A population of values has a normal distribution with $$ \mu=26.1 $$ and $$ \sigma=91.3 $$. A randorn sample of size $$ n=130 $$ is drawn. a. What is the mean of the distribution of sample means? $$ \mu_{\bar{x}}= $$ b. What is the standard deviation of the distribution of sample means? Round your answer to two decimal places. $$ \sigma_{\bar{x}}= $$

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UpStudy AI · Accepted Answer

To solve the problem, we need to use the properties of the sampling distribution of the sample mean.

### Given:
- Population mean ($$ \mu $$) = 26.1
- Population standard deviation ($$ \sigma $$) = 91.3
- Sample size ($$ n $$) = 130

### a. Mean of the distribution of sample means ($$ \mu_{\bar{x}} $$):
The mean of the distribution of sample means is equal to the population mean. Therefore:

$$
\mu_{\bar{x}} = \mu = 26.1
$$

### b. Standard deviation of the distribution of sample means ($$ \sigma_{\bar{x}} $$):
The standard deviation of the distribution of sample means (also known as the standard error) is calculated using the formula:

$$
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
$$

Now, we will calculate $$ \sigma_{\bar{x}} $$:

1. Calculate $$ \sqrt{n} $$:
   $$
   \sqrt{n} = \sqrt{130}
   $$

2. Calculate $$ \sigma_{\bar{x}} $$:
   $$
   \sigma_{\bar{x}} = \frac{91.3}{\sqrt{130}}
   $$

Let's perform the calculations.

First, calculate $$ \sqrt{130} $$ and then use it to find $$ \sigma_{\bar{x}} $$.

Calculating $$ \sqrt{130} $$:

$$
\sqrt{130} \approx 11.4018
$$

Now, substituting this value into the formula for $$ \sigma_{\bar{x}} $$:

$$
\sigma_{\bar{x}} = \frac{91.3}{11.4018} \approx 8.00
$$

### Final Answers:
- a. $$ \mu_{\bar{x}} = 26.1 $$
- b. $$ \sigma_{\bar{x}} \approx 8.00 $$ (rounded to two decimal places)
a. The mean of the distribution of sample means is 26.1. b. The standard deviation of the distribution of sample means is approximately 8.00.

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