Let's solve the problem based on the given data.

Problem Breakdown:

• The population has a normal distribution with:
  • Mean ($\mu$) = 224.9
  • Standard deviation ($\sigma$) = 23.9
• A sample of size $n = 217$ is drawn.

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

For a sample mean distribution, the mean ($\mu_{\bar{x}}$) is the same as the population mean: $\mu_{\bar{x}} = \mu = 224.9$ So, this answer is correct.

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

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

Plugging in the values: $\sigma_{\bar{x}} = \frac{23.9}{\sqrt{217}}$

Calculating: $\sigma_{\bar{x}} \approx \frac{23.9}{14.7309} \approx 1.6220$

Therefore, the standard deviation of the distribution of sample means should be approximately 1.6220 when rounded to four decimal places.

Would you like a detailed breakdown of any steps or have any other questions?

Related Questions:

1. How does increasing the sample size affect the standard error of the mean?
2. What is the relationship between sample size and the shape of the sampling distribution of the mean?
3. How would you calculate the probability of obtaining a sample mean within a specific range?
4. What happens to the standard error if the population standard deviation increases?
5. Can the sample mean be greater than the population mean, and why?

Tip:

Remember, the larger the sample size, the smaller the standard error, which means the sample mean will likely be closer to the population mean.