Understanding why variance of the standard normal distribution equals one intuitively Can anyone explain to me why the variance of the standard normal distribution is 1? I am trying to understand the mechanism behind standardising random variable. While I know minus the variable by the mean is like shifting the graph to make it centre at the origin, I don't know why dividing it by SD makes the variable having SD = 1 as well
 A: The variance of standard normal distribution is $1$ by definition.
Concerning standardizing: if $X$ has a distribution with standard deviation $\sigma_X\neq0$ or equivalently with variance $\sigma_X^2$ then for every constant $c$ (also $c=\mathbb EX$) we have $\mathsf{Var}\left(\frac{X-c}{\sigma_X}\right)=1$ according to the rule:$$\mathsf{Var}(aY+b)=a^2\mathsf{Var}Y$$
Can you deduce this rule yourself?
Applying it on $Y=\frac{X-c}{\sigma_X}$ we get: $$\mathsf{Var}(\sigma_X^{-1}X+(-\sigma_X^{-1}c))=\sigma_X^{-2}\mathsf{Var}X=\sigma_X^{-2}\sigma_X^{2}=1$$
This means that we can write $X=\sigma_XU+\mu_X$ where $U:=\frac{X-\mu_X}{\sigma_X}$ has mean $0$ and variance $1$.
A: Let $X\sim N(\mu,\sigma^2)$ and $Z=\frac{X-\mu}{\sigma}$, then $Z\sim N(0,1)$, because:
$$\mathbb E\left(\frac{X-\mu}{\sigma}\right)=\frac{1}{\sigma}\cdot \mathbb E(X-\mu)=\frac1{\sigma}\cdot \mathbb E(X)-\frac{\mu}{\sigma}=0;\\
\sigma^2\left(\frac{X-\mu}{\sigma}\right)=\frac{1}{\sigma^2}\cdot \sigma^2(X-\mu)=\frac1{\sigma^2}\cdot \sigma^2(X)=1.$$
