QUESTION 9 interval 1: A normally distributed population has a mean of 74 and a standard deviation of 21 . Sample avergaes from samples of size 24 are collected. What would be the upper end of the centered interval that contains $90 \%$ of all possible sample averages? Round to the nearest hundredth

Step 1 ：Given a normally distributed population with a mean of 74 and a standard deviation of 21. Sample averages from samples of size 24 are collected.

Step 2 ：First, calculate the standard error, which is the standard deviation divided by the square root of the sample size. The formula for standard error is \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation and \( n \) is the sample size.

Step 3 ：Substitute the given values into the formula: \( \frac{21}{\sqrt{24}} \) to get the standard error, which is approximately 4.29.

Step 4 ：Next, find the z-score that corresponds to the 95th percentile. The z-score for the 95th percentile is approximately 1.645.

Step 5 ：Finally, calculate the upper end of the interval by adding the product of the z-score and the standard error to the population mean. The formula is \( \mu + Z \times SE \), where \( \mu \) is the population mean, \( Z \) is the z-score, and \( SE \) is the standard error.

Step 6 ：Substitute the values into the formula: \( 74 + 1.645 \times 4.29 \) to get the upper end of the interval, which is approximately 81.05.

Step 7 ：The upper end of the centered interval that contains 90% of all possible sample averages is approximately \(\boxed{81.05}\).