Business Statistics For Contemporary Decision Making 9 th Edition

1. Business Statistics For Contemporary Decision Making9th Edition Ken Black Chapter 5 Discrete Distributions

2. Learning Objectives • Define a random variable in order to differentiate between a discrete distribution and a continuous distribution. • Determine the mean, variance, and standard deviation of a discrete distribution. • Solve problems involving the binomial distribution using the binomial formula and the binomial table. • Solve problems involving the Poisson distribution using the Poisson formula and the Poisson table. • Solve problems involving the hypergeometric distribution using the hypergeometric formula

3. 5.1 Discrete Versus Continuous Distributions A random variable is a variable that contains the outcomes of a chance experiment. • Arandom variable is discrete if the set of all possible values is at most a finite or a countably infinite number of possible values. Examples: • Randomly selecting 25 people who consume soft drinks and determining how many people prefer diet soft drinks • Determining the number of defective items in a batch of 50 items • Counting the number of people who arrive at a store during a five-minute period • Sampling 100 registered voters and determining how many voted for the president in the last election • Discrete distributions have outcomes that generally take on whole number values; you cannot have half a person who prefers diet soft drinks.

4. 5.1 Discrete Versus Continuous Distributions • Arandom variable is continuous if it can take on values at every point over a given interval. Examples: • Sampling the volume of liquid nitrogen in a storage tank • Measuring the time between customer arrivals at a retail outlet • Measuring the lengths of newly designed automobiles • Measuring the weight of grain in a grain elevator at different points of time • Discrete distributions (binomial, Poisson, hypergeometric) are constructed from discrete random variables. • Continuous distributions (uniform, normal, exponential, and others) are constructed from discrete random variables.

5. 5.2 Describing a Discrete Distribution • Ahistogram is the most common graphical way of describing a discrete distribution. • An executive is considering out-of-town business travel for a given Friday. She recognizes that at least one crisis could occur on the day that she is gone and she is concerned about that possibility. Table 5.2 shows a discrete distribution that contains the number of crises that could occur during the day that she is gone and the probability that each number will occur.

6. 5.2 Describing a Discrete Distribution Mean, Variance, and Standard Deviation of Discrete Distributions • The mean or expected value of a discrete distribution is the long run average of occurrences. where long-run average an outcome probability of that outcome • In the long run, the mean or expected number of crises on a given Friday for this executive is 1.15 crises. • However, there will never be exactly 1.15 crises.

7. 5.2 Describing a Discrete Distribution Variance, and Standard Deviation of a Discrete Distribution • The variance of a discrete distribution has the following formula: • The standard deviation is then calculated by taking the square root of the variance: where an outcome probability of a given outcome mean of the distribution

8. 5.2 Describing a Discrete Distribution Variance, and Standard Deviation of a Discrete Distribution

9. 5.3 Binomial Distribution Assumptions of the Binomial Distribution • The experiment involves n identical trials. • Each trial has only two possible outcomes denoted as success or as failure. • Each trial is independent of the previous trials. • The terms p and q remain constant throughout the experiment, where the term p is the probability of getting a success on any one trial and the term q = (1 − p) is the probability of getting a failure on any one trial. • There are many binomial distributions, each characterized by the parameters n (the sample size) and p(the probability of success). • Examples: • Flipping a coin 10 times; probability of heads = .5. • With a known defective rate of 9%, checking 15 products for quality.

10. 5.3 Binomial Distribution Solving a Binomial Problem A survey of relocation administrators by Runzheimer International revealed several reasons why workers reject relocation offers. Included in the list were family considerations, financial reasons, and others. • 4% of the respondents said they rejected relocation offers because they received too little relocation help. • Suppose five workers who just rejected relocation offers are randomly selected and interviewed. • Assuming the 4% figure holds for all workers rejecting relocation, what is the probability that exactly one of the workers rejected the offer because of too little relocation help?

11. 5.3 Binomial Distribution Solving a Binomial Problem A survey of relocation administrators by Runzheimer International revealed several reasons why workers reject relocation offers. Included in the list were family considerations, financial reasons, and others. • 4% of the respondents said they rejected relocation offers because they received too little relocation help. • Suppose five workers who just rejected relocation offers are randomly selected and interviewed. • Assuming the 4% figure holds for all workers rejecting relocation, what is the probability that exactly one of the workers rejected the offer because of too little relocation help?

12. 5.3 Binomial Distribution Solving a Binomial Problem, continued One way to solve this problem would be to figure out all the possible sequences such that there was exactly one worker who rejected the offer due to lack of relocation help (R). • Let T stand for all other reasons. • There are five ways to get a sequence such that there is only one worker who rejects for this reason. • Then, using the special multiplication rule for independent events, the probability of the first sequence is: (.04)(.96)(.96)(.96)(.96) = .03397 Five ways to get only one worker. Probability of each sequence.

13. 5.3 Binomial Distribution Solving a Binomial Problem, continued With five sequences with identical probability, the probability of getting one worker will be A simpler way to determine the number of sequences is to use the combination rule introduced in Chapter 4. • Multiplying the combinations by the probability of each, , gives the binomial formula.

14. 5.3 Binomial Distribution Example: A Gallup survey found that 65% of all financial consumers were very satisfied with their primary financial institution. Suppose that 25 financial consumers are sampled. • If the Gallup survey result still holds true today, what is the probability that exactly 19 are very satisfied with their primary financial institution? • On average, how many very satisfied customers would you expect to get in a sample of 25? • Calculate the mean, (25)(.65) = 16.25. • The average number of very satisfied customers is about 16.

15. 5.3 Binomial Distribution Using the Binomial Table Binomial distributions are described by their sample size and probability, and can be summarized in a table. • Suppose that with n=20 and p=.6, a researcher wants to know the probability of less than 10 successes? Excerpt from the n= 20 table: P(x<10) = .127, found by adding the probability of each number less than 10. Look for correct n and p.

16. 5.3 Binomial Distribution Using the Computer to Product a Binomial Distribution Both Minitab and Excel will print binomial table values or find a binomial probability. • Astudy of bank customers stated that 64% of all financial consumers believe banks are more competitive today than they were five years ago. Suppose 23 financial consumers are selected randomly. What is the probability that ten or fewer believe this? The Minitab (or Excel) output can give the answer directly, or could be used to generate a table for this sample size and probability.

17. 5.3 Binomial Distribution Mean and Standard Deviation of a Binomial Distribution For a binomial distribution: • Example: If 40% of all graduate business students at a large university are women and if random samples of 10 graduate business students are selected many times, how many would you expect to be women? • Calculating the mean, (10)(.4), the expectation is that, on average, four of the 10 students would be women. • The standard deviation would be

18. 5.3 Binomial Distribution Graphing Binomial Distributions • A binomial distribution can be graphed by using all possible x values and their associated probabilities: Peak and skew of the distribution change with the value of p, the probability of success.

19. 5.4 Poisson Distribution Assumptions of the Poisson Distribution • It is a discrete distribution. • It describes rare events. • Each occurrence is independent of the other occurrences. • It describes discrete occurrences over a continuum or interval. • The occurrences in each interval can range from zero to infinity. • The expected number of occurrences must hold constant throughout the experiment. • The Poisson distribution describes the occurrence of rare events. • Examples: • Number of telephone calls per minute at a small business • Number of hazardous waste sites per county in the United States

20. 5.4 Poisson Distribution If a Poisson-distributed phenomenon is studied over a long period of time, a long-run average can be determined, λ. where 0, 1, 2, 3,… long-run average occurrences 2.718282 • The λ value must hold constant throughout a Poisson experiment. • The researcher must be careful not to apply a given lambda to intervals for which lambda changes. • For example, the average number of customers arriving at a Macy’s store during a one-minute interval will vary from hour to hour, day to day, and month to month.

21. 5.4 Poisson Distribution Working Poisson Problems by Formula Example: Suppose bank customers arrive randomly on weekday afternoons at an average of 3.2 customers every 4 minutes. What is the probability of exactly 5 customers arriving in a 4-minute interval on a weekday afternoon? • The lambda for this problem is 3.2 customers per 4 minutes. • The value of x is 5 customers per 4 minutes. • The probability of 5 customers randomly arriving during a 4-minute interval when the long-run average has been 3.2 customers per 4-minute interval is .1140.

22. 5.4 Poisson Distribution What to Do When Intervals Are Different • Lambda and x intervals must be the same. Example: Suppose that on Saturday mornings, a specialty clothing store averages 2.4 customer arrivals every 10 minutes. What is the probability that, on a given Saturday morning, 2 customers will arrive at the store in a 6-minute interval? • λ = 2.4 customers/10 minutes • x = 2 customers/6 minutes • Must change the λ interval, NOT the x interval. • 0.6(2.4 customers) = 1.44 customers, and 0.6(10 minutes) = 6 minutes. • New λ= 1.44 customers/6 minutes • P(x=2) = = .2456

23. 5.4 Poisson Distribution Using the Poisson Tables • For every value of λ, there is a different Poisson distribution, but the distribution is the same for every λ regardless of the interval, so tables can be used. Example: Suppose that during the noon hour in the holiday season, a UPS store averages 2.3 customers every minute and that arrivals at such a store are Poisson distributed. During such a season and time, what is the probability that more than four customers will arrive in a given minute?

24. 5.4 Poisson Distribution Mean and Standard Deviation of a Poisson Distribution • The mean of a Poisson distribution is λ, the long-run average of occurrences. • The variance is also λ, so the standard deviation is . Graphing the Poisson Distribution • The height and skew of the distribution are determined by λ. λ = 6.5 λ = 1.6

25. 5.4 Poisson Distribution Using the Computer to Generate Poisson Distributions • Both Minitab and Excel can generate Poisson tables for any value of λ. Example: One study by the National Center for Health Statistics claims that, on average, an American has 1.9 acute illnesses or injuries per year. If these cases are Poisson distributed, lambda is 1.9 per year. What does the Poisson probability distribution for this lambda look like?

26. 5.4 Poisson Distribution Approximating Binomial Problems by the Poisson Distribution • Binomial problems with large sample sizes and small values of p can be approximated by the Poisson distribution. • If n > 20 and np < 7, the approximation can be used. • Example: The following binomial distribution problem can be worked by using the Poisson distribution: n = 50 and p = .03. What is the probability that x = 4? • First find the mean of the binomial distribution, np, which will be used for the λ value. • λ = (50)(.03) = 1.5 • P(x=4) = = .0471

27. 5.4 Poisson Distribution Approximating Binomial Problems by the Poisson Distribution, continued. • The Poisson approximation gave a probability of .0471 • Using the binomial formula would give a probability of .0459 Binomial Poisson approximation

28. 5.5 Hypergeometric Distribution • Like the binomial distribution, the hypergeometric distribution has two outcomes, success or failure. • Unlike the binomial distribution, the researcher must know the size of the population and the probability of success in the population. Characteristics of the Hypergeometric Distribution: • It is a discrete distribution. • Each outcome consists of either a success or a failure. • Sampling is done without replacement. • The population, N, is finite and known. • The number of successes in the population, A, is known.

29. 5.5 Hypergeometric Distribution • Like the binomial distribution, the hypergeometric distribution has two outcomes, success or failure. • Unlike the binomial distribution, the researcher must know the size of the population and the probability of success in the population. • Should be used instead of binomial when sampling is done without replacement and the sample is greater than or equal to 5% of the population. Characteristics of the Hypergeometric Distribution: • It is a discrete distribution. • Each outcome consists of either a success or a failure. • Sampling is done without replacement. • The population, N, is finite and known. • The number of successes in the population, A, is known.

30. 5.5 Hypergeometric Distribution • Hypergeometric formula where N = population size, n = sample size, A = number of successes in the population, and x is the number of successes in the sample. • Hypergeometric distributions are characterized by three parameters, N, A, and n, so creating tables is nearly impossible. Example: Twenty-four people, of whom eight are women, apply for a job. If five of the applicants are sampled randomly, what is the probability that exactly three of those sampled are women? • Small, finite population • Sample size is 21% of the population • Hypergeometric is the appropriate distribution

31. 5.5 Hypergeometric Distribution • Example, continued. Twenty-four people, of whom eight are women, apply for a job. If five of the applicants are sampled randomly, what is the probability that exactly three of those sampled are women? • A, the number of occurrences in the population, is 8. • After evaluating each counting rule in the formula, • In a sample of 5, there is a 15.81% chance that three will be women.

32. 5.5 Hypergeometric Distribution Using the Computer to Generate Hypergeometric Distribution Probabilities • Both Excel and Minitab can find hypergeometric probabilities for given values of N, A, n, and x. • For the previous example problem, Excel and Minitab gave the following outputs: