Going from a normal distribution to a standard normal distribution with a change of variable If $X$ follows a normal distribution with parameters $\mu$ and $\sigma^2$ show that $Z = (X- \mu)/\sigma$ follows a standard normal distribution. This doesn't seem to intuitive to me. We shift $X$ so that the top of the bell curve is over $0$, that bit makes sense. But then we scale the curve by a scalar, wouldn't this change the area under the curve, making it no longer a pdf since if we integrate $Z$ over the real line we need to get $1$? I guess my intiution is failing because I'm trying to imagine area along an infinite line.
Now I would like to go about proving this using the definition of the normal distribution. The definition I'm working with is $$f(x,\mu,\sigma) = \frac{1}{\sigma \sqrt{2 \pi}}e^{\frac{-(x- \mu)^2}{2 \sigma^2}}$$
I thought the proof would be showing that $$f(\frac{x-\mu}{\sigma},\mu,\sigma) = \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}$$ But this didn't work, I got an ugly expression in the exponent and I couldn't see why the $\sigma$ in the denominator would disappear at the front of the expression.  
How Should I go about proving this is a fundamental way? I think their is probably an easier proof using properties such as how adding and multiplying normal distribution's by scalars modifies it. But what I'm really after is a proof using the normal distribution definition
Thanks in advance.
 A: Write $\displaystyle P\{Z \leq a\} = P\left\{\frac{X-\mu}{\sigma} \leq a\right\}
= P\{X\leq \mu+a\sigma\}=\int_{-\infty}^{\mu+a\sigma}f_X(x)\,\mathrm dx$ and then
make a change of variable in the integral, setting $z=\frac{x-\mu}{\sigma}$,
hopefully not forgetting to change the upper limit appropriately. DO NOT
attempt to actually integrate: just do a change of variables, draw a
box around the integral that you have obtained, and admire the contents
of the box.
After the admiration is over, argue
that the formula
$\displaystyle P\{Z \leq a\} = F_Z(a) = \int_{-\infty}^a f_Z(z)\,\mathrm dz$
allows us to conclude something about the density function
$f_Z(z)$ via the admirable contents of the box you just drew. 
A: as I see you are thinking the area is divided by $$\sigma(standard deviation)$$ .....because Random Variable is divided by $\sigma$,
but $\sigma$ is also multiplied by all Probability function so the area remains same.
Now, for you second query disappearance of $\sigma$
as you Put $$X=Z\sigma+\mu$$ you must put $$dx=\sigma*dz$$(because you changed the Random Variable)
so the $\sigma$ canceled out.
