1. McGraw-Hill/Irwin Copyright © 2013 by The McGraw-Hill Companies, Inc. All rights reserved. A PowerPoint Presentation Package to Accompany Applied Statistics in Business & Economics, 4 th edition David P. Doane and Lori E. Seward Prepared by Lloyd R. Jaisingh

2. 7-2 Continuous Probability Distributions Chapter Contents 7.1 Describing a Continuous Distribution 7.2 Uniform Continuous Distribution 7.3 Normal Distribution 7.4 Standard Normal Distribution 7.5 Normal Approximations 7.6 Exponential Distribution 7.7 Triangular Distribution (Optional) Chapter 7

3. 7-3 Chapter Learning Objectives (LO’s) LO7-1 : Define a continuous random variable. LO7-2: Calculate uniform probabilities. LO7-3: Know the form and parameters of the normal distribution. LO7-4: Find the normal probability for given z or x using tables or Excel. LO7-5: Solve for z or x for a given normal probability using tables or Excel. Chapter 7 Continuous Probability Distributions

4. 7-4 Chapter Learning Objectives (LO’s) LO6: Use the normal approximation to a binomial or a Poisson distribution. distribution. LO7: Find the exponential probability for a given x. LO8: Solve for x for given exponential probability. LO9: Use the triangular distribution for “what-if” analysis (optional). Chapter 7 Continuous Probability Distributions

5. Discrete Variable – each value of X has its own probability P(X).Discrete Variable – each value of X has its own probability P(X). Continuous Variable – events are intervals and probabilities are areas under continuous curves. A single point has no probability.Continuous Variable – events are intervals and probabilities are areas under continuous curves. A single point has no probability. Events as Intervals Events as Intervals Chapter 7 7.1 Describing a Continuous Distribution LO7-1: Define a continuous random variable. LO7-1 7-5

6. Continuous PDF’s: Denoted f (x) Must be nonnegative Total area under curve = 1 Mean, variance and shape depend on the PDF parameters Reveals the shape of the distribution PDF – Probability Density Function PDF – Probability Density Function Chapter 7 7.1 Describing a Continuous Distribution LO7-1 7-6

7. Continuous CDF’s: Denoted F(x) Shows P(X ≤ x), the cumulative proportion of scores Useful for finding probabilities CDF – Cumulative Distribution Function Chapter 7 7.1 Describing a Continuous Distribution LO7-1 7-7

8. Continuous probability functions: Continuous probability functions: Unlike discrete distributions, the probability at any single point = 0.Unlike discrete distributions, the probability at any single point = 0. The entire area under any PDF, by definition,The entire area under any PDF, by definition, is set to 1. is set to 1. Mean is the balance point of the distribution.Mean is the balance point of the distribution. Probabilities as Areas Probabilities as Areas Chapter 7 7.1 Describing a Continuous Distribution LO7-1 7-8

9. Expected Value and Variance Expected Value and Variance Chapter 7 7.1 Describing a Continuous Distribution LO7-1 The mean and variance of a continuous random variable are analogous to E(X) and Var(X ) for a discrete random variable, Here the integral sign replaces the summation sign. Calculus is required to compute the integrals. 7-9

10. 7-10 Characteristics of the Uniform Distribution Characteristics of the Uniform Distribution If X is a random variable that is uniformly distributed between a and b, its PDF has constant height. Denoted U(a, b) Area = base x height = (b-a) x 1/(b-a) = 1 Chapter 7 7.2 Uniform Continuous Distribution LO7-2: Calculate uniform probabilities. LO7-2

11. 7-11 Characteristics of the Uniform Distribution Characteristics of the Uniform Distribution Chapter 7 7.2 Uniform Continuous Distribution LO7-2

12. 7-12 Example: Anesthesia Effectiveness Example: Anesthesia Effectiveness An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes.An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes. X is U(15, 30)X is U(15, 30) a = 15, b = 30, find the mean and standard deviation.a = 15, b = 30, find the mean and standard deviation. Chapter 7 Find the probability that the effectiveness anesthetic takes between 20 and 25 minutes.Find the probability that the effectiveness anesthetic takes between 20 and 25 minutes. 7.2 Uniform Continuous Distribution LO7-2

13. 7-13 P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33% Chapter 7 Example: Anesthesia Effectiveness Example: Anesthesia Effectiveness 7.2 Uniform Continuous Distribution LO7-2

14. 7-14 Characteristics of the Normal Distribution Characteristics of the Normal Distribution Normal or Gaussian (or bell shaped) distribution was named for German mathematician Karl Gauss (1777 – 1855). Defined by two parameters, µ and . Denoted N(µ,  ). Domain is –  < X < +  (continuous scale). Almost all (99.7%) of the area under the normal curve is included in the range µ – 3  < X < µ + 3 . Symmetric and unimodal about the mean. Chapter 7 7.3 Normal Distribution LO7-3: Know the form and parameters of the normal distribution. LO7-3

15. 7-15 Characteristics of the Normal Distribution Characteristics of the Normal Distribution Chapter 7 7.3 Normal Distribution LO7-3

16. 7-16 Normal PDF f (x) reaches a maximum at µ and has points of inflection at µ ± Normal PDF f (x) reaches a maximum at µ and has points of inflection at µ ±  Bell-shaped curve NOTE: All normal distributions have the same shape but differ in the axis scales. Chapter 7 Characteristics of the Normal Distribution Characteristics of the Normal Distribution 7.3 Normal Distribution LO7-3

17. 7-17 Normal CDFNormal CDF Chapter 7 Characteristics of the Normal Distribution Characteristics of the Normal Distribution 7.3 Normal Distribution LO7-3

18. 7-18 Characteristics of the Standard Normal Distribution Characteristics of the Standard Normal Distribution Chapter 7 7.4 Standard Normal Distribution Since for every value of µ and , there is a different normal distribution, we transform a normal random variable to a standard normal distribution with µ = 0 and  = 1 using the formula. LO7-3

19. 7-19 Characteristics of the Standard Normal Characteristics of the Standard Normal Standard normal PDF f (x) reaches a maximum at z = 0 and has points of inflection at +1. Shape is unaffected by the transformation. It is still a bell-shaped curve.Shape is unaffected by the transformation. It is still a bell-shaped curve. Figure 7.11 Chapter 7LO7-3 7.4 Standard Normal Distribution

20. 7-20 Characteristics of the Standard Normal Characteristics of the Standard Normal Standard normal CDFStandard normal CDF Chapter 7 A common scale from -3 to +3 is used. Entire area under the curve is unity. The probability of an event P(z 1 < Z < z 2 ) is a definite integral of f (z). However, standard normal tables or Excel functions can be used to find the desired probabilities. LO7-3 7.4 Standard Normal Distribution

21. 7-21 Normal Areas from Appendix C-1 Normal Areas from Appendix C-1 Appendix C-1 allows you to find the area under the curve from 0 to z. For example, find P(0 < Z < 1.96): Chapter 7LO7-3 7.4 Standard Normal Distribution

22. 7-22 Normal Areas from Appendix C-1 Normal Areas from Appendix C-1 Now find P(-1.96 < Z < 1.96).Now find P(-1.96 < Z < 1.96). Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96).Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96). So, P(-1.96 < Z < 1.96) =.4750 +.4750 =.9500 or 95% of the area under the curve. Chapter 7LO7-3 7.4 Standard Normal Distribution

23. 7-23 Basis for the Empirical Rule Basis for the Empirical Rule Approximately 68% of the area under the curve is between + 1  Approximately 95% of the area under the curve is between + 2  Approximately 99.7% of the area under the curve is between + 3  Chapter 7LO7-3 7.4 Standard Normal Distribution

24. 7-24 Normal Areas from Appendix C-2 Normal Areas from Appendix C-2 Appendix C-2 allows you to find the area under the curve from the left of z (similar to Excel).Appendix C-2 allows you to find the area under the curve from the left of z (similar to Excel). For example,For example, P(Z < -1.96) P(Z < 1.96 P(Z < 1.96) P(-1.96 < Z < 1.96) Chapter 7 LO7-4: Find the normal probability for given z or x using tables or Excel. 7.4 Standard Normal Distribution LO7-4

25. 7-25 Normal Areas from Appendices C-1 or C-2 Normal Areas from Appendices C-1 or C-2 Appendices C-1 and C-2 yield identical results.Appendices C-1 and C-2 yield identical results. Use whichever table is easiest.Use whichever table is easiest. Finding z for a Given Area Finding z for a Given Area Appendices C-1 and C-2 can be used to find the z-value corresponding to a given probability. For example, what z-value defines the top 1% of a normal distribution? This implies that 49% of the area lies between 0 and z which gives z = 2.33 by looking for an area of 0.4900 in Appendix C-1. Chapter 7 7.4 Standard Normal Distribution LO7-4

26. 7-26 Finding Areas by using Standardized Variables Finding Areas by using Standardized Variables So John’s score is 1.57 standard deviations about the mean.So John’s score is 1.57 standard deviations about the mean. Chapter 7 P(X < 86) = P(Z < 1.57) =.9418 (from Appendix C-2)P(X < 86) = P(Z < 1.57) =.9418 (from Appendix C-2) So, John is approximately in the 94 th percentile.So, John is approximately in the 94 th percentile. Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in? That is, what is P(X < 86) where X represents the exam scores?Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in? That is, what is P(X < 86) where X represents the exam scores? 7.4 Standard Normal Distribution LO7-4

27. 7-27 Finding Areas by using Standardized Variables Finding Areas by using Standardized Variables NOTE: You can use Excel, Minitab, TI83/84 etc. to compute these probabilities directly. Chapter 7 7.4 Standard Normal Distribution LO7-4

28. 7-28 Inverse Normal Inverse Normal How can we find the various normal percentiles (5th, 10th, 25th, 75th, 90th, 95th, etc.) known as the inverse normal? That is, how can we find X for a given area? We simply turn the standardizing transformation around: Chapter 7 LO7-5: Solve for z or x for a normal probability using tables or Excel. Solving for x in z = (x − μ)/  gives x = μ + zσ 7.4 Standard Normal Distribution LO7-5

29. 7-29 Inverse Normal Inverse Normal Chapter 7 For example, suppose that John’s economics professor has decided that any student who scores below the 10th percentile must retake the exam. The exam scores are normal with μ = 75 and σ = 7. What is the score that would require a student to retake the exam? We need to find the value of x that satisfies P(X < x) =.10. The z-score for with the 10th percentile is z = −1.28. 7.4 Standard Normal Distribution LO7-5

30. 7-30 Inverse Normal Inverse Normal Chapter 7 The steps to solve the problem are: Use Appendix C or Excel to find z = −1.28 to satisfy P(Z < −1.28) =.10. Substitute the given information into z = (x − μ)/σ to get −1.28 = (x − 75)/7 Solve for x to get x = 75 − (1.28)(7) = 66.03 (or 66 after rounding) Students who score below 66 points on the economics exam will be required to retake the exam. 7.4 Standard Normal Distribution LO7-5

31. 7-31 Inverse Normal Inverse Normal Chapter 7 7.4 Standard Normal Distribution LO7-5

32. 7-32 Normal Approximation to the Binomial Normal Approximation to the Binomial Binomial probabilities are difficult to calculate when n is large.Binomial probabilities are difficult to calculate when n is large. Use a normal approximation to the binomial distribution.Use a normal approximation to the binomial distribution. As n becomes large, the binomial bars become smaller and continuity is approached.As n becomes large, the binomial bars become smaller and continuity is approached. Chapter 7 7.5 Normal Approximations LO7-6: Use the normal approximation to a binomial or a Poisson. LO7-6

33. 7-33 Normal Approximation to the Binomial Rule of thumb: when n  ≥ 10 and n(1-  ) ≥ 10, then it is appropriate to use the normal approximation to the binomial distribution. In this case, the mean and standard deviation for the binomial distribution will be equal to the normal µ and , respectively. Chapter 7 Example Coin Flips Example Coin Flips If we were to flip a coin n = 32 times and  =.50, are the requirements for a normal approximation to the binomial distribution met? 7.5 Normal Approximations LO7-6

34. 7-34 Example Coin Flips Example Coin Flips n  = 32 x.50 = 16 n(1-  ) = 32 x (1 -.50) = 16 So, a normal approximation can be used. When translating a discrete scale into a continuous scale, care must be taken about individual points. Chapter 7 For example, find the probability of more than 17 heads in 32 flips of a fair coin. This can be written as P(X  18). However, “more than 17” actually falls between 17 and 18 on a discrete scale. 7.5 Normal Approximations LO7-6

35. 7-35 Example Coin Flips Example Coin Flips Since the cutoff point for “more than 17” is halfway between 17 and 18, we add 0.5 to the lower limit and find P(X > 17.5).Since the cutoff point for “more than 17” is halfway between 17 and 18, we add 0.5 to the lower limit and find P(X > 17.5). This addition to X is called the Continuity Correction.This addition to X is called the Continuity Correction. At this point, the problem can be completed as any normal distribution problem.At this point, the problem can be completed as any normal distribution problem. Chapter 7 7.5 Normal Approximations LO7-6

36. 7-36 Chapter 7 Example Coin Flips Example Coin Flips P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5) = P(Z > 0.53) = 0.2981 7.5 Normal Approximations LO7-6

37. 7-37 Normal Approximation to the Poisson Normal Approximation to the Poisson The normal approximation to the Poisson distribution works best when is large (e.g., when exceeds the values in Appendix B). Set the normal µ and  equal to the mean and standard deviation for the Poisson distribution. Chapter 7 Example Utility Bills Example Utility Bills On Wednesday between 10A.M. and noon customer billing inquiries arrive at a mean rate of 42 inquiries per hour at Consumers Energy. What is the probability of receiving more than 50 calls in an hour? = 42 which is too big to use the Poisson table. Use the normal approximation with  = 42 and  = 6.48074 7.5 Normal Approximations LO7-6

38. 7-38 Example Utility Bills Example Utility Bills To find P(X > 50) calls, use the continuity-corrected cutoff point halfway between 50 and 51 (i.e., X = 50.5).To find P(X > 50) calls, use the continuity-corrected cutoff point halfway between 50 and 51 (i.e., X = 50.5). At this point, the problem can be completed as any normal distribution problem.At this point, the problem can be completed as any normal distribution problem. Chapter 7 7.5 Normal Approximations LO7-6

39. 7-39 Characteristics of the Exponential Distribution Characteristics of the Exponential Distribution If events per unit of time follow a Poisson distribution, the time until the next event follows the Exponential distribution.If events per unit of time follow a Poisson distribution, the time until the next event follows the Exponential distribution. The time until the next event is a continuous variable.The time until the next event is a continuous variable. Chapter 7 7.6 Exponential Distribution LO7-7: Find the exponential probability for a given x. NOTE: Here we will find probabilities > x or ≤ x. LO7-7

40. 7-40 Characteristics of the Exponential Distribution Characteristics of the Exponential Distribution Probability of waiting more than xProbability of waiting less than or equal to x Chapter 7 7.6 Exponential Distribution LO7-7

41. 7-41 Example Customer Waiting Time Example Customer Waiting Time Between 2P.M. and 4P.M. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls per minute. What is the probability of waiting more than 30 seconds (i.e., 0.50 minutes) for the next call? Set = 2.2 events/min and x = 0.50 min P(X > 0.50) = e – x = e –(2.2)(0.5) =.3329 or 33.29% chance of waiting more than 30 seconds for the next call. Chapter 7 7.6 Exponential Distribution LO7-7

42. 7-42 Example Customer Waiting Time Example Customer Waiting Time P(X > 0.50) P(X ≤ 0.50) Chapter 7 7.6 Exponential Distribution LO7-7

43. 7-43 Inverse Exponential Inverse Exponential If the mean arrival rate is 2.2 calls per minute, we want the 90 th percentile for waiting time (the top 10% of waiting time).If the mean arrival rate is 2.2 calls per minute, we want the 90 th percentile for waiting time (the top 10% of waiting time). Find the x-value that defines the upper 10%.Find the x-value that defines the upper 10%. Chapter 7 LO7-8: Solve for x for given exponential probability. 7.6 Exponential Distribution LO7-8

44. 7-44 Inverse Exponential Inverse Exponential Chapter 7 7.6 Exponential Distribution LO7-8

45. 7-45 Chapter 7 Mean Time Between Events Mean Time Between Events 7.6 Exponential Distribution LO7-8

46. 7-46 Characteristics of the Triangular Distribution Characteristics of the Triangular Distribution Chapter 7 7.7 Triangular Distribution LO7-9: Use the triangular distribution for “what-if” analysis (optional). LO7-9

47. 7-47 Characteristics of the Triangular Distribution Characteristics of the Triangular Distribution Chapter 7 The triangular distribution is a way of thinking about variation that corresponds rather well to what-if analysis in business. It is not surprising that business analysts are attracted to the triangular model. Its finite range and simple form are more understandable than a normal distribution. 7.7 Triangular Distribution LO7-9

48. 7-48 Characteristics of the Triangular Distribution Characteristics of the Triangular Distribution Chapter 7 It is more versatile than a normal, because it can be skewed in either direction. Yet it has some of the nice properties of a normal, such as a distinct mode. The triangular model is especially handy for what-if analysis when the business case depends on predicting a stochastic variable (e.g., the price of a raw material, an interest rate, a sales volume). If the analyst can anticipate the range (a to c) and most likely value (b), it will be possible to calculate probabilities of various outcomes. Many times, such distributions will be skewed, so a normal wouldn’t be much help. 7.7 Triangular Distribution LO7-9