To find the mean and standard deviation of the sampling distribution of sample means, we need to apply the concepts from the Central Limit Theorem.
### Step 1: Mean of the Sampling Distribution of Sample Means
The mean of the sampling distribution of sample means (denoted as [tex]\(\mu_{\bar{x}}\)[/tex]) is the same as the population mean [tex]\(\mu\)[/tex]. Therefore,
[tex]\[
\mu_{\bar{x}} = \mu
\][/tex]
Given that the population mean [tex]\(\mu = 132\)[/tex],
[tex]\[
\mu_{\bar{x}} = 132
\][/tex]
So, the mean of the sampling distribution of sample means is [tex]\(\mu_{\bar{x}} = 132\)[/tex].
### Step 2: Standard Deviation of the Sampling Distribution of Sample Means
The standard deviation of the sampling distribution of sample means is known as the standard error (denoted as [tex]\(\sigma_{\bar{x}}\)[/tex]). It can be calculated using the formula:
[tex]\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\][/tex]
Here, [tex]\(\sigma\)[/tex] is the population standard deviation, and [tex]\(n\)[/tex] is the sample size. Given [tex]\(\sigma = 23\)[/tex] and [tex]\(n = 57\)[/tex],
[tex]\[
\sigma_{\bar{x}} = \frac{23}{\sqrt{57}}
\][/tex]
### Step 3: Calculate and Round the Standard Error
First, compute the square root of [tex]\(57\)[/tex]:
[tex]\[
\sqrt{57} \approx 7.550
\][/tex]
Next, divide the population standard deviation by this value:
[tex]\[
\sigma_{\bar{x}} = \frac{23}{7.550} \approx 3.046
\][/tex]
Rounding to three decimal places, the standard error is [tex]\(\sigma_{\bar{x}} \approx 3.046\)[/tex].
### Final Answer
[tex]\[
\mu_{\bar{x}} = 132
\][/tex]
[tex]\[
\sigma_{\bar{x}} = 3.046
\][/tex]
Therefore, the mean of the sampling distribution of sample means is [tex]\(\mu_{\bar{x}} = 132\)[/tex], and the standard deviation is [tex]\(\sigma_{\bar{x}} = 3.046\)[/tex].
