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Unformatted text preview: Answers to Practice Questions 2 ACTSC 431/831, FALL 2011 1. Let X 1 = 1 / X and X 2 = e X . (a) We have E ( X 1 ) = R 1 x f X ( x ) dx = and E ( X 2 ) = E ( e X ) = M X (1) = 16 / 9 . Thus, E ( Y ) = 0 . 2 E ( X 1 ) + 0 . 8 E ( X 2 ) = 1 . 7767 . (b) The probability is Pr { Y > E ( Y ) } = 0 . 2Pr { X 1 > E ( Y ) } + 0 . 8Pr { X 2 > E ( Y ) } , where Pr { X 1 > E ( Y ) } = Pr { X < 1 / 1 . 7767 2 } = 0 . 3615 and Pr { X 2 > E ( Y ) } = Pr { X > ln1 . 7767 } = 0 . 33108 . Hence, Pr { Y > E ( Y ) } = 0 . 3372. (c) We have E ( X 2 1 ) = R 1 x f X ( x ) dx = 4 and E ( X 2 2 ) = E ( e 2 X ) = M X (2) = 4 . Thus, E ( Y 2 ) = 0 . 2 E ( X 2 1 ) + 0 . 8 E ( X 2 2 ) = 4 . Hence, V ar ( Y ) = 0 . 84334. 2. (a) i. E ( Y A ) = 1 2 ( E ( X 1 ) + E ( X 2 )) = 30 and E ( Y B ) = 1 2 ( E ( X 1 ) + E ( X 2 )) = 30. ii. E ( Y 2 A ) = 1 2 ( E ( X 2 1 ) + E ( X 2 2 )) = 1306 . 5 and thus V ar ( Y A ) = 406 . 5. Further more, V ar ( Y B ) = 1 4 ( V ar ( X 1 ) + V ar ( X 2 )) = 3 .....
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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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