Step 1 :We are given a sample size of 257, a sample mean of 48.8, and a standard deviation of 17.8. We are asked to find the 99.5% confidence interval for the population mean.
Step 2 :The formula for the confidence interval for a population mean is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
Step 3 :Plugging in the given values, we have \(\bar{x} = 48.8\), \(Z = 2.5758293035489004\) (the Z-score for a 99.5% confidence level), \(\sigma = 17.8\), and \(n = 257\).
Step 4 :Calculating the margin of error, we get \(2.860029549429374\).
Step 5 :Subtracting the margin of error from the sample mean, we get the lower bound of the confidence interval, which is \(45.93997045057063\).
Step 6 :Adding the margin of error to the sample mean, we get the upper bound of the confidence interval, which is \(51.66002954942937\).
Step 7 :Rounding to three decimal places, the 99.5% confidence interval for the population mean is \(\boxed{45.940 \leq \mu \leq 51.660}\).