Answer:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex] the deviation
Since the distribution for X is normal then we can conclude that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X} = \mu[/tex]
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]
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:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex] the deviation
Since the distribution for X is normal then we can conclude that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X} = \mu[/tex]
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]
Answer:
A
Step-by-step explanation:
the answer is a on edg
