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Answers to Practice Questions 1 – ACTSC 431/831, FALL 2011 1. (a) F Y ( y ) = Pr { Y y | N = 0 } Pr { N = 0 } + Pr { Y y | 1 N 5 } Pr { 1 N 5 } + Pr { Y y | N > 5 } Pr { N > 5 } = ( 0 , y < 0 , 1 - 0 . 5 e - y/ 100 - 0 . 2 e - y/ 200 , y 0 . (b) The probability is Pr { Y > 500 } = 1 - F Y (500) = 0 . 019786. (c) Pr { 0 < Y < 300 } = F Y (300 - ) - F Y (0) = 0 . 63048. (d) The mean of the total loss is E ( Y ) = 0 . 3 × 0 + 0 . 5 × 100 + 0 . 2 × 200 = 90. (e) The 20th percentile of Y is π 0 . 2 = 0. (f) The 80th percentile of Y is the root of equation: F Y ( y ) = 0 . 8 , y > 0. Solving the equation, we obtain that the 80th percentile of Y is π 0 . 8 = 153 . 87 . 2. (a) Note that 0 Y L 70. F Y L ( y ) = 0 , y < 0 , Pr { X - 80 y } = 80+ y 200 , 0 y < 40 , Pr { 40 + 0 . 75( X - 120) y } = y +50 150 , 40 y < 70 , 1 , y 70 . (b) Note that

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