1. South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering

2. Introduction to Probability & Statistics Data Analysis Industrial Engineering

3. Data Analysis Histograms Industrial Engineering

4. Experimental Data u Suppose we wish to make some estimates on time to fail for a new power supply. 40 units are randomly selected and tested to failure. Failure times are recorded follow:

5. Histogram u Perhaps the most useful method, histograms give the analyst a feel for the distribution from which the data was obtained. u Count observations within a set of ranges u Average 5 observations per interval class

6. Histogram u Perhaps the most useful method, histograms give the analyst a feel for the distribution from which the data was obtained. u Count observations with a set of ranges u Average 5 observations per interval class u Range for power supply data: 0.5-73.8 u Intervals: 0.0 - 10.040.1 - 50.0 10.1 - 20.050.1 - 60.0 20.1 - 30.060.1 - 70.0 30.1 - 40.070.1 - 80.0

7. Histogram u Class Interval 0.0 - 10.0

8. Histogram

9. Histogram u Class Interval 0.0 - 10.0Count = 15

10. Histogram u Class Interval 10.1 - 20.0

11. Histogram

12. Histogram u Class Interval 10.1 - 20.0Count = 11

13. Histogram Frequency Count Class IntervalsFrequency 0.0 - 10.0 15 10.1 - 20.0 11 20.1 - 30.0 6 30.1 - 40.0 3 40.1 - 50.0 2 50.1 - 60.0 2 60.1 - 70.0 0 70.1 - 80.0 1

14. Histogram Class IntervalsFrequency 0.0 - 10.0 15 10.1 - 20.0 11 20.1 - 30.0 6 30.1 - 40.0 3 40.1 - 50.0 2 50.1 - 60.0 2 60.1 - 70.0 0 70.1 - 80.0 1

15. Exponential Distribution fxe x ()   Density Cumulative Mean 1/ Variance 1/ 2 Fxe x ()   1, x > 0

16. Histogram; Change Interval Class IntervalsFrequency 0.0 - 15.0 21 15.1 - 30.0 10 30.1 - 45.0 4 45.1 - 60.0 3 60.1 - 75.0 1

17. Histogram; Change Interval Class IntervalsFrequency 0.0 - 5.0 8 5.1 - 10.0 7 10.1 - 15.0 6 15.1 - 20.0 4 20.1 - 25.0 3 25.1 - 30.0 3 30.1 - 35.0 2 35.1 - 40.0 1 40.1 - 45.0 1 45.1 - 50.0 1 50.1 - 55.0 1 55.1 - 60.0 1

18. Histogram; Change Class Mark Class IntervalsFrequency -5.0 - 5.0 8 5.1 - 15.0 13 15.1 - 25.0 7 25.1 - 35.0 5 35.1 - 45.0 2 45.1 - 55.0 2 55.1 - 65.0 1 65.1 - 75.0 1

19. Class Problem u The following data represents independent observations on deviations from the desired diameter of ball bearings produced on a new high speed machine.

20. Class Problem

22. Histogram u Intervals & Class marks can alter the histogram u too many intervals leaves too many voids u too few intervals doesn’t give a good picture u Rule of Thumb u # Intervals = n/5 u Sturges’ Rule k = [1 + log 2 n] = [1 + 3.322 log 10 n]

23. Class Problem u The following represents demand for a particular inventory during a 70 day period. Construct a histogram and hypothesize a distribution.

24. Class Problem

26. Relative Histogram ClassFreqRel. 0.0 - 10.0 150.375 10.1 - 20.0 110.275 20.1 - 30.0 60.150 30.1 - 40.0 30.075 40.1 - 50.0 20.050 50.1 - 60.0 20.050 60.1 - 70.0 00.000 70.1 - 80.0 10.025

27. Relative Histogram

28. Histogram u Class Excel Exercise

30. South Dakota School of Mines & Technology Data Analysis Industrial Engineering

31. Data Analysis Empirical Distributions Industrial Engineering

32. Empirical Cumulative u Rank Order the data smallest to largest u u Example: Suppose we collect gpa’s on 10 students 3.5, 2.8, 2.7, 3.3, 3.0, 3.9, 2.9, 3.0, 2.4, 3.1 n i xF i 0.5 )(  

33. Empirical Cumulative u RankObs n i xF i 0.5 )(   12.40.05 22.7 0.15 32.80.25 42.90.35 53.00.45 63.00.55 73.10.65 83.30.75 93.50.85 103.90.95

34. Empirical Cumulative xF i )( 2.40.05 2.7 0.15 2.80.25 2.90.35 3.00.45 3.00.55 3.10.65 3.30.75 3.50.85 3.90.95 Obs

35. Time to Failure

36. Exponential Distribution fxe x ()   Density Cumulative Mean 1/ Variance 1/ 2 Fxe x ()   1, x > 0

37. Inventory Data

38. Diameter Errors

39. Normal Distribution

40. Scatter Plots (Paired Data) u Shows the relationship between paired data u Example: Suppose for example we wish to look at state per student expenditures versus achievement results on the Stanford Achievement Test

41. Scatter Plots (Paired Data)

43. South Dakota School of Mines & Technology Data Analysis Industrial Engineering

44. Data Analysis Box Plots Industrial Engineering

45. Box Plots u Problem with empirical is we may simply not have enough data u For small data sets, analysts often like to provide a rough graphical measure of how data is dispersed u Consider our student data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

46. Box Plots u Ranked student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 Min = 2.4Max = 3.9

47. Box Plots u Ranked student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 Min = 2.4Max = 3.9 2.4 3.9

48. Box Plots u Ranked student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 Median = (3.0+3.0)/2 = 3.0 2.4 3.93.0

49. Box Plots u Ranked student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 Median Bottom= (2.7+2.8)/2 = 2.75 2.4 3.93.02.75

50. Box Plots u Ranked student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 Median Top = (3.3+3.5)/2 = 3.4 2.4 3.93.02.753.4

51. Box Plots u Ranked student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 2.4 3.93.02.753.4

52. Fail Time Data u Min = 0.5 u Max = 73.8 u Lower Quartile = 6.1 u Median= 14.3 u Upper Quartile = 26.7 0.5 6.1 14.3 26.7 73.8

53. Class Problem u The following data represents sorted observations on deviations from desired diameters of ball bearings. Compute a box plot.

54. Class Problem -1.27 0.30 1.00 1.90 2.66

56. South Dakota School of Mines & Technology Data Analysis Industrial Engineering

57. Data Analysis Statistical Measures Industrial Engineering

58. Aside: Mean, Variance Mean : Variance:   xpxxdiscrete x (),  22   ()()xpx x

59. Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6  = E[X] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5

60. Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6  2 = E[(X-  ) 2 ] = (1-3.5) 2 (1/6) + (2-3.5) 2 (1/6) + (3-3.5) 2 (1/6) + (4-3.5) 2 (1/6) + (5-3.5) 2 (1/6) + (6-3.5) 2 (1/6) = 2.92

61. Binomial Mean = 1p(1) + 2p(2) + 3p(3) +... + np(n)   xpx x () xnx n x pp xnx n x               )1( )!(! ! 0

62. Binomial Mean = 1p(1) + 2p(2) + 3p(3) +... + np(n)   xpx x () xnx n x pp xnx n x               )1( )!(! ! 0 Miracle 1 occurs = np

63. Binomial Measures Mean : Variance:   xpx x ()  22   ()()xpx x = np = np(1-p)

64. Binomial Distribution 0.0 0.1 0.2 0.3 0.4 0.5 012345 x P(x) 0.0 0.1 0.2 0.3 0.4 0.5 012345 x P(x) n=5, p=.3n=8, p=.5 x 0.0 0.1 0.2 0.3 0.4 0.5 012345678 P(x) n=4, p=.8 0.0 0.1 0.2 0.3 0.4 0.5 024 x P(x) n=20, p=.5

65. Measures of Centrality u Mean u Median u Mode

66. Measures of Centrality u Mean   xpxxdiscrete x (),     xfxdxxcontinuous(), u u Sample Mean    n i i n x X 1

67. Measures of Centrality u Exercise: Compute the sample mean for the student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

68. Measures of Centrality u Exercise: Compute the sample mean for the student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9    n i i n x X 1 10 9.35.33.31.30.30.39.28.27.24.2   = 3.06

69. Measures of Centrality u Failure Data X1.19 

70. Measures of Centrality u Median Compute the median for the student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 0.3 2 0.30.3    X 

71. Measures of Centrality u Mode Class mark of most frequently occurring interval For Failure data, mode = class mark first interval 0.5  X 

72. Measures of Centrality MeasureStudent Gpa Failure Data Mean3.0019.10 Median3.0414.40 Mode --- 5.00

73. Measures of Centrality MeasureStudent Gpa Failure Data Mean3.0019.10 Median3.0414.40 Mode --- 5.00  Sample mean X is a blue estimator of true mean  X  u.b.

74. Measures of Centrality MeasureStudent Gpa Failure Data Mean3.0019.10 Median3.0414.40 Mode --- 5.00  Sample mean X is a blue estimator of true mean  X  E[ X ] =  u.b.

75. Measures of Dispersion u Range u Sample Variance

76. Measures of Dispersion u Range Compute the range for the student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

77. Measures of Dispersion u Range Compute the range for the student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 Min = 2.4Max = 3.9 Range = 3.9 - 2.4 = 1.5

78. Measures of Dispersion u Variance  22   ()()xpx x  22     ()()xfxdx u u Sample variance x 1 1 2 2 2      n nx s n i i

79. Measures of Dispersion u Exercise: Compute the sample variance for the student Gpa data 2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 x 1 1 2 2 2      n nx s n i i

80. Measures of Dispersion u Sample Variance x 1 1 2 2 2      n nx s n i i  185.0 110 06.3103.95 2    

81. Measures of Dispersion u Exercise: Compute the variance for failure time data s 2 = 302.76

82. An Aside u For Failure Time data, we now have three measures for the data s 2 = 302.76 X1.19 

83. An Aside u For Failure Time data, we now have three measures for the data Expontial ?? s 2 = 302.76 X1.19 

84. An Aside u Recall that for the exponential distribution  = 1/  2 = 1/ 2 If E[ X ] =  and E [s 2 ] = s 2, then 1/ = 19.1 s 2 = 302.76 X1.19  0524. ˆ 

85. An Aside u Recall that for the exponential distribution  = 1/ s 2 = 1/ 2 If E[ X ] =  and E [s 2 ] = s 2, then 1/ = 19.1 or 1/ 2 = 302.76 s 2 = 302.76 X1.19  0575. ˆ  0524. ˆ 