Answer:
[tex] \bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [/tex]
The mean is given by:
[tex] \mu_{\bar X}= 20[/tex]
And the standard error would be:
[tex] \sigma_{\bar X} =\frac{15}{\sqrt{36}}= 2.5[/tex]
Step-by-step explanation:
We have the following info given:
[tex] n =36[/tex] represent the random samples taken from an infinite population
[tex] \mu = 20[/tex] represent the mean
[tex]\sigma =15[/tex] represent the standard deviation
Since we don't know the distribution but we know that the sample size is large enough (n>30) in order to apply the central limit theorem and for this case the sample mean have the following distribution:
[tex] \bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [/tex]
The mean is given by:
[tex] \mu_{\bar X}= 20[/tex]
And the standard error would be:
[tex] \sigma_{\bar X} =\frac{15}{\sqrt{36}}= 2.5[/tex]