Question

Suppose a simple random sample of size n is drawn from a large population with mean mu and standard deviation sigma. The sampling distribution of x overbar has mean mu Subscript x overbar equals​______ and standard deviation sigma Subscript x overbar equals​______.

Answer 1

Answer:

$X \sim N(\mu,\sigma)$  

Where $\mu$ the mean and $\sigma$  the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \mu$

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

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$  

Where $\mu$ the mean and $\sigma$  the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \mu$

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

Answer 2

Answer:

A

Step-by-step explanation:

the answer is a on edg