표준정규분포
\begin{align*}
X &\sim N(0,1) \\\\
f_{X}(x) &= \frac {1}{2\pi}
\exp
\bigg\{\frac{(x-\mu)^2}{2\sigma^2}\bigg\}
\end{align*}
표준화 과정
f_X(x) =
\dfrac{1}{\sqrt{2\pi}\sigma}
\exp\bigg(
-\dfrac{(x-\mu)^2}{2\sigma}
\bigg)
\\[20pt]
\int_{-\infty}^{\infty}
\dfrac{1}{\sqrt{2\pi}\sigma}
\exp\bigg(
-\dfrac{(x-\mu)^2}{2\sigma}
\bigg) = 1
\\[20pt]
\sqrt{2\pi}\sigma =
\int_{-\infty}^{\infty}
\exp
\bigg\{ -\frac{(x-\mu)^2 }{2\sigma} \bigg\}
\\[20pt]
따라서 \mu = 0 이고, \sigma = 1 이라면 다음과 같다.
\sqrt{2\pi} =
\int_{-\infty}^{\infty}
\exp
\bigg\{ -\frac{x^2}{2\sigma} \bigg\}
\\[20pt]
f_Z(z) =
\dfrac{1}{\sqrt{2\pi}\sigma}
\exp
\bigg\{ -\frac{x^2}{2\sigma} \bigg\}