You intend to estimate a population mean $\mu$ with the following sample.
$$\text{Unknown environment 'tabular'}$$
You believe the population is normally distributed. Find the $95\mathrm{\%}$ 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).
$$95\mathrm{\%}\text{ C.I. }=$$
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 $\overline{x}\pm z\frac{s}{\sqrt{n}}$, where $\overline{x}$ 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 $\overline{x}$ 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 $\boxed{{\displaystyle (30.23,50.26)}}$.