({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 ({{
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^
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) ,
10.设x1,x2,···,xn为一个样本, ^2=dfrac (1)(n-1)sum _(i=1)^n(({x)_(i)-overline (x))}^2 是
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
.-(x)^2(n) 的简单随机样本, overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) ,则-|||-|E(overl
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-|||-()
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