According to recent research, 38% of Americans have less than $425 in the bank for an emergency. Let \(\hat{p}\) be the proportion of people who have less than $425 in the bank for an emergency in a random sample of 64 Americans. Determine the probability that the proportion of people in a random sample of 64 Americans who have less than $425 in the bank for an emergency is within 5% of the population proportion. Round the solution to four decimal places. \(P(\hat{p}\text{ within 5\% of the population proportion}) = \boxed{} Determine the probability that the proportion of people in a random sample of 64 Americans who have less than $425 in the bank for an emergency is less than the population proportion by 10% or more. Round the solution to four decimal places. \(P(\hat{p}\text{ less than the population proportion by 10\% or more}) = \boxed{}
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38$ be the population proportion. The sample size is $n = 64$. The standard error is $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.38(1-0.38)}{64}} = \sqrt{\frac{0.2356}{64}} \approx 0.0607$ Show more…
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