Finally, we have the so-called univariate normal distribution, or just the normal distribution $$\begin{equation*} p(x)=\frac{1}{b\sqrt{2\pi}}\exp{\left(-\frac{(x-a)^2}{2b^2}\right)} \end{equation*}$$ with probabilities different from zero in the interval $(-\infty,\infty)$. The integral $\int_{-\infty}^{\infty}\exp{\left(-(x^2\right)}dx$ appears in many calculations, its value is $\sqrt{\pi}$, a result we will need when we compute the mean value and the variance. The mean value is $$\begin{equation*} \mu = \int_0^{\infty}xp(x)dx=\frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}x \exp{\left(-\frac{(x-a)^2}{2b^2}\right)}dx, \end{equation*}$$ which becomes with a suitable change of variables $$\begin{equation*} \mu =\frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}b\sqrt{2}(a+b\sqrt{2}y)\exp{-y^2}dy=a. \end{equation*}$$