38 functions of a random variable 101 example 382

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Unformatted text preview: UOUS-TYPE VARIABLES 97 Solution: (a) Since X has the binomial distribution with parameters n and p, the Gaussian approximation yields P [|p − p| ≤ δ ] = P ˆ X − np X −p ≤δ =P n ≈Φδ n p(1 − p) np(1 − p) − Φ −δ n p(1 − p) ≤δ n p(1 − p) = 2Φ δ n p(1 − p) − 1. For n = 1000, δ = 0.02, and p = 0.5, this is equal to 2Φ(1.265) = 0.794 = 79.4%, and for n = 1000, δ = 0.02, and p = 0.1, this is equal to 2Φ(2.108) = 0.965 = 96.5%. 1000 (b) Select δ so that 2Φ δ p(1−p) − 1 = 0.99. Observing from Table 6.1 for the standard normal CDF that 2Φ(2.58) − 1 ≈ 0.99, we select δ so that δ confidence interval for p = 0.5 requires δ = 2.58 1000 p(1−p) 0.5(1−0.5) 1000 = 2.58, or δ = 2.58 p(1−p) 1000 . The 99% = 0.04. 1(0. (c) Similarly, the 99% confidence interval for p = 0.1 requires δ = 2.58 0.10009) = 0.025. (d) The product p(1 − p), and hence the required δ , is maximized by p = 0.5. Thus, if δ = 0.04 as found in part (b), then the confidence interval contains p with probability at least 0.99 (no matter what the value of p is)...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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