A sample of size $n=54$ has sample mean $\bar{x}=54.8$ and sample standard deviation $s=9.5$. Part: $0 / 2$ Part 1 of 2 Construct a $98 \%$ confidence interval for the population mean $\mu$. Round the answers to one decimal place. A $98 \%$ confidence interval for the population mean is $\square<\mu<\square$.

Step 1 ：We are given a sample of size \(n=54\) with sample mean \(\bar{x}=54.8\) and sample standard deviation \(s=9.5\).

Step 2 ：We are asked to construct a 98% confidence interval for the population mean \(\mu\).

Step 3 ：The formula for a confidence interval is \(\bar{x} \pm Z_{\alpha/2} \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z_{\alpha/2}\) is the z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 4 ：The z-score for a 98% confidence interval is approximately 2.33.

Step 5 ：Substituting the given values into the formula, we get \(\bar{x} \pm Z_{\alpha/2} \frac{s}{\sqrt{n}} = 54.8 \pm 2.33 \frac{9.5}{\sqrt{54}}\).

Step 6 ：Calculating the margin of error, we get approximately 3.0121919698058695.

Step 7 ：Subtracting the margin of error from the sample mean, we get the lower bound of the confidence interval, which is approximately 51.78780803019413.

Step 8 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval, which is approximately 57.812191969805866.

Step 9 ：Rounding the lower and upper bounds to one decimal place, we get 51.8 and 57.8 respectively.

Step 10 ：Thus, a 98% confidence interval for the population mean is \(\boxed{51.8}<\mu<\boxed{57.8}\).