,(X)_(n+1))(ngt 1) 取自总体 sim N(mu ,(sigma )^2) , overline (X)=dfrac (1)(n)sum _(i
(B) dfrac (1)(n+1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-(C) dfrac (1)(n)s
12.设总体 sim N(mu ,(sigma )^2), μ,σ^2均未知,则 dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overl
4.设总体 sim N(mu ,(sigma )^2), x1,x2,···,xn为样本,证明 overline (x)=dfrac (1)(n)sum _(i
4.设X1,.,Xn为正态总体 sim N(mu ,(sigma )^2) 的样本,记 ^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(
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) ,
,(X)_(n)), 是取自 sim N((mu )_(1)(sigma )^2) 的样本,-|||-X与Y相互独立,若 dfrac (k{(x))^2}(su
设总体 X sim N(mu, sigma^2), mu, sigma^2 均未知,则 (1)/(n) sum_(i=1)^n (X_i - overline(
(B) (hat {sigma )}^2=dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overline (X))}^2.-|||-(C
设总体X服从正态分布N(mu,sigma^2),其样本为x_1,x_2,...,x_n,x_(n+1),overline(x_n)=(1)/(n)sum_(i=