# 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.

• Compute the mean and variance of $z.$ What do you get? Also, it should be divided by the square root of the variance (the standard deviation), not the variance. Nov 12 '17 at 0:20

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.

• Hello @heropup can you tell me where you took the last formula with the derivative d/dz please? Nov 13 '21 at 14:44

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.