概率统计第三章习题参考解答

1概率统计第三章习题解答1、52}7{,51}6{}5{}4{XPXPXPXP529)(XE2、2914}7{,296}6{,295}5{,294}4{YPYPYPYP29175)(YE3、设X为取到的电视机中包含的次品数，2,1,0,}{3123102kCCCkXPkkX012pk221222922121)(XE4、设X为所得分数5,4,3,2,1,61}{kkXP12,11,10,9,8,7,361}{kkXP1249)(XE5、（1）由}6{}5{XPXP，则ee!6!565解出6，故6)(XE（2）由于11212211)1(66)1(kkkkkkk不是绝对收敛，则)(XE不存在。6、（1）691)()(03dxxexdxxxfXEx（2）525210150)251()()(dxxxxdxxdFXE7、41)1(42)()(105dxxxxdxxxfXE8、2ln23)11(2)()(212dxxxdxxxfXE29、0)1(23)1(23)()(102012dxxxxdxxxxdxxxfXE10、由4,3,2,1,0,)1(}{44kppCkXPkkk)21)(1(4)2(sinpppXE11、R的概率密度为其它,00,1)(axaxf30332416)(6)(adxaxdxxfxVEa12、5641034010329440920010316103)()())((edxxedxxexdxxfxgXgExx13、Y1的分布函数为1,110,)1(10,0)(11111minyyyyyFnY1的概率密度为其它,010,)1()(1111minyynyfn11)1()()(10111111min11ndyynydyyfyYEnYn的分布函数为1,110,0,0)(maxnnnnnnyyyyyFYn的概率密度为其它,010,)(1maxnnnnynyyf1)()(101maxnndynyydyyfyYEnnnnnnnn14、X的分布律为X012pk281528122813Y的分布律为Y012pk2810281528343)(,21)(YEXE143}2,2{4})1,2{}2,1{(2}1,1{1)(YXPYXPYXPYXPXYE41}2,0{)2(})2,1{}1,0{()1(}0,2{2})1,2{}0,1{(1)(YXPYXPYXPYXPYXPYXPYXE3}2,2{10}1,2{8}0,2{6}2,1{7}1,1{5}0,1{3}2,0{4}1,0{2)23(YXPYXPYXPYXPYXPYXPYXPYXPYXE15、143}2,2{2})1,2{}2,1{}1,1{(1)),(min(YXPYXPYXPYXPYXE149}2,2{52}1,2{31}1,1{21}2,0{2})2,1{}1,0{(1))1/((YXPYXPYXPYXPYXPYXPXYE16、10105224)(xxydyxdxXE10105224)(xxydyydxYE101015224)(xxydyxydxXYE17、400}14{)2000(}13{)1000(}11{1000}10{2000)(XPXPXPXPYE622222106.1}14{)2000(}13{)1000(}11{1000}10{2000)(XPXPXPXPYE6221044.1))(()()(YEYEYD18、2)(02222dxexxXEx42022222)(22dxexxXEx22)(,)22())(()()(222XDXEXEXD19、111)1()(kkpppkXE121222)1()(kkppppkXE2221))(()()(ppXEXEXD20、（1）1)(1kkdxxkxXEkk（2）由于dxxx2，则当)(,1XEk时不存在。（3）21222)(kkdxxkxXEkk)2()1())(()()(2222kkkXEXEXD（4）由于dxxx3222，则当)(,2XDk时不存在。21、（1）143)(,43)(,21)(XYEYEXE569)()()(),(YEXEXYEYXCov2827)(,2816)(22YEXE11245))(()()(,289))(()()(2222YEYEYDXEXEXD55)()(),(YDXDYXCovXY（2）152)(,52)(,52)(XYEYEXE752)()()(),(YEXEXYEYXCov51010225124)(xxydyxdxXE1010225124)(xxydyydxYE251))(()()(,251))(()()(2222YEYEYDXEXEXD32)()(),(YDXDYXCovXY752),(2)()()(YXCovYDXDYXD（3）X的分布律为X012pk0.240.380.38Y的分布律为Y012pk0.160.340.534.1)(,14.1)(YEXE8.1}2,2{4})1,2{}2,1{(2}1,1{1)(YXPYXPYXPYXPXYE2724.0)()()(),(YEXEXYEYXCov34.2)(,9.1)(22YEXE5444.0))(()()(,6004.0))(()()(2222YEYEYDXEXEXD4765.0)()(),(YDXDYXCovXY22、1)()(),(YDXDYXCovXY11),(2)()()(YXCovYDXDYXD36)(9)3(YDYD3),(3)3,(YXCovYXCov51)3,(2)3()()3()43(YXCovYDXDYXDYXD23、（1）17))(16)()(8)()(())4((2332222123221XEXEXEXEXEXXXE6（2）3,2,1,31)(,21)(2iXEXEii21)()(4)()(2)()(4)()(4)())2((3231212322212321XEXEXEXEXEXEXEXEXEXXXE24、1032)(xxxdydxXE100)(xxydydxYE100)(xxxydydxXYE0)()()(),(YEXEXYEYXCov0)()(),(YDXDYXCovXY则X,Y不相关。其他,010,2)(xxXxxdyxf其他,010,101,1)(11yyYyydxyydxyf由于),()()(yxfyfxfYX在平面上不是几乎处处成立的，则X,Y不相互独立。25、设否则号盒子号球放入第第,0,1iiXininXPnXPii,,1,11}0{,1}1{niiniiXEXEXX111)()(,