The question in the image provides the following dataThe question in the image provides the following

• Population mean ($\mu$) = 15
• Population standard deviation ($\sigma$) = 25
• Sample size ($n$) = 57
• Population size ($N$) = 167

Let's address each part:

A. Solving for the sample mean ($\bar{X}$):

The sample mean $\bar{X}$ is equal to the population mean ($\mu$) when dealing with a random sample. Hence, $\bar{X} = \mu = 15$

B. Solving for the standard deviation ($\sigma_x$):

To calculate the standard deviation of the sample, we can use the finite population correction (FPC) formula since the sample size is a significant portion of the population: $\sigma_x = \sigma \times \sqrt{\frac{N - n}{N - 1}}$ Substitute the values: $\sigma_x = 25 \times \sqrt{\frac{167 - 57}{167 - 1}} = 25 \times \sqrt{\frac{110}{166}}$ Let’s calculate it:

C. Solving for variance ($\sigma_x^2$):

The variance is the square of the sample standard deviation: $\sigma_x^2 = (\sigma_x)^2$

I will now calculate the numerical values for both B and C.### B. The standard deviation of the sample ($\sigma_x$) is approximately: $\sigma_x \approx 20.35$

C. The variance of the sample ($\sigma_x^2$) is approximately:

$\sigma_x^2 \approx 414.16$

Would you like more details or further clarification on any part of this problem?

Here are 5 related questions for further practice:

1. How would the sample mean change if the population mean were 20?
2. What happens to the standard deviation if the sample size increases to 100?
3. How would you compute the standard error of the sample mean?
4. Can you explain the significance of using finite population correction (FPC)?
5. What is the effect on variance when the sample size approaches the population size?

Tip: Always apply the finite population correction when the sample size is more than 5% of the population size.