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 $n = 241$. $mu_{\bar{x}}=square$ (simplify your answer.)

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 $\mu$. Given $\mu = 80$, so $\mu_{\bar{x}}=\mu$.
$\mu_{\bar{x}} = 80$

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. Here, $\sigma = 28$ and $n = 241$. First, find $\sqrt{n}=\sqrt{241}\approx15.52$. Then $\sigma_{\bar{x}}=\frac{28}{15.52}\approx1.8$.

Answer:

The mean of the sampling - distribution of sample means $\mu_{\bar{x}} = 80$ and the standard deviation $\sigma_{\bar{x}}\approx1.8$