({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 ({{S)_(n)}^2}({sigma )^2}leqslant 1.5)geqslant 0.95 的最小n值.

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

,(x)_(n),(x)_(n+1) 是来自N(μ,σ^2)的样本, overrightarrow ({x)_(n)}=dfrac (1)(n)sum _(i=

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) ,

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_

10.设x1,x2,···,xn为一个样本, ^2=dfrac (1)(n-1)sum _(i=1)^n(({x)_(i)-overline (x))}^2 是

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

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

样本方差 。^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 是总体方差D X的无偏估计。-|||

3.判断题样本方差S^2=(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2是总体Xsim N(mu,sigma^2)中sigm

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^