Ok so we have the cumulative function for the standard normal distribution: $$\Phi(z)=\frac{1}{\sqrt{2\pi}}\int^{z}_{-\infty}e^{-t^2/2}\;dt$$ and the cumulative function for the general normal distribution: $$P(X\leq x)=\frac{1}{\sigma\sqrt{2\pi}}\int^{x}_{-\infty}e^{-(t-\mu)^2/2\sigma^2}\;dt$$
Now I was taught that we could relate the probabilities as $$P(X\leq x)=P\left (Z\leq \frac{x-\mu}{\sigma}\right )=\Phi\left (\frac{x-\mu}{\sigma}\right )$$ $\textit{However this forgets about the factor of $1/\sigma$ that differs from the two explicit formulae.}$
Why may we use this relation if it does not account for scaling factors?
We can do this because if $X$ is normally distributed $(\mu,\sigma^2)$ then $\frac{x - \mu}{\sigma}$ is distributed standard normal, that is mean 0 and variance 1. To see this let's compute $E(\frac{x - \mu}{\sigma})$ = $\frac{1}{\sigma}(E(x) - E(\mu))$ = $\frac{1}{\sigma}(\mu - \mu) = 0$. Now, $Var(\frac{x - \mu}{\sigma}) = \frac{1}{\sigma^2}E[(x - \mu)^2] = \frac{1}{\sigma^2}[E(x^2) - 2\mu(E(x)) + \mu^2)] = \frac{1}{\sigma^2}[E(x^2) - \mu^2] = \frac{\sigma^2}{\sigma^2} = 1.$ $\newline$Where the last piece comes from the fact that $Var(X) = E(X^2) - (E(X))^2$ and $(E(X))^2 = \mu^2$. This approach allows you to avoid using the messier integration formulas and obtain the desired result.