Points: 0 of 1 Save In a random sample of 8 people, the mean commute time to work was 36.5 minutes and the standard deviation was 7.2 minutes. A $98 \%$ confidence interval using the t-distribution was calculated to be $(28.9,44.1)$. After researching commute times to work, it was found that the population standard deviation is 8.5 minutes. Find the margin of error and construct a $98 \%$ confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. The margin of error of $\mu$ is (Round to two decimal placelaras needed.)

Step 1 ：Given that the mean commute time to work is 36.5 minutes, the lower limit of the confidence interval is 28.9 minutes, the standard deviation is 8.5 minutes, and the sample size is 8 people.

Step 2 ：Calculate the margin of error by subtracting the lower limit of the confidence interval from the mean. In this case, the margin of error is \(36.5 - 28.9 = 7.6\) minutes.

Step 3 ：Next, construct a new 98% confidence interval using the standard normal distribution. The z-score for a 98% confidence interval is approximately 2.33.

Step 4 ：Use this z-score, the known standard deviation of 8.5, and the sample size of 8 to calculate the new confidence interval. The standard error is calculated as \(\frac{8.5}{\sqrt{8}} = 3.005\).

Step 5 ：Calculate the new lower limit of the confidence interval as \(36.5 - 2.33 \times 3.005 = 29.51\) and the new upper limit as \(36.5 + 2.33 \times 3.005 = 43.49\).

Step 6 ：Final Answer: The margin of error is \(\boxed{7.60}\). The new 98% confidence interval using the standard normal distribution is \(\boxed{(29.51, 43.49)}\).