Question

a population has a mean $mu = 80$ and a standard deviation $sigma = 28$. find the mean and standard deviation of a sampling distribution of sample means with sample size $n = 241$. $mu_{\bar{x}}=80$ (simplify your answer.) $sigma_{\bar{x}}=square$ (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Recall the formula for 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 $\mu = 80$, so $\mu_{\bar{x}}=\mu = 80$.

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. Here, $\sigma = 28$ and $n = 241$.
First, calculate $\sqrt{n}=\sqrt{241}\approx15.524$.
Then, $\sigma_{\bar{x}}=\frac{28}{\sqrt{241}}=\frac{28}{15.524}\approx1.804$.

Answer:

$\mu_{\bar{x}} = 80$
$\sigma_{\bar{x}}\approx1.804$