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.
Best Answer
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.