方差：
D(X) = (1/n) Σ(i=1->n) ni [xi-E(x)]^2
= (1/n) Σ(i=1->n) ni [xi^2+E^2(x)-2xiE(x)]
= (1/n) {Σ(i=1->n) nixi^2 + Σ(i=1->n) E^2(x) - 2E(x)Σ(i=1->n) xi}
= (1/n) Σ(i=1->n) nixi^2 + (1/n) Σ(i=1->n) E^2(x) - (1/n) 2E(x)Σ(i=1->n) xi
= 均方值 + 均值的平方 - 2倍均值的平方
= 均方值 - 均值的平方
= (1/n) Σ(i=1->n) (nixi^2) - E^2(x) (1)
E(X) = (1/n) Σ(i=1->n) ni xi (2)
n = Σ(i=1->n) ni (3)
