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)^n(({x)_(i)-overline (x))}^2}({sigma )^2} ,则 sim () .-|||-.-|||-(A) ^2(n-1) ; (B)x^2(n);-|||-(C)N(μ,σ^2); (D) (mu ,dfrac ({sigma )^2}(n))

4.设总体 sim N(mu ,(sigma )^2), x1,x2,···,xn为样本,证明 overline (x)=dfrac (1)(n)sum _(i

1.设X1,X2,···,xn来自总体X的样本, (X)=(sigma )^2, overline (X)=dfrac (1)(n)sum _(i=1)^n(X

多选题设总体sim N(mu ,(sigma )^2) X1,X2,···Xn为来自总体X的样本,sim N(mu ,(sigma )^2) X1,X2,···

1.设总体 sim N(M,(sigma )^2) ,μ未知而σ^2已知,X1,X2,···,Xn是X的样本, overline (X)=dfrac (1)(n

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)_(

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

5.11 设(X1,X2,···Xn, _(n)+1) 是正态总体N(μ,σ^2)的样本, overline (X)=-|||-dfrac (1)(n)sum

3.设总体 approx N(mu ,(sigma )^2), X1,X2···Xn是来自该总体的简单随机样本,则-|||-dfrac (1)({sigma )

设X1,X2,···,Xn是正态总体N(μ,σ^2 )的样本,则 dfrac (1)(n)sum _(i=1)^n(({X)_(i)-overline (X))