Similarly, the variance becomes $$ \begin{equation*} \sigma^2 = \frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}(x-\mu)^2 \exp{\left(-\frac{(x-a)^2}{2b^2}\right)}dx, \end{equation*} $$ and inserting the mean value and performing a variable change we obtain $$ \begin{equation*} \sigma^2 = \frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}b\sqrt{2}(b\sqrt{2}y)^2\exp{\left(-y^2\right)}dy= \frac{2b^2}{\sqrt{\pi}}\int_{-\infty}^{\infty}y^2\exp{\left(-y^2\right)}dy, \end{equation*} $$ and performing a final integration by parts we obtain the well-known result \( \sigma^2=b^2 \). It is useful to introduce the standard normal distribution as well, defined by \( \mu=a=0 \), viz. a distribution centered around zero and with a variance \( \sigma^2=1 \), leading to $$ \begin{equation} p(x)=\frac{1}{\sqrt{2\pi}}\exp{\left(-\frac{x^2}{2}\right)}. \tag{5} \end{equation} $$