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 "

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

__ 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