Sure, let's solve the problem step-by-step:
Given information:
1. The population mean [tex]\(\mu = 84.1\)[/tex]
2. The population standard deviation [tex]\(\sigma = 42.7\)[/tex]
3. The sample size [tex]\(n = 18\)[/tex]
### Part (a)
What is the mean of the distribution of sample means?
The mean of the distribution of sample means, also called the expected value of the sample mean ([tex]\(\mu_{\bar{x}}\)[/tex]), is equal to the population mean ([tex]\(\mu\)[/tex]). This is a fundamental property of the sampling distribution of the sample mean.
So,
[tex]\[
\mu_{\bar{x}} = \mu = 84.1
\][/tex]
Therefore, the mean of the distribution of sample means is [tex]\(84.1\)[/tex].
### Part (b)
What is the standard deviation of the distribution of sample means?
The standard deviation of the distribution of sample means, also known as the standard error of the mean ([tex]\(\sigma_{\bar{x}}\)[/tex]), is calculated using the formula:
[tex]\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\][/tex]
where [tex]\(\sigma\)[/tex] is the population standard deviation, and [tex]\(n\)[/tex] is the sample size.
Plugging in the given values:
[tex]\[
\sigma_{\bar{x}} = \frac{42.7}{\sqrt{18}}
\][/tex]
After computing the above expression, we get:
[tex]\[
\sigma_{\bar{x}} \approx 10.06
\][/tex]
Therefore, the standard deviation of the distribution of sample means, rounded to two decimal places, is [tex]\(10.06\)[/tex].
So, summarizing:
a. The mean of the distribution of sample means is [tex]\(84.1\)[/tex].
b. The standard deviation of the distribution of sample means is [tex]\(10.06\)[/tex].
