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

• It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $\sigma =1$ so it's really just a matter of units. – lulu Jan 12 at 11:05
• If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1. – Bermudes Jan 12 at 11:17

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:$$\mathsf{Var}(aY+b)=a^2\mathsf{Var}Y$$
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 $$1$$.
Let $$X\sim N(\mu,\sigma^2)$$ and $$Z=\frac{X-\mu}{\sigma}$$, then $$Z\sim N(0,1)$$, because: $$\mathbb E\left(\frac{X-\mu}{\sigma}\right)=\frac{1}{\sigma}\cdot \mathbb E(X-\mu)=\frac1{\sigma}\cdot \mathbb E(X)-\frac{\mu}{\sigma}=0;\\ \sigma^2\left(\frac{X-\mu}{\sigma}\right)=\frac{1}{\sigma^2}\cdot \sigma^2(X-\mu)=\frac1{\sigma^2}\cdot \sigma^2(X)=1.$$