Question

the annual salary for one particular occupation is normally distributed, with a mean of about $129,000 and a standard deviation of about $20,000. random samples of 30 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 $mu_{\bar{x}}=square$, and the standard deviation is $sigma_{\bar{x}}=square$. (round to the nearest integer as needed. do not include the $ symbol in your answers.)

Explanation:

Step1: Recall 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 $\mu$. Given $\mu = 129000$, so $\mu_{\bar{x}}=\mu = 129000$.

Step2: Recall the formula for the standard deviation of the sampling - distribution of sample means

The standard deviation of the sampling - distribution of sample means (also known as the standard error) is given by the formula $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Given $\sigma = 20000$ and $n = 30$. Then $\sigma_{\bar{x}}=\frac{20000}{\sqrt{30}}\approx\frac{20000}{5.477}\approx3651$.

Answer:

The mean is $\mu_{\bar{x}} = 129000$, and the standard deviation is $\sigma_{\bar{x}}=3651$.