样本方差 D_(n)=(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2 是总体 Xsim N(mu,sigma^2) 中 s^
dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-n-|||-C. sqrt (dfrac
(B) dfrac (1)(n+1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-(C) dfrac (1)(n)s
3.判断题样本方差S^2=(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2是总体Xsim N(mu,sigma^2)中sigm
4.样本X1,X2,···Xn来自总体 sim N(0,1) , overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) ,
样本方差 。^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 是总体方差D X的无偏估计。-|||
.-(x)^2(n) 的简单随机样本, overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) ,则-|||-|E(overl
1.设X1,···,Xn为正态总体N(μ,σ^2)的样本,记 ^2=dfrac (1)(n-1)sum _(i=1)^n(({x)_(i)-overline (
({S)_(n)}^2=dfrac (1)(n-1)sum _(i=1)^n((x)_(i)--|||-(x))^2 是样本方差,试求满足 (dfrac ({{
设X1,X2,···,Xn是正态总体N(μ,σ^2 )的样本,则 dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overline (X))