Question

the annual salary for one particular occupation is normally distributed, with a mean of about $122,000 and a standard deviation of about $25,000. random samples of 35 are drawn from this population, and the mean of each sample is determined. find the mean and standard deviation of the sampling distribution of these sample means. then, sketch a graph of the sampling distribution. the mean is \\( \mu_{\bar{x}}= \\), and the standard deviation is \\( \sigma_{\bar{x}}= \\). (round to the nearest integer as needed. do not include the $ symbol in your)

Explanation:

Step1: Find the mean of the sampling distribution

According to the Central Limit Theorem, for a sampling distribution of sample means, the mean of the sampling distribution \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\).
Given \(\mu = 122000\), so \(\mu_{\bar{x}}=122000\)

Step2: Find the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution (also known as the standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.
Given \(\sigma = 25000\) and \(n = 35\), then \(\sigma_{\bar{x}}=\frac{25000}{\sqrt{35}}\approx\frac{25000}{5.916}\approx4226\)

Answer:

The mean is \(\mu_{\bar{x}} = 122000\), and the standard deviation is \(\sigma_{\bar{x}}\approx4226\)