Converting a normal random variable

Question

According to what I have learnt, a random variable, X, with mean, $\mu$ and standard deviation, $\sigma$, can be converted to the standardised variable, Z, using the formula: $$Z = \frac{X-\mu}{\sigma}$$

First question: Is there any derivation for this formula? Where does it come from?

I have been told that the standard score allows us to calculate the probability of a score occurring within non-standard normal distribution and compare two scores from different normal distributions. The probability within a certain range can be obtained because of this: $$P(a<X<b) = P\left(\frac{a-\mu}{\sigma}<Z<\frac{b-\mu}{\sigma}\right)$$

Second question: How does $Z = \frac{X-\mu}{\sigma}$ makes the area under the curve within the range of a and b for a non-standard normal distribution equal to the area under the curve within the range of $\frac{a-\mu}{\sigma}$ and $\frac{b-\mu}{\sigma}$ for a standard normal distribution?

In my opinion: $$\int_a^\infty\frac{e^{-\frac{(x-\mu)^{2}}{2\sigma^2}}}{\sigma\sqrt{2\pi}}dx = \int_b^\infty\frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}}dz $$

where the equation on the right is for non-standard normal distribution while the equation on the left is for standard normal distribution. Is my assumption that $b=\frac{a-\mu}{\sigma}$ correct? If so, how to prove it?

There is a proof given by my teacher: $$mean = E(Z) = E\left(\frac{X-\mu}{\sigma}\right) = \frac{E(X)-\mu}{\sigma} = \frac{\mu-\mu}{\sigma} = 0$$ $$Var(Z) = Var\left(\frac{X-\mu}{\sigma}\right) = \frac{Var(X)-0}{\sigma^2} = \frac{\sigma^2}{\sigma^2} = 1$$

In this proof, it states that if $Z = \frac{X-\mu}{\sigma}$, $\mu$ = 0 and $\sigma^2$ = 1. How does this proof tell us that $\frac{X-\mu}{\sigma}$ maps any normal distribution X to the Standard Normal distribution Z? And I still don't understand the relationship between this proof and the equivalence of area for a non-standard normal distribution and a standard normal distribution as stated above. Could someone help me? I really appreciate it.

Answer 1

1. You can verify it by writing the pdf of $X$, then calculate the pdf of $Z=\frac{X-\mu}{\sigma}$ using the change of variable formula. $$f_Z(z)=f_X(x)\cdot\left|\frac{dx}{dz}\right|=f_X(\mu+\sigma z)\cdot \sigma=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$$
2. You can make the change of variable $z=\frac{x-\mu}{\sigma}$ in the integral
3. Your teacher considers as known (maybe without proof) that a linear transformation of a normal variable is normal. So he only calculates the mean and variance.

EDIT: if you don't know or haven't learned yet the change of variable formula you can deduce it as following:

$F_z(z)=P(Z\le z)=P\left(\frac{X-\mu}{\sigma}\le z\right)=P(X\le \mu+\sigma z)=F_X(\mu+\sigma z)$

And by differentiation with respect to $z$ (with chain rule):

$f_Z(z)=f_X(\mu+\sigma z)\cdot \sigma$

It is also proved here.