[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)_(2))}(sqrt {{{x)_(3)}^2+({x)_(4)}^2+({x)_(5)}^2}}approx t(3)-|||-__

7.设X1,X2,···,X6是来自总体 sim N(0,1) 的简单随机样本.-|||-(1)当 k= __ _时, =dfrac (k({X)_(1)+(X

5.设X1,X2,X3,X4是总体 sim N(0,1) 的简单随机样本,则 =dfrac (sqrt {3)(X)_(1)}(sqrt {{{X)_(2)}^

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

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]设总体 approx N(mu ,1), (x1,x2,x3)为其样本,-|||-若估计量-|||-mu =dfrac (1)(2)(x)_(1)+df

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

(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

[题目]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 _(i)=(X)_(i)-overlin