Question

a population has a mean $mu = 79$ and a standard deviation $sigma = 20$. find the mean and standard deviation of a sampling distribution of sample means with sample size $n = 266$.
$mu_{overline{x}}=square$ (simplify your answer.)
$sigma_{overline{x}}=square$ (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Find the mean of the sampling distribution of sample means

The mean of the sampling distribution of sample means, \(\mu_{\bar{x}}\), is equal to the population mean \(\mu\).
\(\mu_{\bar{x}}=\mu\)
Since \(\mu = 79\), then \(\mu_{\bar{x}}=79\)

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

The formula for the standard deviation of the sampling distribution of sample means (also known as the standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)
Given \(\sigma = 20\) and \(n = 266\)
\(\sigma_{\bar{x}}=\frac{20}{\sqrt{266}}\)
First, calculate \(\sqrt{266}\approx16.3095\)
Then \(\sigma_{\bar{x}}=\frac{20}{16.3095}\approx1.226\)

Answer:

\(\mu_{\bar{x}} = 79\)
\(\sigma_{\bar{x}}\approx1.226\)