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Dr. Hackney STA Solutions pg 77

Dr. Hackney STA Solutions pg 77 - Second Edition 5-11 5.29...

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Second Edition 5-11 5.29 Let X i = weight of i th booklet in package. The X i s are iid with EX i = 1 and Var X i = . 05 2 . We want to approximate P 100 i =1 X i > 100 . 4 = P 100 i =1 X i / 100 > 1 . 004 = P ( ¯ X > 1 . 004). By the CLT, P ( ¯ X > 1 . 004) P ( Z > (1 . 004 - 1) / ( . 05 / 10)) = P ( Z > . 8) = . 2119. 5.30 From the CLT we have, approximately, ¯ X 1 n( μ, σ 2 /n ), ¯ X 2 n( μ, σ 2 /n ). Since ¯ X 1 and ¯ X 2 are independent, ¯ X 1 - ¯ X 2 n(0 , 2 σ 2 /n ). Thus, we want . 99 P ( ¯ X 1 - ¯ X 2 < σ/ 5 ) = P - σ/ 5 σ/ n/ 2 < ¯ X 1 - ¯ X 2 σ/ n/ 2 < σ/ 5 σ/ n/ 2 P - 1 5 n 2 < Z < 1 5 n 2 , where Z n(0 , 1). Thus we need P ( Z n/ 5( 2)) . 005. From Table 1, n/ 5 2 = 2 . 576, which implies n = 50(2 . 576) 2 332. 5.31 We know that σ 2 ¯ X = 9 / 100. Use Chebyshev’s Inequality to get P ( - 3 k/ 10 < ¯ X - μ < 3 k/ 10 ) 1 - 1 /k 2 . We need 1 - 1 /k 2 . 9 which implies k 10 = 3 . 16 and 3 k/ 10 = . 9487. Thus P ( - . 9487 < ¯ X - μ < . 9487) . 9 by Chebychev’s Inequality. Using the CLT, ¯ X is approximately n ( μ, σ 2 ¯ X ) with σ ¯ X = . 09 = . 3 and ( ¯ X - μ ) /. 3 n(0 , 1). Thus . 9 = P - 1 . 645 < ¯ X - μ . 3 < 1 . 645 = P ( - . 4935 < ¯ X - μ < . 4935) . Thus, we again see the conservativeness of Chebychev’s Inequality, yielding bounds on
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