《社会统计学》课程练习题（1）答案一、略二、（1）对立事件（2）互不相容事件（3）互不相容事件（1）对立事件三、5025Md400200525(元)4025M0400200516.28(元)25182510Q1200200400.00(元)157565Q3600200690.91(元)22QQ3Q1690.91400290.91(元)222(nibi)52600nb32760000ii50924002N10050924N10010050924225.66(元)四、（1）极差R=1529-65=1464（百元）（2）将数据从小到大排序：65921061181221351481741851529101Q1的位置2.7543(101)Q3的位置8.254Q192(10692)0.75102.5(百元)Q3174(185174)0.25176.75(百元)四分互差Q176.75-102.574.25(百元)（3）222(xi)2674xi2495204N10178017.64421.92(百元)N10五、80P(A)0.810020P(B)0.210012P(B/A)0.158012P(A/B)0.62012P(AB)0.1210040P(C)0.410032P(C/A)0.48012P(A/B)0.158032P(AC)0.32100六、4P(A)157P(B)151P(AB)10P(AB)115P(A/B)0.214P(B)107P(AB)115P(B/A)0.375P(A)104471P(AB)P(A)P(B)P(AB)0.633151510七、10100（1）10口井皆产油的概率为：P(10)C100.30.70.00000590010(2)10口井皆不产油的概率为：P(0)C100.30.70.028250010119（3）该公司赢利的机会为：P(2)1C100.30.7C100.30.70.85069八、424P(x2)e482.7182840.14652!九、1.37011.371.371.371.37P(0x1)P(x0)P(x1)ee0.60220!1!十、0.10.100.1Z10.1(Z)(1)1(1)10.84130.1587投资人投资于此种股票保证不亏的概率为：10.15870.8413。0.20.1Z10.1（Z）(1)0.8413收益在20%以上的可能性为：10.84130.1587。十一、15000-10000Z2.5（1）20001(2.5)10.99380.00620.62%9500-10000Z2.52000（2）100(2.5)1(2.5)0.00620.62%(3)P(800012000)P(1Z1)(1)(1)0.8413(10.8413)68.26%(4)x10000()0.902000x100001.32000x26001000012600十二、xtsnt0.025(24)2.06393636[302.0639,302.0639][27.52,32.48]2525十三、xZnZ0.051.650.02250.0225[21.51.65,21.51.65][21.4175,21.5825]99十四、p?PZp?q?nZ0.0052.580.90.10.90.1[0.92.58,0.92.58][0.8226,0.9774]100100十五、此人上班迟到的概率为：0.430%0.620%24%0.620%此人某天迟到而乘地铁的概率为：50%24%十六、np10020%20n(1p)10080%80因此，可将此二项分布用正态分布来近似计算。np20npq10020%80%4P(x15)P(14.5x15.5)14.520x2015.520P()444x20P(1.3751.125)4x20P(1.1251.375)40.91540.86960.04584.58%x10.01060.0250.050.0250.0106x10.00530.05(1.125)0.0053(1.10)0.00530.86430.8696x20.00770.0250.050.0250.0077x20.00390.05(1.375)0.003