Certainly! Let's solve this step-by-step.
### Given Information:
- Population mean ([tex]\(\mu\)[/tex]) = 213.6
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 32.3
- Sample size ([tex]\(n\)[/tex]) = 210
### Goal:
We need to find:
1. The mean of the distribution of sample means ([tex]\(\mu_{\bar{x}}\)[/tex]).
2. The standard deviation of the distribution of sample means ([tex]\(\sigma_{\bar{x}}\)[/tex]).
### Solution:
1. Mean of the distribution of sample means ([tex]\(\mu_{\bar{x}}\)[/tex])
According to the Central Limit Theorem, the mean of the distribution of sample means is equal to the population mean.
[tex]\[
\mu_{\bar{x}} = \mu = 213.6
\][/tex]
Therefore, the mean of the distribution of sample means ([tex]\(\mu_{\bar{x}}\)[/tex]) is:
[tex]\[
\mu_{\bar{x}} = 213.6
\][/tex]
2. Standard deviation of the distribution of sample means ([tex]\(\sigma_{\bar{x}}\)[/tex])
The standard deviation of the distribution of sample means, also known as the standard error, is calculated using the population standard deviation divided by the square root of the sample size:
[tex]\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\][/tex]
Plugging in the given values:
[tex]\[
\sigma_{\bar{x}} = \frac{32.3}{\sqrt{210}}
\][/tex]
Using a calculator to compute this:
[tex]\[
\sigma_{\bar{x}} \approx 2.22539
\][/tex]
Rounding to two decimal places:
[tex]\[
\sigma_{\bar{x}} = 2.23
\][/tex]
### Final Answers:
[tex]\[
\mu_{\bar{x}} = 213.6
\][/tex]
[tex]\[
\sigma_{\bar{x}} = 2.23
\][/tex]
