STAT1301 (Assig2 ans)

STAT1301 (Assig2 ans) - 09/10 THE UNIVERSITY OF HONG KONG...

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09/10 p. 1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1301 Probability and Statistics I Assignment 2 Solution 1. (a) (b) 2 , 1 , 0 , 1 () = = = = = 2 3 1 1 5 2 0 5 1 1 15 1 x x x x x p (c) < < < < = 2 1 2 1 3 2 1 0 15 4 0 1 15 1 1 0 x x x x x x F 2. = = = = = 12 , 6 9 1 4 12 1 30 , 24 , 20 , 18 , 15 , 10 , 8 , 5 , 3 , 2 18 1 36 , 25 , 16 , 9 , 1 36 1 i i i i i p 3. + + = n n m i n i n m i p 1 , n i ,..., 2 , 1 , 0 = 4. (a) () ( ) 55 1 1 10 2 1 1 10 1 = = + + + = = c c x p x L () () ( )( ) 7 6 1 20 1 10 10 55 1 55 1 10 1 2 10 1 = + + × = = = = = x x x x xp X E ( ) 55 4 1 10 10 55 1 55 1 2 2 10 1 3 10 1 2 2 = + × = = = = = x x x x p x X E 6 7 55 2 = = X Var (b) 25 12 1 4 1 3 1 2 1 1 1 1 4 1 = = + + + = = c c x p x 25 48 1 25 12 4 1 4 1 = × = = = = x x x x x xp X E 5 24 4 3 2 1 25 12 1 25 12 4 1 2 4 1 2 2 = + + + × = × = = = = x x x x x p x X E 1136 . 1 25 48 5 24 2 = = X Var (c) ( ) 9 1 1 1 1 1 0 1 1 1 2 2 2 1 1 = = + + + + + = = c c x p x ( ) 0 1 1 1 1 0 0 1 1 1 9 1 2 2 2 1 1 = + × + + × + + × = = = x x xp X E
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09/10 p. 2 () () () ( ) 9 8 1 1 1 1 0 0 1 1 1 9 1 2 2 2 2 2 2 1 1 2 2 = + + + + + = = = x x p x X E 9 8 0 9 8 2 = = X Var (d) () ( ) 9 1 1 ! 3 ! 2 ! 1 1 3 1 = = + + = = c c x p x 9 23 ! 3 3 ! 2 2 ! 1 1 9 1 3 1 = × + × + × = = = x x xp X E 7 ! 3 3 ! 2 2 ! 1 1 9 1 2 2 2 3 1 2 2 = × + × + × = = = x x p x X E 81 38 9 23 7 2 = = X Var 5. 0, 9 8 , 9 52 6. Not fair, 9 5489 7. Let X be the gain in the game. Then ( ) 9 4 2 9 4 10 5 color same ball both 1 . 1 = × × = = = P X P 9 5 9 4 1 1 = = = X P (a) ()( ) 15 1 9 5 1 9 4 1 . 1 = + = X E (b) ()() 75 82 9 5 1 9 4 1 . 1 2 2 2 = + = X E () () () 45 49 225 1 75 82 2 2 = = = X E X E X Var 8. (a) 29 (b) 61.72 (c) Both are telling the truth. 9. 2 1 1 0 + + = N N X E , ( ) 12 1 2 0 1 = N N X Var 10. A p 1 . 0 +
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09/10 p. 3 11. (a) 14 (b) 45 (c) 9 4 12. (a) 3.88% (b) 53.19% (c) 2.06% 13. (a) () ( balls selected in the largest the is n i P i Y P = = ) samples possible of no. total maximum as with samples possible of no. i = = otherwise 0 if 1 1 N i n n N n i (b) According to (a), we have = = = = n N n i n N n i N n i N n i 1 1 1 1 1 . Consider ( ) × =
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STAT1301 (Assig2 ans) - 09/10 THE UNIVERSITY OF HONG KONG...

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