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The given information is:
- Population mean \( \mu = 183.4 \)
- Population standard deviation \( \sigma = 88.3 \)
- Sample size \( n = 201 \)

### Part a) Mean of the distribution of sample means:
The mean of the distribution of sample means (also known as the expected value of the sample means) is equal to the population mean. So,

\[
\mu_{\bar{x}} = \mu = 183.4
\]

### Part b) Standard deviation of the distribution of sample means:
The standard deviation of the distribution of sample means (also known as the **standard error**) is calculated by:

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

Substituting the values:

\[
\sigma_{\bar{x}} = \frac{88.3}{\sqrt{201}}
\]

Let's calculate that.



A normal distributed population has parameters μ = 183.4 and σ = 88.3. If a random sample of size n = 201 is selected, what is the mean of the distribution of sample means? What is the standard deviation of the distribution of sample means?

Solve a statistics problem involving a normal distributed population with a mean of 183.4 and standard deviation of 88.3. Learn how to calculate the mean and standard deviation of the distribution of sample means for a sample size of 201. Suitable for college-level statistics students.

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