In the image, you are asked to solve a problem related to the Central Limit Theorem. Here's a breakdown:

1. Given Information:

   • The population has a normal distribution with:
     • Mean $\mu = 195.1$
     • Standard deviation $\sigma = 77.4$
   • Sample size $n = 107$

2. Questions:

   • What is the mean of the distribution of sample means?

     • According to the Central Limit Theorem, the mean of the sampling distribution of the sample means, denoted by $\mu_{\bar{x}}$, is the same as the population mean: $\mu_{\bar{x}} = \mu = 195.1$

   • What is the standard deviation of the distribution of sample means?

     • The standard deviation of the sampling distribution of the sample means, denoted by $\sigma_{\bar{x}}$, is given by the formula: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{77.4}{\sqrt{107}}$ Let's calculate that: $\sigma_{\bar{x}} \approx \frac{77.4}{10.344} \approx 7.48$ So, the standard deviation of the sample means is approximately $7.48$, rounded to two decimal places as requested.

Final Answers:

• Mean of sample means $\mu_{\bar{x}} = 195.1$
• Standard deviation of sample means $\sigma_{\bar{x}} \approx 7.48$

Would you like more details or have any questions? Here are 5 related questions for practice:

1. What happens to the standard deviation of the sample means if the sample size is increased?
2. If the sample size was 50 instead of 107, how would $\sigma_{\bar{x}}$ change?
3. Explain why the Central Limit Theorem is important when the sample size is large.
4. How does the Central Limit Theorem apply when the population is not normally distributed?
5. Calculate $\sigma_{\bar{x}}$ if the population standard deviation was $90$ and the sample size was $200$.

Tip: As the sample size increases, the standard deviation of the sample means decreases, making the sample mean a more accurate estimate of the population mean.