The formula for normal distribution is quite complicated, it has $\sigma$ in the exponent and in denominator. And no matter what $\sigma$ is, the shape of its pdf is the same (i.e. for example 3 standard deviations lie on the same point of the graph, no matter what $\sigma$ we choose). Could anyone explain how is that possible?
I guess that's why we can use things such as Z-score - the shape is universal, and thus probability as well. We only calculate the distance from the mean expressed in standard deviations.
http://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule
Note if $X\sim N[\mu,\sigma]$, then $\mathbb{P}[\beta\leq X\leq \alpha]=\mathbb{P}[z_\beta\leq z\leq z_\alpha]$ where $z\sim N[0,1]$ and $$z_{\alpha}=\frac{\alpha-\mu}{\sigma}$$ and similarly for $z_\beta$.
Well by definition, we have
$$\mathbb{P}[\beta\leq X\leq \alpha]=\int_{\beta}^\alpha f(x)\,dx=\int_\beta^\alpha \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}\,dx.$$
Now make the $u$-substitution $u=\frac{x-\mu}{\sigma}\Rightarrow \frac{du}{dx}=\frac{1}{\sigma}\Rightarrow dx=\sigma\cdot du$.
Also the limits change $\alpha\rightarrow z_\alpha$ and $\beta\rightarrow z_\beta$ so we have
$$\mathbb{P}[\beta\leq Z\leq \alpha]=\int_{z_\beta}^{z_\alpha}\frac{1}{\sigma \sqrt{2\pi}}e^{-u^2/2}\sigma\,du=\int_{z_\beta}^{z_\alpha}\frac{1}{ \sqrt{2\pi}}e^{-u^2/2}\,du=\mathbb{P}[z_\beta\leq z\leq z_\alpha].$$