均值为 a + b 2 \frac{a + b}{2} 2a+b​, 总数n为 ( b − a ) (b-a) (b−a)

方差= ( x − 均 值 ) 2 n \frac{(x-均值)^2}{n} n(x−均值)2​

所以[a, b]均匀分布的方差为: v a r i a n c e = ∫ a b ( x − a + b 2 ) 2 d x ( b − a ) = 1 ( b − a ) ⋅ 1 3 ( x − a + b 2 ) 3 ∣ a b = 1 ( b − a ) ⋅ 1 3 ⋅ [ ( b − a + b 2 ) 3 − ( a − a + b 2 ) 3 ] = 1 ( b − a ) ⋅ 1 3 ⋅ [ ( b − a 2 ) 3 − ( a − b 2 ) 3 ] = 1 ( b − a ) ⋅ 2 3 ⋅ ( b − a 2 ) 3 = ( b − a ) 2 12 \begin{aligned} 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\bigg\rvert_{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{aligned} variance​=(b−a)∫ab​(x−2a+b​)2dx​=(b−a)1​⋅31​(x−2a+b​)3∣∣∣∣​ab​=(b−a)1​⋅31​⋅[(b−2a+b​)3−(a−2a+b​)3]=(b−a)1​⋅31​⋅[(2b−a​)3−(2a−b​)3]=(b−a)1​⋅32​⋅(2b−a​)3=12(b−a)2​​

不过也可以用 D ( x ) = E ( x 2 ) − E ( x ) 2 D(x)=E(x^2)-E(x)^2 D(x)=E(x2)−E(x)2来算