【#第一文档网# 导语】以下是®第一文档网的小编为您整理的《二项分布方差公式推导复习过程》,欢迎阅读!
二项分布方差公式推导 精品文档 二项分布方差公式推导 若ξ~B(n,p),q=1-p,求证Dξ=npq 1∵Eξ=np, kCnkpq=npCkpq, n11k1kk Cnkpq=np[(k-1)Ckpq+Cn1pq] n12k1=np[(n-1)pCknpq+Cn1pq] 2kn-kk-1n-kkn-kk-1n-kk-1n-kk-2n-kk-1n-k而Dξ=E2(E)2, n-1kn-k∴Dξ=(1×1×Cn1p1q+2×2 Cn2p2qn-2+…+k×k Cnkpq+…+n×nCnpq)(np)n-1nn02 k-=np(1×Cn-10p0q+2Cn-11p1qn-2+3Cn-12p2qn-2+…+k Cn-1k-1p1n-kq 22n+…+nCn-1n-1pn-1q0)-2np Eξ+np(p+q) =np{[0×Cn-10p0q+1Cn-11p1qn-2+2Cn-12p2qn-2+…+(k-1) Cn-1k-1n-1pq+…+(n-1)Cn-1n-1pn-1q0]+(Cn-10p0q+Cn-11p1qn-2+22n-2k-1k-1n-kk-1n-kn-1Cn-1pq+…+Cn-1pq+…+Cn-1pq)}(np)=np[Eη+(p+q)] (np)=np[(n-1)p+1] (np)=np(1-p) =npq . 2n-1n-102 n-12 收集于网络,如有侵权请联系管理员删除 本文来源:https://www.dywdw.cn/1e945f75834d2b160b4e767f5acfa1c7aa0082b6.html