If $X \sim N(\mu, \sigma ^2)$, show that $(X - \mu) / \sigma \sim N(0,1)$ I don't know how to do this. Do I need to use converge in distribution? (I thought this can only been used if $n$ involves)
 A: The transformations in $\frac {X -\mu}{\sigma}$ are just translation and scaling, i.e., this is a change of coordinate from $X$ to $\frac {X-\mu}{\sigma}$ given by the function $X'=\phi(X)=\frac {X-\mu}{\sigma}$. More formally, you can use the Jacobian to study the effect of the change of variables in the density function:
https://en.wikipedia.org/wiki/Integration_by_substitution#Application_in_probability.
You are making a (linear) change of variable in $X$, given by $X':=\frac{X-\mu}{\sigma}$, or in the notation of the link:
$X':=\phi(X)= \frac {X-\mu}{\sigma} $ , so that $\phi^{-1}=\sigma X'+\mu$, and
then $p(X')= p(\phi^{-1}(X))d\phi^{-1}(X) $.
A: Just write out the explicit formula for the gaussian (https://en.wikipedia.org/wiki/Normal_distribution) 
and then perform the transformation. The algebra shouldn't be difficult. 
A: Okay, so we know that $X\sim N(\mu,\sigma)$, looks like this: 
$$
Pr(X<x) = F_X(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(\frac{t-\mu}{\sigma})^2}dt
$$
Next observe that
$$
Pr(\frac{X-\mu}{\sigma}<x)=Pr(X<x\sigma+\mu)=\int_{-\infty}^{x\sigma+\mu}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(\frac{t-\mu}{\sigma})^2}dt
$$
We then let $y = \frac{t-\mu}{\sigma}$, so $ dy = \frac{dt}{\sigma}$ and $y(x\sigma+\mu) = \frac{x\sigma+\mu-\mu}{\sigma}=x$, while $y(-\infty)=-\infty$.
Hence
$$
Pr(\frac{X-\mu}{\sigma}<x)=Pr(X<x\sigma+\mu)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-y^2}dy
$$
Which is the distribution of $N(0,1)$, so $\frac{X-\mu}{\sigma}\sim N(0,1)$
Note that this is a very simple solution, there are much more insightful proofs that you can do, I just feel that for your level this should be sufficient.
As practice try and do the same with the pdf, rather than the cdf as I did.
$\ddot\smile$
A: One straightforward argument is to calculate the moment generating function of $(X - \mu) / \sigma$,
$$
\begin{align}
M_{(X - \mu) / \sigma}(t) &= M_{X - \mu}(t / \sigma) \\
&= M_{X}(t / \sigma) M_{-\mu} (t / \sigma) \\
&= e^{t \mu / \sigma + t^2 / 2} e^{ - t \mu / \sigma} \\
&= e^{t^2 / 2} ,
\end{align}
$$
which is the standard normal m.g.f.
