Question 7 - of 19 Step 2 of 2 $08: 20: 55$ As part of a fuel efficiency study, a large school district wishes to estimate the average weekly mileage for its school buses. A sample of 23 buses is selected at random and the mileages are recorded for that week. The sample mean was found to be 206 miles with a standard deviation of 31 miles. Assume that the population distribution is approximately normal. Step 2 of 2: Construct a $98 \%$ confidence interval. Round your answers to two decimal places as needed. With $98 \%$ confidence the school district can say that the average weekly mileage for all schools buses in the district is between Answer 2 Points

Step 1 ：The problem is asking us to construct a 98% confidence interval for the average weekly mileage of school buses in a district. The data we have is a sample of 23 buses with a mean mileage of 206 miles and a standard deviation of 31 miles.

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 corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：In this case, \(\bar{x} = 206\), \(s = 31\), and \(n = 23\). The Z-score for a 98% confidence interval is approximately 2.33.

Step 4 ：Substituting these values into the formula, we get \(206 \pm 2.33 \frac{31}{\sqrt{23}}\).

Step 5 ：Calculating the above expression, we find that the margin of error is approximately 15.06.

Step 6 ：Subtracting this margin of error from the sample mean, we get the lower bound of the confidence interval, which is approximately 190.94.

Step 7 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval, which is approximately 221.06.

Step 8 ：Final Answer: With 98% confidence the school district can say that the average weekly mileage for all schools buses in the district is between \(\boxed{190.94}\) and \(\boxed{221.06}\) miles.