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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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  • $\begingroup$ 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. $\endgroup$ Nov 12 '17 at 0:20
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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.

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