What happens when we compute the normalized sample mean Zn in this situation?

Question

$\mathcal{N}(\mu, \sigma^2), \quad \sigma=5.$　Follow a normal population.

What happens when we compute the normalized sample mean Zn in this situation?

Answer 1

First of all we have $\overline X_n=\frac{1}{n} \sum\limits_{i=1}^{n}X_i$. The expected value is just $\mu$. And the variance is $Var\left(\overline X_n\right)=\frac{\sigma^2}{n}$, if the $X_i$'s are i.i.d. Therefore we get

$$Z_{n} = \frac{\overline{X}_n- \mu}{\frac{\sigma}{\sqrt n}}$$

This is a standardized random variable, with the mean $0$ and the variance $1$. The sum/mean of normal distributed random variables are normal distributed as well. Thus $Z_n\sim \mathcal N(0,1)$.

We can use $Z_n$ to calculate $P(\overline{X}_n\leq 9)$, for instance. Let say, that $n=25, \mu=7, \sigma=5$. The calculation goes as follows:

$$P(\overline{X}_n\leq 9)=P\left({Z}_n\leq \frac{9- 7}{\frac{5}{\sqrt{25}}}\right)=\Phi(2)$$

$\Phi(z)$ is the cdf of the standard normal distribution, where the values can be looked up in a table, see here. If we look at the table we find that $\Phi(2)=0.97725$. So the probability that the mean of 5 i.i.d normal distributed variables (with mean 7 and standard deviation of 5) is smaller than 9 is approximately $97.7\%$