Converting from the Normal Distribution to the Standard Normal Distribution

Observe $$\mathrm{P}\left[Z \leq z\right] = \mathrm{P}\left[\dfrac{X-\mu}{\sigma} \leq z\right] = \mathrm{P}\left[X \leq \sigma z + \mu \right]\text{.}$$ Substitute this into the cumulative distribution function of the normal distribution to obtain $$\int_{-\infty}^{\sigma z + \mu} \frac{1}{\sigma \sqrt{2\pi} } \mathrm{exp} \left( - \frac{1}{2} \left( \frac{x- \mu}{\sigma} \right)^2 \right) \mathrm{d} x \text{.}$$ We will need to use the substitution $t= \frac{x-\mu}{\sigma}$. Rearranging for $x$ gives $x = \sigma t +\mu$. Differentiating with respect to $t$ gives $\frac{\mathrm{d} x}{\mathrm{d} t} = \sigma$. Substituting this into the above integral yields $$\int_{-\infty}^{z} \frac{1}{\sigma \sqrt{2\pi} } \mathrm{exp} \left( -\frac{1}{2} t^2 \right) \sigma \mathrm{d} t$$ (note how the upper limit has changed as well here - see integration by substitution). Lastly cancelling $\sigma$ gives $$\int_{-\infty}^{z} \frac{1}{\sqrt{2\pi} } \mathrm{exp} \left( -\frac{1}{2} t^2 \right)\mathrm{d} t$$ which is the cumulative distribution function of the standard normal distribution. We will call this function $\Phi(z)$.

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