hw1 - Homework 1 Solutions 6.2. a. The values T0 can have...

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Homework 1 Solutions 6.2. a. The values T 0 can have are 0, 1, 2, 3, and 4. t P ( T 0 = t ) 0 P ( X 1 = 0 ,X 2 = 0) = . 2 2 = . 04 1 P ( X 1 = 0 ,X 2 = 1) + P ( X 1 = 1 ,X 2 = 0) = 2( . 2)( . 5) = . 2 2 . 37 3 . 3 4 . 09 b. By a proposal, E ( T 0 ) = E ( X 1 ) + E ( X 2 ) = 2 μ = 2 . 2, which is twice the population mean. c. By a proposal, V ( T 0 ) = V ( X 1 )+ V ( X 2 ) = 2 σ 2 = . 98, which is twice the population variance. 6.3. The number of successes X has a binomial distribution with parameters n = 10 and p = . 8. It follows that the sampling distribution of the mean X/ 10 is P ( X/ 10 = k ) = P ( X = 10 k ) = ± 10 10 k . 8 10 k (1 - . 8) 10 - 10 k for k = 0, 1 / 10, ... , 9 / 10, 1. 6.5. a. The values X can have are 1, 1.5, 2, 2.5, 3, 3.5, and 4. As in 6.2 above, x P ( X = x ) 1 P ( X 1 = 1 ,X 2 = 1) = . 4 2 = . 16 1.5 P ( X 1 = 1 ,X 2 = 2) + P ( X 1 = 2 ,X 2 = 1) = 2( . 4)( . 3) = . 24 2 . 25 2.5 . 2 3 . 1 3.5 . 04 4 . 01 b. P ( X 2 . 5) = . 16 + . 24 + . 25 + . 2 = . 85 c. The values
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hw1 - Homework 1 Solutions 6.2. a. The values T0 can have...

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