The president of a large university wishes to estimate the average age of the students presently enrolled. From past studies, the standard deviation ó is known to be 2 years. A sample of 50 students is selected and the mean is found to be 23.3 years.

a. Find the 99% confidence interval of the population.
b. Find the 99% confidence interval for the population standard deviation.

a. To find the 99% confidence interval of the population mean, we can use the formula:

CI = X̄ ± Z * (σ / √n)

Where:
- CI is the confidence interval
- X̄ is the sample mean (23.3 years)
- Z is the z-score corresponding to the desired confidence level (99% confidence level corresponds to a z-score of 2.575 approximately)
- σ is the population standard deviation (2 years)
- n is the sample size (50 students)

Substituting the values into the formula:

CI = 23.3 ± 2.575 * (2 / √50)

Calculating the values:

CI = 23.3 ± 2.575 * (2 / 7.0711)

CI = 23.3 ± 0.92

Therefore, the 99% confidence interval for the population mean is (22.38, 24.22).

b. To find the 99% confidence interval for the population standard deviation, we can use the chi-square distribution.

CI = ((n-1)s^2 / χ^2(α/2, n-1), (n-1)s^2 / χ^2(1-α/2, n-1))

Where:
- CI is the confidence interval
- n is the sample size (50 students)
- s is the sample standard deviation (calculated from the sample, if provided)

Since we don't have the sample standard deviation provided in the question, we cannot calculate the confidence interval for the population standard deviation without it.

a. Use the equations from the previous post.

b. Sorry, I don't know how to do this.