数理统计课后题答案完整版.docx

:因为xi ac所以 =—2Sy—mi yi yni1na所以cycyii1Yi122 —、,2因为 Sxcyicyicyyi2acy身高1542-166166-17~78178-182组中值8172176180学生数17解:x mixini1又因为syiy222所以 s, csy 成立115610160141642617212168281768**********.解:变换yi10\ 27*-3-91234mi234121l*_2S—miXixni1122221015616614160********** - miyi8解:将子样值重新排列(由小到大)-4,-,-,-,-,0,0,,,,,, 4 :1n2n2 Xjn2j1EXDX1 n1n1 xigi1mx1 n2x2n2n1 n2n1n2n1n2n1n2i2xi—2xn1 n2Xi xji1 j1n1n2ns1 X;2n12 2nen2S2n1n2ms) n2s:2n1n2□1X1n2x2n1n22n2S2——2X2n2_2 2n1 X1 n2 X2n1 n2—2n1 n1 n2 X1_ 2n1X1 n2X2n1 n2_ 2n1X1 n2X2n1n2—2n2n1 n2 X22n1 n2n1X1——2n〔n2X22n1n2X1X22n2ElnniiD1nixiExiDxi13n所以 Xi\N0,1i1,2,,n2由 分布定义可知14•解:因为Xj#NnXi22服从2分布Yi1所以 Y& :因为XiNN0,1i1,2,,nX1X2X3NN0,3所以一2n2X2同理X10DX1X2X3 143X1X27N0,1、3N2X1X2X3N2132X4X5X6 iv213 '2由于 分布的可加性,故2ne2n2S2— —2mn?X1 X2n1n22n1n22 2山色n2s2n1n2—:Dxii1,2,,:(1)因为X^'N0,2i1,2,,n—L 1n1nnEXE-Xni1ni1n—r 1n1nnDXD-X2Dxi2——:XNUa,bExabDxi-2bai1,2,,n212在此题中xNu1,1EXi0DX1i1,2,,n3辿N0,1所以XiFy1yPY1 yY1~2y_2y""2f2xdx0因为1x2~n22;所以(2)因为 Xi生Nn0,1所以nXiFY2(3)因为所以故 fY,y(4)因为n122所以0,1,2,ny22nn,12 2n2y2nn220,ny0,1ny~2ny—e尹ndx1,2,,n故fY4Xii1xdx1"Fxe1 e、2ny0X^N0,2n _xli1nPY4yFy,y2n"20,11e2y017•解:因为 XNtn存在相互独立的uUN0,1由定义可知X2VNU22Nf1,nX^N0,218解:因为niin1N0,1f21xdx01~21,2,所以n、,mXii 1n mXi2in1Xii1n2nmXixe所以0,xdxxexdx(2)i1iXi0,11,2,,).E(x)Xi2mi1nmnXi2in1n2nmXi2Fn,mk(1x11\k1P)Pk(11\k1P)2': :因为XI2nE2nD22nin12由分布的性质3可知令,2n0,1c•2n所以有P2).其似然函数为L(P)n(1i1P)x1'n4p(1p)inXin1lndlndp解之得L(P)nlnn)ln(limPn-2nc_n•2nt2?dtc_n•:因为总体X服从U,、abE(X)2令E(X)=XS2=lna+b"I-(an(Xii1XaX..3Sa,b)所以a-b)2n!12r!nrD(X)=S2,X):(1)设x1,x2j|xn为样本观察值则似然函数为:L()n(Xi)1,0Xi1,i 1,2,InL()nkininL()dinL解之得:(2)母体E(x)in-1niniiXidlnL()nkinXi解得(k1)in(inXii1n—ninxii1n—n