Question

a population of values has a normal distribution with $mu = 22.8$ and $sigma = 21.5$. a random sample of size $n = 203$ is drawn.

a. what is the mean of the distribution of sample means?
$mu_{\bar{x}}=$

b. what is the standard deviation of the distribution of sample means? round your answer to two decimal places.
$sigma_{\bar{x}}=$

Explanation:

Step1: Find the mean of the distribution of sample means

According to the Central Limit Theorem, for a sample of size \(n\) from a population with mean \(\mu\), the mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\).
\(\mu_{\bar{x}}=\mu\)
Given \(\mu = 22.8\), so \(\mu_{\bar{x}}=22.8\)

Step2: Find the standard deviation of the distribution of sample means

The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is given by the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)
Given \(\sigma = 21.5\) and \(n = 203\)
\(\sigma_{\bar{x}}=\frac{21.5}{\sqrt{203}}\)
First, calculate \(\sqrt{203}\approx14.25\)
Then \(\sigma_{\bar{x}}=\frac{21.5}{14.25}\approx1.51\)

Answer:

a. \(22.8\)
b. \(1.51\)