(B) (n-1)(S)^2+(overline {X)}^2 (C) (S)^2+(overline {X)}^2. (D) dfrac (n-1)(n)(S)^2+(overline {X)}^2.A、AB、BC、CD、D

A)(S^2)/(sigma^2) sim chi^2(n-1) B)(n(overline(X)-mu)^2)/(S^2) sim F(1, n-1)

(B) dfrac (sqrt {n)(overline (X)-mu )}(S)sim t(n-1).-|||-(C) dfrac (sqrt {n)(ove

X_(n)是来自总体N(mu,sigma^2)的样本，overline(X),S^2分别是样本均值和样本方差，则((n-1)S^2)/(sigma^2)sim

(B) ({S)_(x)}^2+({S)_(Y)}^2sim ({x)_(x)}^2(2n-2).-|||-(C) dfrac (overrightarrow

(overline (X))=dfrac ({sigma )^2}(n)-|||-C. (overline (X)-mu )=dfrac ({sigma )^2

5、设X_(1),X_(2),...,X_(n)是正态总体N(mu,sigma^2)的一个样本，S^2=(1)/(n-1)sum_(i=1)^n(X_(i)-o

dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-n-|||-C. sqrt (dfrac

({S)_(n)}^2=dfrac (1)(n-1)sum _(i=1)^n((x)_(i)--|||-(x))^2 是样本方差,试求满足 (dfrac ({{

样本 X_1, X_2, ldots, X_n 来自总体 X sim N(0,1) ， overline(X) = (1)/(n) sum_(i=1)^n X_

样本方差 D_(n)=(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2 是总体 Xsim N(mu,sigma^2) 中 s^