Solution: We are given a population follows a normal distribution with Mean [tex]\mu=64[/tex] and standard deviation [tex]\sigma=25[/tex]
According to central limit theorem, if we have a population with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] and take large samples from the population, then the distribution of samples means will be approximately normally distributed with mean and standard deviation given below:
[tex]\mu_{\bar{x}}=\mu[/tex]
[tex]\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}[/tex]
Therefore, for the given example the distribution of sample means will be approximately normal with mean and standard deviation given below:
[tex]\mu_{\bar{x}}=\mu=64[/tex]
[tex]\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{25}{\sqrt{155}}=2.01[/tex] rounded to 2 decimal places.
