Determine $\mu_{\bar{X}}$ and $\sigma_{\bar{X}}$ from the given parameters of the population and sample size. \[ \mu=87, \sigma=22, n=21 \] \[ \begin{array}{l} \mu_{\bar{x}}=\square \\ \sigma_{\bar{x}}=\square \end{array} \] (Round to three decimal places as needed.) Time Remaining: 01:40:37 Next
Solution
Step 1 ：The problem is asking for the mean and standard deviation of the sample mean. The mean of the sample mean, denoted as \(\mu_{\bar{X}}\), is equal to the population mean, \(\mu\). The standard deviation of the sample mean, denoted as \(\sigma_{\bar{X}}\), is equal to the population standard deviation, \(\sigma\), divided by the square root of the sample size, \(n\).
Step 2 ：Given that \(\mu = 87\), \(\sigma = 22\), and \(n = 21\), we can calculate \(\mu_{\bar{X}}\) and \(\sigma_{\bar{X}}\).
Step 3 ：Since \(\mu_{\bar{X}}\) is equal to \(\mu\), we have \(\mu_{\bar{X}} = 87\).
Step 4 ：\(\sigma_{\bar{X}}\) is equal to \(\sigma\) divided by the square root of \(n\), so \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{21}} \approx 4.801\).
Step 5 ：So, the final answer is \(\mu_{\bar{x}} = \boxed{87}\) and \(\sigma_{\bar{x}} = \boxed{4.801}\).
