Let's solve the problem step by step.

Given:

• Population mean $\mu = 58.4$
• Population standard deviation $\sigma = 9.4$
• Sample size $n = 13$

We need to answer two questions:

Part (a): What is the mean of the distribution of sample means $\mu_{\bar{X}}$?

• Answer: The mean of the distribution of sample means is the same as the population mean.

$\mu_{\bar{X}} = \mu = 58.4$

So, the answer for part (a) is:

$\mu_{\bar{X}} = 58.4$

Part (b): What is the standard deviation of the distribution of sample means $\sigma_{\bar{X}}$?

The standard deviation of the sample means (also called the standard error of the mean) is given by:

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$

Plugging in the values:

$\sigma_{\bar{X}} = \frac{9.4}{\sqrt{13}}$

First, compute $\sqrt{13}$:

$\sqrt{13} \approx 3.6056$

Now divide:

$\sigma_{\bar{X}} = \frac{9.4}{3.6056} \approx 2.61$

So, the answer for part (b) is:

$\sigma_{\bar{X}} = 2.61$

Final Answers:

• (a) $\mu_{\bar{X}} = 58.4$
• (b) $\sigma_{\bar{X}} = 2.61$ (rounded to two decimal places)

Would you like further details or have any questions?
Here are 5 related questions that may help your understanding:

1. What is the importance of the standard error in statistics?
2. How does increasing the sample size affect the standard deviation of sample means?
3. What is the difference between standard deviation and standard error?
4. How does the Central Limit Theorem relate to the distribution of sample means?
5. What happens to the distribution of sample means if the population is not normally distributed?

Tip: The larger the sample size, the smaller the standard error of the mean. This is why larger samples tend to give more precise estimates.