The problem involves finding the mean and standard deviation of the sampling distribution of sample means for a population. Here's how we solve it:

1. Mean of the sampling distribution (μₓ̄):

The mean of the sampling distribution of sample means is the same as the population mean. Therefore:

$\mu_{\bar{x}} = \mu = 137$

2. Standard deviation of the sampling distribution (σₓ̄):

The standard deviation of the sampling distribution of sample means (also called the standard error) is given by the formula:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Where:

• $\sigma = 30$ is the population standard deviation.
• $n = 48$ is the sample size.

Now calculate:

$\sigma_{\bar{x}} = \frac{30}{\sqrt{48}} \approx \frac{30}{6.93} \approx 4.33$

Final answers:

• Mean of the sampling distribution: $\mu_{\bar{x}} = 137$
• Standard deviation of the sampling distribution: $\sigma_{\bar{x}} \approx 4.33$

Would you like more details or have any questions?

Related questions:

1. What is the importance of the Central Limit Theorem in sampling distributions?
2. How does increasing the sample size affect the standard error of the mean?
3. What is the difference between population standard deviation and sampling distribution standard deviation?
4. How would the results change if the sample size were larger, say $n = 100$?
5. How can you estimate a population mean using confidence intervals?

Tip: The larger the sample size, the smaller the standard error, leading to more precise estimates of the population mean.