（概率课后习题答案详解）
董永俊
(2) S2??0.05
n?16
2X?2.125 t15(0.025)?2.1315
???Xi?X151i?115??0.000293? 即S?0.0171
所以?的置信区间为
[X?Snt15(?2),X?Snt15(?2)]?[2.125?0.017116?2.1315,2.125?0.017116?2.1315]?[2.116,2.1406]
7.10解:根据所给的数据计算: S21X?0.14125,
Y?0.1392
?1??Xi?X3i?13?2?0.00000825 S22?1??Yi?Y4i?14??0.00000 522则X 和Y构成的总体的方差为 S所以??2?(m?1)S1?(n?1)S2m?n?222?0.00000 651?置信系数?2?1?0.95?0.05的置信区间为
1m1n]
[X?Y?tm?n?2(?2)S1m?1n,X?Y?tm?n?2(?2)S?=[X?Y?t7(0.025)S14?15,X?Y?t7(0.025)S14?15]
=[-0.002,0.006] 7.11 解:
??pn?1000
??1?0.95?0.05 Z??Z20.025?1.96 Yn?228
Ynn?0.238 则比例p的区间估计为：
（概率课后习题答案详解）
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（概率课后习题答案详解）
董永俊
??Z[p?2?(1?p?)/n,p??pZ??(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 Z??Z20.025??1?0.95?0.05 X?7.5
?1.96
则?的置信区间为：
[X?Z?X/n,X?2Z?X/n]?[7.5?1.967.5/120,7.5?1.967.5/120]?[7.01,7.99]
2
（概率课后习题答案详解）
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