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
({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 ({{
,(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) ,
设总体 X sim N(0,1),(X_1,X_2,...,X_n) 是总体 X 的样本,令 overline(X)=(1)/(n)sum_(i=1)^nX_i
^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2, 则下列属于σ^2的无偏估计量的是-|||-()
,(X)_(n+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) , overline (X)=dfrac (1)(n)sum _(i
样本 X_1, X_2, ldots, X_n 来自总体 X sim N(0,1) , overline(X) = (1)/(n) sum_(i=1)^n X_
令 Y = (1)/(n) sum_(i=1)^n X_i,则A. $\Cov(X_1, Y)= \frac{\sigma^2}{n}$.B. $Cov(X_1