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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=1734427660-hOXYghwQt85ZVNGEzo77bADJR05KlE9W-0-bd117343e551372ae0ade664b6d1d939)
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=1734427660-VCf6nTDfUWmDCQfy3a09uXaRMCXTSeWz-0-8310aa087acef194f9b864f9a75b6cd8)
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=1734427660-QeHu9uTpD8xzUYQ9VQ7K84XnY6BwgMOM-0-8a12d1357cb6091b5920c8be443999ad)
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=1734427660-R9fVjltrjvCX0MT9gblcqcSKMakNju7S-0-9b63ef5e779c05801b6b59ebf11551e6)
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=1734427660-cJsYsZm69yXUv666aTdbyE8WXrQoTVf2-0-713c0950add25a77b57a1619bf3900c8)
(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=1734427660-fKDQQ2VGcSrRw1PVqqg3GRBDxKsOTZMX-0-379edf06b928d1d89a78733ec65b104d)
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=1734427660-nvIzwpNkwYK9g2yfSMqD89cHyWEv5zen-0-c60a6a84530262a2f4b311a19794f082)
故X的分布函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0001.jpg?sign=1734427660-OstsvAezAnbrD7g2cljNSOSkAc09NERw-0-4ae9fdcf5ee41ffa92dfa9daa700194b)
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=1734427660-pdwQGnMfSHHvjygQBlecWmu3UcK6ty6P-0-61bf6aff3595d605dd9a0e64f3ed6fdd)
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=1734427660-lTVrJhG6isZsNeBOLNOzKN8lBCyih2OZ-0-c45cb92c3820aa3844851de16bacc5e7)
解得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0007.jpg?sign=1734427660-meR2U9M1YVbCuZ3og5Up57ZZD03diAyq-0-f3de311d2dcdeb632b65f7c5fa014c77)
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=1734427660-y8WQufUzA9Ih6iVcz7ytv3aOsN5zhsbb-0-aead698a622aada3925ff047a1e06388)
解得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0009.jpg?sign=1734427660-UYjfCtyn0y4ukWIcuf3CdJjsYnX071qa-0-04e3e45d21a8dd34904537ef86e774dc)
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=1734427660-CiC9ZFkEZRdQH76PrBuSUG7sBoA8HkyN-0-e20cb1c690c2d5edaf048f3aaf4e5b45)
由于
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0012.jpg?sign=1734427660-m8iyecN5tnrKno1cJIkUWPl6VlEgL9Uf-0-876e6638e6ec757ae5b5f037b16b6090)
所以有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0013.jpg?sign=1734427660-II86hWYg9sZyPZ3IVeSA4Etf4UicpeHu-0-f8d06b925b6baad09a0898c091c2e76d)
解得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0065_0014.jpg?sign=1734427660-DNY3LnIG4AIhPZm9lL8YQ4AEUE4liwSx-0-d5cf32a43adf1c4c5785ad0e7c8f49f9)
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=1734427660-HAUG0n3ljZmsMl9eSWpdw0h6UAYzshdI-0-c4012ea2ce4d80f748f7808375510436)
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=1734427660-OQWrRtTaAdy9kEnciXUIE9zZJ8g1wCI3-0-e9ae7d5c3a570ea3b099ff70fc20c294)
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=1734427660-MGlt3XV0bcvf37LHtHO8Kzh3uzU6wxC3-0-0cf58ff4ef1a8600521b7fea4544366f)
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=1734427660-YBqTOYNEnr5Lqfrk21svB9uJ7lrixCgy-0-fe01987e9709252f1083a4e031a89420)
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=1734427660-4XJaS1ZvxFNBHqcTJ2Sq8GNkoe4TkaGO-0-e4a288e9d74b0dee36e95dc7c2c70299)
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=1734427660-BWxQ57CLNwARXxFXWhzNy12M01QdkA0V-0-2b62cef1f74e6f29f15bc1386c374154)
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=1734427660-vVct7Ur58bswdXrDzIWOsKhgc4AOFCtO-0-4da4a9c8202d529630b90810e7158d0c)
解得,故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0007.jpg?sign=1734427660-HIntHbFM9jFhlsvbLPFUPSfSsWPT2SxZ-0-31a011b6b1e7706b2c4613654b16a949)
18.设X的概率密度为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0067_0008.jpg?sign=1734427660-urap1mPlUggGtCOQ1GDB3Fk90qsskXeH-0-6935825b6a83f512515e07ff5c83e264)
(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=1734427660-x5d0K3m0Zuh1i4sSmJ9KCdim04RuAL9B-0-3f864a549838a3f886f2d68fda9cbe6d)
(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=1734427660-swokUVbplTOrNO2ihNxGInRlfqQhAhwN-0-74fa88fc62ad213a9ab41acc6db2573d)
19.设X的概率密度为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0002.jpg?sign=1734427660-9u0woVUKXRxGANCnzYOtEPUtYrTkcnGO-0-5341da3fd94a5f9458fb3f404434f4e7)
(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=1734427660-R4fhnBlyIO5maTTrCTtSiSd3YG7vdgId-0-e34e1a568ce829025bf91b4648ce3a5c)
故c=1π.
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0068_0005.jpg?sign=1734427660-z47UNT9yoz1snTloa2jyC8NfMxOdYWX9-0-2297c3fdfcf4a2ac179e061aff8b87c9)
(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=1734427660-B3tyZ3Ylr6sWYYK42f7VIqxLufni9KWL-0-3ba9113eaf70668e89a91a68d6008870)
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=1734427660-6N8nz9nZEq1RdZI8iRIfxZAjpzVoRARu-0-a5107e239fdea6444cf34262171c5d21)
(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=1734427660-S1gSPuyCLCpRyJZ4nP0pGaLWvwHVSEmT-0-164218cb429098c2e1cd6fd782680156)
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=1734427660-K7zGWXU6EYjaYlmaS0Q83yaaRCSqI6MC-0-05dc7daa57cc88c4e0c609e6de20e07a)
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=1734427660-ZYzHTWmDET3MZxwH2Vw7izOKcEf1nZDq-0-173974915439857e82c5a4eeb74812c3)
(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=1734427660-7KlbbL7io9LqtZ3kMZE8S8JjyjaGa1di-0-dd5b706e88a83406d6f9e7772aa90d64)
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=1734427660-KUoCCahaHX47H6hYj38AuMh2p8JV87UA-0-f4b32c65e4057c842e0a2bf7049db007)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0005.jpg?sign=1734427660-Ig7249lUYRvDbAHrYdfUV6BjQFMDLQNF-0-3a0581f88e048367f472ffeaf0aa2dc8)
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=1734427660-DynqXVSKnYuPORxhk67W24slFvW9Ju7Q-0-d0dffcd12af9ce1a91d13db69823e5dd)
(3)当x<0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0008.jpg?sign=1734427660-crrDlqiuEN0P8jEKWDpoHc8IzlvB7Zjo-0-fdc24759f51199553e2530f366dd2bd4)
当x≥0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0009.jpg?sign=1734427660-3u6JMp8IIO4zNZN3wfj2gekoUBPCqyEi-0-52b29c196b25411355a13e6e96735ac6)
故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0069_0010.jpg?sign=1734427660-SvJKJZlEADaaPi9aORYdl5aDYubMP0LV-0-d49e1e73ab0f4510c3fdbe9f9362e6d2)
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=1734427660-5K7F6Cd8W2NyQh2FTBn1pNMtdHKApoom-0-9638b3857347ac1aa27c248c8fa96555)
26.随机变量X~f(x),其中
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0003.jpg?sign=1734427660-0cnRfCTJHKCkoRxN0Kw813abCtvavPef-0-d554cb98876dd6c5c28456b6ec23ea20)
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=1734427660-eVW0ejz72mjW6DJwRkY7zMk0Db5h35Vd-0-96af640ed59cc3cb20924bc9a04b088d)
因此a=π.当0<x<π时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0005.jpg?sign=1734427660-eqCwLu4m6UE1gGpJc3LJ2tZsboWybegG-0-6b52909ed79e5e35682c4dbba03dc08c)
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=1734427660-bhdpFlcBeFTbrVkhaxlIFEe1f5AqRzu6-0-6acc86f1130e8f33175cc7e291e78c30)
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=1734427660-mefgXiQ70458D2UZC9qehU81bjJbRa9G-0-9755d3c8fc03afe7b78a83247a245a38)
故
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0070_0009.jpg?sign=1734427660-6gqxXrW5Rw2CTz5zpecucuI63HlzaT76-0-6adc49f02731be8b305cc3fd21d99e9b)
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=1734427660-oBrPI1SYzZk4aL00UO5T1ZuUxpEEVpZS-0-7d0641e134254593a14adbfe71933214)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0001.jpg?sign=1734427660-zJElEzVDfTGhatbRyXwL41gyagTglLcP-0-5515c94eba406ca889260e1650aabfb8)
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=1734427660-tB0SSLEDgAzYyhsCsYeiC0Yglq6B8XBM-0-8a92fdb121201101056e85709e34c27f)
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=1734427660-hRbtJ35aGBs98zVSxabt3Srdddg8bweK-0-ac615f4d922c864b517e4ed63b054cbc)
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=1734427660-20MmsplsIPc3va8jq07dK9kcFPqSoEIr-0-cb80bc3417079eed62becac9d66dabaf)
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=1734427660-5EBg2VDEt0FnirI2tAVNf43GgPn83Lyi-0-d67e545d76c51a3d9dc49d878b1819e9)
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=1734427660-eVwEKhYF5MNJxuHxqfJO8LKZgWLe9dul-0-ab8c9127e92357dedd3236925ac2f881)
查表得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0071_0007.jpg?sign=1734427660-pgp6Dwm59AN3v8yGF1g1jOUZRIhkvB0b-0-aeaf2704667ef713c90970fa0085c56c)
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=1734427660-SCP4nYBSna4g0QTT1HrcVlbYGEyyZXNp-0-553eb80f84807508a12256fbd73b7872)
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=1734427660-HJ3WhSET1DyiRtJqCdLUzGpPv2Wel3Cl-0-18d51d67c6a38b53a8112ef3a7e59296)
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=1734427660-iZaiWPHxC2wkGTQD3jdtPCNupZkZ6OIU-0-96b0f5c59e57f7574c632193c6c44d90)
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=1734427660-MUIMFzZ8ZMkixjS1oeEdRbqDRmL9SBJX-0-ccf6dac8bec98197c13327148f300eb1)
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=1734427660-l9wXOcNXfXuPBtM0IfiPMsBb7fO6MtFK-0-56544b903d4a6a91a7fb1b25d3f58f01)
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=1734427660-p7sgC2QFbo5eDC7yJUMfxgCphkBMUQTc-0-08b1e848a93fe0fc63b1e9b9d086f7a2)
Y的密度函数为
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0005.jpg?sign=1734427660-nzIGYFpvycmt0FniYUgZ7KapPbpqkFqr-0-a2eba5e6af81761defe7d5005b862037)
当a>0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0006.jpg?sign=1734427660-cYJfizh016mLdYXTUdjsldsYwPrXz8hT-0-f9eaf03a1499494cfa5fec6e589ae571)
当a<0时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0007.jpg?sign=1734427660-JDy4RSDpl0eUh7FbVd0ztTxTCXvRYFPf-0-bd8fa61580282f09d1b553e17341b08e)
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=1734427660-wvQBcKdVBbri6HLC8NbuLk8GLr5fi1ek-0-1e0a8e9f6d40c1fd21e493b8ebad59b8)
因此
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0073_0011.jpg?sign=1734427660-rTcWMAYr1JaUM8n0pd3yEu71d1f2Ds2g-0-7cb160849e43fbf6f5d12727fb8265cf)
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=1734427660-bzhP0JTinTMgVq38iSgo5argGtHtbiMS-0-52fcced84a8b7f40f893d07cab9d95e9)
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=1734427660-qscvlpLCCvMSBj7IR2mg0uoMH86z9Cs4-0-e75cc7a17549998f4482fdca5843af8f)
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=1734427660-i7DeLAVWG9m1wfYljRiKOKazNao08a14-0-07ae82a4321f7513a2c42a07e6930270)
因此有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0009.jpg?sign=1734427660-wpfTeevfC3qHLNjgP2oU9EuoUfAIwVoy-0-fe278bc4d525dc8af9157e75f2bc9e19)
z=x2在[0, +∞)上单调 ,因此有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0074_0011.jpg?sign=1734427660-K4fCx3NShyFodYTTBH9fTRHEUWN0vvDu-0-40877ada7d8a7dbb11fa51bef4bfec53)
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=1734427660-sM3QiZ1xoIrk1qSFfLNcf4kTAOG0BcVn-0-ae4ed424e631898690a0cfb9fce452db)
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=1734427660-t5NncNfFMCwAh5REzSh4SIQGgQXaaHZE-0-7e803a9889402de65a4b6043f99432a1)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0001.jpg?sign=1734427660-Sa7EAAZjnAP8d8LLvFjRGijT58cZr07E-0-320cf15173fc030e1f29ff4c5f2ac6fd)
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=1734427660-G9XJelSQLreiIIJeO3RdDPgBBHnuOcH1-0-cfca20f4010349abf451c9a85fcfe2bc)
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=1734427660-ghS8n3cvVQpdEXI4uCzjoctZX7v5sDaM-0-009e5fa8ab473ebe42715611a82e267b)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0075_0005.jpg?sign=1734427660-8BbQL5QpegCTvzS6s0ulvnCMULie08dg-0-dc6f46d7a80ecdd2d34277bf2d0b1209)
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=1734427660-j4E3JeaYkXa92stL6sRYvHTvUvcJXfTp-0-9492a850affaa15fd31ee95239358d3b)
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=1734427660-Cemsfh6g7UYObXqhD3Ghcrm4xV3AXeEq-0-dc649766908ebc92730ca0189a8064f2)
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=1734427660-sJR5j7HfW8cF9lJ9J8hWuwwUAxZsPlWX-0-092fb5fa3466a33a21eff0a5984e4144)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0076_0001.jpg?sign=1734427660-HhlRcnO1uWuNvg4nRRoOAQfPMICNDK52-0-64beb5d191335872ad6c9028e4687b63)
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=1734427660-HTUHtpfR5IdwYMh2yuJ8CadkvfIwxRfd-0-5b28f131d57eb4db540685c179f63195)
查表得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0076_0003.jpg?sign=1734427660-FdpFuBdhwsO2a7sd64DawrEXdFmjRcJW-0-b943a0bc46592c9d814e1654b6d0a9fa)
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=1734427660-4A09Q1klE16PE8woxCKjmhcDpg98uVBG-0-f5e87211bbaec5827941562002e62c61)
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=1734427660-9TFBn13l2Km4kccrdNdJZRFn88vFwWiQ-0-2c9e306a9cf4f4f3b59176e45d7455ff)
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=1734427660-sbf1ih5n9DHix4qHDn8YksvYfWZVdhUF-0-6b5e1d49b6f0e080a11c0c697de7cbb0)
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=1734427660-WHbp1Rz4dxUXjFRPyYqVfco1uKPYrZPx-0-14d1d3e001a10b7fbf21032c86ae0518)
(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=1734427660-jKlLphbtCOMTNP2aNnsZbEosPNcZDnZ4-0-23316988c71fdf92ed8e63c02dd2741f)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0077_0006.jpg?sign=1734427660-OaQLPRQlMTz4hqwjC9CHYBAHCYUg37Bt-0-3ba53bcfdc8cdc612efa2898f664ab5c)
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=1734427660-h7N0MTuvc7Ul0Tvei7VDMPhtpn4GFXD6-0-88d452840f6b0a197a0f6af5d16e35d5)
所以
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0002.jpg?sign=1734427660-l5VSFHzM20JFYEwAGVrXwQVJvTJfhXIg-0-94636b871933c688fb3baae90c4d3a14)
即
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0003.jpg?sign=1734427660-Rxu487Myzj3Ce5oqXnIOoUGFCORsgQDL-0-f5a4a2da83ec0c51e352858603e96210)
所以有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0004.jpg?sign=1734427660-Jg8GBE3kN4EWpWIdkh0oyO73FNw3xvhC-0-4fb8a622bc4babdce6b2874b11ced22d)
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=1734427660-yPPhSSqdPFEkti6HHH99DjUe3pXSzta2-0-6c70725ca344136c3f68de7e038928a0)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0006.jpg?sign=1734427660-khVGpEAmmrj8hLzCgzcqweN33JKQgE8B-0-1a5354e484e88e73b14ed19e76ded541)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0007.jpg?sign=1734427660-Eg25UxPj9WOZ7rRDN5nIw70GsEUFwvZd-0-2bab4661533941d3d94a5ce5914bf32c)
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0078_0008.jpg?sign=1734427660-4wvCvux7RHYL4bd9rbgpNQwPhfkGhFmQ-0-79079864a8ac2445ad7fde742bf6281c)
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=1734427660-aJ3pBLP7W19tUgjzDSXlFryh3hnvZ59w-0-2ca4d24aa6b16280edcd04494f3232ef)
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=1734427660-e3evD1lvmT5nPgIwZvqFbpouWGwP0TmM-0-ad5bc030defe16c5471de2cb9f3e5f7a)
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=1734427660-btTLXwwt8Oc6pH51y010HgnHN6O8vP66-0-f6e87f48dcffab8caa7e012fecea90f5)
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=1734427660-VYwWax6K1FmDtXENqMnxtEddkySzRiEw-0-c783ec1d01cb9549ab5c4d967a19a012)
所以由全概率公式得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0003.jpg?sign=1734427660-lSo2IllSiF9jm6MVreLJMNeBQeSZN4f1-0-79d60dbd1fc6d6bc9c05a2abaa51190b)
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=1734427660-VSSBhEzeQPOeefXZfRuBElxV9SIgyMy4-0-190bfa88f602118441ef5966e4d7af0f)
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=1734427660-Cjg18pGrBTVK5iNoU3OyHQjPAmhulGqL-0-e3d2f55a62d549fa964281b60e4ed750)
由全概率公式得
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0079_0006.jpg?sign=1734427660-5pgiiZ5nVMS2axgCEXVWGD21p9iSA0iq-0-46ee257eb5dcc1656ab4fb56016e6fe8)
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=1734427660-a8ulNJYRqeNCDOGoWstjtsQa0dNsI7Nr-0-31a55303a4548aef4d70a955460e58c6)
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=1734427660-A8kA15pbZYJuGAOpnl3ZwlLfQm6JUOT1-0-c65f0149305461c895816f4e46c1ae8b)
(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=1734427660-S2EqzxSsJxhTZhnP9JzJytqDj0rv4T89-0-23321581b5b517aae92c2a75bf065a86)
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=1734427660-kZAqrTYhHgg0AsTxZ68709QRhOLKEQG5-0-e6ce0b2adcb03e39444c9bb65c96d179)
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=1734427660-4GEUWKekkvbaFVjuUAtwYuQLmDDlFdIX-0-77f365fe1bb3a9bbd0342d3d2b43882e)
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=1734427660-bo4Fbmn64CM7pafzMhtlcCubf78PkOqv-0-39947128fad549de3f8ca90262b66355)
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=1734427660-Rq7WFElyl0tIkIQoVkyI2jrlyzOF7kID-0-27545c23f2a86dbcacbaee14c2d7881b)
或
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0002.jpg?sign=1734427660-I87B4pO3ncqpJwVeqZjRxFbZYZU7Tu6i-0-351449f8361c7720c5ae00a5281ee658)
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=1734427660-wGsBqB0HxVvdCvPclyt0R1B6qv1gqUkj-0-e437b4b3a46147db397ab2691b0c1d0a)
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=1734427660-fFh3DJu76Y6UHFFFeI9oHQwocIQoEbiy-0-85bd3eafaaac0b044fa1f7ba5e4906d6)
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=1734427660-1ZAqGwP5YDVwb7UGkgqIFTCrJCsLyNvu-0-b9b1efba9388d47286e35cd5155dc919)
当0≤y<k时,有
![](https://epubservercos.yuewen.com/1A97F7/12738942104014106/epubprivate/OEBPS/Images/figure_0081_0006.jpg?sign=1734427660-ekQAKje8zH0FZls3LY8nd29pUlxFJkQ9-0-c806fd674391cf561273a497b63d1aee)
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=1734427660-PFYxSZsfnlw8x08WV9DDG4stxRzW1ojZ-0-839dc9507d4793ac3a88b4b4ed9b9d51)