Question 1, 5.4.3 HW Score: 085, 0 of 7 points Pert2oti2 Points: 0 of 1 Save A population has a mean \( \mu=72 \) and a standard deviation \( \sigma=29 \). Find the mean and standard deviation of a sampling distribution of sample means with sample size \( n=231 \). \( F_{5}=72 \) (Simplify your answer) \( \sigma_{x}=\square \) (Type an integer or decimal rounded to three decimal places as needed.)

To find the mean and standard deviation of a sampling distribution of sample means, we can use the following formulas: 1. Mean of the sampling distribution of sample means: \( \mu_{\bar{x}} = \mu \) 2. Standard deviation of the sampling distribution of sample means: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \) Given: - Population mean: \( \mu = 72 \) - Population standard deviation: \( \sigma = 29 \) - Sample size: \( n = 231 \) 1. Mean of the sampling distribution of sample means: \( \mu_{\bar{x}} = \mu = 72 \) 2. Standard deviation of the sampling distribution of sample means: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{29}{\sqrt{231}} \) Now, we can calculate the standard deviation of the sampling distribution of sample means. Calculate the value by following steps: - step0: Calculate: \(\frac{29}{\sqrt{231}}\) - step1: Multiply by the Conjugate: \(\frac{29\sqrt{231}}{\sqrt{231}\times \sqrt{231}}\) - step2: Multiply the numbers: \(\frac{29\sqrt{231}}{231}\) The standard deviation of the sampling distribution of sample means is approximately 1.90806.