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# assignment3solution - 10/11 THE UNIVERSITY OF HONG KONG...

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10/11 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 y y x x F (c) 0, 5 1 (d) 0.6575 (e) ( ) ( ) y y y f Y 4 1 3 = , 1 0 < y 2. (a) For , 10 > x ( ) ( ) x t dt t x X P x F x x 10 1 10 10 0 1 10 2 = = = = Hence ( ) > = 10 0 10 10 1 x x x x F . (b) ( ) ( ) 3 2 15 10 1 1 15 1 15 = = = F X P (c) ( ) [ ] = = = × = 10 10 10 2 log 10 10 10 x dx x dx x x X E ( ) [ ] = = = × = 10 10 10 2 2 2 10 10 10 dx dx x x X E The random variable X does not have finite mean and finite variance. (d) ( ) [ ] 3246 . 6 10 20 20 10 10 10 2 1 10 2 3 10 2 = = = = × = x dx x dx x x X E (e) For any , 1 0 < < p ( ) p x p x p x F = = = 1 10 10 1 . Therefore ( ) p p F = 1 10 1 , 1 0 < < p . Lower quartile: ( ) 3 40 25 . 0 1 10 25 . 0 1 25 . 0 = = = F x Median: ( ) 20 5 . 0 1 10 5 . 0 1 5 . 0 = = = F x Upper quartile: ( ) 40 75 . 0 1 10 75 . 0 1 75 . 0 = = = F x

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