To solve this problem, we need to calculate the mean and standard deviation of the distribution of sample means for a population with a given mean ($\mu$) and standard deviation ($\sigma$) when a random sample of a specified size ($n$) is taken.

Given Information

• Population mean, $\mu = 89.9$
• Population standard deviation, $\sigma = 12.8$
• Sample size, $n = 212$

Solutions

1. 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. Therefore: $\mu_{\bar{X}} = \mu = 89.9$

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 calculated using the formula: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ Substituting the given values: $\sigma_{\bar{X}} = \frac{12.8}{\sqrt{212}}$ Calculating this, we find: $\sigma_{\bar{X}} \approx \frac{12.8}{14.5602} \approx 0.88$ (rounded to two decimal places).

Final Answers:

• Mean of the distribution of sample means, $\mu_{\bar{X}} = 89.9$
• Standard deviation of the distribution of sample means, $\sigma_{\bar{X}} \approx 0.88$

Would you like further details on the steps or have any questions?

Here are some related questions:

1. What is the concept of the Central Limit Theorem and why is it important here?
2. How would the standard deviation change if the sample size increased to 500?
3. What is the interpretation of a standard error in the context of sampling?
4. How would the calculation change if the population distribution were not normal?
5. What sample size is needed to make the standard error less than 0.5?

Tip: Remember, as the sample size increases, the standard error decreases, making the sample mean a more accurate estimator of the population mean.