Question

a population has a mean $mu = 86$ and a standard deviation $sigma = 18$. find the mean and standard deviation of a sampling distribution of sample means with sample size $n = 81$. $mu_{\bar{x}}=86$ (simplify your answer.) $sigma_{\bar{x}}=square$ (simplify your answer.)

Explanation:

Step1: Recall mean formula for sampling distribution

The mean of the sampling distribution of sample means $\mu_{\bar{x}}$ is equal to the population mean $\mu$. Given $\mu = 86$, so $\mu_{\bar{x}}=86$.

Step2: Recall standard - deviation formula for sampling distribution

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, $\sigma = 18$ and $n = 81$.
$\sigma_{\bar{x}}=\frac{18}{\sqrt{81}}$

Step3: Calculate the value of $\sigma_{\bar{x}}$

Since $\sqrt{81}=9$, then $\sigma_{\bar{x}}=\frac{18}{9}=2$.

Answer:

$\mu_{\bar{x}} = 86$
$\sigma_{\bar{x}} = 2$