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(X)_(i) =-|||-sqrt (dfrac {1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2} 则 () .-|||-(A) overline (X)sim N(0,1); (B) overline (X)sim N(0,1);-|||-(C) sum _(i=1)^n({X)_(i)}^2sim (chi )^2(n-1); (D) sqrt (n)cdot overline (X)/Ssim t(n-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) ,

(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

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

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 (

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

4.设X1,.,Xn为正态总体 sim N(mu ,(sigma )^2) 的样本,记 ^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(

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

(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

=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

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)(