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Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore Information1Chapter 1The Poisson distributionThis chapter introduces a discrete probability distribution which is used for modellingrandom events. When you have completed it you should be able to calculate probabilities for the Poisson distributionunderstand the relevance of the Poisson distribution to the distribution of random events and usethe Poisson distribution as a modelbe able to use the result that the mean and variance of a Poisson distribution are equalbe able to use the Poisson distribution as an approximation to the binomial distribution whereappropriatebe able to use the normal distribution, with a continuity correction, as an approximation to thePoisson distribution where appropriate. in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationCambridge International AS and A Level Mathematics: Statistics 21.1 The Poisson probability formulaSituations often arise where the variable of interest is the number of occurrences ofa particular event in a given interval of space or time. An example is given in Table1.1. This shows the frequency of 0, 1, 2 etc. phone calls arriving at a switchboard in100 consecutive time intervals of 5 minutes. In this case the ‘event’ is the arrival of aphone call and the ‘given interval’ is a time interval of 5 minutes.Table 1.1 Frequency distribution of number of telephone calls in 5-minute intervalsNumber of callsFrequency07112324324 or more0Some other examples are the number of cars passing a point on a road in a time interval of 1 minute, the number of misprints on each page of a book, the number of radioactive particles emitted by a radioactive source in a timeinterval of 1 second.Further examples can be found in the practical activities in Section 1.4.The probability distribution which is used to model these situations is called thePoisson distribution after the French mathematician and physicist Siméon-DenisPoisson (1781–1840). The distribution is deined by the probability formula2) e λP(λx,x!x 0,1, 2,. .This formula involves the mathematical constant e which you may have already met in unit P3.If you have not, then it is enough for you to know at this stage that the approximate value of e is2.718 and that powers of e can be found using your calculator.Check that you can use your calculator to show that e–2 0.135. and e–0.1 0.904.The method by which Poisson arrived at this formula will be outlined in Section 1.2.This formula involves only one parameter, λ. (λ, pronounced ‘lambda’, is the Greekletter l.) You will see later that λ is the mean of the distribution. The notation forindicating that a random variable X has a Poisson distribution with mean λ isX Po(λ). Once λ is known you can calculate P(X 0), P(X 1) etc. There is noupper limit on the value of X.EXAMPLE 1.1.1The number of particles emitted per second by a radioactive source has aPoisson distribution with mean 5. Calculate the probabilities ofa 0,b 1,c 2,d 3 or more emissions in a time interval of 1 second.a Let X be the random variable ‘the number of particles emitted in1 second’. Then X Po(5). Using the Poisson probability formulaλx50with λ 5, P(P() e λ) e 56737.673767 7 0.00674 ,x!0!correct to 3 signiicant igures.Recall that 0! 1 (see P1 Section 8.3). in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationChapter 1: The Poisson distribution51 0.033 6868. 0.0337, correct to 3 signiicant igures.1!52) e 5 0.084 2222. 0.008 42, correct to 3 signiicant igures.c P(2!d Since there is no upper limit on the value of X the probability of 3 ormore emissions must be found by subtraction.) e 5b P() 1 P(() P() P(() 16 737.737 0.033 6868. 0.0848 22. 0.875, correct to 3 significant figures.P(EXAMPLE 1.1.2The number of demands for taxis to a taxi irm is Poisson distributed with, onaverage, four demands every 30 minutes. Find the probabilities ofa no demand in 30 minutes,b 1 demand in 1 hour,c fewer than 2 demands in 15 minutes.a Let X be the random variable ‘the number of demands in a 30-minuteinterval’. Then X Po(4). Using the Poisson formula with λ 4,40P() e 4 0.0183, correct to 3 signiicant igures.0!b Let Y be the random variable ‘the number of demands in a one-hourinterval’. As the time interval being considered has changed from 30minutes to 1 hour, you must change the value of λ to equal the meanfor this new time interval, that is to 8, giving Y Po(8). Using thePoisson formula with λ 8,81P() e 8 0.002 68 , correct to 3 signiicant igures.1!c Again, the time interval has been altered. Now the appropriate valuefor λ is 2.3Let W be the number of demands in 15 minutes. Then W Po(2).P(W2) P() P(1) e 222021 e 20!1!0 406 ,correct to 3 signiicant igures.Here is a summary of the results of this section.The Poisson distribution is used as a model for the number, X, of events in agiven interval of space or time. It has the probability formulaλxP(X x)x e λ,0,1,2,.,x!where λ is equal to the mean number of events in the given interval.The notation X Po(λ) indicates that X has a Poisson distribution with mean λ.Some books use µ rather than λ to denote the parameter of a Poisson distribution. Bothalternatives are referred to in the syllabus. in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationCambridge International AS and A Level Mathematics: Statistics 2Exercise 1A1 The random variable T has a Poisson distribution with mean 3. Calculatea P(T 2),b P(T 1),c P(T 3).2 Given that U Po(3.25), calculatea P(U 3),b P(U 2),c P(U 2).3 The random variable W has a Poisson distribution with mean 2.4. Calculatea P(W 3),b P(W 2),c P(W 3).4 Accidents on a busy urban road occur at a mean rate of 2 per week. Assumingthat the number of accidents per week follows a Poisson distribution, calculate theprobability thata there will be no accidents in a particular week,b there will be exactly 2 accidents in a particular week,c there will be fewer than 3 accidents in a given two-week period.5 On average, 15 customers a minute arrive at the check-outs of a busy supermarket.Assuming that a Poisson distribution is appropriate, calculatea the probability that no customers arrive at the check-outs in a given 10-secondinterval,4b the probability that more than 3 customers arrive at the check-outs in a15-second interval,6 During April of this year, Malik received 15 telephone calls. Assuming that thenumber of telephone calls he receives in April of next year follows a Poissondistribution with the same mean number of calls per day, calculate the probabilitythata on a given day in April next year he will receive no telephone calls,b in a given 7-day week next April he will receive more than 3 telephone calls.7 Assume that cars pass under a bridge at a rate of 100 per hour and that a Poissondistribution is appropriate.a What is the probability that during a 3-minute period no cars will pass underthe bridge?b What time interval is such that the probability is at least 0.25 that no car willpass under the bridge during that interval?8 A radioactive source emits particles at an average rate of 1 per second. Assumethat the number of emissions follows a Poisson distribution.a Calculate the probability that 0 or 1 particle will be emitted in 4 seconds.b The emission rate changes such that the probability of 0 or 1 emission in 4seconds becomes 0.8. What is the new emission rate? in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationChapter 1: The Poisson distribution1.2 Modelling random eventsThe examples which you have already met in this chapter have assumed that thevariable you are dealing with has a Poisson distribution. How can you decide whetherthe Poisson distribution is a suitable model if you are not told? The answer to thisquestion can be found by considering the way in which the Poisson distribution isrelated to the binomial distribution in the situation where the number of trials is verylarge and the probability of success is very small.Table 1.2 reproduces Table 1.1 giving the frequency distribution of phone calls in 1005-minute intervals.Table 1.2 Frequency distribution of number of telephone calls in 5-minute intervalsNumber of callsFrequency07112324324 or more0If these calls were plotted on a time axis you might see something which looked likeFig. 1.3.0510 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120Fig. 1.3 Times of arrival of telephone calls at a switchboardThe time axis has been divided into 5-minute intervals (only 24 are shown) and theseintervals can contain 0, 1, 2 etc. phone calls. Suppose now that you assume that thephone calls occur independently of each other and randomly in time. In order to makethe terms in italics clearer, consider the following. Imagine the time axis is dividedup into very small intervals of width δt (where δ is used in the same way as it is in puremathematics). These intervals are so small that they never contain more than onecall. If the calls are random then the probability that one of these intervals containsa call does not depend on which interval is considered; that is, it is constant. If thecalls are independent then whether or not one interval contains a call has no effect onwhether any other interval contains a call.5Looking at each interval of width δt in turn to see whether it contains a call ornot gives a series of trials, each with two possible outcomes. This is just the kind ofsituation which is described by the binomial distribution (see S1 Chapter 7). Thesetrials also satisfy the conditions for the binomial distribution that they should beindependent and have a ixed probability of success.Suppose that a 5-minute interval contains n intervals of width δt. If there are, onaverage, λ calls every 5 minutes then the proportion of intervals which contain a callwill be equal to λ . The probability, p, that one of these intervals contains ancall is therefore equal to λ . Since δt is small, n is large and λ is small. You can verifynnfrom Table 1.2 that the mean number of calls in a 5-minute interval is 0.37 so the 0.37 distribution of X, the number of calls in a 5-minute interval, is B n ,. n n ) p xq n x , youFinding P(X 0) Using the binomial probability formula P( x can calculate, for example, the probability of zero calls in a 5-minute interval as0P(n n 0.37 0.37 ) 1 .n x n in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationCambridge International AS and A Level Mathematics: Statistics 2In order to proceed you need a value for n. Recall that n must be large enough toensure that the δt intervals never contain more than one call. Suppose n 1000. Thisgives0P( 1000 0.37 0.37 ) 1 010001000 1000 0.690 68. .However, even with such a large number of intervals there is still a chance that oneof the δt intervals could contain more than one call, so a larger value of n would bebetter. Try n 10 000, giving0P( 10 000 0.37 0.37 ) 1 01000010000 10 000 0.69072. .Explore for yourself what happens as you increase the value of n still further. Youshould ind that your answers tend towards the value 0.690 73 . This is equal to e–0.37,which is the value the Poisson probability formula gives for P(X 0) when λ 0.37.n x This is an example of the general result that 1 tends to the value e–x as n tends to n ininity.Provided that two events cannot occur simultaneously, allowing n to tend to ininitywill ensure that not more than one event can occur in a δt interval.Finding P(X 1) In a similar way you can ind the probability of one call in a5-minute interval by starting from the binomial formula and allowing n to increase asfollows.61P( n 0.37 0.3737 ) 1 n 1 n n 1 0.37 0.373 1 n n 1.Putting n 1000,P(0 37 1) 0 37 1 1000 999 0.37 0.690 94. 0.25564. .Putting n 10 000,P( 0 37 1) 0 37 1 10 000 9999 0.37 0.69075. 0.255579. .Again, you should ind that, as n increases, the probability tends towards the valuegiven by the Poisson probability formula,P(.3737 e 0.37 0.25557. .1)Finding P(X 2), P(X 3), etc. You could verify for yourself that similar results areobtained when the probabilities of X 2, 3, etc. are calculated by a similar method.A spreadsheet program or a programmable calculator would be helpful. λ The general result for P(X x) can be derived as follows. Starting with X B n , n , xP(X n λ λ x) 1 x n n n x n(()().(n x 1)) λx λ x 1 x!n n λx n n 2n x λ . 1 x!nnnn in this web service Cambridge University Pressn xn x.www.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationChapter 1: The Poisson distributionn 1 n 2,, etc. tend towards 1. Thennn xnterm 1 λ can be approximated by 1 λ since x, a constant, is negligible comparedn n with n and, as you have seen previously, this tends towards e–λ.Now consider what happens as n gets larger. The fractionsCombining these results givesP()λx λe .x!The assumptions made in the derivation above give the conditions that a set of eventsmust satisfy for the Poisson distribution to be a suitable model. They are listed below.The Poisson distribution is a suitable model for events which occur randomly in space or time, occur singly, that is events cannot occur simultaneously, occur independently, and occur at a constant rate, that is the mean number of events in a given timeinterval is proportional to the size of the interval.7EXAMPLE 1.2.1For each of the following situations state whether the Poisson distributionwould provide a suitable model. Give reasons for your answers.a The number of cars per minute passing under a road bridge between 10 a.m.and 11 a.m. when the trafic is lowing freely.b The number of cars per minute entering a city-centre car park on a busySaturday between 9 a.m. and 10 a.m.c The number of particles emitted per second by a radioactive source.d The number of currants in buns sold at a particular baker’s shop on aparticular day.e The number of blood cells per ml in a dilute solution of blood which hasbeen left standing for 24 hours.fThe number of blood cells per ml in a well-shaken dilute solution of blood.a The Poisson distribution should be a good model for this situation as theappropriate conditions should be met: since the trafic is lowing freelythe cars should pass independently and at random; it is not possible forcars to pass simultaneously; the average rate of trafic low is likely to beconstant over the time interval given.b The Poisson distribution is unlikely to be a good model: if it is a busy daythe cars will be queuing for the car park and so they will not be movingindependently. in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationCambridge International AS and A Level Mathematics: Statistics 2c The Poisson distribution should be a good model provided that the timeperiod over which the measurements are made is much longer than thelifetime of the source: this will ensure that the average rate at whichthe particles are emitted is constant. Radioactive particles are emittedindependently and at random and, for practical purposes, they can beconsidered to be emitted singly.d The Poisson distribution should be a good model provided that thefollowing conditions are met: all the buns are prepared from the samemixture so that the average number of currants per bun is constant; themixture is well stirred so that the currants are distributed at random; thecurrants do not stick to each other or touch each other so that they arepositioned independently.e The Poisson distribution will not be a good model because the blood cellswill have tended to sink towards the bottom of the solution. Thus the averagenumber of blood cells per ml will be greater at the bottom than the top.fIf the solution has been well shaken the Poisson distribution will be asuitable model. The blood cells will be distributed at random and at aconstant average rate. Since the solution is dilute the blood cells will notbe touching and so will be positioned independently.1.3 The variance of a Poisson distribution8In Section 1.2 the Poisson probability formula was deduced from the distribution λ of X B n , n by considering what happens as n tends to ininity. The variance of a Poisson distribution can be obtained by considering what happens to the variance λ of the distribution of X B n , n as n gets very large. In S1 Section 8.3 you met the formula Var(X) npq for the variance of a binomial distribution. Substituting for pand q gives λ λ λ 1 λ 1 .n n n λAs n gets very large the term tends to zero. This gives λ as the variance of thenPoisson distribution. Thus the Poisson distribution has the interesting property thatits mean and variance are equal.Var( ) n For a Poisson distribution X Po(λ)mean µ E(X ) λ,variance σ2 Var(X ) λ.The mean and variance of a Poisson distribution are equal.The equality of the mean and variance of a Poisson distribution gives a simple way oftesting whether a variable might be modelled by a Poisson distribution. The mean ofthe data in Table 1.2 has already been used and is equal to 0.37. You can verify that thevariance of these data is 0.4331. These values, which are both 0.4 to 1 decimal place, are in this web service Cambridge University Presswww.cambridge.org

Cambridge University Press978-1-316-60042-9 — Cambridge International AS and A Level Mathematics: Statistics 2 CoursebookSteve Dobbs , Jane Miller , Julian GilbeyExcerptMore InformationChapter 1: The Poisson distributionsuficiently close to indicate that the Poisson distribution may be a suitable model forthe number of phone calls in a 5-minute interval. This is conirmed by Table 1.4, whichshows that the relative frequencies calculated from Table 1.2 are close to the theoreticalprobabilities found by assuming that X Po(0.37). (The values for the probabilities aregiven to 3 decimal places and the value for P(X 4) has been found by subtraction.)Note that if the mean and variance are not approximately equal then the Poissondistribution is not a suitable model. If they are equal then the Poisson distributionmay be a suitable model, but is not necessarily so.Table 1.4 Comparison of theoretical Poisson probabilities and relative frequenciesfor the data in Table 1.2xFrequencyRelative frequencyP(X x)0710.71e 0.6911230.23–0.370.37 0.256240.04320.02 4Totals–0.37ee0 37e0 370.37 22!470.37 33!06000100119Exercise 1B1 For each of the following situations, say whether or not the Poisson distributionmight provide a suitable model.a The number of raindrops that fall onto an area of ground of 1 cm2 in a periodof 1 minute during a shower.b The number of occupants of vehicles that pass a given point on a busy road in1 minute.c The number of laws in a given length of material of constant width.d The number of claims made to an insurance company in a month.2 Weeds grow on a large lawn at an average rate of 5 per square metre. A particularmetre square is considered and subdivided into smaller and smaller squares. Copyand complete the table below, assuming that no more than 1 weed can grow in asub-division.Number ofsub-divisions100P(a sub-division containsa weed)P(no weeds in a given squaremetre)5 0 051000.95100 0.005 92110 0005 10 0001 000 000100 000 000Compare your answers to the probability of no weeds in a given square metre,given by the Poisson probability formula. in this web service Cambridge University Presswww.cambridge.org