Gaussian Function

This follows $f (z) = e^{- \frac{z^{2}}{2}}$ for $- \infty < z < \infty$
What is $\int e^{- \frac{z^{2}}{2}} \differential z$
- There is no closed form
But $\int_{- \infty}^{\infty} e^{- \frac{z^{2}}{2}} \differential z = \sqrt{2 π}$
Normalize the function, we need it to be $1$ over the support to be a Probability Density Function.
$\int_{- \infty}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}} \differential z = 1$
See that this is starting to resemble the PDF of the Normal Distribution.
So $f (z) = \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}}, - \infty < z < \infty$ .
$E [Z] = 0$ , $Var (Z) = 1$ and $SD [Z] = 1$
$X \sim N (μ, σ^{2})$
So $f_{X} (x) = \frac{1}{\sqrt{2 π} σ} e^{- \frac{(x - μ)^{2}}{2 σ^{2}}}, - \infty < x < \infty$