Assume that a sample is used to estimate a population mean $\mu$. Find the $90 \%$ confidence interval for a sample of size 53 with a mean of 58.5 and a standard deviation of 5.1. Enter your answer as an openinterval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
$90 \%$ C.1. $=$
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Step 1 ：We are given a sample size of 53, a sample mean of 58.5, and a standard deviation of 5.1. We are asked to find the 90% 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 corresponding to the desired confidence level, \(\sigma\) is the standard deviation of the population, and \(n\) is the size of the sample.

Step 3 ：Substituting the given values into the formula, we get \(58.5 \pm 1.645 \frac{5.1}{\sqrt{53}}\).

Step 4 ：Calculating the margin of error, we get approximately 1.152.

Step 5 ：Subtracting and adding the margin of error from the sample mean, we get the confidence interval as \((58.5 - 1.152, 58.5 + 1.152)\).

Step 6 ：Rounding to one decimal place, we get the confidence interval as \((57.3, 59.7)\).

Step 7 ：\(\boxed{The 90\% confidence interval for a sample of size 53 with a mean of 58.5 and a standard deviation of 5.1 is approximately (57.3, 59.7).}\)