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试题答案要点及评分细则
课程名称： 概率论与数理统计 （A卷） 专业年级： 全院相关班级 填写人： 命题组 2015——2016 年第 1 学期 参 考 答 案 及 得 分 要 点 评分标准（得分）
一、填空题（每小题3分，共30分） 1、ABC?ABC?ABC 3、0.88
2、0.8 4、
Ypk0140.30.30.4
ππ?cosxcosy,0?x?,0?y??5、?22
?0,其他?7、72 9、0.4
16、,36
38、t(n?1)
10、(10.0304,10.0696)
二、（本题12分）某保险公司把被保险人分为三类：“谨慎的”，“一般的”，“冒失的”。统计资
料表明，上述三种人在一年内发生事故的概率依次为0.05，0.15和0.30；如果“谨慎的”被保险人占20%，“一般的”占50%，“冒失的”占30%，求： （1）某被保险人在一年内出事故的概率；
（2）现知某被保险人在一年内出了事故，则他是“谨慎的”的概率是多少？ 解 设A1＝{保险人是“谨慎的”}，A2＝{保险人是“一般的”}，A3＝{保险人是“冒失的”}
B? {保险人在一年内出了事故}
则P(A1)?0.2,P(A2)?0.5,P(A3)?0.3,P(B|A1)?0.05,P(B|A2)?0.15,P(B|A3)?0.30 ····· 4分 （1）由全概率公式得 P(B)?P(1A)P(B1|?A)PB?|A)2(A)P(23P(A) P3(B|A)?0.2?0.05?0.5?0.15?0.3?0.3
?0.175 ················································································································ 8分
（2）由贝叶斯公式得 P(AB)?1P(A?0.051)P(B1A)0.2··························································· 12分 ??0.0 ·57P(B)0.175，Barctaxn???x???
三、（本题12分）设随机变量X的分布函数为 F(x)?A?（1）求常数A与B；（2）求P??1?X?1?；（3）求X的概率密度。 解 （1）由分布函数的性质知
第 1 页 共 4 页
limFx(?)x???x???l?iAm?Baxr?c?tAa?naxr?c?tAa?n?2B? B?
10Fx(?) limx???x???l?iAm?B?211故A?,B? ··························································································································· 4分
2??11??11?1 （2）P??1?X?1??F(1)?F(?1)???arctan1????arctan(?1)?? ·············· 8分
?2???2??2 （3）X的概率密度为
?111??······································ 12分 f(x)?F?(x)???arctanx??,???x??? ·2?2????1?x?四、（本题12分）设随机向量(X,Y)的联合概率密度为
?e?y,0?x?y?1 f(x,y) ???0,other（1）求边缘密度函数；（2）判断随机变X,Y是否独立？（3）求P?X?Y?1?. 解 （1）随机变量X,Y的边缘密度函数分别为
fX(x)????1?11?y???xedy,0?x?1?x?,0?x?1 ····························· 3分 f(x,y)dy????ee??0,other?0,other???fY(x)??????y?y?y???0edx,0?y?1?ye,0?y?1 ································ 6分 f(x,y)dx????other?0,?other?0,（2）由于f(x,y)?fX(x)fY(y)，故随机变量X,Y不相互独立。 ·································· 8分 （3）P?X?Y?1??x?y?1??f(x,y)dxdy?dx?1201?xx12 ····································· 12分 e?ydy?1??ee五、(本题12分) 设随机变量X~U?0,1?,
（1）求Y?eX的概率密度函数; （2）求E?Y?,D?Y?. 解 （1）?X~U?0,1? ?
X的概率密度函数为
?1, f(x)???0,0?x?1 other当0?X?1时，1?Y?e 当y?1时FY(y)?P?Y?y??0
当1?y?e时FY(y)?P?eX?y??P?X?lny???当y?e时FY(y)?P?eX?y??1
第 2 页 共 4 页
lny0f(x)dx?lny
即分布函数
y?1?0,?FY(y)??lny,1?y?e
?1,y?e?故Y的密度函数为
?11?y?e? ···································································································· 6分 fY(y)??y,?0,其他?（2）E?Y???????ee1yfY(y)dy??y?dy??dy?e?1. ······················································· 8分
11yee11y2fY(y)dy??y2?dy??ydy??e2?1?
11 ········································ 10分y22E?Y2??????? D(Y)?E?Y2????E?Y????1212e?1?e?1? ???e?1??3?e? ·······························?? 12分22??x??1,0?x?1六、(本题12分)设总体X的概率密度为f(x,?)??,其中??0为未知参
其它?0,数,x1,x2,?,xn是来自总体X的样本，求参数?的最大似然估计。 解 似然函数
?1 L(?)??f(ix?,?)?1?x??2?x?1??i?1n??1??nxn???1?x2x?n???1 ·········································· 4分 x于是对数似然函数为
lnL(?)?nln?????1??lnxii?1n
··························································································· 6分
dlnL(?)nn令???lnxi?0得参数?的最大似然估计为
d??i?1???n?lnxi?1n
i ······················································································································· 12分
七、(本题10分)已知某炼铁厂铁水含碳量服从正态分布N(4.55,0.1082)，现测定9炉铁水,其平均含碳量为4.484.如果估计方案没有改变,可否认为现在生产的铁水平均含碳量仍为4.55(??0.05)? (已知?(1.96)?0.975)
解 这是一个方差?2已知的正态总体,检验期望值??4.55是否可接受. (1) 提出待检假设H0:??4.55 ·
······················································································ 2分
(2) 构造统计量Z?X??0.在H0成立的条件下Z~N(0,1) · ······································· 4分?n(3) 给定的检验水平??0.05,查表确定临界值z?/2?1.96,确定拒域绝为Z?1.96第 3 页 共 4 页
··· 6分