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The normal distribution is perhaps the most important distribution in statistical applications since many measurements have (approximate) normal distributions. One explanation of this fact is the role of the normal distribution in the Central Theorem.
Clearly, $f\left(x\right)>0$ . Let now evaluate the integral: $$I={\displaystyle \underset{-\infty}{\overset{\infty}{\int}}\frac{1}{\sigma \sqrt{2\pi}}\mathrm{exp}\left[-\frac{{\left(x-\mu \right)}^{2}}{2{\sigma}^{2}}\right]}dx,$$ showing that it is equal to 1. In the integral, change the variables of integration by letting $z=\left(x-\mu \right)/\sigma $ . Then,
$$I={\displaystyle \underset{-\infty}{\overset{\infty}{\int}}\frac{1}{\sqrt{2\pi}}{e}^{-{z}^{2}/2}}dz,$$ since $I>0$ , if ${I}^{2}=1$ , then $I=1$ .
Now $${I}^{2}=\frac{1}{2\pi}\left[{\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{e}^{-{x}^{2}/2}}dx\right]\left[{\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{e}^{-{y}^{2}/2}}dy\right],$$ or equivalently,
$${I}^{2}=\frac{1}{2\pi}{\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{\displaystyle \underset{-\infty}{\overset{\infty}{\int}}\mathrm{exp}\left(-\frac{{x}^{2}+{y}^{2}}{2}\right)dxdy}}.$$
Letting $x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta $ (i.e., using polar coordinates), we have
$${I}^{2}=\frac{1}{2\pi}{\displaystyle \underset{0}{\overset{2\pi}{\int}}{\displaystyle \underset{0}{\overset{\infty}{\int}}{e}^{-{r}^{2}/2}rdrd\theta =}}\frac{1}{2\pi}{\displaystyle \underset{0}{\overset{2\pi}{\int}}d\theta =\frac{1}{2\pi}2\pi =1.}$$
The mean and the variance of the normal distribution is as follows:
$$E\left(X\right)=\mu $$ and $$Var\left(X\right)={\mu}^{2}+{\sigma}^{2}-{\mu}^{2}={\sigma}^{2}.$$
That is, the parameters $\mu $ and ${\sigma}^{2}$ in the p.d.f. are the mean and the variance of X .
If the p.d.f. of X is
$$f\left(x\right)=\frac{1}{\sqrt{32\pi}}\mathrm{exp}\left[-\frac{{\left(x+7\right)}^{2}}{32}\right],-\infty <x<\infty ,$$ then X is $N\left(-\mathrm{7,16}\right)$
That is, X has a normal distribution with a mean $\mu $ =-7, variance ${\sigma}^{2}$ =16, and the moment generating function
$$M\left(t\right)=\mathrm{exp}\left(-7t+8{t}^{2}\right).$$
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