must have 6 X y =0 p ( y ) = 1, so 0 . 9 + 6 X y =1 y c = 1 1 c 6 X y =1 y = 0 . 1 1 c 6 × 7 2 = 0 . 1 c = 210 (b) E ( Y ) = 0 × 0 . 9 + 6 X y =1 y y 210 = 1 210 6 X y =1 y 2 = 1 210 6 × 7 × 13 6 = 13 30 0 . 43 E ( Y 2 ) = 0 2 × 0 . 9 + 6 X y =1 y 2 y 210 = 1 210 6 X y =1 y 3 = 1 210 ± 6 × 7 2 ² 2 = 21 10 = 2 . 1 Var( Y ) = E ( Y 2 ) - [ E ( Y )] 2 = 21 10 - ± 13 30 ² 2 = 1721 900 1 . 91 (c) First note that g ( Y ) = ( 0 Y < 1 Y - 1 Y 1 . Therefore, E ( g ( Y )) = 0 × 0 . 9 + 6 X y =1 ( y - 1) y 210 = 1 210 6 X y =1 y 2 - 6 X y =1 y = 1 210 ± 6 × 7 × 13 6 - 6 × 7 2 ² = 1 3 (3) (a) Y Bin(25 , 0 . 7) (b) P ( Y = 12) = ± 25 12 ² 0 . 7 12 0 . 3 13 = 164 14291 0 . 011 (c) P ( Y 13) = 1 -

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Assignment2 - ASSIGNMENT#2 SOLUTIONS(1(a Is not a...

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