1 单选 设(x1,x2,···,xn)是来自两点分布B(1,p)总体的样本,则-|||-当样本容量n充分大时,样本均值 overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) 近似服从的分-|||-布是 () .-|||-

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

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)^

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) ,

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

设X1,X2,···,Xn是正态总体N(μ,σ^2 )的样本,则 dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overline (X))

8.设X1,X2,···,Xn是来自总体N(μ,σ^2 )的简单随机样本,X是样本均值,记 ({S)_(1)}^2=-|||-dfrac (1)(n-1)sum

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

(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

设x1,x2,···,xn是一组样本观测值,x是样本均值,则样本标准差是-|||-(A) sqrt (dfrac {1)(n)sum _(i=1)^n(({x)

4.设总体 sim N(mu ,(sigma )^2), x1,x2,···,xn为样本,证明 overline (x)=dfrac (1)(n)sum _(i