How is $\mathbb{E}(Z) = 0$ and $\operatorname{Var}(Z) = 1$ true when $Z$ is standard normal? For this normal distribution: 
$$Z = {X-\mu\over \sigma}$$
It always says that $\mathbb{E}(Z) = 0$ and $\operatorname{Var}(Z) = 1$, where $\mathbb{E}(X) =\mu$ and $\operatorname{Var}(X) = \sigma^2$. 
I've seen it explained in pictures (an example normal curve with $20\%$ left tail) where $X$ is the outcome, $\mu$ is the mean and $\sigma$ is the standard deviation. 
But how do you prove this?
 A: You can check that for any (nonzero) constant $c$ : $$\begin{align}
E(X+c) &= E(X) + c\\ 
E(X/c) &= E(X)/c\\ 
V(X+c) &= V(X)\\ 
V(X/c) &= V(X)/c^2
\end{align}$$
Then $$E\left({X-\mu\over \sigma}\right) = {E(X) - E(X)\over\sigma}=0$$
And $$V\left({X-\mu\over \sigma}\right) = {V(X)\over\sigma^2} = 1$$
A: Assuming you have $\mathbb{E}(X) = \mu$ and $\operatorname{Var}(x) = \sigma^2$ then we have $$\mathbb{E}(Z) = \mathbb{E}\left(\frac{X - \mu}{\sigma}\right) = \frac{1}{\sigma}\mathbb{E}(X - \mu) = \frac{1}{\sigma}(\mathbb{E}(X) - \mathbb{\mu})) = \frac{1}{\sigma}(\mu - \mu) = 0$$ by linearity of expectation and using the fact that $\mathbb{E}(\mu) = \mu$. 
Similarly we have $$\operatorname{Var}{Z} = \frac{1}{\sigma^2}\left(\operatorname{Var}{X}\right) = 1$$ using the variance laws: in particular that $\operatorname{Var}{(Y + \alpha)} = \operatorname{Var}{Y}$ (which is intuitive: translating a distribution does not change the measure of its spread) and $\operatorname{Var}{\alpha Y} = \alpha^2 \operatorname{Var}{Y}$.
