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A population has a mean $\mu=83$ and a standard deviation $\sigma=14$. Find the mean and standard deviation of a sampling distribution of sample means with sample size $n=49$.
$\mu_{\bar{x}}=83$ (Simplify your answer.)
$\sigma_{\bar{x}}=\square($ Simplify your answer.)
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Step 1 ：The mean of the sampling distribution of sample means is equal to the population mean. So, \( \mu_{\bar{x}} = \mu = 83 \).

Step 2 ：The standard deviation of the sampling distribution of sample means, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size. So, \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{49}} \).

Step 3 ：Calculate \( \sigma_{\bar{x}} \) to get the final value.

Step 4 ：The mean of the sampling distribution of sample means is \( \mu_{\bar{x}} = \boxed{83} \).

Step 5 ：The standard deviation of the sampling distribution of sample means is \( \sigma_{\bar{x}} = \boxed{2} \).