TI-83 Users Guide

Poisson Probability Distributions

Poisson EVENTS: The difference between binomial and Poisson events is often difficult for students to understand. Here are some guidelines, and then some examples:

In a binomial something happens (or does not) in a trial or interval of time <<<>>> in a Poisson an event may occur any integral number of times over an interval of time.

In a binomial the event happens with a given probability per trial <<<>>> in a Poisson an event happens an average number of times (lamda) over a given number of events or trials.

Now here are some examples of Poisson events. Note how they differ from the statements of Binomial events:

1. A page of a textbook has an average of .25 misprints per page? (notice that there may be zero, one, two or more misprints per page)

2. On average 7.2 people arrive at the bank during any given fifteen minute period that they are open for business.

3. The "911" service receives an average of 15 calls per day.

In these problems two things make them appear to be right for using the Poisson distribution. Each involves an event in which the outcomes are assumed to be random and independent of each other. Instead of the probability of the event we have the average number of occurrences over some interval measure (time, pages, etc). The average frequency of the event is usually symbolized by the Greek letter l,(lambda) as in l=7.2 people per fifteen minutes. In the true Poisson distribution, this value represents both the mean of the distribution and the variance.

EVALUATING POISSON PROBABILITIES: The TI-83 allows you to compute the individual probability of a Poisson outcome, or the complete distribution. The primary keys are the "poissonpdf(" command for probability distributions and the "poissoncdf(" command for the cumulative distribution. The keystrokes to access these are [2nd], [VARS/DISTR] followed by "B" or "C" to select between the two, or you may scroll down (or up) with the arrow keys and highlight the one you want. The command structure for the entry is "poissonpdf(l,k) where k represents the number of occurrences for which we wish to find the probability.

Here are several examples and the display to compute them:

A) A page of a textbook has an average of .25 misprints per page? Find the probability of exactly 1 defect on a page. To find the EXACT event, we use the command "poissonpdf(.25,1)" This gives a result of approximately 19%. Notice that this is a very different result than the binomial assumption that would come about by assuming that the probability of a defect on the page is 25%. In fact the Poisson distribution predicts slightly less than a 25% probability that any number of defects (>0) will occur on the page [P(k>0)=.221199. If this seems counter intuitive, keep in mind that there may be MORE than one on some pages, so if the average is only .25, there must be more than 3/4 with NO defects. If we wished to find the probability of two or less defects, we would use the cumulative command "poissoncdf(.25,2)" which gives the result of .9978. We are assured that there is not likely to be many pages with more than two misprints.

B) The "911" service receives an average of 15 calls per day. What is the probability that they receive more than twenty calls in a day. To answer this question we think of the cumulative probability that there are 20 or less. Using the command "poissoncdf(15,20)" produces .917. The probability of getting MORE than twenty is 1-.917 = .083, telling us that on about 1/12 of the days we can expect more than twenty emergency calls.

Poisson Probability Distributions If we were to be in charge of the Rescue service above, we might wish to know what the probability of as many as thirty people, or how many we might expect on no more than 10% of the days or some other probability value. To do this we need to be able to look at wide range of possible outcomes. We can do this with the List functions of the TI-83. Since we want to look at values from zero up to as many as fifty emergency calls a day, we will enter these numbers in List one. We do this by using the sequence command "seq(x,x,0,50,1)->L1" . Now we will put the Poisson probability of each of these numbers of calls in List L2. We simple repeat the command for the probability function using L1 as the value of k, "poissonpdf(15,L1)->L2". The commands are shown in the screen below. Finally, we add the cumulative distribution in L3. We can take a shortcut to do this using the list commands "cumSum". We have shown a screen shot of the results in the STATS window below. Highlighted on the results for twenty.

GRAPHICS AND STATISTICS: Having created a Probability Density Table, it seems only natural to want to Display a Graph of the results. To do so is not too difficult after we have come so far. Press [2nd], [y=/STAT PLOT] to open the plot menu and select Plot 1. Press enter and highlight as shown on the screen at left below. Repeat for Plot2 using the cumulative information in List3 (see middle screen). Now Clear, or turn off, and graph displays on the y= screen, and set the window range and domain (Usually Zoom9 -zoomstat works well). Then press graph to see a frequency distribution polygon. My results for the probability distribution are shown on the screen at right. Both can be shown at once, but lots of detail of the probability distribution are lost.