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)+dfrac (1)(5)(X)_(3) 是μ的无偏估计量;()A、错B、对

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) ;

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)_

9.设总体 sim N(mu ,(sigma )^2) ,X1,X2,···,Nn为来自X的样本.-|||-证明:统计量-|||-.=dfrac ((dfrac

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)

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)_(

[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

5.设x1,x2,···,xn是来自总体 sim N(mu ,(sigma )^2) 的样本,x为样本-|||-均值,令 =dfrac (sum _{i=1)^

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

(2)设样本X1,X2,···,X5来自总体N (0,1), =dfrac (C({X)_(1)+(X)_(2))}({({{X)_(3)}^2+({X)_(4

(2)设样本X1,X2,···,X5来自总体N (0,1), =dfrac (C({X)_(1)+(X)_(2))}({({{X)_(3)}^2+({X)_(4