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Prepared by Lloyd R. Jaisingh

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 . Chapter Contents 7.1 Describing a Continuous Distribution 7.2 Uniform Continuous Distribution

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Prepared by Lloyd R. Jaisingh

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  1. A PowerPoint Presentation Package to Accompany Applied Statistics in Business & Economics, 4th edition David P. Doane and Lori E. Seward Prepared by Lloyd R. Jaisingh

  2. 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) Continuous Probability Distributions Chapter 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. Continuous Probability Distributions Chapter 7

  4. Chapter Learning Objectives (LO’s) LO6:Use the normal approximation to a binomial or a Poisson 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). Continuous Probability Distributions Chapter 7

  5. 7.1 Describing a Continuous Distribution LO7-1 Chapter 7 • 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. LO7-1: Define a continuous random variable. Events as Intervals 7-5

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

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

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

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

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

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

  12. 7.2 Uniform Continuous Distribution LO7-2 Chapter 7 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. • X is U(15, 30) • a = 15, b = 30, find the mean and standard deviation. • Find the probability that the effectiveness anesthetic takes between 20 and 25 minutes.

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

  14. 7.3 Normal Distribution LO7-3 Chapter 7 LO7-3: Know the form and parameters 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.

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

  16. 7.3 Normal Distribution LO7-3 Chapter 7 Characteristics of the Normal Distribution • 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.

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

  18. 7.4 Standard Normal Distribution LO7-3 Chapter 7 Characteristics of the 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.

  19. 7.4 Standard Normal Distribution LO7-3 Chapter 7 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. Figure 7.11

  20. 7.4 Standard Normal Distribution LO7-3 Chapter 7 Characteristics of the Standard Normal • Standard normal CDF • A common scale from -3 to +3 is used. • Entire area under the curve is unity. • The probability of an event P(z1 < Z < z2) is a definite integral of f(z). • However, standard normal tables or Excel functions can be used to find the desired probabilities.

  21. 7.4 Standard Normal Distribution LO7-3 Chapter 7 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):

  22. 7.4 Standard Normal Distribution LO7-3 Chapter 7 Normal Areas from Appendix C-1 • Now find P(-1.96 < 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.

  23. 7.4 Standard Normal Distribution LO7-3 Chapter 7 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

  24. 7.4 Standard Normal Distribution LO7-4 Chapter 7 LO7-4: Find the normal probability for given z or x using tables or Excel. 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). • For example, P(Z < 1.96) P(-1.96 < Z < 1.96) P(Z < -1.96)

  25. 7.4 Standard Normal Distribution LO7-4 Chapter 7 Normal Areas from Appendices C-1 or C-2 • Appendices C-1 and C-2 yield identical results. • Use whichever table is easiest. 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.

  26. 7.4 Standard Normal Distribution LO7-4 Chapter 7 Finding Areas by using Standardized Variables • 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? • So John’s score is 1.57 standard deviations about the mean. • P(X < 86) = P(Z < 1.57) = .9418 (from Appendix C-2) • So, John is approximately in the 94th percentile.

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

  28. 7.4 Standard Normal Distribution LO7-5 Chapter 7 LO7-5: Solve for z or x for a normal probability using tables or Excel. • 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: Solving for x in z = (x − μ)/ gives x = μ + zσ

  29. 7.4 Standard Normal Distribution LO7-5 Chapter 7 • Inverse Normal • 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.

  30. 7.4 Standard Normal Distribution LO7-5 Chapter 7 • Inverse Normal • 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.

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

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

  33. 7.5 Normal Approximations LO7-6 Chapter 7 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. 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?

  34. 7.5 Normal Approximations LO7-6 Chapter 7 Example Coin Flips • n = 32 x .50 = 16n(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. • 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.

  35. 7.5 Normal Approximations LO7-6 Chapter 7 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). • This addition to X is called the Continuity Correction. • At this point, the problem can be completed as any normal distribution problem.

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

  37. 7.5 Normal Approximations LO7-6 Chapter 7 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. 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

  38. 7.5 Normal Approximations LO7-6 Chapter 7 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). • At this point, the problem can be completed as any normal distribution problem.

  39. 7.6 Exponential Distribution LO7-7 Chapter 7 LO7-7: Find the exponential probability for a given x. 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. • The time until the next event is a continuous variable. NOTE: Here we will find probabilities > x or ≤ x.

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

  41. 7.6 Exponential Distribution LO7-7 Chapter 7 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.

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

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

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

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

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

  47. 7.7 Triangular Distribution LO7-9 Chapter 7 Characteristics of the Triangular Distribution • 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.

  48. 7.7 Triangular Distribution LO7-9 Chapter 7 Characteristics of the Triangular Distribution • 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.

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