stat1301hw3sol

stat1301hw3sol - 11/12 THE UNIVERSITY OF HONG KONG...

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11/12 p. 1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1301 Probability and Statistics I Assignment 3 Solution 1. (a) 4 3 (b) () < < + = 1 1 1 1 3 2 4 1 1 0 3 x x x x x x F X (c) 0, 5 1 (d) 0.6575 (e) = y y y f Y 1 4 3 , 1 0 < < y 2. (a) For , 5 > x () ( ) x t dt t x X P x F x x 5 1 5 5 5 5 2 = = = = Hence > = 5 0 5 5 1 x x x x F . (b) ( ) 12 5 12 5 1 1 12 1 12 = = = F X P (c) [] = = = × = 5 5 5 2 log 5 5 5 x dx x dx x x X E = = = × = 5 5 5 2 2 2 5 5 5 dx dx x x X E The random variable X does not have finite mean and finite variance. (d) [ ] 472 . 4 5 10 10 5 5 5 2 1 5 2 3 5 2 = = = = × = x dx x dx x x X E (e) ( ) 12 5 12 hours 12 least at for function can device a = = X P P Let Y be the number of devices out of 8 that can function for at least 12 hours. Then with the assumption of independence among the lifetimes, 12 5 , 8 ~ b Y . Hence 7185 . 0 12 7 12 5 2 8 12 7 12 5 1 8 12 7 1 3 6 2 7 8 = = Y P .
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11/12 p. 2 4. (a) 32 3 (b) () < < + = 2 1 2 2 12 16 32 1 2 0 3 x x x x x x F X (d) 1 (e) No 5. (a) = y t Y dt e y F 2 1 λ For , 0 y y y t y t Y e e dt e y F 2 1 2 1 2 1 = = = For , 0 > y () () + = y t Y Y dt e F y F 0 2 1 0 y y t e e = + = 2 1 1 2 1 2 1 0
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This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.

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stat1301hw3sol - 11/12 THE UNIVERSITY OF HONG KONG...

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