1. 1 1 Slide Discrete Probability Distributions (Random Variables and Discrete Probability Distributions) Chapter 5 BA 201

2. 2 2 Slide RANDOM VARIABLES

3. 3 3 Slide A random variable is a numerical description of the outcome of an experiment. Random Variables A discrete random variable may assume either a finite number of values or an infinite sequence of values. Example: 1, 2, 3, … A continuous random variable may assume any numerical value in an interval or collection of intervals. Example: 1.0, 1.1, 1.2, …

4. 4 4 Slide Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4) JSL Appliances Discrete Random Variable with a Finite Number of Values We can count the TVs sold, and there is a finite upper limit on the number that might be sold (which is the number of TVs in stock).

5. 5 5 Slide Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2,... Discrete Random Variable with an Infinite Sequence of Values We can count the customers arriving, but there is no finite upper limit on the number that might arrive. JSL Appliances

6. 6 6 Slide Random Variables Question Random Variable x Type Family size x = Number of dependents reported on tax return Discrete Distance from home to store x = Distance in miles from home to the store site Continuous Own dog or cat x = 1 if own no pet; = 2 if own dog(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s) Discrete

7. 7 7 Slide DISCRETE PROBABILITY DISTRIBUTIONS

8. 8 8 Slide The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. We can describe a discrete probability distribution with a table, graph, or formula. Discrete Probability Distributions

9. 9 9 Slide The probability distribution is defined by a probability function, denoted by f ( x ), which provides the probability for each value of the random variable. The required conditions for a discrete probability function are: Discrete Probability Distributions f ( x ) > 0  f ( x ) = 1

10. 10 Slide Discrete Probability Distributions JSL Appliances Units Sold (x) Number of Days f(x) 0800.40 1500.25 2400.20 3100.05 4200.10 2001.00 80/200

11. 11 Slide.10.20.30. 40.50 0 1 2 3 4 Values of Random Variable x (TV sales) Probability Discrete Probability Distributions JSL Appliances Graphicalrepresentation of probability distributionGraphicalrepresentation distribution

12. 12 Slide Expected Value The expected value, or mean, of a random variable is a measure of its central location. The expected value is a weighted average of the values the random variable may assume. The weights are the probabilities. The expected value does not have to be a value the random variable can assume. E ( x ) =  =  x f ( x )

13. 13 Slide Variance and Standard Deviation The variance summarizes the variability in the values of a random variable. The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities. Var( x ) =  2 =  ( x -  ) 2 f ( x ) The standard deviation, , is defined as the positive square root of the variance.

14. 14 Slide Expected Value JSL Appliances Units Sold (x) Number of Days f(x)x f(x) 0800.400.00 1500.25 2400.200.40 3100.050.15 4200.100.40 E(x)1.20 expected number of TVs sold in a day 0 * 0.40

15. 15 Slide  Standard deviation of daily sales =  = 1.2884 TVs Variance JSL Appliances xf(x)x f(x) x -  (x -  ) 2 (x -  ) 2 f(x) 00.400.00-1.201.440.576 10.25 -0.200.040.010 20.200.400.800.640.128 30.050.151.803.240.162 40.100.402.807.840.784 Variance of daily sales =  2 1.660

16. 16 Slide Discrete Uniform Probability Distribution The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by a formula. The discrete uniform probability function is f ( x ) = 1/ n where: n = the number of values the random variable may assume the values of the random variable are equally likely

17. 17 Slide PRACTICE DISCRETE PROBABILITY DISTRIBUTIONS

18. 18 Slide Scenario Amount (x) Number of Donors 535 1025 1520 15 255 A non-profit sends donation solicitations with pre-printed amounts. Based on past years, the non-profit knows approximately how many people will donate each amount. Compute the expected value of x, variance, and standard deviation.

19. 19 Slide f(x) Amount (x) Number of Donors f(x)x f(x) 535 1025 1520 15 255 E(x)

20. 20 Slide Variance and Standard Deviation xDonorsf(x)x*f(x) (x -  )(x -  ) 2 (x -  ) 2 f(x) 535 1025 1520 15 255 E(x) 22 

21. 21 Slide BINOMIAL PROBABILITY DISTRIBUTION

22. 22 Slide Binomial Probability Distribution n Four Properties of a Binomial Experiment 3. The probability of a success, denoted by p, does not change from trial to trial. 4. The trials are independent. 2. Two outcomes, success and failure, are possible on each trial. 1. The experiment consists of a sequence of n identical trials. stationarity assumption

23. 23 Slide Binomial Probability Distribution Our interest is in the number of successes occurring in the n trials. We let x denote the number of successes occurring in the n trials.

24. 24 Slide where: x = the number of successes p = the probability of a success on one trial n = the number of trials f ( x ) = the probability of x successes in n trials Binomial Probability Distribution n Binomial Probability Function

25. 25 Slide Binomial Probability Distribution n Binomial Probability Function Probability of a particular sequence of trial outcomes with x successes in n trials Probability of a particular sequence of trial outcomes with x successes in n trials Number of experimental outcomes providing exactly x successes in n trials Number of experimental outcomes providing exactly x successes in n trials

26. 26 Slide Binomial Probability Distribution Evans Electronics Evans Electronics is concerned about a low retention rate for its employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.

27. 27 Slide Binomial Probability Distribution Evans Electronics The probability of the first employee leaving and the second and third employees staying, denoted ( S, F, F ), is given by p (1 – p )(1 – p ) With a 0.10 probability of an employee leaving on any one trial, the probability of an employee leaving on the first trial and not on the second and third trials is given by (0.10)(0.90)(0.90) = (0.10)(0.90) 2 = 0.081

28. 28 Slide Binomial Probability Distribution Evans Electronics Two other experimental outcomes also result in one success and two failures. The probabilities for all three experimental outcomes involving one success follow. Experimental Outcome ( S, F, F ) ( F, S, F ) ( F, F, S ) Probability of Experimental Outcome p (1 – p )(1 – p ) = (0.1)(0.9)(0.9) = 0.081 (1 – p ) p (1 – p ) = (0.9)(0.1)(0.9) = 0.081 (1 – p )(1 – p ) p = (0.9)(0.9)(0.1) = 0.081 Total = 0.243

29. 29 Slide L (0.1) S (0.9) Binomial Probability Distribution 1 st Worker2 nd Worker3 rd Worker x Prob. L (0.1) S (0.9) 3 2 0 2 2 0.0010 0.0090 0.7290 0.0090 1 1 0.0810 S (0.9) L (0.1) S (0.9) L (0.1) S (0.9) L (0.1) 1 Evans Electronics Using a tree diagram S (0.9) L (0.1)

30. 30 Slide Binomial Probability Distribution Let : p = 0.10, n = 3, x = 1 Evans Electronics Using the probabilityfunction probabilityfunction

31. 31 Slide Binomial Probability Distribution E ( x ) =  = np Var( x ) =  2 = np (1  p ) n n Expected Value n n Variance n n Standard Deviation

32. 32 Slide Binomial Probability Distribution E ( x ) = np = 3(.1) = 0.3 employees out of 3 Var( x ) = np (1 – p ) = 3(0.1)(0.9) = 0.27 Expected Value Variance Standard Deviation Evans Electronics

33. 33 Slide PRACTICE BINOMIAL PROBABILITY DISTRIBUTIONS

34. 34 Slide Scenario Kearn Manufacturing (KM) submits bids for sales a number of times each month. Based on an analysis of past bids, the sales director knows that on average 34% of bids result in sales. KM Last week the sales director submitted three bids. The sales director would like to know how likely it is that two of the bids will result in sales. 1. 1. What are n, x (that two bids result in sales), and p. 2. 2. What is f(x)? 3. 3. Draw a tree diagram illustrating this scenario. 4. 4. Compute the expected value of x. 5. 5. Compute the variance and standard deviation.

35. 35 Slide Bids and Sales p? n? x? f(x)? Tree diagram.

36. 36 Slide Bids and Sales Compute E(x), Var(x), and the standard deviation.

37. 37 Slide Binomial as a Discrete Probability Distribution n3 x2 p0.34 pcpc 0.66 E(x)=  1.0200 =n*p Var(x)0.6732 =n*p*(1-p)

38. 38 Slide Binomial as a Discrete Probability Distribution xf(x)x*f(x) (x-  )(x-  ) 2 (x-  ) 2 *f(x) 00.28750.0000-1.02001.04040.2991 10.4443 -0.02000.00040.0002 20.22890.45780.98000.96040.2198 30.03930.11791.98003.92040.1541  1.0200  0.6732 E(x)=  1.0200 =n*p Var(x)0.6732 =n*p*(1-p)

39. 39 Slide POISSON PROBABILITY DISTRIBUTION

40. 40 Slide A Poisson distributed random variable is often useful in estimating the number of occurrences over a specified interval of time or space It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2,... ). Poisson Probability Distribution

41. 41 Slide Examples of a Poisson distributed random variable: the number of knotholes in 14 linear feet of pine board the number of vehicles arriving at a toll booth in one hour Poisson Probability Distribution Bell Labs used the Poisson distribution to model the arrival of phone calls.

42. 42 Slide Poisson Probability Distribution n Two Properties of a Poisson Experiment 2. 2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval. 1. 1. The probability of an occurrence is the same for any two intervals of equal length.

43. 43 Slide n Poisson Probability Function Poisson Probability Distribution where: x = the number of occurrences in an interval f(x) = the probability of x occurrences in an interval  = mean number of occurrences in an interval e = 2.71828

44. 44 Slide Poisson Probability Distribution n Poisson Probability Function In practical applications, x will eventually become large enough so that f ( x ) is approximately zero and the probability of any larger values of x becomes negligible. Since there is no stated upper limit for the number of occurrences, the probability function f ( x ) is applicable for values x = 0, 1, 2, … without limit.

45. 45 Slide Poisson Probability Distribution Mercy Hospital Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening?

46. 46 Slide Poisson Probability Distribution  = 6/hour = 3/half-hour, x = 4 Mercy Hospital Using the probabilityfunction probabilityfunction Note: Example:

47. 47 Slide Poisson Probability Distribution Poisson Probabilities 0.00 0.05 0.10 0.15 0.20 0.25 0 12345678910 Number of Arrivals in 30 Minutes Probability actually, the sequencecontinues: 11, 12, … Mercy Hospital

48. 48 Slide Poisson Probability Distribution A property of the Poisson distribution is that the mean and variance are equal.   =  2

49. 49 Slide Poisson Probability Distribution Variance for Number of Arrivals During 30-Minute Periods  =  2 = 3 Mercy Hospital

50. 50 Slide PRACTICE POISSON PROBABILITY DISTRIBUTION

51. 51 Slide Scenario On weekdays between 11:00 a.m. and 2:00 p.m., Joe ’ s Pizza receives approximately one order every two minutes. 1. 1. What is the expected number of orders per hour? 2. 2. What is the probability three orders will arrive in a five-minute period? 3. 3. What is the probability no orders will arrive in a five- minute period? 4. 4. What is the variance of the number of orders arriving?

52. 52 Slide Joe’s Pizza 1 order every 2 minutes. Average # of orders every 5 minutes? f(3)? f(0)?

53. 53 Slide Joe’s Pizza 1 order every 2 minutes.

54. 54 Slide

55. 55 Slide OVERVIEW

56. 56 Slide Discrete Probability Distributions x takes discrete values. Values for f(x) may be subjectively assigned, assigned based on frequency of occurrence, or uniform. o o If f(x) is uniform, f(x) = 1/n where n is the number of values x may assume. E(x) = μ =  [x * f(x)] Var(x) =  2 =  [(x – μ ) 2 * f(x)]

57. 57 Slide Binomial Probability Distributions Four properties: (1) n identical trials, (2) success or failure, (3) p does not change, and (4) trials are independent. x is the number of successes. E(x) = μ = n * p Var(x) =  2 = n * p * (1-p)

58. 58 Slide Poisson Probability Distributions Number of occurrences over a specified interval of time or in space. (1) Probability of occurrence e is same for any two intervals of equal length and (2) occurrences are independent. x is the number of occurrences in the interval. μ =  2

59. 59 Slide