STAT1301 (Assig3 ans)full

STAT1301 (Assig3 ans)full - 09/10 p. 1 THE UNIVERSITY OF...

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Unformatted text preview: 09/10 p. 1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1301 Probability and Statistics I Assignment 3 Solution 1. (a) ( ) 3 4 3 1 1 1 3 1 1 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − − − ∫ x x dx x Hence 4 3 = c . (b) For , 1 1 < < − x ( ) ( ) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − = − − ∫ 3 2 3 4 3 3 4 3 1 4 3 3 1 3 1 2 x x t t dt t x F x x X ( ) 3 3 2 4 1 x x − + = Hence ( ) ( ) ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ≥ < < − − + − ≤ = 1 1 1 1 3 2 4 1 1 3 x x x x x x F X . (c) ( ) ( ) 4 3 1 1 3 = − = ∫ − dx x x X E (the integrant is an odd function) ( ) ( ) 5 1 5 3 4 3 4 3 1 1 5 3 1 1 4 2 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − = − − ∫ x x dx x x X E ( ) ( ) ( ) 5 1 2 2 = − = X E X E X Var (d) ( ) ( ) ( ) 3 . 7 . 7 . 3 . − − = ≤ < − F F X P ( ) ( ) ( ) ( ) 3 3 3 . 3 . 3 2 4 1 7 . 7 . 3 2 4 1 − − − × + − − × + = 671 . = (e) For , 1 < < y ( ) ( ) ( ) ( ) y X y P y X P y Y P y F Y ≤ ≤ − = ≤ = ≤ = 2 ( ) ( ) y F y F X X − − = ( ) ( ) 2 3 2 3 3 2 4 1 3 2 4 1 y y y y + − − − + = ( 2 3 3 2 1 y y − = ) Hence ( ) ( ) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − = = − y y y y y F y f Y Y 1 4 3 4 3 4 3 ' 2 1 2 1 , . 1 < < y 09/10 p. 2 2. (a) For , 10 > x ( ) ( ) x t dt t x X P x F x x 10 1 10 10 1 10 2 − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = = ≤ = ∫ Hence ( ) ⎪ ⎩ ⎪ ⎨ ⎧ ≤ > − = 10 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 ln 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 < < p ( ) p x p x p x F − = ⇒ = − ⇒ = 1 10 10 1 . Therefore ( ) p p F − = − 1 10 1 , 1 < < p . Lower quartile: ( ) 3 40 25 . 1 10 25 . 1 25 . = − = = − F x Median: ( ) 20 5 . 1 10 5 . 1 5 . = − = = − F x Upper quartile: ( ) 40 75 . 1 10 75 . 1 75 . = − = = − F x (c) ( ) ( ) 3 2 15 hours 15 least at for function can device a = ≥ = X P P Let Y be the number of devices out of 6 that can function for at least 15 hours. Then with the assumption of independence among the lifetimes, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 3 2 , 6 ~ b Y . Hence ( ) 8999 . 3 1 3 2 2 6 3 1 3 2 1 6 3 1 1 3 4 2 5 6 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ≥ Y P . 09/10 p. 3 3. (a) From the definition of F , we know that ( ) ( ) u u F F ≥ for all ( ) 1 , ∈ u . Since F is a non- decreasing function, we have ( ) ( ) ( ) ( ) ( ) x F u x F u F F x u F ≤ ⇒ ≤ ⇒ ≤ ( ( ) ( ) u u F F ≥ ) On the other hand, since ( ) u F is the minimum...
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This note was uploaded on 05/04/2011 for the course STAT 1301 taught by Professor Smslee during the Spring '08 term at HKU.

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STAT1301 (Assig3 ans)full - 09/10 p. 1 THE UNIVERSITY OF...

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