Use the following sample to estimate a population mean $\mu$.
\[
\begin{array}{|c|}
\hline 73.4 \\
\hline 81.7 \\
\hline 65.9 \\
\hline 63.2 \\
\hline 64.6 \\
\hline
\end{array}
\]
Assuming the population is normally distributed, find the $90 \%$ confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places.
\[
90 \% \text { C.I. }=
\]

Step 1 ：Given a sample of data: \([73.4, 81.7, 65.9, 63.2, 64.6]\)

Step 2 ：Calculate the sample mean (\(\bar{x}\)): \(\bar{x} = \frac{73.4 + 81.7 + 65.9 + 63.2 + 64.6}{5} = 69.76\)

Step 3 ：Calculate the sample standard deviation (\(s\)): \(s = \sqrt{\frac{(73.4-69.76)^2 + (81.7-69.76)^2 + (65.9-69.76)^2 + (63.2-69.76)^2 + (64.6-69.76)^2}{5-1}} = 7.75\)

Step 4 ：Find the z-score corresponding to a 90% confidence level: \(z = 1.645\)

Step 5 ：Substitute these values into the formula for the confidence interval: \(\bar{x} \pm z \frac{s}{\sqrt{n}}\)

Step 6 ：Calculate the lower bound of the confidence interval: \(64.06 = 69.76 - 1.645 \times \frac{7.75}{\sqrt{5}}\)

Step 7 ：Calculate the upper bound of the confidence interval: \(75.46 = 69.76 + 1.645 \times \frac{7.75}{\sqrt{5}}\)

Step 8 ：\(\boxed{\text{Final Answer: The 90% confidence interval for the population mean is } (64.06, 75.46)}\)