Let's solve the question based on the given information.

Given:

• Population mean ($\mu$) = 146.7
• Population standard deviation ($\sigma$) = 52.3
• Sample size ($n$) = 112

(a) 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: $\mu_{\bar{x}} = \mu = 146.7$

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

The standard deviation of the sample mean (standard error) is calculated as: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Substituting the values: $\sigma_{\bar{x}} = \frac{52.3}{\sqrt{112}}$

First, calculate $\sqrt{112}$: $\sqrt{112} \approx 10.583$

Now, calculate: $\sigma_{\bar{x}} = \frac{52.3}{10.583} \approx 4.94$

Final Answers:

• (a) $\mu_{\bar{x}} = 146.7$
• (b) $\sigma_{\bar{x}} = 4.94$ (rounded to two decimal places)

Would you like a detailed breakdown of any steps?

Here are 5 related questions for further practice:

1. What is the effect of increasing the sample size $n$ on $\sigma_{\bar{x}}$?
2. If the population standard deviation ($\sigma$) were smaller, how would $\sigma_{\bar{x}}$ change?
3. How would the answers change if the sample size were halved?
4. Why is the mean of the sample means equal to the population mean?
5. What assumptions are necessary for this calculation to hold?

Tip: Larger sample sizes reduce the standard error, leading to more precise sample mean estimates.