A simple random sample of size $n$ is drawn from a population that is normally distributed. The sample mean, $\bar{x}$, is found to be 114 , and the sample standard deviation, $\mathrm{s}$, is found to be 10 . (c) Construct a $95 \%$ confidence interval about $\mu$ if the sample size, $n$, is 28 . Lower bound: ; Upper bound: (Use ascending order. Round to one decimal place as needed.)

Step 1 ：We are given a simple random sample of size $n = 28$ drawn from a normally distributed population. The sample mean, $\bar{x}$, is found to be 114, and the sample standard deviation, $s$, is found to be 10.

Step 2 ：We are asked to construct a $95\%$ confidence interval about $\mu$. The formula for the confidence interval for a population mean, given a sample, is $\bar{x} \pm Z \frac{s}{\sqrt{n}}$.

Step 3 ：In this formula, $\bar{x}$ is the sample mean, $Z$ is the Z-score which corresponds to the desired confidence level (for a 95% confidence level, the Z-score is approximately 1.96), $s$ is the sample standard deviation, and $n$ is the sample size.

Step 4 ：Substituting the given values into the formula, we get $114 \pm 1.96 \frac{10}{\sqrt{28}}$.

Step 5 ：Calculating the margin of error, we get approximately 3.7.

Step 6 ：Subtracting this margin of error from the sample mean, we get the lower bound of the confidence interval, which is approximately 110.3.

Step 7 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval, which is approximately 117.7.

Step 8 ：Thus, the $95\%$ confidence interval for the population mean $\mu$ is approximately \(\boxed{(110.3, 117.7)}\).