14.2 THREE IMPORTANT DISCRETE PROBABILITY MODELS

Transcription

1 14.2 THREE IMPORTANT DISCRETE PROBABILITY MODELS In Section 14.1 the idea of a discrete probability model was introduced. In the examples of that section the probability of each basic outcome of the experiment was explicitly stated. In the example of the fair die, each side was given a probability of 1 6. Discrete probability models (that

2 is, discrete distributions ) used commonly by statisticians, such as the Poisson, binomial, and hypergeometric distributions, all use a formula to assign a probability to each basic outcome. In Chapter 8, examples concerning theoretical probabilities for the binomial and Poisson distributions were introduced, but the formulas were not given. Here we will learn the formulas. Poisson Distribution From Chapter 8 recall the data concerning the Prussian army and the number of soldiers killed by horses. The data gave the number of units (of a total of 280) in which a certain number of people had died in a particular year: Number of soldiers kicked to death Number of units Proportion ormore 0 0 The theoretical probabilities can be computed using the following formula, the probability law for the Poisson distribution: e px () x! e x! Here e, sometimes called Euler s number, is a famous number in mathematics used in problems involving continuously compounded interest, the natural logarithm, and so on. For practical purposes, e suffices in most problems. There are two other parts to this formula that need to be explained. First, for a positive integer x, the notation x! is read as xfactorial and is given by x! x( x 1)( x 2)( x 3) (2)(1) For example, 3! , and 4! We define 0! 1, which is a property sometimes needed. Second, the value in the formula stands for the theoretical mean, or in the current example, the theoretical mean number of soldiers kicked to death per year in a unit. This is what is referred to as a parameter for the x x

3 distribution. The Poisson distribution, like most other distributions used by statisticians, is actually a family of distributions, and its parameter (for some distributions, parameters) specifies exactly which distribution we are working with. In each application where the Poisson model fits the data well, a different choice of will be appropriate. In the case of the Prussian soldier data, provides a probability model that fits the data best. That is, we assume that the theoretical average number of soldiers kicked to death in a unit per year is Plugging this value into the Poisson formula, we come up with the following theoretical probabilities for the different values: p (0 deaths in a unit) p (1 death in a unit) p (2 deaths in a unit) p (3 deaths in a unit) p (4 deaths in a unit) p (5 deaths in a unit) We could continue calculating these probabilities forever, for 6 deaths, 7 deaths, and so on. For example, the theoretical probability that 10 deaths occur in a unit is a small number ( ), very close to 0. But still it is greater than 0, so that event is possible. The sample space in the case of a Poisson-distributed random variable is any number 0, 1, 2, 3, 4,... Note that the theoretical and obtained proportions are very close. Note also that for practical purposes we can ignore the probabilities for large integers because they are all approximately 0. In general, recall from Chapter 8 that the Poisson distribution can be used in any situation in which we are given a rate of occurrence that is, when we know that a certain event occurs, on average, a given number of times in a given time period, provided the three conditions of Section 8.2 hold. Crudely, these conditions are that the event occurrence rate is stationary over time, that events occur independently of each other, and that events occur one at a time. Any situation in which one of these three conditions is seriously violated should not be considered to be well described by a Poisson distribution. In the case of the Prussian army data, we used the Poisson distribution with a theoretical mean of deaths due to horses in a unit per year. Other examples are the number of phone calls arriving per hour and the number of customers entering a store per minute. In ecology one often studies the population of a certain organism per square mile, or per square meter, as when a biologist counts crab holes on an ocean beach.

4 Once we can arrive at a numerical value for the theoretical mean (which is often estimated from the data), the Poisson distribution formula can be used to give the theoretical probability that the event occurs any given number of times in the time interval. For example, if we concluded that the theoretical mean number of phone calls arriving in an hour is 5.5, we could calculate the probability that we will receive exactly 3 (or any other integer) phone calls in an hour. The Binomial Distribution Another distribution first introduced in Chapter 8 is the binomial distribution. The formula for the theoretical probabilities was not given for this distribution either. However, we did learn how to compute the theoretical binomial probabilities in some simple situations. Recall that the binomial distribution deals with situations that have two outcomes per trial, referred to as success and failure (these terms are used merely for convenience; they do not necessarily imply success or failure in any real sense). Recall the four conditions of Section 8.1 that must hold if we are to use the binomial distribution to model a situation: there must be two possible outcomes per trial, a fixed number of trials (not a random number of trials), the same probability of success in every trial, and independence of outcomes for different trials. If these conditions hold, then the probability of xsuccesses in ntrials is as follows: n x x n x px ( ) p(1 p) where x 0, 1,...,n Here p, the parameter of the binomial distribution, is the probability of success on one trial. The sample space for a binomial random variable is the set of every integer between 0 and n, inclusive. There is yet another new notation in this formula that needs to be explained. The notation n x n x is read as nchoose x. We sometimes also call this the number of combinations of size x selected from n items. This notation is related to the factorial, just introduced, through the following definition: n! x!( n x)! Here 0! 1 is needed when x 0 or x n.

5 Example 14.7 On any given day, the probability that Jill will make it to class on time is 75%. Assume that her attendance behaviors on different days are independent. What is the probability that she will make it on time on exactly four days in a five-day school week? Solution Because the four assumptions necessary for using the binomial distribution hold in this case, we will use the binomial distribution to solve this problem. In this case n 5, since there are five trials. The probability of success, or of Jill s making it to class on time, is 0.75, so p p (4 successes in 5 trials) (0. 75) (0. 25) What is the probability that she will make it to class on time all five days? p (5 successes in 5 trials) (0. 75) (0. 25) Thus Jill will come late to class at least one day a week in about 76% of the weeks in the term. This is because p(fewer than 5 successes) 1 p(5 successes) The Hypergeometric Distribution Now we will introduce a new discrete distribution, the hypergeometric distribution. We will introduce it through an example from quality control, which is perhaps its main area of application. Suppose a manufacturer ships 50 units of its product to a store, and 10 of the units are defective. If an individual comes along and buys seven of these units (choosing them at random), what is the probability that exactly two of them will be defective? This is the type of situation that can be modeled using the hypergeometric distribution. In general, the hypergeometric distribution is used when you have a total of n elements and this total of n is divided into two distinct groups, which we will label, for convenience, defectives and nondefectives, even if the context of the problem is not one of defective versus nondefective objects. The number of defectives is denoted a, and the number of nondefectives is denoted b. Note that a b must equal n. If rdraws are made without replacement, as was the case when the seven units were purchased in the

6 example, the probability of getting exactly x elements from the group of defectives is as follows: b x r x px ( defectives out of the rdraws) a n r This is the function for assigning probabilities given by the hypergeometric distribution. It can be used for any value of xsuch that 0 x n, though of course x a is a practical constraint, too: there cannot occur more sampled defectives than exist in the population. Continuing the example of the defective units: Likewise, p (exactly 2 defectives) p (exactly 1 defective) p (none defective) Finally, we can conclude, based on the additivity of probabilities of exclusive events, discussed in the previous section, that p(1 defective or less) p(0 defectives) p(1 defective) The hypergeometric distribution has important applications in the field of quality control. As seen in the above example, the distribution can be used to model situations in which we are interested in finding out how many defectives one can expect when drawing units from a larger group of a known defective rate. In statistics one often infers from the number of defectives observed in a random sample from the population something about the proportion of defectives in the population. In this manner one can assess whether a manufacturing process is in control in the sense of

7 SECTION 14.2 EXERCISES producing, on average, an acceptably small percentage of defectives. The hypergeometric distribution is the probability backbone of this theory. More generally, the hypergeometric distribution can be used whenever we are looking at one large group that can be divided into two subgroups on the basis of a certain characteristic and we sample randomly without replacement. Using the hypergeometric distribution, we can determine the probability of pulling out a certain number from the first group when we take a sample of a specific size and are sampling without replacement. The use of formulas such as these we have introduced for the Poisson, binomial, and hypergeometric distributions greatly simplifies the task of coming up with probabilities for certain outcomes. The five-step method was our original method for the calculation of probabilities of this type, but of course in that case we were only obtaining estimates of the probabilities accurate estimates if we had a large enough number of trials, of course. Now we have formulas that give us the exact theoretical probabilities and are also much simpler to use and less time consuming. The disadvantage, though, is that in order to compute probabilities involving the statistic of interest, we must logically deduce, from the original model given in the first two steps of the five-step method, which theoretical probability distribution describes the statistic of interest. In the case of the Poisson distribution, there are three conditions that must be met, and in the case of the binomial there are four. The five-step method is much more general, allowing one to estimate probabilities for almost any situation in which we are able to set up a model and decide what constitutes a trial and what the statistic of interest is. We can therefore use the five-step method when the statistic of interest is neither binomial, Poisson, or hypergeometric! Note: In each of the three models, it can be shown that the sum of the probabilities px ( ) over all possible x s equals 1. (Try it in special cases, such as n 3, p 1/3, for the binomial.) 1. What are the following expressions equal to? a. Explain why the number of telephone calls a. 4! b. 2! arriving per minute follows a Poisson dis- c. 5! d. 0! tribution. 2. What are the following expressions equal to? b. What is the value of the parameter, the 5 Hint: 5 choose 2 is the same as theoretical mean? 2. a. (5 choose 2) b. (4 choose 1) c. What is the probability that the secretary c. (2 choose 2) d. (4 choose 2) does not receive any calls in a minute? d. What is the probability that the secretary 3. A secretary is working in an office where receives one call in a minute? telephone calls arrive at a rate of 2.3 per mine. What is the probability that the secretary ute throughout the day. receives three or more calls in a minute? ( Hint: Recall that p( A) 1 p(not A).)

8 4. In a box there are five doughnuts, and Alice a. Explain why the number of lines the stock takes two of them at random. Out of the five goes up in five days follows a binomial doughnuts, three are cream-filled, and two are distribution. What is a success in this not. case? a. Explain why the number of cream-filled b. What is the value of p, the probability of doughnuts Alice obtains follows a hyper- success on one trial? geometric distribution. c. What is the probability that in five days, b. What are the values of n, a, b, and r in the the stock goes up on every day? hypergeometric formula in this case? d. What is the probability that in five days, c. What is the probability that Alice obtains the stock goes up on exactly four days? exactly one cream-filled doughnut in the 8. In a group of four used cars at an auction, two she takes out of the box? two have previously been involved in serious d. What is the probability that both of the accidents. A car dealer will buy two of the doughnuts Alice pulls out are cream-filled? cars chosen at random. Use the hypergeometric 5. During rush hour, a certain intersection sees distribution to model the number of cars an average of 5.1 cars every minute. Use the purchased that have been in accidents. Poisson distribution to model the number of a. What are the values of nab,,, and r in this cars arriving at the intersection per minute. case? a. What is the value of, the theoretical mean? b. What is the probability that the dealer picks b. What is the probability that exactly two both of the cars that have been in accidents? cars arrive during a given minute? c. What is the probability that the dealer picks c. What is the probability that exactly four exactly one of the cars that have been in cars arrive during a given minute? accidents? Note: You will need a calculator! 9. There are six multiple-choice questions on a 6. Refer to the situation given in Exercise 5. test, and each one has five choices. Paul has a. How many cars would you expect every no idea as to what the correct answers are, two minutes on average? so he guesses on each one. Use the binomial b. Using your result from part (a), what is distribution to model the number of correct the probability that there will be exactly answers. 7 cars in a two-minute period? Hint: Is it a. What is the value of n, the number of true that the number of cars arriving in a trials, and the value of p, the probability of two-minute period can be modeled by the success? Poisson distribution?) b. What is the probability that Paul guesses c. What is the probability that there will be none of the six questions correctly? exactly 12 cars in a three-minute period? c. What is the probability that Paul guesses exactly two of the questions correctly? 7. A certain stock has a 60% chance of going up d. What is the probability that Paul gets very and a 40% chance of going down on any given lucky and guesses all six questions corday. Assume that each day is independent of rectly? any other (so whatever happened yesterday does not effect what happens today). For additional exercises, see page 735.