武汉大学2012概率统计期末考试B

武汉大学2012——2013第一学期概率统计B 试题

（54学时A ）

学院____________________专业______________学号____________姓名________________

一、（14分）某系有三个班，1班有24位同学，其中12人是特长生；2班有20位同学，

其中8人是特长生；3班有26位同学，其中8人是特长生；现从此70个同学中任找一个同学；（1）求他是特长生的概率？（2）若他是特长生，求他来自1班的概率？（3）若每班任找一人组成三人队参加数模竞赛，求此队的三人全是特长生的概率？二、（12分）某真菌的寿命（单位：小时）在区间（0,5）服从均匀分布；

（1）求其寿命大于3小时的概率？

（2）观测3个此类真菌，求恰有2个寿命大于3小时的概率？三、（14分）若随机变量(,)X Y 的联合概率密度为

(,)0

k f x y ?=?? 221

x y +≤其他；k 为常数。

⑴求随机变量X 和Y 的边沿概率密度();()x y f x f y ； ⑵X 和Y 是否独立？（3

）求Z =

四、（12分）设,A B 为随机事件，111

(),(),()432

P A P B A P A B =

|=|=,设,X Y 分别表示一次实验中,A B 发生的次数。求：（1）二维随机变量（,X Y ）的联合概率分布。（2）,X Y 的相关系数ρ。

五、（12分）若一批种子的发芽率为0.8，分别用切比雪夫不等式和中心极限定理估计这样

的种子10000粒发芽数在7800——8200之间的概率。（标准正态分布的分布函数用

()x Φ表示）

六、（12分）若X 1…Xn 是来自正态总体2

(,)N m s 的样本，X 是样本均值，

1()1n i i S X X n ==--∑是样本方差。 ⑴ 求2

S 的期望和方差。（2）选取常数,a b ，使得X b

t a

-=服从(1)t n -分布。七、（12分）若总体在区间(1,)q 服从均匀分布，X 1…Xn 是其样本，

（1）求q 的矩估计和极大似然估计。（2）判别他们的无偏性。并将不是无偏估计的估计化为无偏估计。八、（12分）据报道：12月7日，全球有感地震18次，其中6级以上2次，某专家说：全球每年发生6级以上地震大约250次，标准差约为60次，最近9年，测得6级以上地震年平均约274.0次，问：可否认为最近9年地球的6级以上地震次数有大幅增加？（0.05α=）（假设地震次数近似服从正态分布2

(,)N m s ，数据为计算方便有改动）已知： (1.65)0.95,(1.96)0.975Φ=Φ= ，其中()x Φ为标准正态分布的分布函数。

武汉大学2012——2013第一学期概率统计B 试题参考答案

（54学时A ）

学院____________________专业______________学号____________姓名________________

三、（14分）某系有三个班，1班有24位同学，其中12人是特长生；2班有20位同学，

其中8人是特长生；3班有26位同学，其中8人是特长生；现从此70个同学中任找一个同学；（1）求他是特长生的概率？（2）若他是特长生，求他来自1班的概率？（3）若每班任找一人组成三人队参加数模竞赛，求此队的三人全是特长生的概率？

解：（1）282

705P =

= ………………………………….6’ （2）123

287P == ……………………………………………10’

（3）12884

24202665

P ==

…………………………………..14’ 四、（12分）某真菌的寿命（单位：小时）在区间（0,5）服从均匀分布；（1）求其寿命大于3小时的概率？

（2）观测3个此类真菌，求恰有2个寿命大于3小时的概率？

解：（1）5

0.45

P dx =

……………………………….6’ （2）设3个真菌中寿命大于3小时的个数为X ，则(3,0.4)X B 所以 2

3(2)0.40.60.288P X C ===或

125

。 …………12’ 三、（14分）若随机变量(,)X Y 的联合概率密度为

(,)0k f x y ?=??

221

x y +≤其他；k 为常数。

⑴求随机变量X 和Y 的边沿概率密度();()x y f x f y ； ⑵X 和Y 是否独立？

（3）求Z =

解：显然，1

k π

（1

）()(,)0X f x f x y dy +∞

-∞

=??

11x -<<其他；

()(,)0Y f y f x y dx +∞

-∞

==??

-11y <<其他； ………………..8’

（2）显然，X 和Y 不独立。 ……………………….10’

（3）设()h z

是一非负连续函数，

(,)()2R h f x y dxdy h z zdz =???；

所以，Z =2()0z z f z ?=??

z ≤≤其他。………14’

四、（12分）设,A B 为随机事件，111

(),(),()432

P A P B A P A B =

|=|=,设,X Y 分别表示一次实验中,A B 发生的次数。求：（1）二维随机变量（,X Y ）的联合概率分布。（2）,X Y 的相关系数ρ。解：（1）1

()()()12

P AB P A P B A =|=

,()1()()6P AB P B P A B ==|

所以 1

{1,1}()12P X Y P AB ====,1{1,0}()()()6

P X Y P AB P A P AB ====-=

1{0,1}()()()12P X Y P AB P B P AB ====-=

,2{0,0}3

P X Y === 故（,X Y ）的联合概率分布为

..6’

(2)

11135,,,,46121636

EX EY EXY DX DY =

==== 故 1

(,)24

COV X Y EXY EXEY =-=,

,X Y 的相关系数ρ …………………………………………………..12’ 五、（12分）若一批种子的发芽率为0.8，分别用切比雪夫不等式和中心极限定理估计这样

的种子10000粒发芽数在7800——8200之间的概率。（标准正态分布的分布函数用

()x Φ表示）

解：用X 表示10000粒种子的发芽数，则(10000,0.8)X B

8000,1600EX DX ==;

所以，由切比雪夫不等式

{78008200}{200}10.96200

P X P X EX ≤≤=|-|≤≥-

= ……………6’ 由中心极限定理

8200800078008000

{78008200}(

)()2(5)114040

P X --≤≤=Φ-Φ=Φ-=…….12’

六、（12分）若X 1…Xn 是来自正态总体2

(,)N m s 的样本，X 是样本均值，

1()1n

i i S X X n ==--∑是样本方差。 ⑴ 求2

S 的期望和方差。（2）选取常数,a b ，使得X b

t a

-=服从(1)t n -分布。解：（1）因为

222

(1)

(1)n S n χσ

-- ，

又2

(1)1,(1)2(1)E n n D n n χχ-=--=- 所以 2

ES DS n σσ==- …………………….6’ （2）a b μ=

= ……………………………………………………12’

七、（12分）若总体在区间(1,)q 服从均匀分布，X 1…Xn 是其样本，

（1）求q 的矩估计和极大似然估计。（2）判别他们的无偏性。并将不是无偏估计的估计化为无偏估计。

解：（1）矩估计

令 1

EX X θ+==,得 121X θ

=- 再求极大似然估计

似然函数 1

(1)()0

L θθ??-=???

121,,...n X X X θ<<其他

所以，q 的极大似然估计 212=max{,,...}n X X X θ …………………..6’ （2）因为 1

()E θθ=，所以，矩估计 121X θ=+是无偏估计。而 2

()1

n E n θθθ+=≠+，故 q 的极大似然估计 212=max{,,...}n

X X X θ不是q 的无偏估计。可将其化为无偏估计 122

(1)max{,,...}1'=n n X X X n

θ+-。 ……….12’

八、（12分）据报道：12月7日，全球有感地震18次，其中6级以上2次，某专家说：全球每年发生6级以上地震大约250次，标准差约为60次，最近9年，测得6级以上地震年平均约274.0次，问：可否认为最近0年地球的6级以上地震次数有大幅增加？（0.05α=）（假设地震次数近似服从正态分布2

(,)N m s ，数据为计算方便有改动）已知： (1.65)0.95,(1.96)0.975Φ=Φ= ，其中()x Φ为标准正态分布的分布函数。解：由题意，做假设，01:250,:250H H μμ=>

274.0,9,60.0X n σ===，0.05α= …………….4’

检验统计量 u =

拒接域为 1.65u > …………………………………………8’

计算得 1.2u =，不落在拒接域内。所以，接受0H ，即认为最近9年地球的6级以上地震次数没有大幅增加。………………………………………………..12’

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