概率统计第2章课件.ppt

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1、2.1 Random VariablesExample 1 Toss coin:1HTRSExample 2 Test the life in years of light bulbs:Definition 2.1Let S be the sample space associated with a particular experiment.A single-valued function X assigning toevery element a real number,X(),is called a random variable.Denoted by X.2In general,Def

2、inition 2.1Let S be the sample space associated with a particular experiment.A single-valued function X assigning to every element a real number,X(),is called a random variable.Denoted by X.and x,y,zrepresent a real number.we use X,Y,Z.represent a random variableNotice that RX is always a set of rea

3、l numbers.Definition 2.2 The set of all possible values of X is called the rangespace of X and is denoted by RX.3Definition 2.2 The set of all possible values of X is called the rangespace of X and is denoted by RX.Notice that RX is always a set of real numbers.For above example,4For above example,5

4、Random variable could take different values depending on different random experiments.Because the experiment results show up randomly the random variable could take values depending on certain kind of probability.(2)The way to take the values for random variable obeys kind of probability rule.Random

5、 variable is kind of function,but it is essentially different to the other general functions.The later kind of functions are defined on real number set while random variables are defined on the sample space whose elements would not all be real numbers.2.NotesNotes(1)Random variable is different to t

6、he common function6The concept of random event is contained within the concept of random variable,which is more extended.From another point of view we can say random event is to search the random phenomena by a static method while random variable is to do so by a dynamic way.(3)The relationship betw

7、een random event&variable7Categories of random variableCategories of random variableDiscrete Observe the number displayed on a rolling dice.Possible values for a random variable X:Random VariableContinuouse.g.11,2,3,4,5,6.Non-discreteOthers(1)Discrete if the number of values a random variable could

8、take is finite or countable infinite then this variable is called discrete random variable.89e.g.2 Let X be a random variable representing“The number of shootings as one shoots continuously until the target is shot.”.Then the possible values X could take:e.g.3 If the probability for one shooter to s

9、hoot the target is 0.8,now he has shot 30 times and let X be the random variable to represent“The number of shootings that are shot on the target”,Then the possible values X could take:10e.g.2 Random variable X represents“The measuring errors for some machine parts”.Then the values X could take is (

10、a,b).e.g.1 Random variable X represents“The length of life for a lamp”.(2)Continuous If all possible values a random variable could take will fully fill in an interval on the axis,this variable will be call a continuous random variable.Then the values X could take is11Summary2.Two ways to classify r

11、andom variable:discrete、continuous.1.Probability theory quantitatively examines the inherent pattern of random phenomena,thus in order to effectively search into random phenomena,we must quantify random events.When representing some non-numerical random event with numbers，the concept of random varia

12、ble is established.Therefore,random variable is defined as a special function in the sample space.2.2 Discrete random variables Definition 2.3With each possible outcome,we associate aNumbercalled the probability of xiThe numbersmust satisfy(i)(ii)A random variable X is said to be a discrete randomva

13、riable if its range space is either finite or countablyinfinite,i.e.12The numbersmust satisfy(i)(ii)Definition 2.4 The function p is called the probability mass functioncalled the probability distribution of X.(pmf)and the collection of pairs13Example 2.1Solution:Let X be a random variable whose val

14、ues xare the possible numbers of defectivecomputers Purchased by the school.Then x can be any of the numbers 0,1,and 2.A shipment of 8 similar microcomputers to a retailoutlet contains 3 that are defective.a random purchase of 2 of these computers,find theprobability distribution for the number of d

15、efectives.If a school makes14Solution:Let X be a random variable whose values xare the possible numbers of defectivecomputers Purchased by the school.Then x can be any of the numbers 0,1,and 2.Thus the probability distribution of X is x 0 1 2p(x)152.3 Some Important Discrete Probability Distribution

16、sUniformWe have a finite set of outcomeswhich has the same probability of occurring(equally likely outcomes).X is said to have a Uniform distribution and weeach ofWriteSo16X is said to have a Uniform distribution and weWritebulb,Example 2.2When a light bulb is selected at random from a boxthat conta

17、ins a 40-watt bulb,a 60-watt bulb,a 75-wattand a 100-watt bulb.FindSolution:each element of the sample space occurs withprobability 1/4.Therefore,we have a uniform distribution:17Solution:each element of the sample space occurs withprobability 1/4.Therefore,we have a uniform distribution:Example 2.3

18、When a die is tossed,S=1,2,3,4,5,6.P(each element of the sample space)=1/6.Therefore,we have a uniform distribution,with 18Example 2.3When a die is tossed,S=1,2,3,4,5,6.P(each element of the sample space)=1/6.Therefore,we have a uniform distribution,with Bernoulli trialA Bernoulli trial is an experi

19、ment which has twoLet p=P(success),q=P(failure)(q=1-p).success and failure.possible outcomes:19The pmf of X is Bernoulli trialA Bernoulli trial is an experiment which has twoLet p=P(success),q=P(failure)(q=1-p).success and failure.possible outcomes:or 20Bernoulli materialBinomialeach of which must r

20、esult in either a success with Consider a sequence of n independent Bernoulli trialsprobability of p or a failure with probability q=1-p.Let X=the total number of successes in these n trialso that X is said to have a Binomial distribution with parametersP(the total number of x successes)=n and p and

21、 we write XBin(n,p)or Xb(x;n,p)Special case,21X is said to have a Binomial distribution with parametersn and p and we write XBin(n,p)or Xb(x;n,p)Special case,when n=1,we have We write B(n,p)b(1,p)22Binomial Distribution23The probability that a certain kind of component will survive a given shock tes

22、t is 3/4.Find the probability that exactly 2 of the next 4 components tested survive.Example 2.4 Assuming that the tests are independent andSolution:p=3/4 for each of the 4 tests,we obtainExample 2.5The probability that a patient recovers from a rare blooddisease is 0.4.this disease,survive,If 15 pe

23、ople are known to have contractedwhat is the probability that(a)at least 10(b)from 3 to 8 survive,and(c)exactly 5 survive?24The probability that a patient recovers from a rare blood disease is 0.4.If 15 people are known to have contracted this disease,what is the probability that(a)at least 10 survi

24、ve,(b)from 3 to 8 survive,and(c)exactly 5 survive?Example 2.5Solution：Let X=the number of people that survive.(a)(b)25(c)(b)Example 2.6(a)The inspector of the retailer randomly picks 20 itemsA large chain retailer purchases a certain kind ofelectronic device from a manufacturer.indicates that the de

25、fective rate of the device is 3%.The manufacturer26from a shipment.(b)Suppose that the retailer receives 10 shipments in aA large chain retailer purchases a certain kind of electronic device from a manufacturer.The manufacturer indicates that the defective rate of the device is 3%.Example 2.6(a)The

26、inspector of the retailer randomly picks 20 items Solution：(a)Denote by X the number of defectiveDevices among the 20.Then this X follows a b(x;20,0.03).will be at least one defective item among these 20?What is the probability that thereshipments containing at least one defective device?shipment.mo

27、nth and the inspector randomly tests 20 devices perWhat is the probability that there will be 3Hence27 Solution：(a)Denote by X the number of defectiveDevices among the 20.Then this X follows a b(x;20,0.03).Hence(b)Assuming the independence from shipment to Therefore,shipment and denoting by Y.Y=the

28、number of shipments containing at least onedefective.Then Yb(y;10,0.4562).28(b)Assuming the independence from shipment to Therefore,shipment and denoting by Y.Y=the number of shipments containing at least onedefective.Then Yb(y;10,0.4562).PoissonThe pmf of a random variable X which has a Poissondist

29、ribution with parameter is given by 29PoissonThe pmf of a random variable X which has a Poissonand we writedistribution with parameter is given by 30Poisson material31Number of telephone ringsNumber of traffic accidentNumber of customers at receptionEarthquakeVolcanic EruptionMass floodingPoisson di

30、stribution32单击图形播放单击图形播放/暂停暂停ESCESC键退出键退出Binomial distribution Poisson distribution33During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4.What is the probability that 6 articles enter the counter in a given millisecond?Example 2.7

31、Solution：Using the Poisson distribution with x=6 and,We find from Table 1 thatExample 2.8Ten is the average number of oil tankers arriving eachday a certain port city.The facilities at the port can34handle at most 15 tankers per day.Example 2.8Ten is the average number of oil tankers arriving each d

32、ay a certain port city.The facilities at the port canSolution：Let X be the number of tankers arriving eachday.Then,What is the probability that on a given day tankershave to be turned away?using Table,we have 352.4 Cumulative Distribution FunctionsDefinition 2.5 The cumulative distribution function(

33、cdf)of the random variable X is defined to beand is denoted by F(x).Properties of F(x):(i)F is non-decreasing.(ii)i.e.i.e.if36Properties of F(x):(i)F is non-decreasing.i.e.if(ii)(iii)F is right continuous.(iv)F(x)is defined for all real numbers x.The cdf of a discrete random variable X is a stepfunc

34、tion with jumps at thei.e.37The cdf of a discrete random variable X is a stepfunction with jumps at the38e.g.39e.g.Example 2.9The pmf of X is40The pmf of X isFind:1)The Cumulative distribution function of X.2)X -1 2 3Solution:1)-1 2 3Example 2.941Solution:1)-1 2 32)422)-1 0 1 2 31432.5 Continuous Ra

35、ndom VariablesDefinition 2.6 X is a continuous random variable if there exists aThe function f is called the probability density function with the property that for every subset of realnonnegative function f defined for all real xnumbers B(pdf)of X.Properties of the pdf(i)44Properties of the pdf(i)T

36、his follows by setting B=(ii)145This follows from(iii)If we let B=a,b thenThis follows from(iv)This follows from461(iv)This follows fromNotes(a)If X is continuous then F(x)is continuous.Also,(b)P(aXb)represents thebetween x=a and x=b.area under the graph of f f(x)=F(x)at all point where F is continu

37、ous.47f(x)=F(x)(c)The meaning of density function:i.e.The probability that X is in a small intervalis approximately equal to f(x)times the width of theinterval(d)For any specified value of X,say x0,we have.(b)P(aXb)represents thebetween x=a and x=b.area under the graph of f 48i.e.The probability tha

38、t X is in a small intervalis approximately equal to f(x)times the width of theinterval(d)For any specified value of X,say x0,we have.Hence,thenif X is continuous then the probabilities49 Example 2.10 Let X be a continuous r.v.with.pdf FindSolution:502.6 Some Continuous Probability DistributionsUnifo

39、rm(or rectangular)DistributionA uniform random variable X on the interval(a,b)hasprobability density function(pdf)We write XU(a,b).51Cdf:52Example 2.11 Suppose that a large conference room for a certain company can be reserved for no more than 4 hours,However,the use of the conference room is such t

40、hatNormal DistributionThe pdf of a Normal random variable,X,withis given byparametersboth long and short conferences occur quite often.In fact,it can be assumed that length X of a conference has a uniform distribution on the interval 0,4.(a)What is the probability density function?(b)What is the pro

41、bability that any given conferencelasts at least 3 hours?53Normal DistributionThe pdf of a Normal random variable,X,withis given byparametersWe write54Geometric Characteristics of the density for Normal Distribution 555657Cdf:58 Normal Distribution is one of the most important andCommonly observed d

42、istribution.For example,measuring uncertainty,human physical characteristics such as height,weight,etc Measurements for manufactured products made under sameConditions,i.e.length,diameter,mass,height,etc,all seem to obey normal distribution.Application and Backgroundof the Normal Distribution Gaussi

43、an material59In the special case when and thedistribution is called the standard Normal distribution.i.e.We write60For a standard Normal random variable Xdistribution is called the standard Normal distribution.i.e.We write61Example 2.12Given a standard normal distribution，find the areaunder the curv

44、e that lies(a)to the right of z=1.84(b)between z=-1.97 and z=0.86.Example 2.13 Given a standard normal distribution，find the valueof k such that62Figure 2.10 Areas for Example2.16(a)P(Zk)=0.3015，Example 2.13 Given a standard normal distribution，find the value of k such thatA Theorem:IfThenand(b)P(kZ

45、-0.18)=0.4197.63For a Normal random variable XA Theorem:IfThenExample 2.1464Example 2.14Given a random variable X having a normaldistribution with，Find theprobability that X assumes a value between 45 and 62.Example 2.15Given that X has a normal distribution withand，find the probability that X assum

46、es avalue greater than 362.Exponential DistributionThe pdf of an Exponential random variable withParameter is given by65Exponential DistributionThe pdf of an Exponential random variable withParameter is given byWe write66 Some components or devices have a life span that obeys the exponential distrib

47、ution.For example,life span of mobile devices,electric power devices,animals,and others all obey this exponential distribution.Application and BackgroundCdf:67If 5 of these components are installed in differentSuppose that a system contains a certain type of component whose time in years to failure

48、is given by T.The random variable T is modeled nicely by the Example 2.16 exponential distribution with mean time to failure Systems what is the probability that at least 2 are stillfunctioning at the end of 8 years?682.7 Functions of Random Variablesa)when the r.v.x is discreteThe pmf of Y takes on

49、 a given value,say yj,isSuppose the r.v.x has the probability distribution belowExample 2.17xi 0 1 2 3 4 5P(xi)p(0)p(1)p(2)p(3)p(4)p(5)Let Y=(X-2)2,Find the pmf of the r.v.Y.69Suppose the r.v.x has the probability distribution belowExample 2.17xi 0 1 2 3 4 5P(xi)p(0)p(1)p(2)p(3)p(4)p(5)Let Y=(X-2)2,

50、Find the pmf of the r.v.Y.Solution:0 1 4 9 p(2)p(1)+p(3)p(0)+p(4)p(5)The cdf of Y isb)when the r.v.X is continuous70calculated by takingThe cdf of Y isb)when the r.v.X is continuousOnce the cdf FY(y)is obtained,Example 2.18 Suppose XU(0,1)and let Y=X2,Find the cdf and pdf of Y.suppose X is continuou