1 / 51

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). Continuous Probability Distributions.

ruana
Télécharger la présentation

Continuous Probability Distributions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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

  2. 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

  3. 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

  4. 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 7-4

  5. 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 7-5

  6. 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 7-6

  7. 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. 7-7

  8. 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. 7-8

  9. 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

  10. 7.2 Uniform Continuous Distribution LO7-2 Chapter 7 Characteristics of the Uniform Distribution

  11. 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.

  12. 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%

  13. 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.

  14. 7.3 Normal Distribution LO7-3 Chapter 7 Characteristics of the Normal Distribution

  15. 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.

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

  17. 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.

  18. 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

  19. 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.

  20. 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):

  21. 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.

  22. 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

  23. 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)

  24. 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.

  25. 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.

  26. 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.

  27. 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

  28. 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.

  29. 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.

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

  31. The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. a. What proportion of brook trout caught will be between 12 and 18 inches in length?  b. The first quartile for the lengths of brook trout would be:  c. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers?  The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10 percent of the citizens would need at least how many minutes to complete the form? 

  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. A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine). The normal approximation of the probability of reaching at least 20 customers is:  n = π = µ = n π = σ = √nπ(1 - π) = x = For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors?  n = π = µ = n π = σ = √nπ(1 - π) = x =

  38. 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

  39. 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.

  40. In a T-F exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points?  n = π = µ = n π = σ = √nπ(1 - π) = x = A guesser would have a 50 percent chance of a correct answer, so we set π = .50. There are n = 100 questions, so we calculate μ = nπ = (100)(.50) = 50 and σ = [nπ(1 - π)]1/2 = [100(.50)(1 - .50)]1/2 = 5. Use x = 59.5 (with the continuity correction) and calculate the binomial P(X ≥ 60) ≈ P(z ≥ 1.90) using z = (x - μ)/σ = 1.90.

  41. 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.

  42. 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

  43. 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.

  44. 7.6 Exponential Distribution LO7-7 Chapter 7 Example Customer Waiting Time P(X ≤ 0.50) P(X > 0.50)

  45. If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of: waiting more than 0.5 hour for the next arrival is:  waiting less than 0.5 hour for the next arrival is: If arrivals occur at a mean rate of 2.6 events per minute, the exponential probability of waiting more than 1.5 minutes for the next arrival is: 

  46. 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%.

  47. 7.6 Exponential Distribution LO7-8 Chapter 7 Inverse Exponential

  48. 7.6 Exponential Distribution LO7-8 Chapter 7 Mean Time Between Events

  49. 7.7 Triangular Distribution LO7-9 Chapter 7 LO7-9: Use the triangular distribution for “what-if” analysis (optional). Characteristics of the Triangular Distribution

  50. 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.

More Related