CHAPTER
10

The Poisson Probability Distribution

In This Chapter

• Describe the characteristics of a Poisson process
• Calculate probabilities using the Poisson equation
• Use the Poisson probability tables
• Use Excel to calculate Poisson probabilities
• Use the Poisson equation to approximate the binomial equation

Now that we have mastered the binomial probability distribution, we are ready to move on to the next discrete theoretical distribution, the Poisson. This probability distribution is named after Simeon Poisson, a French mathematician who developed the distribution during the early 1800s.

The Poisson distribution is useful for calculating the probability that a certain number of events will occur over a specific period of time. We could use this distribution to determine the likelihood that 10 customers will walk into a store during the next hour or that 2 car accidents will occur at a busy intersection this month. So let’s grab some crêpes and croissants and learn about some French math.

Characteristics of a Poisson Process

In Chapter 9, we defined a binomial experiment as counting the number of successes over a specific number of trials. The result of each trial is either a success or a failure. A Poisson process counts the number of occurrences of an event over a period of time, area, distance, or any other type of measurement.

Rather than being limited to only two outcomes, the Poisson process can have any number of outcomes over the unit of measurement. For instance, the number of customers who walk into a local convenience store during the next hour could be zero, one, two, three, or so on. The random variable for the Poisson distribution would be the actual number of occurrences–in this case, the number of customers arriving during the next hour.

The mean for a Poisson distribution is the average number of occurrences that would be expected over the unit of measurement. For a Poisson process, the mean has to be the same for each interval of measurement. For instance, if the average number of customers walking into the store each hour is 11, this average needs to apply to every one-hour increment.

Another characteristic of a Poisson process is that the number of occurrences during one interval is independent of the number of occurrences in other intervals. In other words, if six customers walk into the store during the first hour of business, this would have no effect on the number of customers arriving during the second hour.

The last characteristic of a Poisson process is that the intervals don’t overlap. For example, when counting the number of customers walking into the store in one hour periods, the one-hour periods cannot overlap with each other. We can count the number of customers arriving between 9 to 10 A.M. and 10 to 11 A.M. and so forth, but we cannot use the 9:30 to 10:30 A.M. period because this overlaps with the other intervals.

Examples of random variables that may follow a Poisson probability distribution include the following:

• The number of cars that arrive at a tollbooth over a specific period of time
• The number of typographical errors found in a manuscript
• The number of students who are absent in my Monday morning statistics class
• The number of professional football players who are placed on the injured list each week

Now that you understand the basics of a Poisson process, let’s move into probability calculations.

The Poisson Probability Distribution

If a random variable follows a pattern consistent with a Poisson probability distribution, then we can calculate the probability of a certain number of occurrences over a given interval. To make this calculation, we need to know the average number of occurrences for the event over this interval. To demonstrate the use of the Poisson probability distribution, I’ll use this example.

The following story is true, but the names have not been changed because nobody in this story is innocent. Each year, Brian, John, and Bob make a golf pilgrimage to Myrtle Beach, South Carolina. On their last night one particular year, they were browsing through a golf store. Brian somehow convinced Bob to purchase a used, fancy, brand-name golf club that he swore he absolutely had to have in order to reach his full potential as a golfer. Even used, this club cost more than any Bob had purchased new, but teenagers have this special talent that allows them to disregard any rational adult logic when their minds are made up.

Early the following morning, they packed their bags, checked out of the hotel, and drove to their final round of golf, which Bob had cleverly planned to be along their route back home. On the first tee, Brian pulled out his new, used prize possession and proceeded to hit a “duck hook,” which is a golfer’s term for a ball that goes very short and very left, often into a bunch of trees never to be seen again. Bob smiled nervously at Brian and tried to convince himself that he’d be fine on the next hole. After hitting duck hooks on holes two, three, and four, Bob found himself physically restraining Brian from throwing his new, used prized possession into the lake.

After their round was over, Bob drove back to Myrtle Beach to return the club, adding an hour to what would have been a 10-hour car ride. At the golf store, the woman cheerfully said she would take the club back, but she needs … the receipt. Now Bob vaguely remembered putting the receipt someplace “special” just in case he would need it, but after packing, checking out, and playing golf, he would have had a better chance of discovering a cure for cancer than remembering where he had put that piece of paper.

Not being one to give up easily, Bob marched back to the car and started unpacking everything. After a short while, the same woman walked out to tell Bob the store would gladly refund his money without the receipt if he would just pack up his things and put them back in the car. A very powerful technique that Bob discovered when he needed to return something without a receipt, simply take along some dirty clothes in a suitcase and spread them out in the parking lot in front of the store. It works like a charm.

Anyway, let’s assume that the number of tee shots that Brian normally hits that actually land in the fairway during a round of golf is five. The fairway is the area of short grass where the people who have designed this nerve-wracking game intended your tee shot to land. We will also assume that the actual number of fairways that Brian “hits” during one round follows the Poisson distribution.

We can now use the Poisson probability distribution to calculate the probability that Brian will hit x number of fairways during his next round, as follows:

where:

x = the number of occurrences of interest over the interval

λ = the mean number of occurrences over the interval

e = the mathematical constant ≈ 2.71828

P(x) = the probability of exactly x occurrences over the interval

We can now calculate the probability that Brian will hit exactly seven fairways during his next round. With λ = 5, the equation becomes this:

In other words, Brian has slightly more than a 10 percent chance of hitting exactly seven fairways.

We can also calculate the cumulative probability that Brian will hit no more than two fairways using the following equations:

There is a 12.46 percent chance that Brian will hit no more than two fairways during his next round.

In the previous example, the mean of the Poisson distribution happened to be an integer (5). However, this doesn’t have to always be the case. Suppose the number of absent students for my Monday morning statistics follows a Poisson distribution, with the average being 2.4 students. The probability that there will be three students absent next Monday is as follows.

Looks like I need to start taking roll on Mondays!

There’s one more cool feature of the Poisson distribution: the variance of the distribution is the same as the mean. In other words:

σ² = λ

This means that there are no nasty variance calculations like the ones we dealt with in previous chapters for this distribution.

Poisson Probability Tables

Just like the binomial distribution, the Poisson probability distribution has a table that allows you to look up the probabilities for certain mean values. You can find the Poisson distribution table in Appendix B of this book. The following is an excerpt from this appendix with the probabilities from our Myrtle Beach example underlined.

The probability table is organized by values of λ, the average number of occurrences. Notice that the sum of each block of probabilities for a particular value of λ adds to 1.

As with the binomial tables, one limitation of using the Poisson tables is that you are restricted to using only the values of λ that are shown in the table. For instance, the previous table would not be useful for λ = 4.5. However, other statistics books might contain Poisson tables that are more extensive than the one in Appendix B.

The Poisson distribution for λ = 5 is shown graphically in the following histogram. The probabilities in Figure 10.1 are taken from the last column in the previous table.

Figure 10.1

Poisson probability distribution.

Note that the most likely numbers of occurrences for this distribution are four and five.

Here’s another example. Let’s assume that the number of car accidents each month at a busy intersection that Bob used to pass on his way to work follows the Poisson distribution with a mean of 1.8 accidents per month. What is the probability that three or more accidents will occur next month? You can express this as:

P(x ≥ 3) = P(x = 3) + P(x = 4) + P(x = 5) + + P(x = ∞)

Technically, with a Poisson distribution, there is no upper limit to the number of occurrences during the interval. You’ll notice from the Poisson tables that the probability of a large number of occurrences is practically zero. Because we cannot add all the probabilities of an infinite number of occurrences, we need to use the complement rule (Do you remember this from Chapter 6? It all comes back!), which is:

P(x ≥ 3) = 1 – P(x <3)

because:

P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + + P(x = ∞) = 1.0

Therefore, to find the probability of three or more accidents, we’ll use the following:

P(x ≥ 3) = 1 – [P(x = 0) + P(x = 1) + P(x = 2)]

Using the probabilities underlined in the following Poisson table (I seem to have misplaced my calculator), we have this:

Values of λ

P(x ≥ 3) = 1 – (0.1653 + 0.2975 + 0.2678)

P(x ≥ 3) = 1 – 0.7306 = 0.2694

There is almost a 27 percent chance that three or more accidents will occur in this intersection next month.

Using Excel to Calculate Poisson Probabilities

You can also conveniently calculate Poisson probabilities using Excel. The built-in POISSON.DIST function has the following characteristics:

POISSON.DIST(x, λ, cumulative)

where:

cumulative = FALSE if you want the probability of exactly x occurrences

cumulative = TRUE if you want the probability of x or fewer occurrences

For instance, Figure 10.2 shows the POISSON.DIST function being used to calculate the probability that there will be exactly two accidents in the intersection next month.

POISSON.DIST function in Excel for exactly x occurrences.

Cell A1 contains the Excel formula =POISSON.DIST(2,1.8,FALSE) with the result being 0.2678. This probability is underlined in the previous table.

Excel will also calculate the cumulative probability that there will be no more than two accidents in the intersection, as shown in Figure 10.3.

Figure 10.3

POISSON.DIST function in Excel for no more than x occurrences.

Cell A6 contains the Excel formula =POISSON.DIST(2,1.8,TRUE) with the result being 0.7306, a probability that we saw in the last calculation and which is also the sum of the underlined probabilities in the previous table.

One benefit of using Excel to determine Poisson probabilities is that you are not limited to the values of λ shown in the Poisson table in Appendix B. Excel’s POISSON.DIST function allows you to use any value for λ.

Using the Poisson Distribution as an Approximation to the Binomial Distribution

I don’t know about you, but when I have two ways to do something, I like to choose the one that’s less work. If you don’t agree with me, feel free to skip this material. If you do, read on!

We can use the Poisson distribution to calculate binomial probabilities under the following conditions:

• when the number of trials, n, is greater than or equal to 20, and
• when the probability of a success, p, is less than or equal to 0.05.

The Poisson formula would look like this:

where:

n = the number of trials

x = the number of successes

p = the probability of a success

You might be asking yourself at this moment just why you would want to do this. The answer is because the Poisson formula has fewer computations than the binomial formula, and, under the stated conditions, the distributions are very close to one another.

Just in case you are from Missouri (the “Show Me” state), I’ll demonstrate this point with an example. Suppose there are 20 traffic lights in a town and each has a 3 percent chance of not working properly (a success) on any given day. What is the probability that exactly 1 of the 20 lights will not work today? This is a binomial experiment with n = 20, x = 1, and p = 0.03. From Chapter 9, we know that the binomial probability is this:

The Poisson approximation is as follows:

Because: np = (20)(0.03) = 0.6

Even if you’re from Missouri, I think you would have to agree that the Poisson calculation is easier, and the two results are very close. So my advice to you is to use the Poisson equation if you’re faced with calculating binomial probabilities with n ≥ 20 and p ≤ 0.05.

This concludes our discussion of discrete probability distributions. We hope you’ve had as much fun with these as we’ve had!

Practice Problems

1. The number of rainy days per month at a particular town follows a Poisson distribution with a mean value of six days. What is the probability that it will rain four days next month?

2. The number of customers arriving at a particular store follows a Poisson distribution with a mean value of 7.5 customers per hour. What is the probability that five customers will arrive during the next hour?

3. The number of pieces of mail that I receive daily follows a Poisson distribution with a mean value of 4.2 per day. What is the probability that I will receive more than two pieces of mail tomorrow?

4. The number of employees who call in sick on Monday follows a Poisson distribution with a mean value of 3.6. What is the probability that no more than three employees will call in sick next Monday?

5. The number of spam e-mails that I receive each day follows a Poisson distribution with a mean value of 2.5. What is the probability that I will receive exactly one spam e-mail tomorrow?

6. Historical records show that 5 percent of people who visit a particular website purchase something. What is the probability that exactly 2 people out of the next 25 will purchase something? Use the Poisson distribution to estimate this binomial probability.

The Least You Need to Know

• A Poisson process counts the number of occurrences of an event over a period of time, area, distance, or any other type of measurement.
• The mean for a Poisson distribution is the average number of occurrences that would be expected over the unit of measurement and has to be the same for each interval of measurement.
• The number of occurrences during one interval of a Poisson process is independent of the number of occurrences in other intervals.
• The intervals of a Poisson process don’t overlap.
• If x is a Poisson random variable, the probability of x occurrences over the interval of measurement is
• If the number of binomial trials is greater than or equal to 20 and the probability of a success is less than or equal to 0.05, then you can use the equation for the Poisson distribution to approximate the binomial probabilities.