A population has a mean \( \mu=82 \) and a standard deviation \( \sigma=6 \). Find the mean and standard deviation of a sampling distribution of sample means with sample size \( n=36 \). \( \mu_{\bar{x}}^{-}=\square \) (Simplify your answer.)

The mean of the sampling distribution of sample means, denoted as \( \mu_{\bar{x}} \), is equal to the population mean \( \mu \). Thus, in this case, \( \mu_{\bar{x}} = 82 \). The standard deviation of the sampling distribution of sample means, known as the standard error (SE), is calculated using the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \). Given \( \sigma = 6 \) and \( n = 36 \), we find: \[ \sigma_{\bar{x}} = \frac{6}{\sqrt{36}} = \frac{6}{6} = 1. \] So, \( \mu_{\bar{x}} = 82 \) and \( \sigma_{\bar{x}} = 1 \).