Question

a population has a mean $mu = 71$ and a standard deviation $sigma = 14$. find the mean and standard deviation of a sampling distribution of sample means with sample size $n = 49$. $mu_{\bar{x}}=square$ (simplify your answer.)

Explanation:

Step1: Recall mean of sampling distribution property

The mean of the sampling - distribution of sample means $\mu_{\bar{x}}$ is equal to the population mean $\mu$.
$\mu_{\bar{x}}=\mu$

Step2: Substitute the population mean value

Given $\mu = 71$, so $\mu_{\bar{x}}=71$.

Step3: Recall standard deviation of sampling distribution formula

The standard deviation of the sampling - distribution of sample means (also known as the standard error) is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size.

Step4: Substitute the given values

Given $\sigma = 14$ and $n = 49$. First, calculate $\sqrt{n}=\sqrt{49}=7$. Then $\sigma_{\bar{x}}=\frac{14}{7}=2$.

Answer:

The mean of the sampling - distribution of sample means $\mu_{\bar{x}} = 71$ and the standard deviation of the sampling - distribution of sample means $\sigma_{\bar{x}}=2$.