发布时间 : 星期三 文章《概率论与数理统计》第三版-科学出版社-课后习题答案更新完毕开始阅读b228736c3b68011ca300a6c30c2259010202f326
7.8
解;因为Var(??)?E(c??(1?c)??)?c?1?(1?c)?2 ?212222要使Var(??)最小则对Var(??)关于c求一阶导并令其等于零可得
?)22?Var(??2c?1?2(1?c)?2?0 ?c解得 c??2?1??2222
因为对Var(??)关于
2?)Var(??c求二阶导可得 ?2?1?2?2?c2222?0
故当c??2?1??222时Var(??)达到最小。
7.9 解(1)根据题意和所给的数据可得
??0.05,n?16,Z??Z0.025?1.96,??0.01,X?2.125
222?nZ?2?0.01?1.96?0.0049
162所以?的置信区间为
[X??Zn?2,X???]?[2.125?0.0049,2.125?0.0049]?[2.1201,2.1299] Zn2(2) ??0.05 n?16 X?2.125 t15(0.025)?2.1315
S2115??15i?1?Xi?X??0.000293? 即S?0.0171
2
所以?的置信区间为
[X?S?S?0.01710.0171(),X?()]?[2.125??2.1315,2.125??2.1315]?[2.116,2.1406] tt151522nn16167.10解:根据所给的数据计算: X?0.14125, Y?0.1392
2S2113??3i?1?Xi?X?214?0.00000825 S2??4i?12?Yi?Y??0.0000052
22则X 和Y构成的总体的方差为 S2?(m?1)S1?(n?1)S2m?n?2?0.0000065
所以?1??2置信系数??1?0.95?0.05的置信区间为
?11?11[X?Y?tm?n?2()S?,X?Y?tm?n?2()S?]
2mn2mn=[X?Y?t7(0.025)S1111?,X?Y?t7(0.025)S?] 4545=[-0.002,0.006]
7.11 解: n?1000 ??1?0.95?0.05 Z??Z0.025?1.96 Yn?228
2??Yn?0.238 则比例pn??Z?[p2p的区间估计为：
?(1?p?)/n,p??Z?p?(1?p?)/n]?[0.238?1.960.238(1?0.238)/1000,0.238?1.960.238(1?0.238)/1000]p2
=[0.202,0.254]
7.12 解：根据题意有，n?120 ??1?0.95?0.05 X?7.5
Z??Z20.025?1.96
则?的置信区间为：
[X?Z?X/n,X?Z?X/n]?[7.5?1.967.5/120,7.5?1.967.5/120]?[7.01,7.99]
22