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Probability, Central Limit Theorem?
Y~B(n,p), n = 1000. Use the normal approximation to the binomial to approximate P(|Y/n - p| ≤ 0.05) and compare with the bound using Chebychev's inequality. If Y = 700, determine the corresponding confidence interval for p.
1 Answer
- cidyahLv 74 years agoFavorite Answer
P(|Y/n - p| ≤ 0.05) >= 1- sigma^2 / (0.05)^2
Assume p^=0.5
sigma^2 = p^(1-p^)/n = (0.5)(0.5)/1000 =0.00025
1- sigma^2 / (0.05)^2 = 1- 0.00025 / (0.05)^2 = 0.9
P(|Y/n - p| ≤ 0.05) >= 0.9
with n=700
sigma^2 = p^(1-p^)/n = (0.5)(0.5)/700 =0.000357
1- sigma^2 / (0.05)^2 = 1- 0.000357 / (0.05)^2 = 0.9
P(|Y/n - p| ≤ 0.05) >= 0.857
We need the sample proportion and the sample size to determine the confidence interval for P.
