Question 1, 5.4.3 HW Score: 0\%, 0 of 7 points Pat 1 of 2 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 \( n=231 \). \[ \mu_{x}^{\prime}=\square \text { (Simplífy your answer.) } \]

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 (\( \mu_{x}^{\prime} \)): \[ \mu_{x}^{\prime} = \mu \] 2. Standard deviation of the sampling distribution of sample means (\( \sigma_{x}^{\prime} \)): \[ \sigma_{x}^{\prime} = \frac{\sigma}{\sqrt{n}} \] Given: - Population mean (\( \mu \)) = 72 - Population standard deviation (\( \sigma \)) = 29 - Sample size (\( n \)) = 231 1. Mean of the sampling distribution of sample means (\( \mu_{x}^{\prime} \)): \[ \mu_{x}^{\prime} = \mu = 72 \] 2. Standard deviation of the sampling distribution of sample means (\( \sigma_{x}^{\prime} \)): \[ \sigma_{x}^{\prime} = \frac{\sigma}{\sqrt{n}} = \frac{29}{\sqrt{231}} \] 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 mean of the sampling distribution of sample means is 72, and the standard deviation of the sampling distribution of sample means is approximately 1.90806.