dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-n-|||-C. sqrt (dfrac
,(X)_(n+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) , overline (X)=dfrac (1)(n)sum _(i
(B) (hat {sigma )}^2=dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overline (X))}^2.-|||-(C
,(X)_(n+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) . overline (X)=dfrac (1)(n)sum _(i
,(x)_(n),(x)_(n+1) 是来自N(μ,σ^2)的样本, overrightarrow ({x)_(n)}=dfrac (1)(n)sum _(i=
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) ,
.-(x)^2(n) 的简单随机样本, overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) ,则-|||-|E(overl
5.11 设(X1,X2,···Xn, _(n)+1) 是正态总体N(μ,σ^2)的样本, overline (X)=-|||-dfrac (1)(n)sum
1.设X1,X2,···,xn来自总体X的样本, (X)=(sigma )^2, overline (X)=dfrac (1)(n)sum _(i=1)^n(X
({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 ({{