Th e Poisson distribution is a probability distribution that describes a pattern of behavior when some event occurs at varying, random intervals over a continuum of time, length, or space. Such a pattern of behavior is called a Poisson process.
For example, if we graphed the number of muggings in Chickasaw, Oklahoma, on a time dimension, we might find the following (where the Xs represent muggings):
Th e number of potholes per foot on South Flood Street in Normal, Oklahoma, also follows a Poisson process, as shown in the following diagram.
In theory, each of these continuums can be divided into equal segments such that no more than one event (pothole, mugging) occurs in each segment. Every equal segment of Flood Street would either have a pothole or not have a pothole.
In addition, for the process to be a Poisson process, the occurrence or nonoccur- rence of any event must not aff ect the probability that other events will occur.
Th at is, the fact that a pothole occurs at 621 South Flood does not aff ect whether or not a pothole occurs at 527 South Flood.
Up to this point, the process fi ts the description of a Bernoulli process: An event either occurs (success) or it does not (failure), and events are independent of each other. To determine the probability of any event when using the Bernoulli process, we need to know the probability of the event (p), the number of successes (r), and the number of trials (n). In the present situation, we could empirically determine the probability of a mugging occurring in Chickasaw during any time period (p). In this case, Chickasaw might experience .3 mugging per hour. Note that probabilities are based on time, length, or space. Th e number of successes (r)
X X XXX X X
12:00 a.m.
12:00 p.m.
XX XX XXXX
0 feet
X
X X
500 feet
X
could also be determined. We can count the number of muggings in any period.
We do not, however, know the number of trials. How many potential muggings were there in Chickasaw? Clearly this information cannot be determined. Th e major diff erence between a Poisson process and a Bernoulli process, then, is that the number of trials is not known in a Poisson process.
Th e use of the Poisson distribution can best be illustrated with an example.
Th e Elazar Rape Crisis Center wants to staff its crisis hot line so that a center member will always be present if a rape victim calls. Th e Crisis Center operates between the hours of 6:00 p.m. and 6:00 a.m. By examining the past records of the center, the director has determined that the mean number of rapes per hour is .05. Th is information can be determined from police records.
Th e fi rst question is, What is the probability that no rapes will occur on any given night? Th e steps in using the Poisson probability distribution are as follows:
Step 1: Adjust the mean to refl ect the number of hours (length or space) you want to consider. In the present example, the mean number of rapes in an hour is .05. You want to know the mean number of rapes in 12 hours. Because these events are independent, the mean for a 12-hour period is .6 (.05 3 12). Th is new mean adjusted to the period of interest is called l, the Greek letter lambda (some texts refer to it as lt or “lambda t”).
Step 2: To determine the probability of zero rapes, you turn to the Poisson distribution tables (Table 2 in the Statistical Tables). For those of you who feel that using a table is cheating, exact probabilities can be calcu- lated with the following formula:
y5 lxe2 l x!
where l is the probability of the event, x is the number of occurrences, e is a constant (2.71828), and y is the probability of x occurrences. If you value your time, turn to Table 2. Scanning across the top of the table, fi nd l equal to .6. You should fi nd the following information:
X l
.6
0 .5488
1 .3293
2 .0988
3 .0198
4 .0030
5 .0004
6 .0000
Th is table gives a Poisson probability distribution for l 5.6. It tells you that if l 5.6, the probability of no rapes in a single night is .5488.
Th e probability of exactly one rape is .3293. Th e probability of exactly two rapes is .0988, and so on.
How might this information be used in a management setting? Suppose that the director believes that (based on past experience) assisting a rape victim takes about 4 hours. One crisis center staff member remains with the victim for this period to assist with the medical treatment and police procedures. If the center had one staff member and two rapes occurred in a 4-hour period, then no one would be at the center to assist the second victim. Th e director wants to avoid this situation. How many staff members should be at the center?
Step 1: Adjust the mean to the time period. Th e mean is .05 for 1 hour, and you want to know about 4-hour periods (the amount of time a staff member will be away from the center). Th erefore, l is .2 1.05342. Step 2: Find the probabilities in the Poisson distribution table for 0, 1, 2, 3, 4,
and 5 rapes when l is .2.
Number of Rapes Probability
0 .8187
1 .1637
2 .0164
3 .0011
4 .0001
5 .0000
To make the staffi ng decision, the director uses the preceding table. If 0 or 1 rape occurs in a 4-hour period, one staff member is suffi cient. Th is will occur in 98% of the 4-hour periods 1.81871.16375.98242.
Because the director wants to minimize the time during which the crisis center cannot respond, the probability is interpreted as follows: If the crisis center staff s only one employee, this will be inadequate 1.8% of the time periods 112.9824 5.0176, or about 1.8%). Th e 12-hour day is equivalent to three 4-hour periods, so a rape with no one pres- ent will occur on 5.4% 11.8%332 of the days (or about once every 3 weeks). Th e director considers this risk unacceptable.
If two staff members are stationed at the crisis center, the probability of suffi cient staff is .9988. Translating this into days when a rape will occur with no one at the crisis center 1.0012335.0036, or .36%) reveals that the staff will be inadequate .36% of the days, or once every 278 days (36 times in 10,000 days). Th e staff director feels that this risk is tolerable and decides to staff two persons.
(Note: Staffi ng three persons would increase the probability of suffi cient staff to .9999. Th is means inadequate staff problems would occur once every 10,000 days.)
You may have noticed that the Poisson probability tables give values only for l less than or equal to 20. Poisson tables for larger l values are not presented because the normal distribution table provides fairly accurate probabilities when l is greater than 20. Th e normal distribution of this magnitude has a mean of l and a standard deviation equal to the square root of l. For a l value of 30, there- fore, you would use the normal table with a mean of 30 and a standard deviation of 5.5 1!3055.52.