Assume that a sample is used to estimate a population mean $\mu$. Find the $95 \%$ confidence interval for a sample of size 58 with a mean of 44.4 and a standard deviation of 17.2. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. \[ 95 \% \text { C.I. }= \]

Step 1 ：Given that the sample size \(n = 58\), the sample mean \(\bar{x} = 44.4\), and the standard deviation \(\sigma = 17.2\).

Step 2 ：The z-score for a 95% confidence interval is approximately \(z = 1.96\).

Step 3 ：The formula for a confidence interval is given by \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\).

Step 4 ：Substitute the given values into the formula to find the margin of error, which is \(4.426602464566966\).

Step 5 ：Subtract the margin of error from the sample mean to find the lower bound of the confidence interval, which is \(39.97339753543303\).

Step 6 ：Add the margin of error to the sample mean to find the upper bound of the confidence interval, which is \(48.82660246456697\).

Step 7 ：\(\boxed{\text{The 95% confidence interval for a sample of size 58 with a mean of 44.4 and a standard deviation of 17.2 is approximately (39.973, 48.827).}}\)