Chapter 5, part C

V. Poisson Probability Distribution Siméon Denis Poisson (1781-1840), a French mathematician, developed the Poisson to estimate the probability of a number of occurrences over a specified interval of time or space. It’s said that the French Army used Poisson’s methods to predict the incidence of soldiers’ death by mule kicks. Tres debonair, no?

A. Properties of a Poisson Experiment 1. The probability of an occurrence is the same for any two intervals of equal length. 2. The occurrence or nonoccurrence in any interval is independent of the same in any other interval. Example: Experiment is the number of cars through a drive-thru window in 1 hour. Can you see how this experiment satisfies the conditions?

B. Poisson Probability Function f(x) is the probability of exactly x occurrences in an interval.  is the expected value or mean # of occurrences in an interval e is the ln(1) or approximately 2.71828.

C. Example with time intervals At a drive up window at a local bank, past experience tells us that the mean number of cars in a 15 minute period is =10 cars. x is the random variable, # of cars in a 15 minute span. What does this have to do with mules?

Problems What is the probability of exactly 7 cars in any 15 minute span? How would this be useful?

Problems Find the probability of 1 car in 5 minutes. First we need to convert  =10 from a 15 minute span to a 5 minute span. If  =10 cars in 15 minutes,  =10/3=3.33 cars in 5 minutes.

D. Example with distance intervals Suppose a pipeline needs 1 repair every 100 miles, per year ( =1 per 100 miles). If 1000 miles of pipe are built, what is the probability that there will need to be 5 repairs in a year? If  =1 per 100 miles,  =10 per 1000 miles.

VI. Hypergeometric Probability Distribution This discrete probability distribution is related to the binomial, but the trials are not independent. This means the probability of a success changes from trial to trial.

Probability function The hypergeometric is used to calculate the probability of x successes in a sample n, selected from a population N, without replacement. r: # of successes in population N N-r: # of failures

An example Coke and Pepsi are constantly running taste tests to determine consumer preferences. Suppose that in a group of 10 people, 6 prefer Coke and 4 prefer Pepsi. If we randomly select 3 at a time, answer the following: a. What is the probability that 2 prefer Coke? b. What is the probability that 2 or 3 prefer Pepsi?

Solutions a. N=10, r=6, n=3, x=2. Make sure you understand where these values come from. Practice!

Solutions B. Calculating the probability that 2 or 3 prefer Pepsi is the same as the probability that 1 or 0 will prefer Coke. P(Pepsi majority) = f(0) + f(1) = .30 + .0333 = .3333 In other words, a group of 3 consumers will have a Pepsi majority 1/3 of the time and a Coke majority 2/3 of the time.