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三、习题解答
(A)
1.已知随机变量X服从0-1分布,并且P{X≤0}=0.2,求X的概率分布.
解 X只取0与1两个值,P{X=0}=P{X≤0}-P{X<0}=0.2, P{X=1}=1-P{X=0}=0.8.
2.一箱产品20件,其中有5件优质品,不放回地抽取,每次一件,共抽取两次,求取到的优质品件数X的概率分布.
解 X可以取0,1,2三个值.由古典概型概率公式可知
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0063_0003.jpg?sign=1738893989-Q4B5Mwgb7cerGb4NP3FxFAKf1DlDD084-0-df0fa24ae348680b3ade7558a298b6f8)
依次计算得X的概率分布如下表所示:
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0063_0004.jpg?sign=1738893989-b1zvuqHzxllq8177caeVu6eTCbWtZOul-0-da5a581e9311f974abffcd198b197e7c)
3.上题中若采用重复抽取,其他条件不变,设抽取的两件产品中,优质品为X件,求随机变量X的概率分布.
解 X的取值仍是0,1,2.每次抽取一件取到优质品的概率是1/4,取到非优质品的概率是3/4,且各次抽取结果互不影响,应用伯努利公式有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0063_0005.jpg?sign=1738893989-NHrG3QBDb06aYIUhUB804ELMJX16QLCq-0-c56cc4306081d1cb1008728d6d265dfd)
4.第2题中若改为重复抽取,每次一件,直到取到优质品为止,求抽取次数X的概率分布.
解 X可以取1,2, …可列个值.且事件{X=m}表示抽取m次前m-1次均未取到优质品且第m次取到优质品,其概率为.因此X的概率分布为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0063_0007.jpg?sign=1738893989-KfkVVkvgRjk2CzncKXRDEWZq7gWALTlH-0-c03558c425486e28f5cc76b1b6a495bd)
5.盒内有12个乒乓球,其中9个是新球,3个为旧球,采取不放回抽取,每次一个直到取得新球为止,求下列随机变量的概率分布:
(1)抽取次数X;
(2)取到的旧球个数Y.
解 (1)X可以取1,2,3,4各值,则有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0064_0001.jpg?sign=1738893989-DNHT8YcCGgbN0G4kkOJXHgyXWxsUXGfT-0-2a9bf5bf2d2507d172729feb95717d2c)
(2)Y可以取0,1,2,3各值.
P{Y=0}=P{X=1}=0.75,
P{Y=1}=P{X=2}≈0.2045,
P{Y=2}=P{X=3}≈0.0409,
P{Y=3}=P{X=4}≈0.0045.
6.上题盒中球的组成不变,若一次取出3个,求取到的新球数目X的概率分布.
解 X可以取0,1,2,3各值,则有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0064_0002.jpg?sign=1738893989-znFFkEM60Lijx7yWx77qFWfR9c545FML-0-f13c588a2f22e842829041d686d7686d)
7.将3人随机地分配到5个房间去住,求第一个房间中人数的概率分布和分布函数.
解 用X表示第一个房间中的人数,则其可能的取值为0,1,2,3.分别算得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0064_0003.jpg?sign=1738893989-ouYPWONsyODk4MLdVEqRcIQX8ebrrl93-0-32c6b930dd6fd3f3ed5bc4c00d5599d0)
故X的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0001.jpg?sign=1738893989-U4YKssVYBNAhRHNelYMAczxHhpxsbq55-0-732e4b3c6a5f1b807a4aa1e63573116e)
8.袋中装有n个球,分别编号为1,2, …, n,从中任取k(k≤n)个,求取出的k个球最大编号的概率分布.
解 用X表示k个球的最大编号,则X可能的取值为k, k+1, …, n.考虑随机事件{X=l},总样本点数为,若k个球的最大编号是l,编号是l的球一定被取出,剩下k-1个球从编号为1,2, …, l-1的l-1个球中取,共
种取法,所以随机事件{X=l}所包含的样本点数为
,由古典概型概率公式得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0005.jpg?sign=1738893989-2pv18wzCNkLP8HiivlcFVm4wdLyXw9OX-0-7d98889ecee0643c3b35251f600079aa)
9.已知P{X=n}=pn, n=2,4,6, …,求p的值.
解 由题意可知
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0006.jpg?sign=1738893989-8RHkl6FKfxd2IoDYT8khoRoe9d82qrm0-0-cac285376bbf120086b294e1497e52f2)
解得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0007.jpg?sign=1738893989-dZ52ufyc3Y5fLhgchI4AYvfOEwfXQFms-0-11a22832a15d4c3ad504734bd255a55a)
10.已知P{X=n}=cn, n=1,2, …,100,求c的值.
解 由题意可知
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0008.jpg?sign=1738893989-zL5DLvVtqi7CrhRNpj1LhitEGmftzyCU-0-3d5b07352118ea1afb28e36c4013df31)
解得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0009.jpg?sign=1738893989-fmc4iSoSf1lGraFsBJXJ2x6BZH7gHOeu-0-227d0681c9a45df756d8774977fbf266)
11.已知 , …,且λ>0,求常数c.
解 由题意知
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0011.jpg?sign=1738893989-5YBnd0hFFn3ikxZ4mN7ngURF0dcyaec3-0-02a490ecc8732dec60d335cc47f269f1)
由于
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0012.jpg?sign=1738893989-zV3twuFvqS0Rhlo5nHGoCkcIxtSgiLKz-0-8b4af4e3dd62a6d0c00c4a7f3ccb8742)
所以有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0013.jpg?sign=1738893989-K9zD5zFZGyZUsPl2XHdzOyfSMjJyeAe2-0-02a41fe08375f71492608b532e0a2e60)
解得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0014.jpg?sign=1738893989-9bmzXTsNn1Of6bmIz2ajH4c0f6bNEmqe-0-a80b452029d03a351bd3dbbe0370f164)
12.某人任意抛硬币10次,写出出现正面次数的概率分布,并求出现正面次数不小于3及不超过8的概率.
解 用X表示抛10次出现正面的次数,则X可能的取值为0,1,2, …,10.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0066_0001.jpg?sign=1738893989-XPPHFhttdKAVTsbTwtI9YFXMtQU6sSuy-0-06f1b14bdae296685d1388fdc81d5e38)
13.甲、乙二人轮流投篮,甲先开始,直到有一人投中为止,假定甲、乙二人投篮的命中率分别为0.4及0.5,求:
(1)二人投篮总次数Z的概率分布;
(2)甲投篮次数X的概率分布;
(3)乙投篮次数Y的概率分布.
解 设事件Ai(i=1,3,5, …)表示“在第i次投篮中甲投中”, Bj(j=2,4,6, …)表示“在第j次投篮中乙投中”,且A1, B2, A3, B4, …相互独立.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0066_0002.jpg?sign=1738893989-GeXgbjaPdRV0Ic7Tn9PFH84wXNR2Bigq-0-e24ecc87aa3dc291303cd9d078d72621)
14.一条公共汽车路线的两个站之间,有四个路口处设有信号灯,假定汽车经过每个路口时遇到绿灯可顺利通过,其概率为0.6,遇到红灯或黄灯则停止前进,其概率为0.4,求汽车开出站后,在第一次停车之前已通过的路口信号灯数目X的概率分布(不计其他因素停车).
解 X可以取0,1,2,3,4,分别得到
P{X=0}=0.4, P{X=1}=0.6 × 0.4=0.24,
P{X=2}=0.6 2 × 0.4=0.144,
P{X=3}=0.6 3 × 0.4=0.0864,
P{X=4}=0.6 4=0.1296.
15.已知
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0066_0003.jpg?sign=1738893989-pU6BrSzNBTsWHc6UoARR9pnG1GztrU1Q-0-303779e4a9f9dbe618b4932a60369661)
问f(x)是否为密度函数.若是,确定a的值;若不是,说明理由.
解 如果f(x)是密度函数,则f(x)≥0,因此a≥0,但是,当a≥0时,
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0001.jpg?sign=1738893989-OaTcJ0HjBAnXwTjhBPkc0lSrP3jykVSb-0-41242eef428558aca8b3616c33b918de)
由于不等于1,因此f(x)不是密度函数.
16.某种电子元件的寿命X是随机变量,概率密度为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0003.jpg?sign=1738893989-GEN23JKxJTTrBzxYUMZlmk1BgRsxaF0k-0-e543a4622dc90e20fede9e175ff90485)
3个这种元件串联在一个线路上,计算这3个元件使用了150h后仍能使线路正常工作的概率.
解 串联线路正常工作的充分必要条件是3个元件都能正常工作.而3个元件的寿命是3个相互独立同分布的随机变量,因此若用事件A表示“线路正常工作”,则
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0004.jpg?sign=1738893989-sVgAn6BEgvlt7esseQ80gaTvwhq9PVVs-0-2b2c27fc684fa956335e6e174eae5d04)
17.设随机变量X~f(x), f(x)=Ae-|x|.试确定系数A并计算P{|X|≤1}.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0005.jpg?sign=1738893989-8fkZrT5iWZTsyGETjlzPU1TEB43CEHXg-0-fbfb0ef8f9231c3a67fa2dfe45dbd5da)
解得,故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0007.jpg?sign=1738893989-u9Jpno3bVI8btN3SP7CainIZsCbyK8pX-0-8cb18186c25274836fb07369a7d00e46)
18.设X的概率密度为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0008.jpg?sign=1738893989-ClqVeUnkzmIkyetbwZrbqZegQYnvRnRQ-0-4b6f96a43d6ab1a51eae7c1c1ac24c8d)
(1)确定常数c;(2)计算; (3)写出分布函数.
解(1)解得
.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0011.jpg?sign=1738893989-mTKsQulxC7e1N0vcY7HAms86rnFX54HF-0-e29cfefa35d1d95a6071553db4311722)
(3)当x≤-1时,F(x)=0;当x≥1时,F(x)=1;当-1<x<1时,
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0001.jpg?sign=1738893989-NJsQUBUtwuNRRt8tQhknsqIzOLleHDOY-0-29b5e9774a1ccac9025a315a558598d8)
19.设X的概率密度为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0002.jpg?sign=1738893989-JTjNc16p7qfy4L2BG8p5pQdG2c2A81cK-0-4214b1ac6e8b383857e8e58b842947b4)
(1)确定常数c;(2)计算; (3)写出分布函数.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0004.jpg?sign=1738893989-kiiVmjYfz2FrZhkKIizFXI5vgXYcTkI4-0-fd0c9d3eb2e58dc00e4180797cf41d04)
故c=1π.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0005.jpg?sign=1738893989-Jsbw3PYAZFLak8MNyxYXoqKqv0wW7s1v-0-d6f2c0597115f6cb748aec7b854c9668)
(3)当x≤-1时,F(x)=0;当x≥1时,F(x)=1;当-1<x<1时,
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0006.jpg?sign=1738893989-7QyTfPYDJEajlcZF0jyXwL7w0aUuehej-0-88b8d5a7d40f2c861523d2bd6a3fb959)
20.设连续型随机变量X的分布函数F(x)为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0007.jpg?sign=1738893989-punWyNSRO2gHlkYUUIVRwt90kMc5USf4-0-e7186088616daf18441906ddc211b534)
(1)确定系数A; (2)计算P{0≤X≤0.25}; (3)求概率密度f(x).
解(1)连续型随机变量X的分布函数是连续函数,F(1)=F(1-0),则有A=1.
(2)P{0≤X≤0.25}=F(0.25)-F(0)=0.5.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0008.jpg?sign=1738893989-JG3NaqUlDtHzBVbFRtASNDvgwlfRY7lh-0-e383ca2c3b9cfe0b955aed35d72f9f2c)
21.随机变量X的分布函数F(x)为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0009.jpg?sign=1738893989-9u4lR4ecjQmLaWuuHQ7CD5wt4P9sXfYO-0-74477a0906333b5e776da40ea734ecff)
试确定常数A的值并计算P{0≤X≤4}.
解 由F(2+0)=F(2),可得,故A=4,且
P{0≤X≤4}=P{0<X≤4}=F(4)-F(0)=0.75.
22.设X的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0001.jpg?sign=1738893989-R55CCuEWYtI1nnrRwSuQYwdrrgs0nf0f-0-26071c27f9cafab1f59da84fe54ab66b)
(1)确定常数A; (2)计算P{|X|<2}; (3)求概率密度f(x).
解 (1)由F(0+0)=F(0),可得0=A-1,故A=1.
(2)P{|X|<2}=F(2)-F(-2)=1-e-4.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0002.jpg?sign=1738893989-EXZ3IirZFpFhqmpWOuLP37ZRHX7dbUQT-0-ec50ad46529a0ae4ee0a4ee1f7f73759)
23.设X的分布函数为
F(x)=A+Barcta nx, -∞ <x<+∞.
(1)确定常数A, B; (2)计算P{|X|<1}; (3)求概率密度f(x).
解 (1) ,可得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0004.jpg?sign=1738893989-qBDrjafkoTLxPTjrLxQiE9ZXhE4NuWSy-0-36648b6bdc9c54f09f94735adc1b3ba4)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0005.jpg?sign=1738893989-Wxs7q1UGc8fawkInRJTMryl2zgeRfbsI-0-b8884606927462061d48b5d07e57b4c9)
24.设X的概率密度为
f(x)=Ae-|x|, -∞<x<+∞.
(1)确定常数A; (2)求分布函数f(x); (3)计算X落在(0,1)内的概率.
解 (1)由第17题,有.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0007.jpg?sign=1738893989-ZHsgKy6vImPNIw0FnKuvLJQ8VxohdoHT-0-1822df1ea4c471cae10a5bf1bc7c80ef)
(3)当x<0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0008.jpg?sign=1738893989-eqeAv11zz5h7b1YqxPmz5Iz5EUr7en6M-0-932b8b321519054d7a752df56944f8e2)
当x≥0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0009.jpg?sign=1738893989-SdfV7fvzg7UzGA7fyDdWbStEHKNwCRCY-0-f8f66537caf4bcf77dcac1c8860a88cc)
故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0010.jpg?sign=1738893989-2tPrDL3N35LAaOD00H7yTmgniOjdSw5j-0-750f900f2e75339a598a88fc01b863fd)
25.随机变量 ,试确定A的值并求分布函数F(x).
解 ,因此
,所求分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0002.jpg?sign=1738893989-ZtzqMg68P3Jeiioam7HnGlP4ZdCt7cQj-0-2aa06c584db474b685633d264cd9039a)
26.随机变量X~f(x),其中
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0003.jpg?sign=1738893989-wZYgLIs32IluFN5ah4IT7ILr6YapeONC-0-aefdfb83536f986eb36204534c935aa9)
试确定a的值并求分布函数F(x).
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0004.jpg?sign=1738893989-LBDlnEC7Ln1PWi8oZS4EjtFLXaC5YlzA-0-e9c5a08ce7ee6f950628bfdd97f472b7)
因此a=π.当0<x<π时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0005.jpg?sign=1738893989-SpkLZ6Z0FSxabJip1D2KI8jvEu9bru8q-0-33bd13a7b9d1b05b75aba43e784efc8f)
27.随机变量X的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0006.jpg?sign=1738893989-2v3qVAMeRTYidT5mp3rbsjkWkcU4wZzh-0-e7728ffab56e6bbeae1847db66cc83d2)
求X的概率密度,并计算
解 当x≤0时,X的概率密度为f(x)=0;当x>0时,
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0008.jpg?sign=1738893989-5hpDwMEYXlRrXVQN2tfKlMc5iJ2FY0F7-0-41c0cfa8136264817530cc9355f69106)
故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0009.jpg?sign=1738893989-grO54nsHhprLpM9mkUDlA7pSKTL1Zf7h-0-2f4e73463b637be61410600020aefa77)
28.某公共汽车站,每隔8分钟有一辆汽车通过,乘客到达汽车站的任一时刻是等可能的,求乘客到达汽车站后候车时间不超过3分钟及至少5分钟的概率.
解 用X表示乘客到达汽车站后候车时间,则X~U(0,8).X的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0010.jpg?sign=1738893989-IQQAahkKK1GfGOhKKH7JoIIHcXkh31k2-0-ad1a1cd41a34081619232e7e0518acdf)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0001.jpg?sign=1738893989-w5E1vxFnTLe9zQtirg169cabgh3RIowe-0-beca62fdf7a72dacaf1b6b1e68047883)
29.设ξ~U(0,10),求方程x 2+ξx+1=0有实根的概率.
解 ξ的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0002.jpg?sign=1738893989-Iahq8GCCZY8w4vDAi6Le1qN9Dn3fV3dl-0-947e20b9fccab432b6f4879a625bcdb1)
方程x2+ξx+1=0当ξ2-4≥0时有实根,故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0003.jpg?sign=1738893989-LK0JAZ1fjRLPeJQJTKHAquZKKoZDaLiB-0-517b4edbe650d5c642f0a9d946ddc088)
30.一批产品中有15%的次品,逐个进行返样抽取检查,共抽取20个样品,问取出的20个样品中最可能有几个次品,并求相应的概率.
解 用X表示抽取20个样品中的次品的件数,由于(20+1)×0.15=3,则取出的20个样品中最可能有3个次品,且
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0004.jpg?sign=1738893989-SsNjSEMDyI5mB0sQY61vt6sGOMJNXQRo-0-8fa8544c6d4f9cf1e7ba9abef62b71cc)
31.在1000件产品中含有15件次品,现从中任取6件产品,分别求其中恰含有2件次品和不含次品的概率.
解 用X表示抽取的6件产品中次品的件数,次品率为0.015,故X近似地服从二项分布B(6,0.015),
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0005.jpg?sign=1738893989-l8E7Ye4EhcZQmJvUVqvPcdNWqKymIoL3-0-d200659f2212643f95125469b227f72c)
32.电话交换台每分钟接到呼唤的次数服从泊松分布P(3),求一分钟内接到4次呼唤、不超过5次呼唤和至少3次呼唤的概率.
解 用X表示每分钟接到的呼唤次数,则X服从泊松分布P(3),即有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0006.jpg?sign=1738893989-TvyoabrqMwOW7hP9sNF4SPZdAk4qL7K4-0-2d7c5fb12c0663e7609c9b4c75f7ef2b)
查表得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0007.jpg?sign=1738893989-7FielT7MdsRKOvtnkyUqbVY5E3Ru0yv4-0-815405bdd891b59e5db2c40fb63f2297)
33.设书籍中每页的印刷错误服从泊松分布,经统计发现在某本书上,有1个印刷错误的页数与有2个印刷错误的页数相同,求任意检验4页,每页上都没有印刷错误的概率.
解 设一页书上印刷错误为X,4页中没有错误的页数为Y,依题意有
P{X=1}=P{X=2},
即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0008.jpg?sign=1738893989-e1v8OClGBBuUwYhANdTd6aKCG44iqjzn-0-8485aa43c2d268e989cb5fef48885f46)
解得λ=2,即X服从λ=2的泊松分布.
每页上没有印刷错误的概述是
p=P{X=0}=e-2,
显然Y~B(4, e-2),故
P{Y=4}=p4=e-8.
34.每个粮仓内老鼠数目服从泊松分布,若已知一个粮仓内,有1只老鼠的概率为有2只老鼠的概率的2倍,求粮仓内无鼠的概率.
解 设X为粮仓内老鼠数目,依题意有
P{X=1}=2P{X=2},
即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0072_0001.jpg?sign=1738893989-wFLq1DbSAPiY7pZdgQn8BvKSGySTDiZm-0-3a72a547dd78b85a1561a27fc52b1d41)
解得λ=1, P{X=0}=e-1.
35.上题中条件不变,求10个粮仓内有老鼠的粮仓不超过2个的概率.
解 接上题,设10个粮仓中有老鼠的粮仓的数目为Y,则Y~B(10, p),其中
p=P{X>0}=1-P{X=0}=1-e-1, q=e-1.
P{Y≤2}=P{Y=0}+P{Y=1}+P{Y=2}=e-8(36e-2-80e-1+45).
36.随机变量X服从参数为0.7的0-1分布,求X2, X2-2X的概率分布.
解 X2仍服从0-1分布,且
P{X2=0}=P{X=0}=0.3,
P{X2=1}=P{X=1}=0.7.
X2-2X的取值为-1与0,则
P{X2-2X=0}=P{X=0}=0.3,
P{X2-2X=-1}=1-P{X=0}=0.7.
37.设X的概率分布为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0072_0002.jpg?sign=1738893989-GOTJAQVefgUMhSGhCxTP4ido29lgxaFT-0-60a43471d9991b272432c4937aabfa0b)
求3X+2和2X2-1的概率分布.
解 P{3X+2=-1}=P{X=-1}=0.1,
P{3X+2=2}=P{X=0}=0.2,
P{3X+2=5}=P{X=1}=0.3,
P{3X+2=17}=P{X=5}=0.4.
P{2X2-1=1}=P{X=-1}+P{X=1}=0.4,
P{2X2-1=-1}=P{X=0}=0.2,
P{2X2-1=49}=P{X=5}=0.4.
38.从含有3件次品的12件产品中任取3件,设其中次品数为X,求2X+1的概率分布.
解 X可能的取值为0,1,2,3,则有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0001.jpg?sign=1738893989-bYlplxHMV1V0ZOM2VsaICls6d7AhpPl7-0-ee7334dded462f458e3db0a2ff9cbd89)
39.已知 , Y=lgX,求Y的概率分布.
解 Y的取值为±1, ±2, …,则
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0003.jpg?sign=1738893989-J2lHYrHme5df2qBJNGywNUCk4lApBYxU-0-eee4b0ca47b5847e015afc7e562c4bc9)
40.X服从[a, b]上的均匀分布,Y=aX+b, (a≠0),求证Y也服从均匀分布.
证明 X的密度函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0004.jpg?sign=1738893989-NwjN8kopaCDsJSJ9UQ6VkDGsi3gndeV9-0-416a9344fe05563366301d3404ee510b)
Y的密度函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0005.jpg?sign=1738893989-kwBeJsut3pvBif04dHKoNQjWEtAmTWL6-0-07e692757f6daf4a29d418c0df82273c)
当a>0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0006.jpg?sign=1738893989-pVwBaNYRWmpIuYj6pb1wkWWIYEWdIRpz-0-a7d3f17482137dd184644066ad16f97e)
当a<0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0007.jpg?sign=1738893989-WXXkIvO7dGhUf0eN7lO9q21rXSjLvh5Y-0-d3286b109a2a3491e6f92c0ebc0307a0)
41.随机变量服从 上的均匀分布,Y=cosX,求Y的概率密度.
解 显然y=cosx在 上单调,在(0,1)上有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0010.jpg?sign=1738893989-MqfRSxscJswnSLj1JjjHKhBl1vpQzqt4-0-6c2b5ec6a5435aa93d56a371ec344de0)
因此
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0011.jpg?sign=1738893989-4uuadh3j1gK1yzvbjGla9omYGBzMqvYd-0-24430ca4b1b3a3f51aadd502d629062a)
42.随机变量服从(0,1)上的均匀分布,Y=eX, Z=|lnX|,分别求随机变量Y与Z的概率密度fY(y)及fZ(z).
解 y=ex在(0,1)内单调,x=lny可导,且,因此有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0002.jpg?sign=1738893989-As8d0D3vFeqBXkpnnbH3bWeXf3fUaROZ-0-6c0d69f7ddab9d813d1cb9e5effd90b2)
在(0,1)内,lnx<0, |lnx|=-lnx单调,且 ,因此有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0004.jpg?sign=1738893989-JOIRwOQMBNxRECPk4OstrgA25ABOQji6-0-44751c30d97a29802b9e2b2fd33cbded)
43.设X服从参数λ=1的指数分布,求的概率密度fY(y)及Z=X 2的概率密度fZ(z).
解 在[0, +∞)上单调,x=y2(0≤y<+∞),
.根据题意有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0008.jpg?sign=1738893989-kJzPjTs86zM2BVqd11IucgmfnMeBpIBr-0-6b88edaba91bbe1cbce2705f9f0c119b)
因此有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0009.jpg?sign=1738893989-LsLU16D6qKE2ZPtehEoZ0yCNvkPxTCoN-0-4490a97f4747ba5ccb4759444fa60260)
z=x2在[0, +∞)上单调 ,因此有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0011.jpg?sign=1738893989-2IbqxoPRSYomH8rMvrNylHemI6oXDqla-0-3191793c0b6bc3c696560a8fb48561dd)
44.随机变量X~f(x),当x≥0时 ,分别计算随机变量Y与Z的概率密度fY(y)及fZ(z).
解 由于y=arctanx是单调函数,其反函数是 在
内不恒为零,因此,当
时,有
在x>0时也是x的单调函数,其反函数为
,因此当z>0时,
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0016.jpg?sign=1738893989-Ra0P9sg4qKGSdCgH8M3KXqsShn867zxG-0-c4d62017a6c5a3e8d80defffc146e476)
即Y服从区间 上的均匀分布.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0020.jpg?sign=1738893989-rj8ZT5Eaa3abYuSd5XAZjswfNxyIWJ3o-0-db36466d24bc21aff0f830cbde9d0cb0)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0001.jpg?sign=1738893989-J4PplYENwNqUTIUZuQBKp4Dc5V9pa9Y2-0-aa1e3598a93b4a043b9f80d1d4d0c109)
即与X同分布.
45.一个质点在半径为R、圆心在原点的圆的上半圆周上随机游动.求该质点横坐标X的概率密度fX(x).
解 如图2.1所示,设质点在圆周位置为M,弧的长记为L,显然L是一个连续型随机变量,L服从[0, πR]上的均匀分布.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0004.jpg?sign=1738893989-ONUa5RLPpuZfgWSsOTKGLishvlAu433E-0-ed6a37b422fd4bdd4c8e0bda90152a5f)
M点的横坐标X也是一个随机变量,它是弧长L的函数,且
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0006.jpg?sign=1738893989-ZCNd30ckjiqxob94d8ctMpMBQ0Cvgbpr-0-0bcf2bdd89812c42b77635d6322a0c94)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0005.jpg?sign=1738893989-jdxlIPKa0Z1tjFoc6aUUolyWfLyar5cA-0-d0c0bbeda77df86da291e3497a995e18)
图 2.1
函数x=Rcos(l/R)是l的单调函数(0<l<πR),其反函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0007.jpg?sign=1738893989-UI4avxgG5jrT09fIdtmE8YThFQb3V8Hb-0-bd41e543ba46179bf5ef65bd65efed41)
当-R<x<R时,l'x≠0,此时有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0008.jpg?sign=1738893989-CzdVOSIRFKEXqAtWOP8AkENWiHfS7aoD-0-86c7f275ef638035a0f13adc6ffef01d)
当x≤-R或x≥R时,fX(x)=0.
46.设X~N(3,4),求:
(1)P{X≤2.5}; (2)P{X>1.3}; (3)P{1≤X≤3.5};
(4)P{|X|>2.8}; (5)P{|X|<1.6}; (6)P{X-2>5}.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0009.jpg?sign=1738893989-8X2TM1fD9x0dPCjv2VfU0MEV6sLYA7FJ-0-7723cacc74766f09c6f6f0fc06d13500)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0076_0001.jpg?sign=1738893989-MPzFkcxtM5cg6SqMXEdrqjmZNnNCuliT-0-068c4d4bfd529ca1e08c6fb074e69f42)
47.随机变量X~N(μ, σ2),若P{X<9}=0.975, P{X<2}=0.062,试计算μ和σ2的值并求P{X>6}.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0076_0002.jpg?sign=1738893989-agMFO2lXc2xI3XdjkiM8quJ2Y1t979HF-0-157f150c1175ae018bcf3d5881d23a91)
查表得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0076_0003.jpg?sign=1738893989-YOoiCiVlhUaqlPbsvNVS17BFxZCoUai3-0-63a820a350e8f5b49b4924457ed81f98)
解关于μ和σ的方程组,得
μ=5.08,σ=2.
故
P{X>6}=1-P{X≤6}=1-Φ(0.46)=0.328.
48.已知随机变量X~N(10,22), P{|X-10|<c}=0.95, P{X<d}=0.023,试确定c和d的值.
解 查表得
.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0076_0006.jpg?sign=1738893989-pXLxnT7zF3wwzGKVAeb84XPNhJr1n1Sh-0-7bae7deab98348b10ab619523486fd72)
查表得 .
49.假定随机变量X服从正态分布N(μ, σ2),确定下列各概率等式中a的数值:
(1)P{μ-aσ<X<μ+aσ}=0.9;
(2)P{μ-aσ<X<μ+aσ}=0.95;
(3)P{μ-aσ<X<μ+aσ}=0.99.
解
(1)2Φ(a)-1=0.9, Φ(a)=0.95, a=1.64;
(2)2Φ(a)-1=0.95, Φ(a)=0.975, a=1.96;
(3)2Φ(a)-1=0.99, Φ(a)=0.995, a=2.58.
50.设X~N(160, σ2),如要求X落在区间(120,200)内的概率不小于0.8,则应允许σ最大为多少?
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0077_0001.jpg?sign=1738893989-c1NEXj7yOudzfGn3TEIytZHCJqGURJC0-0-187c7fcc6c7b0d2bdae84e5930de1b4e)
查表得Φ(1.28)≈0.9,于是可得.故σ最大约为31.
51.设一节电池使用寿命X~N(300,352),求:
(1)使用250h后仍有电的概率;
(2)满足关系式P{|X-300|<d}=0.9的数值d;
(3)满足关系式P{X>c}=P{X<c}的数值c.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0077_0003.jpg?sign=1738893989-IlGgb4xd7yXFX1dVL8vKMBneDzenfw7v-0-467f3311a67138712baba520f20f0ccd)
52.设某班有40名同学,期末考试成绩X~N(375,81),假设按成绩评定奖学金,一等奖学金评4人,二等奖学金8人,问至少得多少分才能得到一、二等奖学金?
解 假设分别至少得分为a和b,才能得到一、二等奖学金.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0077_0004.jpg?sign=1738893989-NZGxIE8ChXwP7VguR2AyIVocjsg1XGgs-0-eede7afa6570c1fc630de45c04883cf5)
(B)
1.设随机变量X的概率密度为f(x),且f(-x)=f(x).F(x)是X的分布函数,则对任意实数a,有( ).
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0077_0005.jpg?sign=1738893989-yA2DU215TyZuBDz4qu32ZVuhlr9lyFms-0-4894b16b45e5f0eabdd4748c6e20dab9)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0077_0006.jpg?sign=1738893989-1Y6ZvG7ERROu8CStSUGRmBdD3uWgwTcq-0-35fbceeddb79a3ed899ec8082609b3a0)
C.F(-a)=F(a);
D.F(-a)=2F(a)-1.
解 ,
由于
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0001.jpg?sign=1738893989-uYjJFGkbrfIuEC04M8DWpwUYCmR9ZkI2-0-34c59ff6ef1ffa7b5e8ec521f0321613)
所以
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0002.jpg?sign=1738893989-hZ3dCyKV7FLVPUhUaHtjoPEpUyCs1vxC-0-b8802dc396b7788266827d8b57361e01)
即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0003.jpg?sign=1738893989-4RVzIhIpjboSMs9S0VDrtm3Co9puSlUF-0-ec526bc041250ba7839b6eb4186c7103)
所以有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0004.jpg?sign=1738893989-8umKm0liohjMm3JteF2hrC6LFQM4olum-0-bb4e9f87c07670ee8669ae8bb6b8120b)
B为正确答案.
2.设F1(x)与F(x2)分别为随机变量X1与X2的分布函数,为使F(x)=aF1(x)-bF2(x)是某一随机变量的分布函数,a, b的值应取( ).
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0005.jpg?sign=1738893989-2APYsoQio3DMcK5FaFASz3sDNwFZS67z-0-fdc20f88cd38b3e7168f9120d1af9d9b)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0006.jpg?sign=1738893989-v4ft7T5opMoQYkqKGFzVFYIPZ8KO82XY-0-35fe8cc2bf4d51b52fdec531695efd38)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0007.jpg?sign=1738893989-7VB7hJWIKXPIZnkhCtWTrAEyELCh7Fb2-0-125ebb5560041482ce026abb1e6b513a)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0008.jpg?sign=1738893989-ds2knfiovtptLXk180hI35ADShIQLe5h-0-f0975bfd18351e77ecece8798c4d12c8)
解 由分布函数的性质,应有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0009.jpg?sign=1738893989-GhztjxsMxSOjq1lwXf3Cf7shDi9Es0S1-0-fb4a7e8e25a01168e6ca4fceec5bfd6e)
所以A为正确答案.
3.设随机变量X服从正态分布N(μ, σ2),则随σ的增大,概率P{|X-μ|<σ}( ).
A.单调增大;
B.单调减小;
C.保持不变;
D.增减不定.
解 由正态分布的标准化变换得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0010.jpg?sign=1738893989-hCxGr0obTZRbTvWZYhX6l3N9D14r74Y0-0-713f836d895b86e3d174304cbd0d9a44)
所以概率P{|X-μ|<σ}的大小与σ无关.C正确.
4.设随机变量X服从正态分布N(μ1, θ21),随机变量Y服从正态分布N(μ2, θ22),且P{|X-μ1|<1}>P{|Y-μ2|<1},则必有( ).
A.θ1<θ2;
B.θ1>θ2;
C.μ1<μ2;
D.μ1>μ2.
解 因为θi>0(i=1,2),由正态分布的标准化变换有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0011.jpg?sign=1738893989-MxEzrfQbKVBxdQOtCdKjbG1StYubpTnp-0-6af773adedeb6d74c8e46ed72d7ed0e4)
所以A正确.
5.从数1,2,3,4中任取一个数,记为X,再从1,2, …, X中任取一个数,记为Y,求P{Y=2}.
解 显然,随机变量X能取1,2,3,4这4个值,由于事件{X=1}, {X=2}, {X=3},{X=4}构成完备事件组,且,则有条件概率
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0002.jpg?sign=1738893989-Sy6BEdoEiZLrhcrjazuQ3lR7t2inrXh2-0-26e2ec5855c89ca411c71ce529dbb61f)
所以由全概率公式得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0003.jpg?sign=1738893989-iSOOKpgW2h1afNTMZXun1uGr8yGp5o3A-0-eb01ec34d3ae4684560b57b0376a27be)
6.设在一段时间内进入某一商店的顾客人数X服从参数为λ的泊松分布,每个顾客购买某种商品的概率为p,并且每个顾客是否购买该种商品相互独立,求进入商店的顾客购买该种商品的人数Y的概率分布.
解 由题意得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0004.jpg?sign=1738893989-kfENKORrJSqpQ8sklXUDOmKMygJvOrb2-0-f0698e4dbee47b342172b0bd25c1b446)
设购买某种物品的人数为Y,在进入商店的人数X=m的条件下,随机变量Y的条件分布为二项分布B(m, p),即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0005.jpg?sign=1738893989-H20urid3KiQaY72QkXQVJgpL7nej9x7U-0-4b6063d66a7dbbdd0256cbc6ab6264b2)
由全概率公式得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0006.jpg?sign=1738893989-QsZownqedpeWHbOH3ccvudczc9KHl6mT-0-f145c26365e18289586459121079e1a6)
7.设X是只取自然数为值的离散随机变量.若X的分布具有无记忆性,即对任意自然数n与m,都有
P{X>n+m|X>m}=P{X>n},
则X的分布一定是几何分布.
解 由无记忆性知
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0007.jpg?sign=1738893989-VyHih4En1r7FDvPVoHwB2AnJWeu1W1uN-0-f6cfc03e2b9d7290bf598c9967177a08)
或
P{X>n+m}=P{X>n}·P{X>m}.
若把n换成n-1仍有
P{X>n+m-1}=P{X>n-1}·P{X>m}.
上两式相减可得
P{X=n+m}=P{X=n}·P{X>m}.
若取n=m=1,并设P{X=1}=p,则有
P{X=2}=p(1-p).
若取n=2, m=1,可得
P{X=3}=P{X=2}·P{X>1}=p(1-p)2.
若令P{X=k}=p(1-p)k-1,则由归纳法可推得
P{X=k+1}=P{X=k}·P{X>1}=p(1-p)k, k=0,1, …,
这表明X的分布就是几何分布.
8.假设一大型设备在任何长为t的时间内发生故障的次数N(t)服从参数为λt的泊松分布.(1)求相继两次故障之间时间间隔T的概率分布;(2)求在设备已经无故障工作8小时的情况下,再无故障工作8小时的概率Q.
解 发生故障的次数N(t)是一个随机变量,且N(t)服从参数为λt的泊松分布,即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0080_0001.jpg?sign=1738893989-QJXSiVfMkK5mVGq5zb69nJkd252HM8EX-0-2e563f3f85d0b0390c61ad12117a71be)
(1)相继两次故障之间时间间隔T是非负连续型随机变量,所以,当t<0时,分布函数为F(t)=P{T≤t}=0;当t≥0时,{T>t}与{N(t)=0}等价,于是有
F(t)=P{T≤t}=1-P{T>t}=1-P{N(t)=0}=1-e-λt,
即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0080_0002.jpg?sign=1738893989-Tn6aLGwX7z7fvTXPoaS95ebW1u6s94sS-0-40ea8d52e99a695aa41a2f6046b02873)
因此,随机变量T服从参数为λ的指数分布.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0080_0003.jpg?sign=1738893989-opYjlF7ktoLpnL0rOtnMo3BjqqP3UkZz-0-efc1e08fec3b4fd56d3ec4497a4c23b3)
9.设随机变量X的概率密度为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0080_0004.jpg?sign=1738893989-uypYmoEOr3snRpmHL9Eimm6AEgqmz5pj-0-17ae368cf569ac9b126021aa7b0b96bc)
F(x)是x的分布函数,求随机变量Y=F(X)的分布函数G(y).
解 对X的概率密度积分得X的分布函数
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0080_0005.jpg?sign=1738893989-2CsbVy1jH6dS0mSJGvHkGRRyukoQwa1I-0-e0e6f6a277510afd5586f0156399236a)
当y≤0时,有
G(y)=P{Y≤y}=P{F(X)≤y}=0,
当y≥1时,有
G(y)=P{Y≤y}=P{F(X)≤y}=1,
当0<y<1时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0001.jpg?sign=1738893989-NDJ68u34QcW78zC2wPdCZIupksD14Scx-0-d9e8e8a8476027a4f41a3bb51ea1b3d5)
或
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0002.jpg?sign=1738893989-bZzgTCFbEWVIGHgRWXiHNDdLmFplB00f-0-668da15b45f9d93a1c11ac4e9235012f)
于是,Y=F(X)的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0003.jpg?sign=1738893989-ADSUy4vMkCMc97x6wXZkrjQWGmpIiEZz-0-d56696b1c29d034ae59693a135fe3995)
即Y=F(X)服从区间[0,1]上的均匀分布.
10.假设随机变量X服从参数为λ的指数分布,求随机变量Y=min{X, k}的分布函数(k为一常数,k>0).
解 由题设条件
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0004.jpg?sign=1738893989-w5vGJtcI3wKu3RuiKuAVMA7sC7z04Hrf-0-353f2505d9c1715f60682df462c9fa99)
所以
FY(y)=P{Y≤y}=P{m in{X, k}≤y}.
当y<0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0005.jpg?sign=1738893989-cjSYTG4DJJmWUTjF7MvjKwXn1r4CLYsY-0-7e4d8e8a53099644959ed8e622a02f26)
当0≤y<k时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0006.jpg?sign=1738893989-021h1A8llHIukPjEWsx545QUXMXEBAul-0-2bdc92a8b4a569ece89407a21856401a)
当y≥k时,有
FY(y)=P{Y≤y}=P{m in{X, k}≤y}=1.
所以Y的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0007.jpg?sign=1738893989-hG4yu6NlNM9gOSXz0C6N3sOw1QO7Bfzq-0-5c7d3c0a4ce14bfc09f903eee3cc92cd)