Answer:
![\bar X \sim N(\mu, (\sigma)/(√(n)))](https://img.qammunity.org/2021/formulas/mathematics/college/4bvte95qymxyikwf6tc010pimabbzegclr.png)
The mean is given by:
![\mu_(\bar X)= 20](https://img.qammunity.org/2021/formulas/mathematics/college/y510chzcbdqdbz3bgu26qqfpyja0hlrjpv.png)
And the standard error would be:
![\sigma_(\bar X) =(15)/(√(36))= 2.5](https://img.qammunity.org/2021/formulas/mathematics/college/ri880b4xpvvxt0d8r3fvgz3o80fcjch5ga.png)
Explanation:
We have the following info given:
represent the random samples taken from an infinite population
represent the mean
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:
![\bar X \sim N(\mu, (\sigma)/(√(n)))](https://img.qammunity.org/2021/formulas/mathematics/college/4bvte95qymxyikwf6tc010pimabbzegclr.png)
The mean is given by:
![\mu_(\bar X)= 20](https://img.qammunity.org/2021/formulas/mathematics/college/y510chzcbdqdbz3bgu26qqfpyja0hlrjpv.png)
And the standard error would be:
![\sigma_(\bar X) =(15)/(√(36))= 2.5](https://img.qammunity.org/2021/formulas/mathematics/college/ri880b4xpvvxt0d8r3fvgz3o80fcjch5ga.png)