Assume that a sample is used to estimate a population mean $\mu$. Find the $99.5 \%$ confidence interval for a sample of size 257 with a mean of 48.8 and a standard deviation of 17.8 . Enter your answer as a tri-linear inequality accurate to 3 decimal places.

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 \(\boxed{45.940 \leq \mu \leq 51.660}\).