管理统计学之随机变量教材

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1、管理统计学管理统计学第五章第五章2：随机变量及其重要的离散：随机变量及其重要的离散概率分布概率分布Random Variables and Some Important Discrete Probability Distributions学习目标学习目标 Learning Objectivesn1.计算离散随机变量的期望值和方差nCompute the Expected Value&Variance of Discrete Random Variablesn2.描述二项、泊松和超几何概率分布nDescribe the Binomial,Poisson,&Hypergeometric Proba

2、bility Distributionsn3.介绍两个随机变量的协方差与和的思想Introduce the Ideas of Covariance and Sum of Two Random VariablesRandom Variables(R.V.)using variables denoting events Properties:choosing value randomly,the probability of choosing value is definite.being divided into continuous variables and discrete variabl

3、es often being denoted by capital letter X,Y.Random Variables and Their Distributions The distribution of discrete r.v.X x1 x2 .xn Pi P1 P2 .Pn Writing like P(X=xi)=formula is also ok.For example:P(X=i)=The Distribution of R.V.n1ii1P!iei (n may be,Pi0)PX xxx(|)!e-The distribution of continuous r.v.T

4、he distribution function:F（x）=P（Xx）It can describe discrete r.v.too.0F（x）1，F（x）is nondecreasing as x increases F（）=?，F（-）=?The distribution density function:f（x）f（x）0，F（x）=,F（x）=f（x）.xdt)t(fThe Distribution of R.V.=1.dt)t(fThe geometrical meaning of density f(x)If X is a continuous r.v.,then P(X=a)=

5、0,so F(x)=P(Xa)=1-F(a)P(aXb)=F(b)-F(a)For example:Which of the following formulas are impossible?P(X=6)=0.8 P(X=10)=1.3 F(3.2)=1.6 f(1.5)=0.8 F1(3)=0.5 F1(4)=0.4 f2(3)=1.6 f2(4)=1.0Random Variables and Their Distributions离散随机变量离散随机变量 Discrete Random Variablen1.用数值表达的一个事件n掷两个硬币出现的反面次数n观察到0,1,2个反面n2.离

6、散随机变量 n全部整数(0,1,2,3 等等)n通过计数得到离散随机变量的例子离散随机变量的例子 Examples试验试验随机变随机变量量可能值可能值打100个促销电话销售数目0,1,2,.,100检查70个收音机残品数目0,1,2,.,70回答33个问题正确数目0,1,2,.,33在收费站数车从 11:00 到 1:00到达车辆数目0,1,2,.,离散概率分布离散概率分布Discrete Probability Distributionn1.列出所有可能的 Xi,P(Xi)数对nXi=随机变量的取值(可能的结果)nP(Xi)=与每一取值相应的概率n2.互斥(没有重叠)n3.完备(没有遗漏)n

7、4.0 P(Xi)1n5.P(Xi)=1离散概率分布的例子离散概率分布的例子 Examplen概率分布n取值,Xi 概率,P(Xi)n01/4=.25n12/4=.50n21/4=.25 事件:掷两个硬币.出现反面的次数.?1984-1994 T/Maker Co.图示离散概率分布图示离散概率分布n(0,.25),(1,.50),(2,.25)列表列表表格表格图形图形等式等式#反面反面f(Xi)次数次数P(Xi)01.2512.5021.25P xnxnxppxnx()!()!()1.00.25.50012XP(X)离散随机变量概括度量离散随机变量概括度量 Summary Measuresn1

8、.期望值Expected Valuen概率分布的均值n所有可能值的加权平均n=E(X)=Xi P(Xi)n2.方差 Variancen对均值的方差的加权平均n2=E(Xi(Xi P(Xi)思考题 Thinking Challengen扔两个硬币。你感兴趣的是反面出现的次数。反面的次数，作为一个随机变量，它的期望值和标准差是什么?1984-1994 T/Maker Co.离散概率分布函数离散概率分布函数 Probability Distribution n1.模型的类型n代表一些内在的现象n2.数学公式 n3.代表离散随机变量n4.学会求出准确的概率P Xxxx(|)!e-离散概率分布离散概率分

9、布 Discrete Probability Distribution 离散离散 概率分布概率分布泊松分布泊松分布超几何分布超几何分布二项分布二项分布二项分布二项分布 Binomial Distributionn1.n 次观察(试验)的样本中成功的次数n2.15次旋转轮盘赌转盘中红色出现的次数n3.一批五件产品中残品的个数n4.33个问题的考试中正确回答的个数n5.100个进入店内的顾客中买东西的顾客个数二项分布特征二项分布特征 Binomial Distribution Propertiesn1.两种不同的抽样方法n无限无限总体无放回无放回n有限有限总体有放回有放回n2.n 次相同的试验序列

10、n3.每次试验有两个可能的结果n成功(所希望的结果)或者失败n4.每次试验的概率恒定n5.每次试验互相独立 二项分布的概率分布函数二项分布的概率分布函数 Probability Distribution FunctionP(X=x|n,p)=给定 n 和 p，X=x的概率n=样本 数量p=成功的概率 x=在样本中成功的次数 (X=0,1,2,.,n)P Xxn pnx nxppxn x(|,)!()!()1二项分布示例二项分布示例事件:投掷一个硬币4次。记录反面的个数。出现三次反面的概率是多少?P Xxn pnx nxppP Xxn x(|,)!()!()(|,.)!()!.(.)13 4 5

11、43 4351 534 3 .25二项分布的特征二项分布的特征Characteristicsn=5 p=0.1n=5 p=0.5均值均值标准差标准差E Xnpnpp()()1.0.2.4.6012345XP(X).0.2.4.6012345XP(X)泊松分布泊松分布 Poisson Distributionn1.在一单位区间内发生的事件的次数n单位可以是单位可以是n时间,长度,面积,空间n2.例子n在20分钟内惠顾的顾客数目n每年美国罢工的次数n每一批录像机中的残品数目泊松过程泊松过程 Poisson Processn1.事件概率不变n平均60次/小时就是在60个 1分钟的区间内1次/分钟.n

12、2.每个区间发生一个事件n不能同时到达n3.事件的发生相互独立n一个人的到达不会影响到另一个人的到达?1984-1994 T/Maker Co.泊松概率分布函数泊松概率分布函数 Probability Distribution FunctionP(X=x|)为给定 时 X=x 的概率=成功事件数的期望(均值)e=2.71828(自然对数的底)x=每单位时间、空间中成功的次数P Xxxx(|)!e-泊松分布特征泊松分布特征 Characteristics=0.5=6均值均值标准差标准差iiNiE XX P X()()1.0.2.4.6012345XP(X).0.2.4.60246810XP(X)

13、泊松分布应用示例泊松分布应用示例 Examplen顾客以每小时72人的速度到达.n3分钟之内达到4名顾客的概率是多少?n72每小时=1.2 每分钟 =3.6 每三分钟 P XxxP Xx(|)!(|.).!ee =0.1912-3.64 36(3 6)44泊松分布题解泊松分布题解:求出求出 Finding n75 单词/分=(75 单词/分)(60 分/小时)n =4500 单词/小时n 6 错误/小时=6 错误/4500 单词n =0.00133 错误/单词n在一个255个单词的段落中(区间):n =(0.00133 错误/单词)(255 单词)n =0.34 错误/255个单词的段落 超几

14、何分布超几何分布 Hypergeometric Distributionn1.单一抽样方法n有限有限 总体 无无 放回n2.n 次相同试验的序列n3.每次试验有两个后果n成功(所希望的后果)或者失败n4.恒定试验概率n5.试验之间相互依赖超几何分布超几何分布 Hypergeometric DistributionP(X)=给定 n,N,和 A 时 X=x 的概率n=样本容量N=总体容量 A =总体中成功的次数x=样本中成功的次数 (X=0,1,2,.,n)PX)()()AX()N-An-XNn超几何分布特征超几何分布特征 Characteristics均值均值标准差标准差E Xnp 对于对于

15、p=A/N npp()()1NnN1超几何分布的例子超几何分布的例子一个班有10个孩子，其中4 人患了感冒。试计算从该班抽取一个包括3 个孩子的样本中有2 个孩子患感冒的概率?0.30X)42(10-4 3-210 3()()()P协方差协方差 Covariance=(Xi XYiYP(Xi Yi)I=1N度量两个随机变量的相关程度Measures the Joint Variability of Two Random VariablesXY两个随机变量的和两个随机变量的和 Sum of Two Random VariablesE(X+Y)=E(X)+E(Y)Var(X+Y)=2 两个随机变量

16、的和的期望值两个随机变量的和的方差2 YXY2XX+YExample:The probability that one unit normally uses water is 0.75.Compute the probability:that there are 4 days that use water normally in the recent 6 days(using water independently everyday).Example:20 machines work independently.It has been known that the probability th

17、at every machine breaks down in one hour is 0.01.Compute the probability:There are 2 breaking down in 20 machines in one hour.If there are two persons who are looking after machines,compute the probability that at least one person is idle.Thinking ChallengeAbout the approximation relationship among

18、the Hypergeometric,Binomial and Poisson Distribution (1)When N is large and n is small(n/N0.1),the Binomial Distribution can approximate the Hypergeometric Distribution（we order P=M/N）(2)When n is large and p is small,the Poisson can approximate the Binomial Distribution（we take=np）。）。（n100，p0.01,np

19、=1，0.1np10)Special FocusqThe Binomial Distribution is used for determining the probability that there have x times success in a n times experiment,it has the properties as follows:1.The experiment consists of n same sequences.2.Every experiment only has two outcomes:success and failure.3.Every exper

20、iment has the same successful probability p.4.Every experiment is independent of others.qThe Poisson Distribution is used for determining the probability that an event occurs x times in a section of time or space.It needs to suppose that 1.In any two equal intervals,the probability that the event occurs one time is equal.2.In different intervals,whether the event occurs independently.Binominal Distribution vs.Poisson Distribution