確率密度関数と分散

$$\begin{array}{rl}⟨X⟩& =\int xf(x)\mathrm{d}x\\ & ={\int }_0^{5}x\frac{1}{5}\mathrm{d}x\\ & =\frac{1}{5}{\left[\frac{x^2}{2}\right]}_0^5\\ & =\frac{5}{2}\\ ⟨X^2⟩& =\int x^{2}f(x)\mathrm{d}x\\ & ={\int }_0^5{x}^2\frac{1}{5}\mathrm{d}x\\ & =\frac{1}{5}{\left[\frac{x^3}{3}\right]}_0^5\\ & =\frac{25}{3}\end{array}$$
これより、分散$\sigma$は
$$\begin{array}{rl}\sigma & =⟨X^2⟩–⟨X{⟩}^2\\ & =\frac{25}{3}–{\left(\frac{5}{2}\right)}^2\\ & =\frac{25}{12}\end{array}$$
と求まる。