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) ,X1,X2,···X5sim N(0,1) ,X1,X2,···X5sim N(0,1) ,X1,X2,···X5sim N(0,1) ,X1,X2,···X5

设总体sim N(0,1) X1,X2,···,X6是来自总体X的简单随机样本sim N(0,1) X1,X2,···,X6分别为样本均值和样本方差sim N(

设样本X1,X2,X3,X4,X5来自总体N(0,1),设样本X1,X2,X3,X4,X5来自总体N(0,1),设样本X1,X2,X3,X4,X5来自总体N(0

总体sim N(0,1), X1,X2,X3,X4,X55,X6,X7,X8为来自总体sim N(0,1), X1,X2,X3,X4,X55,X6,X7,X8的

设x1,x2···,x5为来自总体x1,x2···,x5的样本，x1,x2···,x5，x1,x2···,x5，x1,x2···,x5与x1,x2···,x5分

设X1,X2,···X16为来自总体 sim N(2,(sigma )^2)设X1,X2,···X16为来自总体 sim N(2,(sigma )^2)（

22.设X1,X2,···,x6是来自总体N(0,1 )的样本,则 _(1)sim ()-|||-(A)N(0,1) (B)N(0,6) (C)x^2(1) (

多选题设总体sim N(mu ,(sigma )^2) X1,X2,···Xn为来自总体X的样本,sim N(mu ,(sigma )^2) X1,X2,···

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) ,

(2)设样本X1,X2,···,X5来自总体N (0,1), =dfrac (C({X)_(1)+(X)_(2))}({({{X)_(3)}^2+({X)_(4

(2)设样本X1,X2,···,X5来自总体N (0,1), =dfrac (C({X)_(1)+(X)_(2))}({({{X)_(3)}^2+({X)_(4