The grade point averages (GPA) for 12 randomly selected college students are shown on $2.3 \quad 3.3 \quad 2.8$ the right. Complete parts (a) through (c) below. $\begin{array}{lll}1.6 & 0.9 & 4.0\end{array}$ Assume the population is normally distributed. $\begin{array}{lll}2.5 & 1.4 & 3.8\end{array}$ $\begin{array}{lll}0.3 & 2.3 & 3.1\end{array}$ (a) Find the sample mean. $\bar{x}=2.36$ (Round to two decimal places as needed.) (b) Find the sample standard deviation. $s=1.14$ (Round to two decimal places as needed.) (c) Construct a $95 \%$ confidence interval for the population mean $\mu$. A $95 \%$ confidence interval for the population mean is (Round to two decimal places as needed.)

Step 1 ：The sample mean \(\bar{x}\) is given as 2.36.

Step 2 ：The sample standard deviation \(s\) is given as 1.14.

Step 3 ：The sample size \(n\) is 12 as there are 12 students.

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

Step 5 ：We can calculate the margin of error using the formula \(z \frac{s}{\sqrt{n}}\). Substituting the given values, we get a margin of error of approximately 0.645.

Step 6 ：We can then calculate the lower and upper bounds of the confidence interval by subtracting and adding the margin of error from the sample mean respectively. This gives us a lower bound of approximately 1.715 and an upper bound of approximately 3.005.

Step 7 ：Final Answer: A 95% confidence interval for the population mean is \(\boxed{(1.71, 3.01)}\)