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.
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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
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