1. 1 STAT 500 – Statistics for Managers STAT 500 Statistics for Managers

2. 2 STAT 500 – Statistics for Managers Agenda for this Session Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications

3. 3 STAT 500 – Statistics for Managers Agenda for this Session# 3 Part 2 Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications

4. 4 STAT 500 – Statistics for Managers Any random variable whose only possible values are 0 and 1 is called a Bernoulli random variable. The outcome 0 is called failure and the outcome 1 is called success. Probability of success is: p Examples: Tossing a coin (Head/Tail), War between two countries (Victory/Defeat), Taking an examination (pass/Fail), bidding for a contract (win/lose) Parameters:

5. 5 STAT 500 – Statistics for Managers Agenda for this Session# 3 Part 2 Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications

6. 6 STAT 500 – Statistics for Managers Binomial random variable is the number of successes in many trials of Bernoulli experiment. Formulation of Business Problem Models using Binomial Distribution  Quality control problems in which units are good or bad, processes are either in control or out of control.  Sales personnel either make a sale or fail to make a sale whenever a customer walks in to buy a car.  Auditors sample clients’ accounts to determine if any errors exist. At each trial an error exists or does not exist

7. 7 STAT 500 – Statistics for Managers Let  p be the probability of success in each trial of the experiment.  n be the total number of trials.  x be the number of successes out of n trials  P(x) be the probability of x successes

8. 8 STAT 500 – Statistics for Managers Parameters of the Binomial Distribution:

9. Using Excel To determine P(x) Obtain this panel in Excel by choosing Insert, Function, Statistical, BINOMDIST

10. Using Binomial tables at the end of the book To determine the P(x), the probability of x successes out of n trials with p, the probability of success in each trial: 1)Locate the appropriate table corresponding to the number of trials, n 2)Locate the probability P(x) by going west to east (horizontally) for x, and North to South for p 3) P(x) = 0.1323 for n = 5, x = 3, and p = 0.3

11. 11 STAT 500 – Statistics for Managers Agenda for this Session# 3 Part 2 Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications

12. 12 STAT 500 – Statistics for Managers The Poisson random variable is a special class of a binomial random variable for which n is large and unknown, p is small and unknown, but np is known. It is the mean rate, lambda. Formulation of Business Problem Models using Poisson Distribution  The number of earthquakes during some fixed time span.  The number of wars per year  The number of deaths in a given period of time for the policyholders of a life insurance company.  The number of misprints on the page of a book  The number of wrong telephone numbers that are dialed in a day

13. 13 STAT 500 – Statistics for Managers Parameters of the Poisson Distribution: Usually lambda, the average rate is expressed in units per time (e.g., 10 trains per hour) or units per space interval (e.g., Number defective spots per square yard of cloth)

14. Using Excel To determine P(x) Obtain this panel in Excel by choosing Insert, Function, Statistical, POISSON

15. 15 STAT 500 – Statistics for Managers A note on Cumulative Distribution may be in Order. Let us say, we are interested in determining the probability of x <= 2, we will go by P (x <= 2) = P (x = 0) + P(x = 1) + P (x = 2). In the case of both Poisson and Binomial distributions, we will get the values of P (x = 0), P(x = 1), and P (x = 2) and then add them up. Alternatively, there are tables available (known as cumulative tables) that would give the cumulative values directly, i.e., P (x <= n). In Excel, you can generate the cumulative probabilities by setting the cumulative logical variable to TRUE.

16. 16 STAT 500 – Statistics for Managers Agenda for this Session# 3 Part 2 Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications

17. Geometric Distribution A geometric random variable differs from other types of random variables in one crucial aspect--its value is the number of trials that are performed until the first success occurs, where the first successful trial is included in that number.  Probability Formula: Let x be the number of trials required to get the first success Let p be probability of success in one trial P(x) is the probability of getting the first success at the x th trial

18. Parameters of a Geometric Random Variable Example: Keep buying lottery tickets until you win.

19. 19 STAT 500 – Statistics for Managers Agenda for this Session# 3 Part 2 Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications

20. Application # 1 The “Witches of Wall Street” are people who predict whether a stock will go up or down using tarot cards, astrological readings or other supernatural means. One such “witch” claims she can correctly predict a daily increase or decrease in price for any stock 60% of the time. In order to test her, a broker selects five stocks at random from the New York Stock Exchange and asks the “witch” to predict whether the price will increase or decrease the next day. Assume that the five predictions are made independently. If the “witch's” claim is true, a) what is the probability that she correctly predicts an increase or decrease in all 5 stocks? b) what is the probability that she makes an incorrect prediction for all five stocks? c) Suppose you observe four incorrect predictions by the “witch” do you believe her claim?

21. Application # 1 (a) p, the probability of success in each trial = 0.6 n, the number of trials = 5 x, The number of successes = 5 P (x = 5) = =BINOMDIST(5,5,0.6,FALSE) = 0.0778 (b) p, the probability of success in each trial = 0.6 n, the number of trials = 5 x, The number of successes = 0 P (x = 0) =BINOMDIST(0,5,0.6,FALSE) = 0.01024 (c) p, the probability of success in each trial = 0.6 n, the number of trials = 5 x, The number of successes = 1 P (x = 1) =BINOMDIST(1,5,0.5,FALSE) = 0.0768. There is About 8% chance that she can come out with 4 wrong

22. Application # 2 An airplane has 18 seats, but 20 reservations, because from past experience the airline knows that there is a 10% chance that someone who holds a ticket will not show up. What is the probability that the airline will have to bump someone? p, the probability of passenger showing up = 0.9 n, the number of trials = 20 A passenger is bumped when the number of passengers Showed exceeds 18, i.e., 19 or 20 P (x = 19 or x = 20) = P (x = 19) + P (x = 20) BINOMDIST(19,20,0.9,FALSE)+ BINOMDIST(20,20,0.9,FALSE) =0.2702 + 0.1216 = 0.3918

23. Application # 3 You are taking a multiple choice examination with 20 questions. Each question has 5 choices. What is the probability that a person choosing to answer questions at random would get no more than 5 questions correct? What is the probability that a person choosing to answer questions at random would get 10 or more questions correct? What is the probability that a person choosing to answer questions at random would get at least 8 questions correct?

24. Application # 3 (a) p, the probability of success in each trial = 0.2 n, the number of trials = 20 x, The number of successes = no more than 5 P (x <= 5) = =BINOMDIST(5,20,0.2,True) = 0.8042 (b) p, the probability of success in each trial = 0.2 n, the number of trials = 20 x, The number of successes = 10 or more P (x >= 10) = 1 – P (x <=9) = 1 - BINOMDIST(9,20,0.2,True) = 1 – 0.9974 = 0.0026 (c) p, the probability of success in each trial = 0.2 n, the number of trials = 20 x, The number of successes = at least 8, i.e., 8 or more P (x >= 8) = 1 – P (x <=7) = 1 - BINOMDIST(7,20,0.2,True) = 1 – 0.9679 = 0.0321

25. Application # 4 A firm bills its accounts on a 2% discount for payment in ten days and the full amount after 30 days. In the past, 30% of all invoices have been paid within ten days. During the first week of October, the firm sends out 10 invoices. If you assume independence, what is the probability that: No one takes the discount? The number of trials, n = 10; number of successes: 0 Probability of success, p = 0.3 P(x = 0) = BINOMDIST(0,10,0.3,False) = 0.0282 At least one takes the discount? P(x = 0) = 0.0282 P (x >= 1) = 1 – P(x = 0) = 1 – 0.0282 = 0.9718

26. Application # 4 Every one takes the discount? P(x = 10) = BINOMDIST (10,10,0.3,False) = Very Small, Close to zero. Exactly three take the discount? P(x = 3) = BINOMDIST(3,10,0.3,False) = 0.2668 At most three take the discount? P(x <=3) = P (x = 0) + P (x = 1) + P(x = 2) + P (x = 3) P (x <= 3) = BINOMDIST(3,10,0.3,True) = 0.6496 At least four take the discount? P(x >=4) = 1 – P (x <=3) = 1 – 0.6496 = 0.3504

27. Application # 5 The average number of thread defects in standard bolt of cloth produced by New York Weavers Inc. is 6. Assume the thread distribution of thread defects follows Poisson distribution. What is the probability that a bolt of cloth will have exactly five defects? Lambda = Average Defect Rate = 6 per bolt of cloth. P (x = 5) = POISSON (5,6,False) = 0.1606 What is the probability that a half-bolt cloth will have four defects? Lambda = Average Defect Rate = 3 per half-bolt of cloth. P (x = 4) = POISSON (4,3,False) = 0.1680

28. Application # 5 Airline passengers arrive randomly and independently at the passenger screening facility at a major international airport. The mean arrival rate is 12 passengers per minute. What is the probability of no arrivals in a 1-minute period? Lambda = Average arrival Rate = 12 passenger per minute. P (x = 0) = POISSON (0,10,False) = Very Small, close to zero. What is the probability of no arrivals in a 15-second period? Lambda = Average arrival Rate = 3 passenger per 15- second. P (x = 0) = POISSON (0,3,False) = 0.0498 What is the probability of at least one arrival in a 15-second period? P (x >= 1) = 1 – P (x = 0) = 1 – 0.0498 = 0.9502

29. 29 STAT 500 – Statistics for Managers Agenda for this Session# 3 Part 2 Bernoulli Random Variables Binomial Distribution Poisson Distribution Geometric Distribution Applications