分别求μ和σ的双侧置信水平0.90的置信区间。
设_(xy)=sum _(i=1)^4((x)_(i)-overline (x))((y)_(i)-overline (y)),_(xy)=sum _(i=1)
=2-|||-4.设X1,X2,···,x3是来自总体 approx N(1,4) 的简单随机样本, overline (X)=dfrac (1)(n)sum
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) ,
(B) dfrac (1)(n+1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-(C) dfrac (1)(n)s
设X_(1),X_(2),...,X_(10)是来自正态总体N(0,1)的简单随机样本,则统计量Y=(1)/(4)(sum_(i=1)^4X_(i))^2+(1
4.设总体 sim N(mu ,(sigma )^2), x1,x2,···,xn为样本,证明 overline (x)=dfrac (1)(n)sum _(i
55单选题设 _(xy)=sum _(xarrow {y)_(4)}((x)_(i)-overline (x))((y)_(i)-overline (y)) ,
3.设X_(1),...,X_(10)为来自标准正态总体Xsim N(0,1),Y_(1)=7sum_(i=1)^3X_(i)^2,Y_(2)=3sum_(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)_(
对某一距离进行4次独立测量,得到的数据为(单位:米):15.51,15.47,15.50,15.52 由此计算出 s = sqrt((1)/(n-1) sum_