Let's go through the problem step-by-step.

1. Mean of the distribution of sample means ($\mu_{\bar{x}}$):

   The mean of the distribution of sample means is the same as the population mean ($\mu$). $\mu_{\bar{x}} = \mu = 22.6$ So, $\mu_{\bar{x}} = 22.6$.

2. Standard deviation of the distribution of sample means ($\sigma_{\bar{x}}$):

   The standard deviation of the sample means, also called the standard error, is calculated using the formula: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ where:

   • $\sigma = 29.4$ (population standard deviation)
   • $n = 86$ (sample size)

   Plugging in the values: $\sigma_{\bar{x}} = \frac{29.4}{\sqrt{86}} \approx \frac{29.4}{9.2736} \approx 3.17$

   So, $\sigma_{\bar{x}} \approx 3.17$.

Final Answers:

1. $\mu_{\bar{x}} = 22.6$
2. $\sigma_{\bar{x}} \approx 3.17$

Would you like any further explanation on these concepts?

Here are 5 related questions to deepen your understanding:

1. What is the significance of the Central Limit Theorem in relation to the distribution of sample means?
2. How does increasing the sample size affect the standard error?
3. If the population distribution were normal, how would that influence the sample mean distribution?
4. What happens to the mean and standard error if the population mean changes but the sample size remains constant?
5. How would you calculate a confidence interval for the sample mean in this case?

Tip: Remember that as sample size increases, the standard error decreases, meaning the sample mean becomes a more accurate estimate of the population mean!