Assume that a sample is used to estimate a population mean $\mu$. Find the $90 \%$ confidence interval for a sample of size 71 with a mean of 62.1 and a standard deviation of 9.6. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.
\[
90 \% \text { C.I. }=
\]
The answer should be obtained without any preliminary rounding.

Step 1 ：Given that the sample size (n) is 71, the sample mean (\(\bar{x}\)) is 62.1, and the standard deviation (\(\sigma\)) is 9.6.

Step 2 ：We are asked to find the 90% confidence interval. The Z-score (Z) for a 90% confidence level is 1.645. This value can be found in a Z-table or using a statistical calculator.

Step 3 ：The formula for the confidence interval is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\).

Step 4 ：Substitute the given values into the formula: 62.1 \(\pm\) 1.645 * \(\frac{9.6}{\sqrt{71}}\).

Step 5 ：Calculate the margin of error: 1.645 * \(\frac{9.6}{\sqrt{71}}\) = 1.874165594619733.

Step 6 ：Subtract the margin of error from the sample mean to get the lower bound of the confidence interval: 62.1 - 1.874165594619733 = 60.22583440538027.

Step 7 ：Add the margin of error to the sample mean to get the upper bound of the confidence interval: 62.1 + 1.874165594619733 = 63.974165594619734.

Step 8 ：Final Answer: The 90% confidence interval for a sample of size 71 with a mean of 62.1 and a standard deviation of 9.6 is \(\boxed{(60.226, 63.974)}\).