为什么二项分布的方差公式是npq?

证明:方差D(ξ)=E(ξ^2)-[E(ξ)]^2 =0^2×C(0,n)q^n+1^2×C(1,n)pq^(n-1)+……+n^2×C(n,n)p^n-(np)^2 =np[C(0,n-1)q^(n-1)+2C(1,n-1)pq^(n-2)+……+nC(n-1,n-1)p^(n-1)]-(np)^2 (化简方法同前) =np{[C(1,n-1)pq^(n-2)+2C(2,n-1)p^2q^(n-3)+……+(n-1)C(n-1,n-1)p^(n-1)] +[C(0,n-1)q^(n-1)+C(1,n-1)pq^(n-2)+……+C(n-1,n-1)p^(n-1)]}-(np)^2 =np{(n-1)p[C(0,n-2)q^(n-2)+C(1,n-2)pq^(n-3)+……+C(n-2,n-2)p^(n-2)] +(q+p)^(n-1)}-(np)^2 =np{(n-1)p(q+p)^(n-2)+(q+p)^(n-1)}-(np)^2 =np[(n-1)p+1]-(np)^2 =np(1-p)=npq。(因为p+q=1) +nC(n-1,n-1)p^(n-1)]}-(np)^2 =npq[C(0,n-2)q^(n-2)+C(1,n-2)pq^(n-3)+……+C(n-2,n-2)p^(n-2)] +n^2p^n-(np)^2 =npq(q+p)^(n-2)+n^2p^n-n^2p^2 =npq。