To solve this problem, let’s address each part step by step.

1. Mean of the Distribution of Sample Means ($\mu_{\bar{x}}$):

   • In a normal distribution, the mean of the distribution of sample means is the same as the population mean.
   • Given that the population mean ($\mu$) is 175.7, we have: $\mu_{\bar{x}} = \mu = 175.7$

2. Standard Deviation of the Distribution of Sample Means ($\sigma_{\bar{x}}$):

   • The standard deviation of the distribution of sample means, also known as the standard error, is given by: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
   • Here, $\sigma = 31.4$ and $n = 84$, so: $\sigma_{\bar{x}} = \frac{31.4}{\sqrt{84}}$

   Let’s calculate this:

   $\sigma_{\bar{x}} = \frac{31.4}{\sqrt{84}} \approx \frac{31.4}{9.165} \approx 3.42$

So, the answers are:

• $\mu_{\bar{x}} = 175.7$
• $\sigma_{\bar{x}} \approx 3.42$

Would you like further details on the calculations or concepts here?

Here are some additional questions to expand on this topic:

1. How does sample size affect the standard deviation of the sample mean distribution?
2. Why is the mean of the sample mean distribution the same as the population mean?
3. How would these values change if the sample size were increased to 100?
4. What does the Central Limit Theorem tell us about the distribution of sample means?
5. How would the calculations change if the population were not normally distributed?

Tip: The standard deviation of the sample mean distribution decreases as the sample size increases, making larger samples more precise for estimating the population mean.