[a, b]均匀分布方差

均值为 $\frac{a+b}{2}$, 总数n为$(b-a)$

方差= $\frac{(x-均值{)}^2}{n}$

所以[a, b]均匀分布的方差为:
$$\begin{array}{rl}variance& =\frac{{\int }_a^b(x-\frac{a+b}{2}{)}^{2}dx}{(b-a)}\\ & =\frac{1}{(b-a)}\cdot \frac{1}{3}(x-\frac{a+b}{2}{)}^3{|}_a^b\\ & =\frac{1}{(b-a)}\cdot \frac{1}{3}\cdot [(b-\frac{a+b}{2}{)}^3-(a-\frac{a+b}{2}{)}^3]\\ & =\frac{1}{(b-a)}\cdot \frac{1}{3}\cdot [(\frac{b-a}{2}{)}^3-(\frac{a-b}{2}{)}^3]\\ & =\frac{1}{(b-a)}\cdot \frac{2}{3}\cdot (\frac{b-a}{2}{)}^3\\ & =\frac{(b-a{)}^2}{12}\end{array}$$

不过也可以用$D(x)=E(x^2)-E(x{)}^2$来算