Here we will consider sampling a normal distribution   $n$   mutually independent times, and consider the density function for mean of the sample.   That is, let   $X$   be a normal random variable with density function   $\displaystyle{ Z_{\mu,\sigma} }\,$ , and let   $\displaystyle{ X_1 }\, $ , $\,\displaystyle{ X_2 }\, $ , … , $\,\displaystyle{ X_n }$   be   $n$   mutually independent samples of   $X\;$ .   Then
$$ \bar{X} =\textstyle{\frac{1}{n}}\left( X_1 + \cdots + X_n \right) $$
will be the random variable denoting the mean of the sample.

For the purposes of determining the density function of   $\bar{X}\,$ , we will first determine its moment generating function.

We already know how to compute the moment generating function for a linear combination of independent random variables.   That is, we have seen that if   $\displaystyle{ X_1 }\, $ , … , $\,\displaystyle{ X_n }$   are mutually independent, then
$$
M_{a_1\, X_1 +\cdots +a_n\, X_n}(t) =M_{a_1\, X_1}(t) \cdot M_{a_2\, X_2}(t) \cdot \, \cdots\, \cdot M_{a_n\, X_n}(t)
=M_{X_1}(a_1\, t) \cdot M_{X_3}(a_2\, t)\cdot \, \cdots \, \cdot M_{X_n}(a_n\, t) \;\text{.}
$$
Thus
$$
M_{\bar{X}}(t) =M_{\frac{1}{n}X_1}(t)\cdot M_{\frac{1}{n}X_2}(t)\cdot\,\cdots\,\cdot M_{\frac{1}{n}X_n}(t)
=M_{X_1}\left(\frac{t}{n}\right)\cdot M_{X_2}\left(\frac{t}{n}\right)\cdot\,\cdots\,\cdot M_{X_n}
\left(\frac{t}{n}\right) \;\text{.}
$$
Since the random variables   $\displaystyle{ X_1 }\, $ , … , $\,\displaystyle{ X_n }$   all share the same density function, this gives
$$ M_{\bar{X}}(t) =M_{X}\left( \frac{t}{n}\right)^n \;\text{.} $$

We have already determined the moment generating function for the normal random variable   $X$   with density   $\displaystyle{ Z_{\mu,\sigma} }$   to be
$$ M_X(t) ={\rm e}^{\mu\, t +\frac{1}{2}\,\sigma^2\, t^2} \;\text{.} $$
This allows us to say that
$$
M_{\bar{X}}(t) =\left( {\rm e}^{\mu\frac{t}{n} +\frac{1}{2}\,\sigma^2\left(\frac{t}{n}\right)^2 } \right)^n ={\rm e}^{\mu t +\frac{1}{2}\,\left(\frac{\sigma^2}{n}\right)\, t^2 } \;\text{.}
$$

The special thing about this is that the resulting moment generating function looks a lot like that of   $\displaystyle{ Z_{\mu,\sigma} }\;$ .   In fact, it is precisely the moment generating function of   $\displaystyle{ Z_{\mu,\frac{\sigma}{\sqrt{n}}} }\;$ .   We have already noted that the moment generating function for a random variable completely determines the probability density function of that random variable.   Consequently   $\bar{X}$   has density   $\displaystyle{ Z_{\mu,\frac{\sigma}{\sqrt{n}}} }\;$ .   That is, $\,\bar{X}$   is also a normal random variable, with the same mean as   $X\,$ , but a smaller standard deviation   –   i.e.   $\displaystyle{ \frac{\sigma}{\sqrt{n}} }$   rather than   $\sigma\;$ .

Here are graphs of the density functions for   $\displaystyle{ Z_{1,2} }\,$ , $\displaystyle{ Z_{0,1} }\,$ , and   $\displaystyle{ Z_{-1,\frac{1}{2}} }\;$ .   Notice how the density function with smaller standard deviation has graph much more concentrated about the mean.

The above observation regarding the average of a sample from a normal random variable can be seen as a generalization of an easier fact.

Sum of Independent Normal Random Variables

Fact:   Let   $\displaystyle{ X_1 }$   and   $\displaystyle{ X_2 }$   be two independent random variables, with means   $\displaystyle{ \mu_1 }$   and   $\displaystyle{ \mu_2 }\,$ , and variances   $\displaystyle{ \sigma_1^2 }$   and   $\displaystyle{ \sigma_2^2 }\,$ , respectively.   Then
$$ X_1 + X_2 $$
is a normal random variable, with mean   $\displaystyle{ \mu_1 +\mu_2 }$   and variance   $\displaystyle{ \sigma_1^2 +\sigma_2^2 }\;$ .

This can be seen by following the logic above.   The moment generating functions for   $\displaystyle{ X_1 }$   and   $\displaystyle{ X_2 }$   are
$$
M_{X_1} ={\rm e}^{\mu_1 t +\frac{1}{2}\sigma_1^2 t^2} \qquad \text{and} \qquad
M_{X_2} ={\rm e}^{\mu_2 t +\frac{1}{2}\sigma_2^2 t^2} \,\text{,}
$$
respectively.   Thus
$$
\begin{array}{rl}
M_{X_1 +X_2} & =M_{X_1} M_{X_2} \\ & \\
& ={\rm e}^{\mu_1 t +\frac{1}{2}\sigma_1^2 t^2} {\rm e}^{\mu_2 t +\frac{1}{2}\sigma_2^2 t^2} \\ & \\
& ={\rm e}^{\left(\mu_1 +\mu_2\right) t +\frac{1}{2} \left(\sigma_1^2 +\sigma_2^2\right) t^2}
\end{array}
$$
is the moment generating function for $\displaystyle{ Z_{\mu_1 +\mu_2 , \sqrt{\sigma_1^2 +\sigma_2^2}} }\;$ .   That is, $\displaystyle{ X_1 +X_2 }$   is a normal random variable with mean   $\displaystyle{ \mu_1 +\mu_2 }$   and variance   $\displaystyle{ \sigma_1^2 +\sigma_2^2 }\;$ .

Here are a couple of computational examples.

Example: A standard normal random variable, $\,X$   (with density   $\displaystyle{ Z_{0,1} }\,$ ), takes values in the interval   $[-1.96, 1.96]\,$ , symmetric about its mean, with   $95\%$   probability.   If   $\bar{X}$   is the sample mean of twenty-five samples of   $X\,$ , find the interval symmetric about its mean within which   $\bar{X}$   takes   $95\%$   of its values.

Since   $X$   and   $\bar{X}$   both have the same mean, $\,\displaystyle{ \mu_X =\mu_{\bar{X}} =0 }\,$ , $\,\bar{X}$   has density function   $\,\displaystyle{ Z_{0,\frac{1}{\sqrt{25}}} =Z_{0,0.2} }\;$ .   This takes   $95\%$   of its values in the symmetric interval   $\left[ -\frac{1.96}{5}, \frac{1.96}{5} \right] =[-.392, .392]\;$ .

Example: If   $X$   is standard normal random variable, how large must a sample be to ensure that   $\bar{X}$   takes   $95\%$   of its values in   $[-.15,.15]\;$ ?