Engineering Statistics Chapter 2 Special Variables 2B Poisson Distribution

Rare Events In many statistical problems, an event which seldom happens is considered. The rareness is of course relative. Even though people who die of cancer because of smoking in Malaysia is in tens of thousands annually, compared to the population, it is still a rare event. In such cases, it is found that a distribution using exponential function will model the probabilities of the event based on the mean number of occurrences.

Poisson Distribution If the mean number occurrences X of a rare event for a certain period is, then X is said to follow the Poisson distribution if P(X = r) = e - r /r! We represent a Poisson distribution with a mean by P( ). Since is the average of a number of event, its value can be a non-integer. In addition, there is no top limit for the value of X. This means that, even though may be 5, 3, 1, even 0.2, the values for X can range from 0 to .

First example 1.On the average, 2.5 cars are found parking illegally on a busy road during a traffic check. Assume this number follows the Poisson distribution. Find the probability (i)No illegally-parked car is found during a round by a police; (ii)Up to 3 cars are found illegally parked on the road during a check. Solution: Let I represent the number of illegally parked car during a check on the road. Then we assume I~P(2.5) (i)P(I=0) = e /0! = (ii)P(I  3) = e e /1! + e /2! + e /3! = = (4 d.p).

Using Tables Just as for binomial distributions, statistical tables for Poisson distributions are constructed to assist in calculations. As Poisson distributions have only one parameter each, the tables are simpler. The UTM tables for Poisson distributions are also cumulative from below, showing P(X  r) for each r. Your experience in using the binomial distribution tables should make it easier to read the Poisson distribution tables.

Example 2 The marriage bureau records an average of 5.5 marriages each week. Assume this number follows the Poisson distribution. What is probability (i)5 marriages are registered this week? (ii)5 or more marriages are recorded this week? Solution: Let M represent the number of marriages for a week. Then M~P(5.5) (i)P(M=5) = e /5! = (ii)P(M  5) = 1 – P(M  4) = 1 – (table) =

Multiple Intervals If the mean number of occurrences of an event is for a certain period, then it makes sense that the mean number for double the period would be 2, for triple the period 3 and so on. For example, if on the average, 3.2 voters vote in a minute, then for 2 minutes, the average should be 6.4, and so on. Hence if X~P( ) for a period of 5 minutes, then Y~P(2 ) if Y represent the number of events for double the period.

Example 3 On the average, there are 1.2 traffic lights on 1 km of a city road. Assume this number follows the Poisson distribution. What is the probability we find (i)5 traffic lights or less on a stretch of road 5 km long? (ii)More than 3 traffic lights on a road 3 km long? Solution: In this case, the ‘period’ is actually a distance. The theory remains the same as for time interval. (i)Let L5 represent the number of traffic lights on a 5 km long road. Then L5~P(6).  P(L5  5) = (ii)Let L3 represent the number of traffic lights on a 3 km long road. Then L3~P(3.6).  P(L3>3) = 1 – P(L3  3) = 1 – =

Example 4 – sub-intervals It is estimated that the number of critical cases received at the emergency ward averages 4.8 per hour. Assume this number follows the Poisson distribution. What is the probability (i)No critical case is received in 10 minutes? (ii)At least 3 critical cases come in during a half-hour interval? Solution: Just as multiple intervals yield multiples for average, the reverse is true too. Thus (i)Let C 1/6 represent the number of critical cases in 10 minutes, then C 1/6 ~P(0.8). So P(C 1/6 =0)=e -0.8 = (ii)Let C ½ represent number of critical cases in ½ hour, then C ½ ~P(2.4). So P(C ½  3) = 1 – P(C ½  2) = 0..

Combining Poisson Distributions Unlike the binomial distributions, given X 1 ~P( 1 ) and X 2 ~P( 2 ), we can sum the two variables: X 1 + X 2 ~P( ). For example if the weekly mean number of accidents on route A is 2.2 and that on route B (separate from and independent of A) is 1.5 then the combined mean number is 3.7 and A~P(2.2) and B~P(1.5)  A+B~P(3.7). Unfortunately, there is still no short-cut for finding probabilities of the type P(A>B) or P(A>B+2) and similar events. For such problems, only computer programs specially designed to solve them will reduce the tedium of works needed.

Example 5 The number of passengers reporting sick during a domestic flight averages 0.8; for international flight, 1.2. Two planes, one on domestic route and the other international, just landed at an airport. What is the probability 4 passengers are sick? Solution: To solve this problem without combining the two variables, you need to find P(D=0), … P(D=4) and P(I=0), … P(I=4), then combine them to give the answer as P(D=0)  P(I=4) + P(D=1)  P(I=3) + P(D=2)  P(I=2) + P(D=3)  P(I=1) + P(D=4)  P(I=0). Combining, we have D+I~P(2). So P(D+I=4) = e /4! = You might like to try the first method and compare your result with this.

Interpolation again Just as for binomial distributions, the Poisson table gives probabilities for only a small set of. There will be problems when is not exactly one of those given in the tables. As usual, you may decide to carry out separate calculations to obtain accurate answers. Alternately, interpolations will provide sufficiently good answers when the problem demands fast answers.

Example 6 – interpolation example The mean number of goats getting choked by palm leaves in a integrated oil palm plantation is 1.8 per day. During a week, what is the probability (i)At least 10 goats are choked on palm leaves? (ii)8 to 15 goats are choked on palm leaves? Solution: Let G represent the number of choking goats for a week (7 days). Then G~P(12.6). We set G1~P(12) and G2~P(13)

Example 6 - solution P(G1  10) = 1 – P(G1  9) = 1 – = P(G2  10) = 1 – P(G2  9) = 1 – = Hence P(G  10) = ( – )  0.6 = (ii) P(8  G1  15) = P(G1  15) – P(G1  7) = – = P(8  G2  15) = P(G2  15) – P(G2  7) = – = Hence P(8  G  15) = ( – )  0.6 =

Compound Problems As shown in the part on binomial distribution, the result of an analysis may be used as the basis for another problem. The next few examples combine the Poisson distribution with binomial distribution.

Example 7 (i)On the average, a police station receives 5.2 reports of lost ID cards per day. What is the probability the number of loss reports exceeds 8, the critical number for a red day? (ii)The station tracks the records for 10 days. What is the probability it has more than 3 red days during the period?

Example 7 (Solution) (i) Let L represent the number of lost cards. L~P(5.2). P(L>9) = 1–P(L  9) = (ii) Let R represent the number of red days in 10 days. R~Bin(10, k) P(R>3) = 1–P(R  3) =

Example 8 It is estimated that on the average, there are 2.2 floods a year in a district in Trengganu. (i)What is the probability a district suffers more than 5 floods for a certain year? (ii)A district getting 5 floods or more in a year will be allocated special subsidies in the next financial years. There are 12 districts in Trengganu. What is the probability at least 2 districts will get such assistance for the next year?

Example 8 (Solution) (i)F=number of floods in a year. F~P(2.2). P(F  5) = 1–P(F  4) = r (ii) S=districts getting subsidies. S~Bin(12, r) P(S  2)