Standardizing Normally Distributed Random Variables Having the normally distributed Random variables. We can normalize it and use table values in order to calculate probability of some event.
The standardization takes formula
$$z = \frac{ X - \text{expected value} }{ \text{variance}}$$
It is told that by doing this, we are forcing our variable $X$ to have expected value $0$ and variance $1$. However why is that? Why by doing steps above we force the distribution to behave like that?
Thanks for help.
 A: No, your formula to standardize a normal random variable is not correct.
The formula should be $$Z = \frac{X - \mu}{\sigma},$$ where $\sigma$ is the standard deviation, not the variance, which is $\sigma^2$.
Recall that for $X \sim \operatorname{Normal}(\mu,\sigma^2)$, with mean $\mu$ and variance $\sigma^2$, the probability density function is $$f_X(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/(2\sigma^2)}.$$  Then $Z = (X-\mu)/\sigma$ has PDF $$f_Z(z) = f_X(\sigma z + \mu) \left|\frac{d}{dz}[\sigma z + \mu] \right| = \frac{1}{\sqrt{2\pi} \sigma} e^{-(\sigma z + \mu - \mu)^2/(2\sigma^2)} \sigma = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}.$$  Therefore $Z$ has mean $0$ and variance $1$; i.e., it is standard normal.
A: Let $\mu$ be the mean, that is $\mathbb{E}[X]=\mu$ and $\sigma^2$ be the variance, that is $\operatorname{Var}{X}=\sigma^2$.
Notice that $$z=\frac{x-\mu}{\sigma}$$
There is a mistake in your standardization, we do not divide by the variance but the standard deviation. 
$$\mathbb{E}[Z]=\frac{1}{\sigma}(\mathbb{E}[X]-\mu)$$
$$\operatorname{Var}[Z]=\frac{1}{\sigma^2}(\operatorname{Var}[X-\mu])$$
Are you able to complete the proof?
Remark: You might also want to prove perhaps using mgf that $Z$ is a normal distribution too.
A: You can get to the same result of @heropup and @MatteoMenghini, starting from the cumulative distribution functions.
$$
\begin{equation}
Z := \frac{X-\mu_x}{\sigma_x}
\end{equation}
$$
The transformation is monotonically increasing, since $\frac{d}{dx}\left(\frac{x-\mu_x}{\sigma_x}\right)=1/\sigma_x>0$, then we have that
$$
\begin{align}
F_z(z) &= P(Z\leq z) \stackrel{\mathscr{S}}{=} P\left(\frac{X-\mu_x}{\sigma_x}\leq z\right)
\\&=
P(X \leq \sigma_x z + \mu_x) \stackrel{\mathscr{S}}{=} P(X \leq x)
\\&=
F_x(x)
\end{align}
$$
substituting
$$
F_z(z) \stackrel{\mathscr{S}}{=} F_x(\sigma_x z + \mu_x) \label{Fz}
$$
differentiating
$$
\begin{align}
\frac{d}{dz} F_z(z) &= \frac{d}{dz} F_x(\sigma_x z + \mu_x)
\\
f_z(z) &= \sigma_x f_x(\sigma_x z + \mu_x)
\\
f_z(z) &= \sigma_x f_x(x) \label{fz}
\end{align}
$$
