1. ©2003 Thomson/South-Western 1 Chapter 6 – Continuous Probability Distributions Slides prepared by Jeff Heyl, Lincoln University ©2003 South-Western/Thomson Learning™ Introduction to Business Statistics, 6e Kvanli, Pavur, Keeling

2. ©2003 Thomson/South-Western 2 Probability for a Continuous Random Variable Figure 6.1 2060 X Curve describing the population Area = P(20 < X < 60)

3. ©2003 Thomson/South-Western 3 Properties of a Normal Distribution  Continuous random variable  Symmetrical in shape (bell shaped)  The probability of any given range of numbers is represented by the area under the curve for that range  Probabilities for all normal distributions are determined using the standard normal distribution

4. ©2003 Thomson/South-Western 4 Histogram for Lightbulb Life X Relative frequency 280300320340360380400420440460480500520 Figure 6.2

5. ©2003 Thomson/South-Western 5 Distribution of Lightbulb Life Figure 6.3 µ = 400 450 X Inflection point This distance is  = 450 - 400 = 450 - 400 = 50 = 50 ||||||||| P

6. ©2003 Thomson/South-Western 6 Probability Density Function for Normal Distribution f(x) = e - 1  2π 12121212 x - µ 

7. ©2003 Thomson/South-Western 7 Normal Curves with Unequal Means and Equal Standard Deviations Relative frequency FemalesMales Average female height Average male height Height || Figure 6.4

8. ©2003 Thomson/South-Western 8 Normal Curves with Equal Means and Unequal Standard Deviations Figure 6.5 Relative frequency Company A Company B Average age in both companies Age

9. ©2003 Thomson/South-Western 9 Area Under the Normal Curve Area =.5 Total area = 1 X Area =.5 µ Figure 6.6

10. ©2003 Thomson/South-Western 10 Normal Curve for Lightbulbs Figure 6.7 X  = 50 |300 350 400 450 500 360

11. ©2003 Thomson/South-Western 11 Standard Normal Curve Z  = 1 |-2 |11|111 20 0 µ = 0 (µ - 2  ) (µ -  ) (µ +  ) (µ + 2  ) Figure 6.8

12. ©2003 Thomson/South-Western 12 Standard Normal Curve Figure 6.9 Z Area =.4474 1.62 0

13. ©2003 Thomson/South-Western 13 Determining the Probability for a Standard Normal Random Variable P(Z > 1.62) =.5 -.4474 =.0526 Area =.5 01.62 Z Figure 6.10

14. ©2003 Thomson/South-Western 14 P(Z < 1.62) =.5 +.4474 =.9474 Determining the Probability for a Standard Normal Random Variable A 1 =.5 01.62 A 2 =.4474 Z Figure 6.11

15. ©2003 Thomson/South-Western 15 Determining the Probability for a Standard Normal Random Variable P(1.0 < Z < 2.0)= P(0 < Z < 2.0) - P(0 < Z < 1.0) =.4772 -.3413 =.1359 A 1 =.3413 02 A 2 =.4772 Z 1 Figure 6.12

16. ©2003 Thomson/South-Western 16 Determining the Probability for a Standard Normal Random Variable P(-1.25 < Z < 1.15)= P(-1.25 < Z < 0) + P(0 < Z < 1.15) = A 1 + A 2 =.3944 +.3749 =.7693 A 1 =.3944 01.15 A 2 =.3749 Z -1.25 Figure 6.13

17. ©2003 Thomson/South-Western 17 Determining the Probability for a Standard Normal Random Variable These areas are the same 01.25 Z -1.25 Figure 6.14 P(-1.25 < Z < 1.15)= P(-1.25 < Z < 0) + P(0 < Z < 1.15) = A 1 + A 2 =.3944 +.3749 =.7693

18. ©2003 Thomson/South-Western 18 Determining the Probability for a Standard Normal Random Variable P(Z < -1.45)= P(Z < 0) - P(-1.45 < Z < 0) =.5 -.4265 =.0735 A 1 =.5 -.4265 0 A 2 =.4265 Z -1.25 Figure 6.15

19. ©2003 Thomson/South-Western 19 Determining the Probability for a Standard Normal Random Variable Area =.5 -.03 0 z Area =.03 Z Figure 6.16 P(Z ≥ z) =.03 =.5 -.03 =.47

20. ©2003 Thomson/South-Western 20 Determining the Probability for a Standard Normal Random Variable P(Z ≤ z) =.2 =.5 -.2 =.3 Area =.5 -.2 0z Area =.2 Z Figure 6.17

21. ©2003 Thomson/South-Western 21 Areas Under Any Normal Curve Figure 6.18 µ = 0,  = 50 Y -120-100-80-60-40-20020406080100120 

22. ©2003 Thomson/South-Western 22 Areas Under Any Normal Curve Figure 6.19 µ = 0,  = 1 Y -2.4-2.0-1.6-1.2-.8-.40.4.81.21.62.02.4 

23. ©2003 Thomson/South-Western 23 Areas Under Any Normal Curve Figure 6.20A P(X < 360) = ? X = lifetime of Everglo bulb 400360 Same area

24. ©2003 Thomson/South-Western 24 Areas Under Any Normal Curve Figure 6.20B From Table A.4, this area is.2881. So shaded area=.5 -.2881 =.2119 Z =standard normal 0-.8 Same area

25. ©2003 Thomson/South-Western 25 Interpreting Z  In Example 6.2 Z = - 0.8 means that the value 360 is.8 standard deviations below the mean  A positive value of Z designates how may standard deviations (  ) X is to the right of the mean (µ)  A negative value of Z designates how may standard deviations (  ) X is to the left of the mean (µ)

26. ©2003 Thomson/South-Western 26 ATM Example Figure 6.21A  = $625 X $2000$3700$5000

27. ©2003 Thomson/South-Western 27 ATM Example Figure 6.21B X-2.7202.08 Area =.4812 Area =.4967

28. ©2003 Thomson/South-Western 28 Policyholder Lifetimes Figure 6.22 57.4 57.4 – 61.8 61.8 – 66.2 66.2 – 70.6 70.6 – 75.0 75.0 – 65 X =age at death (years) (1)(2)  = 4.4 yr 70

29. ©2003 Thomson/South-Western 29 Policyholder Lifetimes Figure 6.23 Z A 1 =.1064 A 1 + A 2 = P(Z > -.27) =.1064 +.5 =.6064 A 2 =.5 -.27

30. ©2003 Thomson/South-Western 30 Policyholder Lifetimes Figure 6.24 Z A 2 =.5 -.3051 =.1949 A 1 =.3051 (Table A.4).86

31. ©2003 Thomson/South-Western 31 Policyholder Lifetimes Figure 6.25 Z A 3 = A 2 - A 1 =.4772 -.3051 =.1721.862.00 A 1 =.3051 A 2 =.4772

32. ©2003 Thomson/South-Western 32 Everglo Lightbulbs After how many hours will 80% of the Everglo bulbs burn out? Example 6.5 A 1 =.5 A 2 =.3 A 1 + A 2 =.8 400 x0x0x0x0 X =lifetime of Everglo bulb Figure 6.26A P(X < x 0 ) =.8

33. ©2003 Thomson/South-Western 33 Everglo Lightbulbs After how many hours will 80% of the Everglo bulbs burn out? Example 6.5 A 1 =.5 A 2 =.3.84 Figure 6.26B P(X < x 0 ) =.8 P(Z <.84) =.5 +.2995 =.7995 .8 x 0 - 400 50 =.84 x 0 - 400 = (50)(.84) = 42 x 0 = 400 + 42 = 442

34. ©2003 Thomson/South-Western 34 Bakery Example Figure 6.27A 35 x0x0x0x0 X =demand for French bread (loaves) Area =.9 µ = 35  = 8

35. ©2003 Thomson/South-Western 35 Bakery Example Figure 6.27B 1.28 Z A 1 =.5 A 2 =.4 A 1 + A 2 =.9

36. ©2003 Thomson/South-Western 36 Empirical Rule 1.Approximately 68% of the data should lie between X - s and X + s 2.Approximately 95% of them should lie between X - 2s and X + 2s 3.Approximately 99.7% of them should lie between X - 3s and X + 3s These numbers are generated directly from Table A.4

37. ©2003 Thomson/South-Western 37 Empirical Rule Figure 6.28 1 Z A 1 =.3413 A 2 =.3413 A 1 + A 2 =.3413 +.3413 =.6826

38. ©2003 Thomson/South-Western 38 Determining Areas and Values With Excel Figure 6.29A

39. ©2003 Thomson/South-Western 39 Determining Areas and Values With Excel Figure 6.29B

40. ©2003 Thomson/South-Western 40 Allied Manufacturing Figure 6.30

41. ©2003 Thomson/South-Western 41 Allied Manufacturing Figure 6.31A

42. ©2003 Thomson/South-Western 42 Allied Manufacturing Figure 6.31B

43. ©2003 Thomson/South-Western 43 Normal Approximation to the Binomial Distribution Poisson approximation: Use when n > 20 and np ≤ 7 Normal approximation: Use when np > 5 and n(1 - p) > 5

44. ©2003 Thomson/South-Western 44 Normal Approximation to the Binomial Distribution = Solution to 1 = Solution to 2 A normal curve with µ = 6 and  = 1.732 X 0123456789101112 4.5 5.5 Figure 6.32

45. ©2003 Thomson/South-Western 45 How to Adjust for Continuity If X is a binomial random variable with n trials and probability of success = p, then: 4.Be sure to convert a probability to a ≥ before switching to the normal approximation 3.P(a ≤ X ≤ b)  P ≤ Z ≤ b +.5 - µ  a -.5 - µ  1.P(X ≤ b)  P Z ≤ b +.5 - µ  2.P(X ≥ a)  P Z ≥ a -.5 - µ 

46. ©2003 Thomson/South-Western 46 Continuous Uniform Distribution The probability of a given range of values is proportional to the width of the range µ = a + b 2 b - a 12 12  = = = =

47. ©2003 Thomson/South-Western 47 Continuous Uniform Distribution Relative frequency Content of cup (fluid ounces) 678 Figure 6.33

48. ©2003 Thomson/South-Western 48 Continuous Uniform Distribution Figure 6.34 X =amount of soda (ounces) 6 |77|777 8 12 – Total area = (2) = 1 12

49. ©2003 Thomson/South-Western 49 Continuous Uniform Distribution Figure 6.34 X =amount of soda (ounces) a = 6 | µ = 7 b = 8 – Total area = (b - a) = (2)(.5) = 1 1 b - a =.5 1 b - a

50. ©2003 Thomson/South-Western 50 Continuous Uniform Distribution Figure 6.36 X = amount of soda (ounces) 6 |77|777 8 |6.5 7.5.5.5 – Area= (8 - 7.5)(.5) =.25

51. ©2003 Thomson/South-Western 51 Continuous Uniform Distribution Figure 6.37 X = amount of soda (ounces) 6 |77|777 8 |6.5 7.5.5.5 – Area= (7.5 - 6.5)(.5) =.5

52. ©2003 Thomson/South-Western 52 Exponential Distribution  Time between arrivals to a queue - time between people arriving at a line to check out in a department store (people, machines, or telephone calls may wait in a queue)  Lifetime of components in a machine

53. ©2003 Thomson/South-Western 53 Exponential Distribution If the random variable Y, representing the number of arrivals over a specified time period T, follows a Poisson distribution, then X, representing the time between successive arrivals, will be an exponential random variable P(X ≥ x 0 ) = e -Ax 0 µ = 1/A  = 1/A

54. ©2003 Thomson/South-Western 54 Exponential Distribution Total area = 1.0 0 X Figure 6.38

55. ©2003 Thomson/South-Western 55 Exponential Distribution Figure 6.39 0 X A A –A A – x0x0x0x0 Area= P(X ≥ x 0 ) = e -Ax 0

56. ©2003 Thomson/South-Western 56 Exponential Distribution Figure 6.40 0 A A –A A –.5 Area= e -(4)(.5) = e -2 =.135 X =time between arrivals

57. ©2003 Thomson/South-Western 57 Exponential Distribution Figure 6.41 Area = 1 - e -(.001)(1000) = 1 - e -1 =.632 0 X =battery lifetime 1000.001.001 – LongLife

58. ©2003 Thomson/South-Western 58 Exponential Distribution Figure 6.41 Area = 1 - e -(.001)(365) = 1 - e -.365 =.306 0 X =battery lifetime 1000.001.001 – LongLife