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A normally distributed population has parameters [tex]\mu = 99.1[/tex] and [tex]\sigma = 79.2[/tex]. If a random sample of size [tex]n = 238[/tex] is selected:

a. What is the mean of the distribution of sample means?

[tex]\mu_z = \square[/tex]

b. What is the standard deviation of the distribution of sample means? Round to two decimal places.

[tex]\sigma_{z} = \square[/tex]

Answer :

Sure! Let's break down the steps to find the mean and the standard deviation of the distribution of sample means.

a. Mean of the distribution of sample means

The mean of the distribution of sample means, also known as the sampling distribution of the sample mean, is the same as the mean of the population. This principle comes from the Central Limit Theorem, which states that the sampling distribution of the sample mean will have the same mean as the original population.

Given:
- Population mean ([tex]\(\mu\)[/tex]) = 99.1

Therefore, the mean of the distribution of sample means ([tex]\(\mu_z\)[/tex]) is:

[tex]\[
\mu_z = 99.1
\][/tex]

b. Standard deviation of the distribution of sample means

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

[tex]\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\][/tex]

Where:
- [tex]\(\sigma\)[/tex] is the population standard deviation
- [tex]\(n\)[/tex] is the sample size

Given:
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 79.2
- Sample size ([tex]\(n\)[/tex]) = 238

Plug these values into the formula:

[tex]\[
\sigma_{\bar{x}} = \frac{79.2}{\sqrt{238}}
\][/tex]

Calculating this gives approximately 5.13, when rounded to two decimal places.

So, the standard deviation of the distribution of sample means ([tex]\(\sigma_z\)[/tex]) is:

[tex]\[
\sigma_z = 5.13
\][/tex]

In conclusion:
- The mean of the distribution of sample means is [tex]\(\mu_z = 99.1\)[/tex].
- The standard deviation of the distribution of sample means is [tex]\(\sigma_z = 5.13\)[/tex].

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