6.设总体Xsim N(mu,sigma^2)，X_(1),X_(2),...,X_(20)为其样本，S^2=(1)/(19)sum_(i=1)^20(X_(i)-overline(X))^2为样本方差，求P0.4sigma^2leq S^2leq 2sigma^2.

30 总体Xsim N(mu,sigma^2)，x_(1),x_(2),...,x_(n)为其样本，overline(x)=(1)/(n)sum_(i=1)^n

设总体 X sim N(mu, sigma^2), X_(1), X_(2), ..., X_(n) 为来自总体X的简单随机样本，则 sum_(i=1)^n((

1.6 总体X-N(mu,sigma^2)，x_(1),x_(2),...,x_(n)为其样本，bar(x)=(1)/(n)sum_(i=1)^nx_(i)，s

设X_(1),X_(2)...,X_(n)是来自总体X的样本，则(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2为(）.A.

设（X_(1),X_(2),...,X_(10),X_(11)）是来自于正态总体Xsim N(mu,sigma^2)的样本，bar(X)=(1)/(n)sum_

12.设x_(1),x_(2),...,x_(n),x_(n+1)是来自N(mu,sigma^2)的样本，overline(x)_(n)=(1)/(n)sum_

4.设X_(1),X_(2)...,X_(n)是来自正态总体N(mu,sigma^2)的样本，试求样本方差S^2=(1)/(n-1)sum_(i=1)^n(X_

设X_(1),X_(2),...,X_(n)为总体Xsim N(mu,sigma^2)的样本，证明hat(mu)_(1)=(1)/(2)X_(1)+(2)/(3

设X_(1),X_(2),...,X_(n)为总体X的简单样本，则样本均值overline(X)=(1)/(n)sum_(i=1)^nX_(i).A. 对B.

设X_(1),X_(2),...,X_(n)是来自总体X的样本，则(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2是（）A.