A population has a mean $\mu=133$ and a standard deviation $\sigma=27$. Find the mean and standard deviation of the sampling distribution of sample means with sample size $n=60$.
The mean is $\mu_{\bar{x}}=\square$, and the standard deviation is $\sigma_{\bar{x}}=$
(Round to three decimal places as needed.)
Answer
Final Answer: The mean of the sampling distribution of sample means is \(\mu_{\bar{x}} = \boxed{133}\), and the standard deviation is \(\sigma_{\bar{x}} = \boxed{3.486}\).
Step 1 ：The mean of the sampling distribution of sample means is equal to the mean of the population. So, \(\mu_{\bar{x}} = \mu = 133\).
Step 2 ：The standard deviation of the sampling distribution of sample means, also known as the standard error, is equal to the standard deviation of the population divided by the square root of the sample size. In this case, \(\sigma = 27\) and \(n = 60\).
Step 3 ：We calculate the standard deviation of the sampling distribution of sample means as \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\).
Step 4 ：Substituting the given values, we get \(\sigma_{\bar{x}} = \frac{27}{\sqrt{60}}\).
Step 5 ：Calculating the above expression, we get \(\sigma_{\bar{x}} = 3.486\) (rounded to three decimal places).
Step 6 ：Final Answer: The mean of the sampling distribution of sample means is \(\mu_{\bar{x}} = \boxed{133}\), and the standard deviation is \(\sigma_{\bar{x}} = \boxed{3.486}\).
