Question

a population has a mean $mu = 142$ and a standard deviation $sigma = 30$. find the mean and standard deviation of the sampling distribution of sample means with sample size $n = 46$. the mean is $mu_{\bar{x}}=square$, and the standard deviation is $sigma_{\bar{x}}=square$. (round to three decimal places as needed.)

Explanation:

Step1: Recall the mean of sampling - distribution of sample means

The mean of the sampling - distribution of sample means $\mu_{\bar{x}}$ is equal to the population mean $\mu$. So, $\mu_{\bar{x}}=\mu$.
Given $\mu = 142$, then $\mu_{\bar{x}}=142$.

Step2: Recall the standard deviation of 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 $\sigma = 30$ and $n = 46$, then $\sigma_{\bar{x}}=\frac{30}{\sqrt{46}}$.
Calculate $\sqrt{46}\approx6.782$, and $\sigma_{\bar{x}}=\frac{30}{6.782}\approx4.423$.

Answer:

The mean is $\mu_{\bar{x}} = 142$, and the standard deviation is $\sigma_{\bar{x}}\approx4.423$.