(B) dfrac (1)(n+1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-(C) dfrac (1)(n)s
12.设x_(1),x_(2),...,x_(n),x_(n+1)是来自N(mu,sigma^2)的样本,overline(x)_(n)=(1)/(n)sum_
({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+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) , overline (X)=dfrac (1)(n)sum _(i
dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-n-|||-C. sqrt (dfrac
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) ,
17.设x_(1),x_(2),...,x_(n),x_(n+1)是来自N(mu,sigma^2)的样本,又设overline(x)_(n)=(1)/(n)su
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_
设X1,X2,···, _(n)(ngeqslant 2) 为来自总体N(μ,1)的简单随机样本,-|||-记 overline (x)=dfrac (1)(n
2.设(X1,···,xn,xn+1)是取自正态总体N (μ,σ^2)的样本,求 =(x)_(n+1)-dfrac (1)(n+1)sum _(i=1)^n+1