You intend to estimate a population mean $\mu$ with the following sample. \begin{tabular}{|r|} \hline 45.4 \\ \hline 48.3 \\ \hline 39.3 \\ \hline 5.4 \\ \hline 55.8 \\ \hline 43.5 \\ \hline 51.3 \\ \hline 57.8 \\ \hline 46.6 \\ \hline 38 \\ \hline 11.3 \\ \hline \end{tabular} You believe the population is normally distributed. Find the $95 \%$ confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place). \[ 95 \% \text { C.I. }= \] 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 of 11 data points: 45.4, 48.3, 39.3, 5.4, 55.8, 43.5, 51.3, 57.8, 46.6, 38, 11.3. We are asked to find the 95% confidence interval for the population mean, assuming the population is normally distributed.

Step 2 ：The formula for the confidence interval is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score which corresponds to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：First, we calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\). The sample mean is approximately 40.24545454545454 and the sample standard deviation is approximately 16.949062725493917.

Step 4 ：For a 95% confidence level, the z-score is approximately 1.96.

Step 5 ：Substituting these values into the formula, we find the confidence interval to be approximately (30.22938260078375, 50.261526490125334).

Step 6 ：Rounding to two decimal places, the final answer is \(\boxed{(30.23, 50.26)}\).