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
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) ,
4.设总体 sim N(mu ,(sigma )^2), x1,x2,···,xn为样本,证明 overline (x)=dfrac (1)(n)sum _(i
[题目]-|||-设总体X服从N(1,9),X1,···X 9为X的样本,-|||-则有 ()-|||-A、 dfrac (overline {X)-1}(1)
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
(8)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)su
1.设总体 sim N(M,(sigma )^2) ,μ未知而σ^2已知,X1,X2,···,Xn是X的样本, overline (X)=dfrac (1)(n
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
5.设x1,x2,···,xn是来自总体 sim N(mu ,(sigma )^2) 的样本,x为样本-|||-均值,令 =dfrac (sum _{i=1)^
(8)设X1,X2,···,X9是来自正态总体N(μ,σ^2)的样本, _(1)=dfrac (1)(6)sum _(i=1)^6(X)_(i) , _(2)=