Question

a population has a mean $mu = 168$ and a standard deviation $sigma = 23$. find the mean and standard deviation of the sampling distribution of sample means with sample size $n = 51$.

the mean is $mu_{overline{x}}=square$, and the standard deviation is $sigma_{overline{x}}=square$.
(round to three decimal places as needed.)

Explanation:

Step1: Find the mean of the sampling distribution

According to the central limit theorem, the mean of the sampling distribution of sample means ($\mu_{\bar{x}}$) is equal to the population mean ($\mu$).
$\mu_{\bar{x}}=\mu$
Given $\mu = 168$, so $\mu_{\bar{x}}=168$.

Step2: Find the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution of sample means ($\sigma_{\bar{x}}$) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size.
Given $\sigma = 23$ and $n = 51$, then $\sigma_{\bar{x}}=\frac{23}{\sqrt{51}}$.
Calculate $\sqrt{51}\approx7.141$, so $\sigma_{\bar{x}}=\frac{23}{7.141}\approx3.221$.

Answer:

The mean is $\mu_{\bar{x}} = 168$, and the standard deviation is $\sigma_{\bar{x}}=3.221$.