Solutions to Problem Set 6

# Solutions to Problem Set 6 - Solutions to Problem Set 6...

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Solutions to Problem Set 6 March 5, 2012 Problem 1 a. E ( X ) = N i = 1 x i p ( x i ) = 9. b. V ( X ) = E ( X 2 ) - ( E ( X ) 2 ) = 104 . 7 - 81 = 23 . 7 Problem 2 In such experiment there are ± 6 2 = 15 ² possible pairs. Because we draw simulta- neously and later on we only care about the sum, the order of numbers does not matter. The sample space is given by {( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 2 , 6 ) , ( 3 , 4 ) , ( 3 , 5 ) , ( 3 , 6 ) , ( 4 , 5 ) , ( 4 , 6 ) , ( 5 , 6 )} . Each pair will be realized with probability 1 / 15 because of the random sampling. 1

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a. The random variable X that represents the sum of the two tokens is given by X P ( X ) 3 (( 1 , 2 )) 1 / 15 4 (( 1 , 3 )) 1 / 15 5 (( 1 , 4 ) , ( 2 , 3 )) 2 / 15 6 (( 1 , 5 ) , ( 2 , 4 )) 2 / 15 7 (( 1 , 6 ) , ( 2 , 5 ) , ( 3 , 4 )) 3 / 15 8 (( 2 , 6 ) , ( 3 , 5 )) 2 / 15 9 (( 3 , 6 ) , ( 4 , 5 )) 2 / 15 10 (( 4 , 6 )) 1 / 15 11 (( 5 , 6 )) 1 / 15 b. E
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## This note was uploaded on 04/05/2012 for the course STATS V3100018.0 taught by Professor Giuseppearbia during the Spring '12 term at NYU.

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Solutions to Problem Set 6 - Solutions to Problem Set 6...

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