A simple random sample of size $n=13$ is drawn from a population that is normally distributed. The sample mean is found to be $\bar{x}=50$ and the sample standard deviation is found to be $s=10$. Construct a $90 \%$ confidence interval about the population mean. The lower bound is The upper bound is (Round to two decimal plak as needed.)

Step 1 ：We are given a simple random sample of size \(n=13\) drawn from a population that is normally distributed. The sample mean is \(\bar{x}=50\) and the sample standard deviation is \(s=10\). We are asked to construct a \(90\%\) confidence interval about the population mean.

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

Step 3 ：In this case, we have \(\bar{x} = 50\), \(s = 10\), and \(n = 13\). The z-score for a \(90\%\) confidence interval is approximately \(1.645\).

Step 4 ：Substituting these values into the formula, we get the margin of error as \(4.56240911395251\).

Step 5 ：Subtracting the margin of error from the sample mean, we get the lower bound of the confidence interval as \(45.43759088604749\).

Step 6 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval as \(54.56240911395251\).

Step 7 ：Rounding to two decimal places, the final answer is: The \(90\%\) confidence interval about the population mean is approximately \(\boxed{[45.44, 54.56]}\).