A population has a mean $\mu=72$ and a standard deviation $\sigma=6$. Find the mean and standard deviation of a sampling distribution of sample means with sample size $n=36$. $\mathrm{n}=36$.
$\mu_{x}=\square$ (Simplify your answer.)

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Step 1 ：The mean of a sampling distribution of sample means is equal to the mean of the population from which the samples were drawn. Therefore, the mean of the sampling distribution, denoted as \(\mu_{x}\), is equal to the population mean \(\mu\), which is 72.

Step 2 ：The standard deviation of a sampling distribution (often called the standard error) is equal to the standard deviation of the population divided by the square root of the sample size. Therefore, the standard deviation of the sampling distribution, denoted as \(\sigma_{x}\), is equal to the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\), which is \(\sqrt{36}\).

Step 3 ：Let's calculate these values.

Step 4 ：\(\mu = 72\)

Step 5 ：\(\sigma = 6\)

Step 6 ：\(n = 36\)

Step 7 ：\(\mu_x = 72\)

Step 8 ：\(\sigma_x = 1.0\)

Step 9 ：I now know the final answer.

Step 10 ：Final Answer: The mean of the sampling distribution of sample means is \(\boxed{72}\) and the standard deviation of the sampling distribution of sample means is \(\boxed{1}\).