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

Given:

• $\mu = 15$ (population mean),
• $\sigma = 25$ (population standard deviation),
• $n = 57$ (sample size),
• $N = 167$ (population size).

We are asked to solve for:

A. Sample mean ( $\bar{X}$ ):

For a random sample, the sample mean $\bar{X}$ is usually the same as the population mean $\mu$, unless otherwise indicated. Therefore: $\bar{X} = \mu = 15$

B. Standard deviation of the sample mean ( $\sigma_{\bar{X}}$ ):

The standard deviation of the sample mean, also called the standard error (SE), can be calculated using the finite population correction (FPC) when dealing with a finite population size.

The formula for the standard error of the mean with FPC is: $\sigma_{\bar{X}} = \sigma \times \sqrt{\frac{1}{n}} \times \sqrt{\frac{N - n}{N - 1}}$ Substituting the given values: $\sigma_{\bar{X}} = 25 \times \sqrt{\frac{1}{57}} \times \sqrt{\frac{167 - 57}{167 - 1}}$ Let's compute this.

C. Variance of the sample mean ( $\sigma^2_{\bar{X}}$ ):

The variance of the sample mean is the square of the standard deviation of the sample mean: $\sigma^2_{\bar{X}} = (\sigma_{\bar{X}})^2$

I will now compute parts B and C.B. The standard deviation of the sample mean ( $\sigma_{\bar{X}}$ ) is approximately 2.70.

C. The variance of the sample mean ( $\sigma^2_{\bar{X}}$ ) is approximately 7.27.

Would you like any further details or clarifications?

Here are 5 related questions for further practice:

1. How does increasing the sample size $n$ affect the standard deviation of the sample mean?
2. What is the difference between population standard deviation and sample standard deviation?
3. How is the standard error affected if the population is considered infinite?
4. How would you calculate the confidence interval for the sample mean?
5. What is the significance of the finite population correction factor?

Tip: The standard error decreases as the sample size increases, meaning larger samples provide more accurate estimates of the population mean.