概率论与数理统计第三课后答案习题.doc

leading cadres awareness of right in place, study the party Constitution and party rules, series of speeches can make proper effect. Party cadres to two lead by example, to lead by example we must change our mind, recognizing that two is important. First, learn the Communist Party Constitution Party rules, learning learning series important speech by General Secretary can enhance theory. With the third revolution the rise of rapid changes in our life are feeling, the Communists should adhere to the theory of confidence will continue to learn from the voices of the times, the times, in turn, will have new requirements for leading party cadres. Two is the most basic meaning of members all mastering the core theory and the most advanced weapons theory, complement the spirit of calcium. Secondly, the Communist Party Constitution Party rules, learning learning series important speech by General Secretary be able to firmly build the ideological foundation. Under the impact of multiple values, the two is to help cultivate independent judgment in numerous miscellaneous multiple concepts, so that the majority consensus of party members, the important magic weapon of the party with the resonance frequency. Finally, the Constitution of the Communist Party, party rules, learning learning series important speech can stand crowds, General Secretary position. Now, some grass-roots work in the the old way didnt work, hard way cannot, the new approach would not phenomenon, sometimes due to a mass of party members and cadres not understanding, does not meet. In fact, the mass convincing and identityBehind are likely to be party members and leading cadres themselves ignoring beliefs held, resulting in lack of persuasion and sense of identity. Two is in fact guaranteed party cadres work, an important prerequisite for convincing the masses. Bacon said it well: practical men can handle individual matters, but looking at the whole operation globally, but only man can do to knowledge. Giving up two effective, must first raise awareness of leading cadres of party members, as party members learn real responsibility to know to music , good changes, in parallel with the ground, do not forget to also answer the antenna, draw catches on meteorology, which sit between heaven and Earth which melds together the dreams of practicing Communist. Members cadres should in two learn a do in the based post do contribution in all members in the carried out learn Constitution Party rules, and learn series speech, do qualified members learning education, this is following party of mass line education practice activities and three strict three real topic education zhihou, deepening party education of and once important practice, is promoted three strict three real topic education from key minority to all members expand, and from concentrated education to regular education extends of important initiatives. Vast numbers of party members and cadres in the course of two, should be第四章 随机变量的数字特征1. 甲、乙两台自动车床，生产同一种零件，生产1000件产品所出的次品数分别用x，h表示，经过一段时间的考察，知x，h的分布律如下： x 0 1 2 3 h 0 1 2 p 0.7 0.1 0.1 0.1 p 0.5 0.3 0.2试比较两台车床的优劣。解：因为Ex=00.7+10.1+20.1+30.1=0.6; Eh=00.5+10.3+20.2=0.7。故就平均来说，甲机床要优于乙机床。2. 连续型随机变量x的概率密度为 又知Ex=0.75，求k, a之值 。 解：首先由密度函数性质知；又 Ex=0.75，即有 ；由上述两式可求得k=3, a=2。3.已知随机变量x的分布律为x-1023p1/81/43/81/4求Ex，E(3x-2)，Ex2，E(1-x)2 。解：Ex=(-1)(1/8)+0(1/4)+2(3/8)+3(1/4)=11/8; Ex2=(-1)2(1/8)+02(1/4)+22(3/8)+32(1/4)=31/8; E(1-x)2=(1-(-1)2(1/8)+(1-0)2(1/4)+(1-2)2(3/8)+(1-3)2(1/4)=17/8或者， E(1-x)2=E(1-2x+x2)=1- (E2x)+Ex2=17/8。4. 若x的概率密度为。求(1)Ex，(2)Ex2 。解：(1)中因e-|x|为偶函数，x为奇函数，故xe-|x|为奇函数，且积分区间关于原点对称，该积分又绝对收敛，事实上故 Ex=0。 (2)。5. 轮船横向摇摆的随机振幅x的概率密度为 求(1)确定系数A；(2)遇到大于其振幅均值的概率是多少？解：(1)由密度函数性质知,即 (2) ， 。6. 一个仪器由两个主要部件组成，其总长度为此二部件长度之和，这两个部件的长度x和h为两个相互独立的随机变量，其分布律如下表： x910 11 h 6 7 p 0.30.5 0.2 p 0.4 0.6试求E(x+h)，E(xh)。解：因为 Ex=90.3+100.5+110.2=9.9，Eh=60.4+70.6=6.6，故 E(x+h)=Ex+Eh=9.9+6.6=16.5；又x和h为两个相互独立的，因此有E(xh)=ExEh=9.96.6=65.34。7. 已知(x，h)的联合概率密度为 试求E(x2+h2)。解：E(x2+h2)=。8. 一民航送客车载有20位旅客自机场开出，旅客有10个车站可以下车，如到达一个车站没有旅客下车就不停车。以x表示停车的次数，求Ex (设每位旅客在各个车站下车是等可能的，并设各旅客是否下车是相互独立的)。解：引入随机变量 易知，现在求Ex 由题设，任一游客在第i站不下车的概率为9/10，因此，20位游客都不在第i站下车的概率为(9/10)20，在第i站下车的概率为1-(9/10)20。也就是 Px i=0=(9/10)20， Px i=1=1-(9/10)20()，因此， Ex i=1-(9/10)20()。故Ex=E(次)9. 圆的直径用x度量，而x且在a，b上服从均匀分布，试求圆的周长和圆的面积的数学期望和方差。解：由于x服从a，b上的均匀分布，因此x的分布密度为 而圆的周长L=px，圆的面积A=px2/4，故有 EL=E(px)=pEx=， DL=D(px)=p2Dx=； EA=px2/4=，又 =，因此 DA=EA2-(EA)2= = 10. 设随机变量x，h相互独立，其概率密度分别为： ， 试求E(xh)，D(x+h)。解：因为 ， ， ， ，又x与h是独立的，故有 E(xh)=ExEh=11=1； D(x+h)=Dx+Dh=。11. 设随机变量x与h相互独立，且Ex=Eh=0，Dx=Dh=1，求E(x+h)2 。解： E(x+h)2= E(x2+2xh+h2)= Ex2+2E(xh)+Eh2，又x与h相互独立，因此 E(xh)= ExEh，而Dx=，同理 故有 E(x+h)2=E(x2+2xh+h2)= Ex2+2 ExEh+Eh2 =+2 ExEh+=1+1=2。12. 若连续型随机变量的概率密度是 且已知Ex=0.5，Dx=0.15，求系数a, b , c 。解：因为，即有 又Ex=0.5，故 又Ex=0.5，Dx=0.15，因而Ex2=0.4，因此 解、组成的方程组，解得a=12，b=-12，c=3。13. 设随机变量x有分布函数 求E(2x+1)，D(4x) 。解：先求x的分布密度函数 故 ， ,因此。从而有 E(2x+1)=2Ex+1=，D(4x)=16Dx=。14. 证明：当k=Ex时，E(x-k)2的值最小，且最小值为Dx。解：E(x-k)2=E(x-Ex)+(Ex-k)2= E(x-Ex)2+2E(x-Ex)(Ex-k)+E(Ex-k)2 = E(x-Ex)2+E(Ex-k)2=Dx+ E(Ex-k)2 Dx。即当k= Ex时，E(x-k)2取得最小值Dx。15. 如果x与h相互独立，不求出(xh)的分布，直接用x的分布和h的分布能否计算出D(xh)，怎样计算？解：因为x与h相互独立，故D(xh)=E(xh)2- E(xh)2= E(x2h2)-(ExEh)2 = Ex2Eh2)-(Ex)2(Eh)2。16. 一台仪器有10个独立工作的元件组成，每一个元件发生故障的概率为0.1，试求发生故障的元件数的方差。解：引入随机变量易知， ， ，故 x。17. 设随机变量x服从瑞利(Rayleigh)分布，其概率密度为 求Ex，Dx。解： = 。18. 若x1，x2，x3为相互独立的随机变量，且 试求： 的数学期望和方差。解：， ,故 。19.设二维随机变量(x，h)的联合分布律为 hx-101-11/81/81/801/801/811/81/81/8计算rxh，并判断x与h是否独立。证明：由题得(x, h )的边际分布律各为x-101h-101pi.3/82/83/8p.j3/82/83/8 pijpi.p.j，(i,j=1,2,3)故x与h不独立。又 ，即x与h不相关。20. 设二维随机变量(x，h)的联合概率密度为： 试验证x和h是不相关的，但x和h并不相互独立。解：先求fx(x),fh (y): 同理 显然，f(x, y)fx(x)fh (y)，故x与h不独立。 又 故 ，即x与h不相关。21. 设随机变量(x，h)的联合概率密度为： 求：Ex，Eh，Cov(x，h)。解： 22 . 设有随机变量x和h，已知Dx=25，Dh=36，r xh=0.4，计算D(x+h)，D(x-h)。解：由于 故 D(X+Y)=61+24=85， D(X-Y)=61-24=37。23. 证明：当x，h不相关时，有： (1)E(xh)=ExEh (2)D( xh)=Dx+Dh。证明：(1)因为 ，由题知x，h是不相关的，故r xh=0，因此，有E(xh)=ExEh。(2)D(xh)=E(xh)2-E(xh)2=Ex22xh+h2-(Ex)22(Ex)(Eh)+(Eh)2 = Ex2-(Ex)2+Eh2-(Eh)22(Ex)(Eh)2(Ex)(Eh)=Dx+Dh。24. 设(x，h)在。试求r xh。解：因为(x，h)在，故联合密度为 。25. 设(x，)的联合概率密度为 证明：x与h不独立，但x2与h2独立。解：x与h的边际概率密度为 同理 显然， f(x, y)fx(x)fh (y)，故x与h不独立。令 ，则当z0时，；当0z1时， ；当z1时， ,故 类似地可求得h1的分布密度函数为 令(x1，h1)的分布函数为F(z, w)，则有当z0,或w0，易知 F(z, w)=0；当0z1，0 w1时， ；当z1，0 w1时， ；当0z1，w1时， ；当z1,或w1， F(z, w)=1；故(x1，h1)的联合分布密度函数为 因此有 ，即x1，h1是相互独立的。26. 设x1，x2为独立的随机变量，且都服从N(0，2)，记 。解： 而 , 。27. 设随机变量x服从指数分布，其概率密度为 试求k阶原点矩E(xk) 。解：E(xk)= 非谓语动词是历年高考英语的重要考点之一，也是较难掌握的难点之一。它贯穿于英语学习和考试过程的始终。但是，只要认真分析、透彻理解、看透本质、准确把握，就一定能在高考中运筹帷幄，游刃有余。based on their actual, striving to be learning and done markers. Strengthening theoretical study, belief and faith. Theory is an action guide, only theoretical knowledge in place, to unity, action can consciously. Constitution is the fundamental law of the party, party rules are party must follow the General rules. Studying and implementing the party Constitution Party rules and a series of important speech, General Secretary, is a guide to persisting and developing socialism with Chinese characteristics, is a basic work of strengthening the partys construction, but also each and every Communist Party Members obligations and solemn responsibility. These important content, for each party, must know would believe will do both. WeGeneral Secretary of the party Constitution, a series of remarks with, and hold, grasp the real learning that basis, based on the Constitution speak to issue, in accordance with the series changed the problem. Second, disciplined in line, to become law-abiding model. Is not no rules. If there is no iron discipline, party members and cadres will become a mess, it will lose its fighting strength and vitality, purity and naturally much less advanced. By two education, party members and cadres to keep the bottom line, not the more red lines, consciously practice the three Suns and consciously act according to the Constitution, party rules, remains a political force, to perform party oath, a tree of its own image. Three is the loyal, dare to play and play to party members. Learning is the Foundation, is the key to development is the goal. Two main objective is to let the majority of the play to the exemplary role of party members and cadres, party Constitution, party rules to examine. To use new criteria to guide our behavior. Through education, the ideal and conviction, keep in mind that membership, learn, learn, learn go play a vanguard and exemplary role, the courage to play as, and always maintain the pioneer, pioneering and enterprising spirit, and co