ck 4.2 Part 1 of 2 Points: 0 of 1 Save K A population has a mean $\mu=80$ and a standard deviation $\sigma=8$. Find the mean and standard deviation of a sampling distribution of sample means with sample size $n=64$. $\mu_{\dot{-}}=\square$ (Simplify your answer.) Clear all Check answer
Solution
Step 1 ：A population has a mean of \( \mu = 80 \) and a standard deviation of \( \sigma = 8 \). We are asked to find the mean and standard deviation of a sampling distribution of sample means with sample size \( n = 64 \).
Step 2 ：The mean of the sampling distribution of sample means is equal to the mean of the population. Therefore, \( \mu_{\dot{-}} = \mu = 80 \).
Step 3 ：The standard deviation of the sampling distribution of sample means is given by the formula \( \sigma_{\dot{-}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the population and \( n \) is the sample size.
Step 4 ：Substituting the given values into the formula, we get \( \sigma_{\dot{-}} = \frac{8}{\sqrt{64}} \).
Step 5 ：Solving the above expression, we find that \( \sigma_{\dot{-}} = 1.0 \).
Step 6 ：Therefore, the mean and standard deviation of the sampling distribution of sample means with sample size \( n = 64 \) are \( \mu_{\dot{-}} = 80 \) and \( \sigma_{\dot{-}} = 1.0 \) respectively.
Step 7 ：\(\boxed{\mu_{\dot{-}} = 80, \sigma_{\dot{-}} = 1.0}\)
