stat2601hw3sol.pdf - 13/14 Spring THE UNIVERSITY OF HONG...

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13/14 Spring p. 1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2601 Probability and Statistics I Assignment 3 Solution 1. (a) 2 1 1 1 1 2 1 2 1 2 x x dx x Hence 2 k . (b) For 2 1 x ,   x x x x t t dt t x F x x X 2 1 1 2 1 2 2 1 2 1 2 1 1 2 Hence   2 1 2 1 1 2 1 0 2 x x x x x x F X . (c) 3 . 1 8 . 1 8 . 1 3 . 1 F F X P 3 . 1 1 3 . 1 2 8 . 1 1 8 . 1 2 2 2 5726 . 0 (d) 6137 . 1 2 log 2 3 log 2 1 1 2 2 1 2 2 1 2 x x dx x x X E 3 8 2 3 2 2 2 1 1 2 2 1 3 2 1 2 2 1 2 2 2 x x dx x dx x x X E 0626 . 0 2 log 2 3 3 8 2 X Var (e) 5 . 0 1 2 5 . 0 5 . 0 2 5 . 0 5 . 0 x x x F 0 4 9 4 5 . 0 2 5 . 0 x x 8 17 9 4 2 4 4 4 9 9 2 5 . 0 x The root 1 8 17 9 is rejected as it lies out of the support. Therefore the median of X is 6404 . 1 8 17 9 5 . 0 x .
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