x1，x2是取自总体N(μ，1)(μ未知)的样本。hat (mu )_(1)=dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) ;hat (mu )_(2)=dfrac (1)(3)(X)_(1)+dfrac (1)(2)(X)_(2) ;hat (mu )_(3)=dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)。针对这三个估计量，下列说法正确的是( )。 hat (mu )_(1)=dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) ;hat (mu )_(2)=dfrac (1)(3)(X)_(1)+dfrac (1)(2)(X)_(2) ;hat (mu )_(3)=dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)

设sim N(mu ,1), X1、X2、X来自总体样本, hat (mu )=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+

4.设总体 sim N(mu ,(sigma )^2), x1,x2,x3为来自X的样本 hat (mu )=dfrac (1)(4)(x)_(1)+b(x)_

样本X1,X2,X3来自总体X,若 hat (mu )=dfrac (1)(3)(X)_(1)+-|||-(X)_(2)+dfrac (1)(2)(X)_(3)

(D) (hat {mu )}_(4)=dfrac (1)(3)(X)_(1)+dfrac (3)(4)(X)_(2)-dfrac (1)(12)(X)_(3)

[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

3.设x1,x2,x3是取自N(μ,1)的样本,以下μ的四个估计量中最有效的是 () .-|||-(A) (hat {mu )}_(1)=dfrac (1)(5

9.设x1,x2,x3是来自总体X的样本,若 (X)=u (未知), hat (mu )=dfrac (1)(2)(x)_(1)-a(x)_(2)+3a(x)_

设X1,X2为来自正态总体N(μ,σ^2 )的样本,下列不是μ的无偏估计量的是-|||-A (mu )_(1)=dfrac (2)(3)(X)_(1)+dfra

... +({X)_(n)}^2)-|||-;(5) (mu )^2+dfrac (1)(3)((X)_(1)+(X)_(2)+(X)_(3))-|||-;(6

10.设总体 sim N(mu ,(sigma )^2) ,X1,X2,X3是来自X的样本,则当 a= __ _时, mu =dfrac (1)(7)(X)_(