1.设总体 approx N(1,9), X1,X2,···,x9是来自总体X的一个样本,则 ()-|||-(A) dfrac (overline {X)-1}(1)approx N(0,1) (B) dfrac (overline {X)-1}(9)approx N(0,1) (C) dfrac (overline {X)-1}(sqrt {3)}approx N(0,1) (D) dfrac (overline {X)-1}(3)approx N(0,1)

[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)

5、设 sim N(1,9) ,X1,X2,···X9为X的样本,记 overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i)

9.设总体服从正态分布N(μ,1),且μ未知,设-|||-X1,X2,···,Xn为来自该总体的一个样本,记-|||-overline (X)=dfrac (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

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) ,

设总体sim N(0,1) ,X1,X2,···X5为来自总体的样本,令sim N(0,1) ,X1,X2,···X5则下列说法错误的是( )sim N(0,1

[题目]设x1,x2,...,x5是总体 approx N(0,1) 的简单-|||-随机样本,则当 k=_ 时, =dfrac (k({x)_(1)+(x)_

6.设X1,X2,···,Xn为来自正态总体X的一个样本,且 E(X)=-1 ,-|||-.((X)^2)=4 .则样本均值X服从 ()-|||-(A)N(0,

(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)=

=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