Let's break down the problem based on the information provided:

Given:

• Population mean $\mu = 60$
• Population standard deviation $\sigma = 15$
• Sample size $n = 25$

We are asked to find:

• The sample mean $\bar{X}$
• The standard deviation of the sampling distribution $\sigma_{\bar{X}}$
• The variance of the sampling distribution $\sigma_{\bar{X}}^2$

A. Solve for the sample mean $\bar{X}$:

The sample mean $\bar{X}$ is equal to the population mean $\mu$ since the sample is taken from the population: $\bar{X} = \mu = 60$

B. Solving for the standard deviation of the sample mean (standard error) $\sigma_{\bar{X}}$:

The standard deviation of the sample mean, also known as the standard error, is given by: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ Substitute the values: $\sigma_{\bar{X}} = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3$

C. Solving for the variance of the sample mean $\sigma_{\bar{X}}^2$:

The variance is simply the square of the standard error: $\sigma_{\bar{X}}^2 = (\sigma_{\bar{X}})^2 = 3^2 = 9$

Summary of Results:

• Sample mean $\bar{X}$: $60$
• Standard deviation $\sigma_{\bar{X}}$: $3$
• Variance $\sigma_{\bar{X}}^2$: $9$

Would you like further details or explanations on any of the steps? Here are five related questions for practice:

1. How does increasing the sample size $n$ affect the standard error $\sigma_{\bar{X}}$?
2. If the population standard deviation $\sigma$ was 20 instead of 15, how would the results change?
3. What is the relationship between the variance of the sample mean and the population variance?
4. Why is the sample mean equal to the population mean in this case?
5. How does sampling distribution theory apply to real-world statistical problems?

Tip: The larger the sample size, the smaller the standard error of the sample mean, making estimates of the population mean more precise.