Chapter 8 Binomial and Geometric Distribu7ons

Transcription

1 Chapter 8 Binomial and Geometric Distribu7ons 8.2 Geometric Distributions Children s cereals sometimes contain prizes. Imagine that packages of Chocolate- Coated Sugar Bombs contain one of three baseball cards: Mark McGwire, Sammy Sosa, or Barry Bonds. Shak wanted to get a Sammy Sosa card and had to buy eight boxes until getting his desired card. Shak feels especially unlucky. Should Shak consider himself especially unlucky? On average, how many boxes would a person have to buy to get the Sammy Sosa card? In this activity, you will become familiar with the geometric distribution, including the shape of the distribution and how to find its mean. 1

2 1. Roll your die: If the side with 1 or 2 dots lands up, this will represent the event of buying a box of Chocolate- Coated Sugar Bombs and ge@ng a Sammy Sosa card. If one of the other sides lands on top, roll again. Count the number of rolls unbl you get a 1 or 2. Add to the class histogram showing the number of rolls it took to get your first Sammy Sosa card. (Do this 10 Bmes and record each on the class histogram so that we have more data.) Describe the shape of this distribubon. What was the average number of boxes purchased to get a Sammy Sosa card? EsBmate the chance that Shak would have to buy eight or more boxes to get his card. What assumpbons did you make in this simulabon about the distribubon of prizes? Do you think they are reasonable ones? 1. Roll your die: 2. Doubles: In some games, a player must roll doubles before conbnuing, such as when in jail in Monopoly. Use a pair of dice or random numbers to simulate rolling a pair of dice. Count the number of rolls unbl you get doubles. Add your informabon to the class histogram of the number of rolls our class required to roll doubles. (Do this 10 Bmes, too.) Describe the shape of the distribubon. What was the average number of rolls required? 2

3 1. Roll your die: 2. Doubles: 3. Wai5ng Time Distribu5on: In steps 1 & 2, you constructed a waibng- Bme distribubon using simulabon. Now construct a theorebcal waibng- Bme distribubon for ge@ng a different cereal prize. Boxes of Post s Coca Pebbles recently contained one of four endangered animal sbckers: a parrot, an African elephant, a Bger, or a crocodile. Suppose 4096 children want a sbcker of a parrot. How many of them would you expect to get a parrot in the first box of Cocoa Pebbles they buy? What assumpbons are you making? How many children do you expect will have to buy a second box? How many of them do you expect will get a parrot in the second box? 1. Roll your die: 2. Doubles: 3. WaiBng Time DistribuBon: 4. Fill in the following table: Number of Boxes Purchased to get first Parrot Sticker 1 20 Number of Children Make a histogram of the theorebcal waibng- Bme distribubon. The height of each bar of the histogram is what proporbon of the height of the bar to its lea? What is the average number of boxes purchased? 3

4 Wrap-Up 1. Describe the shape of a waiting-time (geometric) distribution for a given probability p of success on each trial. Will the first bar in a waiting-time distribution always be the highest? Why or why not? The height of each bar is what proportion of the height of the bar to its left? 2. What are some other situations that can be modeled by a waiting-time distribution? Geometric Settings In a binomial setting, the number of trials n is fixed and the binomial random variable X counts the number of successes. In other situations, the goal is to repeat a chance behavior until a success occurs. These situations are called geometric settings. Defini7on: A geometric se;ng arises when we perform independent trials of the same chance process and record the number of trials un;l a par;cular outcome occurs. The four condi;ons for a geometric se?ng are B I T S Binary? The possible outcomes of each trial can be classified as success or failure. Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial. Trials? The goal is to count the number of trials un;l the first success occurs. Success? On each trial, the probability p of success must be the same. 4

5 Geometric Random Variable In a geometric setting, if we define the random variable Y to be the number of trials needed to get the first success, then Y is called a geometric random variable. The probability distribution of Y is called a geometric distribution. Definition: The number of trials Y that it takes to get a success in a geometric setting is a geometric random variable. The probability distribution of Y is a geometric distribution with parameter p, the probability of a success on any trial. The possible values of Y are 1, 2, 3,. Note: Like binomial random variables, it is important to be able to distinguish situations in which the geometric distribution does and doesn t apply! Example: The Birthday Game Ms. Raskin is planning to give you 10 problems for homework. As an alternative, you can agree to play the Birthday Game. Here s how it works. A student will be selected at random from the class and asked to guess the day of the week on which Ms. Raskin s best friend was born. If the student guesses correctly, the class will have only one homework problem. If the student guesses the wrong day of the week, Ms. Raskin will once again select a student from the class at random. The chosen student will try to guess the day of the week on which a different one of Ms. Raskin s many friends was born. If this student gets it right, the class will have two homework problems. The game continues until a student correctly guesses the day on which one of Ms. Raskin s many friends was born. Ms. Raskin will assign a number of homework problems that is equal to the total number of guesses made by members of the class. Are you ready to play the Birthday Game? 5

6 Example: The Birthday Game The random variable of interest in this game is Y = the number of guesses it takes to correctly identify the birth day of one of your teacher s friends. What is the probability the first student guesses correctly? The second? Third? What is the probability the k th student guesses correctly? Verify that Y is a geometric random variable. B: Success = correct guess, Failure = incorrect guess I: The result of one student s guess has no effect on the result of any other guess. T: We re counting the number of guesses up to and including the first correct guess. S: On each trial, the probability of a correct guess is 1/7. Calculate P(Y = 1), P(Y = 2), P(Y = 3), and P(Y = k) P(Y =1) =1/7 P(Y = 2) = (6 /7)(1/7) = P(Y = 3) = (6 /7)(6/7)(1/7) = NoBce the pabern? Geometric Probability If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3,. If k is any one of these values, P(Y = k) = (1 p) k 1 p Example: Monopoly In the board game Monopoly, one way to get out of jail is to roll doubles. How likely is it that someone in jail would roll doubles on his first, second, or third attempt? If this was the only way to get out of jail, how many turns would it take, on average? The random variable of interest in this game is Y = the number of attempts it takes to roll doubles once. What is the probability of rolling doubles the first attempt? The second? Third? What is the probability of rolling doubles on the k th attempt? Verify that Y is a geometric random variable. B: Success = roll doubles, Failure = not rolling doubles I: The result of one roll has no effect on other rolls. T: We re counting the number of rolls up to and including the first doubles roll. S: On each trial, the probability of a correct guess is 1/6. Calculate P(Y = 1), P(Y = 2), P(Y = 3), and P(Y = k) P(Y =1) =1/6 P(Y = 2) = (5 /6)(1/6) = P(Y = 3) = (5 /6)(5/6)(1/6) = NoBce the pabern? Geometric Probability If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3,. If k is any one of these values, P(Y = k) = (1 p) k 1 p 6

7 Mean of a Geometric Distribution The table below shows part of the probability distribution of Y. We can t show the entire distribution because the number of trials it takes to get the first success could be an incredibly large number. y i p i Shape: The heavily right-skewed shape is characteristic of any geometric distribution. That s because the most likely value is 1. Center: The mean of Y is µ Y = 7. We d expect it to take 7 guesses to get our first success. Spread: The standard deviation of Y is σ Y = If the class played the Birth Day game many times, the number of homework problems the students receive would differ from 7 by an average of Mean (Expected Value) of Geometric Random Variable If Y is a geometric random variable with probability p of success on each trial, then its mean (expected value) is E(Y) = µ Y = 1/p. Geometric Probability on the Calculator geometpdf(n,p,k) computes P(Y = k) geometcdf(n,p,k) computes P(Y < k) TI-83/84: These commands are found in the distributions menu (2 nd /VARS) TI-89: These commands are found in CATALOG under Flash Apps For the Birthday Game with probability of success p = 1/7 on each trial: P(Y = 10) = geometpdf(1/7, 10) = To find P(Y < 10) = geometcdf(1/7, 10) =

8 Summary In this section, we learned that ü A binomial setting consists of n independent trials of the same chance process, each resulting in a success or a failure, with probability of success p on each trial. The count X of successes is a binomial random variable. Its probability distribution is a binomial distribution. ü The binomial coefficient counts the number of ways k successes can be arranged among n trials. ü If X has the binomial distribution with parameters n and p, the possible values of X are the whole numbers 0, 1, 2,..., n. The binomial probability P(X of observing = k) = n p k (1 successes p) n k in n trials is k Summary In this section, we learned that ü The mean and standard deviation of a binomial random variable X are µ X = np σ X = np(1 p) ü The Normal approximation to the binomial distribution says that if X is a count having the binomial distribution with parameters n and p, then when n is large, X is approximately Normally distributed. We will use this approximation when np 10 and n(1 - p) 10. 8

9 Summary In this section, we learned that ü A geometric setting consists of repeated trials of the same chance process in which each trial results in a success or a failure; trials are independent; each trial has the same probability p of success; and the goal is to count the number of trials until the first success occurs. If Y = the number of trials required to obtain the first success, then Y is a geometric random variable. Its probability distribution is called a geometric distribution. ü If Y has the geometric distribution with probability of success p, the possible values of Y are the positive integers 1, 2, 3,.... The geometric probability that Y takes any value is P(Y = k) = (1 p) k 1 p ü The mean (expected value) of a geometric random variable Y is 1/p. 9