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.)

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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:
  $$\frac{29\sqrt{231}}{\sqrt{231}	imes \sqrt{231}}$$
- step2: Multiply the numbers:
  $$\frac{29\sqrt{231}}{231}$$
The standard deviation of the sampling distribution of sample means is approximately 1.90806.
The mean of the sampling distribution of sample means is 72, and the standard deviation is approximately 1.908.

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