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Unformatted text preview: ANSWERS TO PRACTICES QUESTIONS 5 ACTSC 431/831, FALL 2011 1. We have F S ( x ) = Pr { S x } = 1 625 x 4 , x 5 , E ( S ) = Z 5 x 2500 x 5 dx = 20 3 , V ar ( S ) = Z 5 x 2 2500 x 5 dx ( E ( S )) 2 = 50 9 . Thus, Pr { S (1 + ) E [ S ] } = Pr n S E [ S ] + q V ar ( S ) o = 0 . 95 gives = 0 . 58606 and = 1 . 6576. 2. We have E ( S ) = 1 . 6 and V ar ( S ) = 7 . 94667 . So, the probability is Pr { S 1 . 6 + 7 . 94667 } = Pr { S 9 } = Pr { N = 0 } + X n =1 Pr { X 1 + + X n 9 } Pr { N = n } . Note that Pr { X 1 9 } = 1, Pr { X 1 + X 2 9 } = 1 Pr { X 1 + X 2 > 9 } = 8 / 9, Pr { X 1 + X 2 + X 3 9 } = 1 / 27, and Pr { X 1 + + X n 9 } = 0 for n = 4 , 5 ,.... Thus, Pr { S 1 . 6 + 7 . 94667 } = Pr { S 9 } = 0 . 9778. 3. E [( S 2) + ] = E ( S ) E ( S 2) = 1 . 65 Pr { S = 1 }  2 Pr { S 2 } = 1 . 65 . 45 2(1 . 45) = 0 . 1 ....
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This note was uploaded on 02/08/2012 for the course ACTSC 431 taught by Professor Laundriualt during the Fall '09 term at Waterloo.
 Fall '09
 laundriualt

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