=2-|||-4.设X1,X2,···,x3是来自总体 approx N(1,4) 的简单随机样本, overline (X)=dfrac (1)(n)sum _(i=1)^4(X)_(i) =sum _(i=1)^n(({x)_(i)-overline (x))}^2, 则统计景-|||-12(X5-X)^2 服从 __ -

4.设X1,X2,···,Xn为来自正态总体 sim N(0,1) 的简单随机样本, overline (X)=dfrac (1)(n)sum _(i=1)^n

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) ,

(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

(16)设X1,X2,···,xn为来自标准正态总体X的简单随机样本,记 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

7.设X1,X2,···, _(n)(ngeqslant 2) 为来自正态总体N(μ,1)的简单随机样本,若 overline (X)=-|||-dfrac (

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

3.设总体 approx N(mu ,(sigma )^2), X1,X2···Xn是来自该总体的简单随机样本,则-|||-dfrac (1)({sigma )

设X1,X2,···, _(n)(ngt 2) 为来自总体N(0,1)的简单随机样-|||-本,X为样本均值,记 overline (X)=dfrac (1)(