STAT1301 (Assig2 ans)full

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

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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. Sample space: {} BO BB WO WB WW , , , , = Ω Note that the outcomes are not equally likely with associated probabilities: () 3 1 45 15 2 10 2 6 = = = WW P , 5 2 45 3 6 = × = WB P 15 2 45 1 6 = × = WO P , 15 1 45 2 3 = = BB P , 15 1 45 1 3 = × = BO P The random variable X is the following function: 2 = WW X , 1 = WB X , ( ) 0 = WO X , ( ) 0 = BB X , 1 = BO X (a) The the possible values of X are –1, 0, 1, 2. (b) The probability mass function is given by { } 15 1 1 = = = BO P X P { } 5 1 15 1 15 2 , 0 = + = = = BB WO P X P { } 5 2 1 = = = WB P X P { } 3 1 2 = = = WW P X P The cumulative distribution function is () ( ) < < < < = = 2 if 1 2 1 if 3 2 1 0 if 15 4 0 1 if 15 1 1 if 0 x x x x x x X P 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
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09/10 p. 2 3. For , the sample should contain the largest i numbers and should not contain the ( i +1) th largest number (otherwise ). Hence the remaining i X = i X > i n numbers should be selected from {} 1 ,..., 3 , 2 , 1 + i n m . Therefore the pmf of X is given by () + + = = n n m i n i n m i X 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 ( ) 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
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09/10 p. 3 5. () ( )( ) 0 9 4 1 9 1 0 9 4 1 1 1 = × + × + × = = = x x xp X E () () ( ) 9 8 9 4 1 9 1 0 9 4 1 2 2 2 1 1 2 2 = × + × + × = = = x x p x X E () ( ) 9 52 4 0 3 9 8 2 4 3 2 4 3 2 2 2 = + × × = + = + X E X E X X E 6. Let X be the gain from the game. Obviously X takes on values -20, -10, 22, 82. The probability mass function of X is given by ( ) 2 1 even 10 = = = P X P , ( ) 6 1 36 6 7 is total 22 = = = = P X P ( ) 18 1 36 2 11 is total 82 = = = = P X P , 18 5 18 1 6 1 2 1 1 20 = = = X P ()( ) 3 7 18 5 20 18 1 82 6 1 22 2 1 10 = × + × + × + × = X E The game is not fair. For the variance of the gain, 3 1846 18 5 20 18 1 82 6 1 22 2 1 10 2 2 2 2 2 = × + × + × + × = X E () () () 9 5489 3 7 3 1846 2 2 2 = = = X E X E X Var .
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STAT1301 (Assig2 ans)full - 09/10 THE UNIVERSITY OF HONG...

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