A population has a mean $\mu=88$ and a standard deviation $\sigma=18$. Find the mean and standard deviation of a sampling distribution of sample means with sample size $n=36$.
\[
\mu_{\bar{x}}=\square(\text { Simplify your answer.) }
\]

Step 1 ：The mean of a sampling distribution of sample means, often denoted as \(\mu_{\bar{x}}\), is equal to the population mean, \(\mu\). So in this case, \(\mu_{\bar{x}}\) should be 88.

Step 2 ：The standard deviation of a sampling distribution of sample means, often denoted as \(\sigma_{\bar{x}}\), is equal to the population standard deviation, \(\sigma\), divided by the square root of the sample size, \(n\). So in this case, \(\sigma_{\bar{x}}\) should be \(\frac{18}{\sqrt{36}}\).

Step 3 ：Simplify \(\sigma_{\bar{x}}\) to get the final answer. \(\sigma_{\bar{x}} = \frac{18}{\sqrt{36}} = 3.0\)

Step 4 ：Final Answer: The mean of the sampling distribution of sample means is \(\boxed{88}\) and the standard deviation of the sampling distribution of sample means is \(\boxed{3}\).