# 6 POISSON DISTRIBUTIONS

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1 6 POISSON DISTRIBUTIONS Chapter 6 Poisson Distributions Objectives After studying this chapter you should be able to recognise when to use the Poisson distribution; be able to apply the Poisson distribution to a variety of problems; be able to approximate the binomial distribution by a suitable Poisson distribution. 6.0 Introduction This distribution is introduced through the Activity below. Activity 1 Vehicle survey The survey should be carried out on a motorway or dual carriageway, well away from any obstacles that would prevent free flow of traffic such as road works, roundabouts or traffic lights. (A useful, but not essential, piece of apparatus is something that will 'beep' at the end of each minute. Some watches with alarms can be set to go off every minute, or perhaps your chess club has a lightning buzzer. It is something that can be made without much difficulty by an electronics student.) Note the number of vehicles that pass you in one minute and also the number of lorries. Collect readings for 100 minutes in a period which is either wholly in or completely outside any 'rush hour' surge of traffic that might exist. For each of your distributions calculate: (a) mean, E( X) (b) variance, V( X) where X is ' number of vehicles (or lorries) passing in one minute'. (If several people are involved then the far carriageway can be studied separately. The 'number of red cars' is another example of a variable that could be examined.) 115

2 Chapter 6 Poisson Distribution The Poisson distribution,which is developed in the next section, is of particular use when the number of possible occurrences of an event is unlimited. Possible examples are when describing the number of: (a) (b) (c) flaws in a given length of material; accidents on a particular stretch of road in a week; telephone calls made to a switchboard in one day. 6.1 Developing the distribution Surveys of the type undertaken in Activity 1 are important for transport planners who have to make decisions about road building schemes. For example, in considering the need for an extra lane on a dual carriageway, a survey of the number of lorries over several days gave the number passing a point per minute during the evening rush hour, as in the table opposite. Here is an example of a random variable X, 'the number of lorries per minute', which is certainly going to produce a discrete probability distribution, but each one minute trial will have many possible outcomes. This cannot be a binomial distribution since, in theory at least, the possible values of X are unlimited. Now the mean of the data is given by x = No. of lorries per minute Frequency ( x ) ( f ) whilst the variance is s 2 = So allowing a little for experimental error, it seems that the distribution has its mean equal to its variance. A relationship between succeeding frequencies can be seen by dividing consecutive data = = = =

3 Chapter 6 Poisson Distributions The initial probability, P( X =0) = , can then be used to calculate the others. P( X =1) = 5 1 P( X =0) P( X =2) = 5 2 P( X =1) = P( X =0) P( X =3) = 5 3 P( X =2) = P( X =0) P( X = 4) = P( X =0). Hence the probability distribution can be written P( X = n) = 5 n nn 1 ( ) P( X =0) = 5n n! P ( X =0 ), using factorials. Since the sum of the probabilities is one, putting p = P( X =0), it follows that 1 = p + 5p + 52 p + 53 p 3! + 54 p 4! +... = p ! ! +... p = e 5. = pe 5, since e 5 = , The exponential number e is a very important number in advanced mathematical analysis and can be found on all scientific calculators. The exponential function, e x, takes the form e x = 1 + x + x 2 + x3 3! +... and it is this result that has been used above with x =

4 Chapter 6 Poisson Distribution Activity 2 Telephone calls Many school and college switchboards are computerised and can provide a print-out of calls over any period. Other switchboards in local firms may also be able to help. Study the number of incoming calls in, for example, ten minute periods, during a time of day avoiding lunch and other breaks. Look at the results for several days. Calculate the mean and variance of your distribution and try to fit a Poisson distribution to your figures. Activity 3 As an alternative or additional practical to Activity 2, study the number of arrivals of customers at a post office in two minute intervals. The length of the time interval may well be shortened in the case of a large and busy site. The key parameter in fitting a Poisson distribution is the mean value, usually denoted by λ. This is the average number of occurrences in the specified period (e.g. cars passing in a minute). The probability distribution is then given by: x P( X = x) e λ λ e λ λ 2 e λ λ 3 e λ 3! λ 4 e λ 4!... In general, if X is a Poisson distribution, then P( X = x) = λx e λ x! (x = 0, 1, 2,...) and this is denoted by X ~ Po( λ ). The Poisson distribution was first derived in 1837 by the French mathematician Simeon Denis Poisson whose main work was on the mathematical theory of electricity and magnetism. The distribution arises when the events being counted occur (a) (b) independently; such that the probability that two or more events occur simultaneously is zero; 118

5 Chapter 6 Poisson Distributions (c) (d) randomly in time or space; uniformly (that is, the mean number of events in an interval is directly proportional to the length of the interval). Example If the random variable X follows a Poisson distribution with mean 3.4, find P X = 6 ( ). Solution This can be written more quickly as: if X ~ Po 3. 4 P( X = 6). Now P( X =6) = e λ λ 6 6! ( ) find = e 3.4 ( 3. 4) 6 6! ( mean, λ =3.4) = = (to 3 d.p.). Example The number of industrial injuries per working week in a particular factory is known to follow a Poisson distribution with mean 0.5. Find the probability that (a) (b) in a particular week there will be: (i) less than 2 accidents, (ii) more than 2 accidents; in a three week period there will be no accidents. Solution Let A be 'the number of accidents in one week', so A ~ P 0 (0.5). (a) (i) P( A<2) = P( A 1) = (from tables in Appendix 3 (p257), to 4 d.p.) or, from the formula, P( A<2) = P( A=0) + P( A=1) = e e ! 119

6 Chapter 6 Poisson Distribution = 3 2 e (ii) P( A>2) = 1 P( A 2) or 1 P A=0 = ( from tables) = ( to 4 d. p. ) [ ( ) + P( A=1) + P( A=2) ] ( ) 2 = 1 e e e = 1 e 0.5 ( ) = e (b) P (0 in 3 weeks) = ( e 0.5 ) Could the number of vehicles on a single carriageway road passing a fixed point in some time interval be expected to follow a Poisson distribution? Exercise 6A Give answers to 3 significant figures. 1. If X ~ Po( 3), find: (a) P( X =2) (b) P( X =3) (c) P( X 5) (d) P( X <3). 2. If X ~ Po λ and P( X =5). ( ) and P( X = 4) = 3P X =3 ( ), find λ 3. If X ~ Po( λ ) and P( X =0) = 0.323, find the value of λ to two decimal places and use this to calculate P( X =3). 4. Investigate whether or not the following figures might result from a Poisson variable. x P( X = x) *5. If X ~ Po( 2), Y~Po( 3) and Z ~ Po( 5), find: (a) P( X+Y =0) (b) P( X+Y =1) (c) P( Z=0) (d) P( Z=1) (e) P( X + Y 2) (f) PZ 2 ( ) 120

7 Chapter 6 Poisson Distributions 6.2 Combining Poisson variables Activity 4 The number of telephone calls made by the male and female sections of the P.E. department were noted for fifty days and the results are shown in the table opposite. The number of calls by men are given first in each pair of numbers. Investigate the distributions of the numbers of calls made: (a) (b) (c) by the male staff; by the female staff; in total each day i.e = 2, = 4, etc. 0, 2 2, 2 6, 0 3, 5 1, 2 2, 2 1, 1 2, 2 1, 1 2, 3 7, 0 1, 4 3, 6 2, 3 3, 0 4, 1 5, 1 4, 3 5, 4 6, 4 1, 0 2, 3 3, 2 3, 3 6, 1 2, 3 2, 2 2, 1 3, 5 5, 3 4, 3 4, 2 3, 4 4, 3 3, 1 3, 1 3, 3 4, 4 5, 4 2, 1 5, 6 1, 2 2, 2 1, 2 3, 3 4, 2 0, 5 4, 4 2, 2 4, 1 Number of telephone calls ( male, female ) Activity 5 Study the numbers of cars and lorries from your survey in Activity 1and look at the distributions of the numbers of: (a) cars (b) lorries (c) all vehicles. Now consider the result of combining two independent Poisson variables A ~ Po( 2) and B ~ Po( 3). Define the new distribution C = A + B. What can you say about C? You know that P( A=0) = e 2, P( A=1) = e 2 2, P( A=2) = e 2 2 2,... P( B=0) = e 3, P( B=1) = e 3 3, P( B=2) = e 3 3 2,... This gives PC=0 ( ) = P( A=0) P( B=0) = e 2 e 3 = e

8 Chapter 6 Poisson Distribution PC=1 ( ) = P( A=0) P( B=1) + P( A=1) P( B=0) = e 2 3e e 2 e 3 = 5e 5, and PC=2 ( ) = P( A=0) P( B=2) + P( A=1) P( B=1) + P( A=2) P( B=0) = e 2 9e 3 + 2e 2 3e 3 + e e 3 = e = 25e 5 = e What sort of distribution do these results indicate? In general, if A ~ Po a random variables, then C = ( ) and B ~ Po( b) are independent ( A + B) ~ Po( a + b). You have seen this result illustrated in a special case above. The proof requires a good working knowledge of the binomial expansion and is set as an optional activity below. *Activity 6 By noting that ( ) = P( A=i) P( B= n i) PC=n n i=0 and that ( a + b) n = n i=0 n i ai b n i prove that C ~ Po( a + b). Example The number of misprints on a page of the Daily Mercury has a Poisson distribution with mean 1.2. Find the probability that the number of errors (a) on page four is 2; (b) on page three is less than 3; (c) on the first ten pages totals 5; (d) on all forty pages adds up to at least

9 Chapter 6 Poisson Distributions Solution Let E be 'the number of errors on one page', so that E ~ Po( 1. 2). (a) P E =2 ( ) = e 1.2 ( 1. 2) (b) P E <3 or ( ) = P ( E 2 ) , from tables, P( E <3) = P( E =0) + P( E =1) + P( E =2) = e e ( 1. 2)2 e 1.2 = e 1.2 ( ) = 2.92e 1.2 = (to 4 d.p.). (c) Let E 10 be 'the number of errors on 10 pages', then E 10 ~ Po( 12), as E 10 = E + E E, and E 10 ~ Po( ) = Po( 12). Hence P( E 10 = 5) = e ! (d) Similarly E 40 ~ Po( 48). P( E 40 3) = 1 P( E 40 2) = 1 e 48 + e e = e (to 3 d.p.). Tables may well be a time saving device, as was true with the binomial distribution. 123

10 Chapter 6 Poisson Distribution Activity 7 A firm has three telephone numbers. They all receive numbers of calls that follow Poisson distributions, the first having a mean of 8, the second 4 and the third 3 in a period of half an hour. Find the probability that (a) (b) (c) (d) the second and third lines will receive a total of exactly six calls in half an hour; the firm will receive at least twelve calls in half an hour; line two will receive at most six calls in one hour; line one will receive no calls in 15 minutes. Example A shop sells a particular make of video recorder. (a) Assuming that the weekly demand for the video recorder is a Poisson variable with mean 3, find the probability that the shop sells (i) at least 3 in a week, (ii) at most 7 in a week, (iii) more than 20 in a month (4 weeks). Stocks are replenished only at the beginning of each month. (b) Find the minimum number that should be in stock at the beginning of a month so that the shop can be at least 95% sure of being able to meet the demands during the month. Solution (a) Let X be the demand in a particular week. Thus X ~ Po( 3) and, using the Poisson tables in the Appendix, (i) PX 3 ( )=1 PX 2 ( ) = (Note that the tables give the = cumulative probabilities) (ii) PX 7 ( )=0.9881, from tables. (iii) If Y denotes the demand in a particular month, then Y ~ Po( 12)and PY>20 ( ) = 1 PY 20 ( ) = =

11 Chapter 6 Poisson Distributions (b) You need to find the smallest value of n such that From the tables, PY n ( ) PY 17 ( )=0.9370, PY 18 ( )= So the required minimum stock is 18. A very important property of the Poisson distribution is that if X ~ Po( λ ), then EX ( )=V( X)=λ. That is, both the mean and variance of a Poisson distribution are equal to λ. To show that E( X)=λ, note that, by definition, ( ) = xp( X = x) E X all x = 0 e λ +1 ( λe λ )+2 λ2 e λ + 3 = λ e λ 1+ λ + λ2 + λ3 3! + K = λ e λ e λ (using the result from page 117) = λ λ3 e λ which proves the result. The proof of V( X)=λ follows in a similar but more complicated way. 3! +K *Activity 8 If X ~ Po( λ ) show that V( X)=λ. Exercise 6B 1. Incoming telephone calls to a school arrive at random times. The average rate will vary according to the day of the week. On Monday mornings, in term time there is a constant average rate of 4 per hour. What is the probability of receiving (a) 6 or more calls in a particular hour, (b) 3 or fewer calls in a particular period of two hours? During term time on Friday afternoons the average rate is also constant and it is observed that the probability of no calls being received during a particular hour is What is the average rate of calls on Friday afternoons? (AEB) 125

12 Chapter 6 Poisson Distribution 2. Write down two conditions which need to be satisfied in order to use the Poisson distribution. The demand at a garage for replacement windscreens occurs randomly and at an average rate of 5 per week. Determine the probability that no more than 7 windscreens are demanded in a week. The windscreen manufacturer uses glass which contains random flaws at an average rate of 48 per 100 m 2. A windscreen of area 0.95 m 2 is chosen at random. Determine the probability that the windscreen has fewer than 2 flaws. A random sample of 5 such windscreens is taken. Find the probability that exactly 3 of them contain fewer than 2 flaws. (AEB) 3. A garage uses a particular spare part at an average rate of 5 per week. Assuming that usage of this spare part follows a Poisson distribution, find the probability that (a) exactly 5 are used in a particular week, (b) at least 5 are used in a particular week, (c) exactly 15 are used in a 3-week period, (d) at least 15 are used in a 3-week period, (e) exactly 5 are used in each of 3 successive weeks. If stocks are replenished weekly, determine the number of spare parts which should be in stock at the beginning of each week to ensure that on average the stock will be insufficient on no more than one week in a 52 week year. (AEB) 4. A shopkeeper hires vacuum cleaners to the general public at 5 per day. The mean daily demand is 2.6. (a) Calculate the expected daily income from this activity assuming an unlimited number of vacuum cleaners is available. The demand follows a Poisson distribution. (b) Find the probability that the demand on a particular day is: (i) 0 (ii) exactly one (iii) exactly two (iv) three or more. (c) If only 3 vacuum cleaners are available for hire calculate the mean of the daily income. A nearby large store is willing to lend vacuum cleaners at short notice to the shopkeeper, so that in practice she will always be able to meet demand. The store would charge 2 per day for this service regardless of how many, if any, cleaners are actually borrowed. Would you advise the shopkeeper to take up this offer? Explain your answer. (AEB) 6.3 The Poisson distribution as an approximation to the binomial Despite having tables and powerful calculators, it is often difficult to make binomial calculations if n, the number of experiments, becomes very large. In these circumstances, it is easier to approximate the binomial by a Poisson distribution. Activity 9 Birthday dates Obtain the dates of birth of the students in your college or school from official records or by running a survey. Record the number of people having a birthday on each date (omitting 29th February). Draw up a frequency table. 1 Use the theoretical probability to work out the expected 365 frequencies of 0, 1, 2,... people from the n people considered, 126

13 Chapter 6 Poisson Distributions who have birthdays on the same date, according to the binomial 1 distribution B n, 365. Using a mean number of people per day, n, calculate the expected frequencies from the Poisson 365 n distribution Po 365. Compare the observed and two sets of expected frequencies. Poisson's first work on the distribution that bears his name arose from considering the binomial distribution and it was derived as an approximation to the already known binomial model. When might a Poisson distribution give probabilities close to those of a binomial distribution? If X ~ Bn,p ( ) Po( λ ), then the means and variances of the two distributions must be about the same. This gives mean : np=λ variance : npq=λ. So, if p is small ( for example p = then in the Activity above), q 1 and n = λ p. So when n is very large and p is very small, binomial probabilities may be approximated by Poisson probabilities with λ = np. Normally, for this approximation it is required that n 50 and p 0.1 The approximation improves as n, p 0. Example A factory produces nails and packs them in boxes of 200. If the probability that a nail is substandard is 0.006, find the probability that a box selected at random contains at most two nails which are substandard. 127

16 Chapter 6 Poisson Distribution 15. A polytechnic offers a short course on advanced statistical methods. As the course involves a large amount of practical work only 8 places are available. Advertising starts two months before the course and if, at the end of one month, 3 or fewer places have been taken the course is cancelled. If 4 or more places have been taken by the end of one month the course is run regardless of the number of applications received in the second month. If the number of applications per month follows a Poisson distribution with mean 3.6, and places are allocated on a first come first served basis, what is the probability that at the end of one month the course will be (a) cancelled, What is the probability that (c) a place will be available at the start of the second month, (d) the course will run with 8 students? If the course is offered on four separate occasions, what is the probability that it will (e) run once and be cancelled three times, (f) run with 8 students on 2 or more occasions? (AEB) (b) full? 130

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