south dakota school of mines & technology introduction to probability & statistics...

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School of Mines & School of Mines & TechnologyTechnology

Introduction to Introduction to Probability & Statistics Probability & Statistics

Industrial EngineeringIndustrial Engineering

Introduction to Introduction to Probability & StatisticsProbability & Statistics

Data AnalysisData Analysis



HistogramsHistograms


Experimental DataExperimental Data 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:

2.7 25.8 19.6 4.5 0.56.4 18.3 41.6 5.8 73.813.9 32.2 27.7 5.1 12.034.9 21.0 10.2 46.1 37.914.9 24.1 1.0 29.8 3.37.1 59.9 9.4 12.9 7.911.1 2.1 16.0 22.5 8.63.8 51.8 1.6 17.1 14.7

HistogramHistogram Perhaps the most useful method,

histograms give the analyst a feel for the distribution from which the data was obtained.

Count observations within a set of ranges Average 5 observations per interval class

HistogramHistogram Perhaps the most useful method, histograms give

the analyst a feel for the distribution from which the data was obtained.

Count observations with a set of ranges Average 5 observations per interval class

Range for power supply data: 0.5-73.8

Intervals: 0.0 - 10.0 40.1 - 50.010.1 - 20.0 50.1 - 60.020.1 - 30.0 60.1 - 70.030.1 - 40.0 70.1 - 80.0

HistogramHistogram

Class Interval 0.0 - 10.0 Count = 15

2.7 25.8 19.6 4.5 0.56.4 18.3 41.6 5.8 73.813.9 32.2 27.7 5.1 12.034.9 21.0 10.2 46.1 37.914.9 24.1 1.0 29.8 3.37.1 59.9 9.4 12.9 7.911.1 2.1 16.0 22.5 8.63.8 51.8 1.6 17.1 14.7

HistogramHistogram

Class Interval 10.1 - 20.0Count = 11

2.7 25.8 19.6 4.5 0.56.4 18.3 41.6 5.8 73.813.9 32.2 27.7 5.1 12.034.9 21.0 10.2 46.1 37.914.9 24.1 1.0 29.8 3.37.1 59.9 9.4 12.9 7.911.1 2.1 16.0 22.5 8.63.8 51.8 1.6 17.1 14.7

HistogramHistogramClass Intervals

Frequency 0.0 - 10.0 1510.1 - 20.0 1120.1 - 30.0 630.1 - 40.0 340.1 - 50.0 250.1 - 60.0 260.1 - 70.0 070.1 - 80.0 1

Power Supply Failure Times

0

5

10

15

20

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Time Class

Fre

qu

en

cy

Exponential Exponential DistributionDistribution

f x e x( ) Density

Cumulative

Mean 1/

Variance 1/2

F x e x( ) 1

, x > 0

Exponential Life

0.0

0.5

1.0

0 0.5 1 1.5 2 2.5 3

Time to Service

De

ns

ity

Histogram; Change Histogram; Change IntervalInterval

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

Change of Interval

0

5

10

15

20

25

0-15 15-30 30-45 45-60 60-75

Failure Times

Fre

qu

en

cy

Histogram; Change Histogram; Change IntervalInterval

Class Intervals Frequency 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

Change of Interval

0

2

4

6

8

10

Failure Times

Fre

qu

en

cy

Histogram; Change Class Histogram; Change Class MarkMark

Class IntervalsFrequency

-5.0 - 5.0 8 5.1 - 15.0 1315.1 - 25.0 725.1 - 35.0 535.1 - 45.0 245.1 - 55.0 255.1 - 65.0 165.1 - 75.0 1

Change of Class Mark

02468

101214

-5 -5

5-15

15-25

25-35

35-45

45-55

55-65

65-75

Failure Times

Fre

qu

en

cy

Class ProblemClass Problem The following data represents independent

observations on deviations from the desired diameter of ball bearings produced on a new high speed machine.

Deviations from desired diameters of ball bearings2.31 0.94 1.70 1.00 -0.161.49 2.48 2.58 0.19 1.712.10 1.97 0.56 2.28 1.180.30 0.59 0.38 0.01 1.590.48 1.15 0.77 0.31 1.631.71 0.95 2.29 -0.12 0.440.19 0.45 1.55 0.89 2.440.00 -0.51 0.27 0.60 2.200.66 2.36 2.29 0.21 2.30-1.27 1.03 1.55 1.90 1.301.01 0.24 -0.54 2.66 0.22

Class ProblemClass Problem

Diameter Errors

0

5

10

15

Error

Fre

qu

en

cy

Class ProblemClass Problem

Diameter Error

02468

1012

Error

Fre

qu

en

cy

HistogramHistogram

Intervals & Class marks can alter the histogram too many intervals leaves too many voids too few intervals doesn’t give a good picture

Rule of Thumb # Intervals = n/5 Sturges’ Rule

k = [1 + log2n] = [1 + 3.322 log10n]

Class ProblemClass Problem The following represents demand for a particular

inventory during a 70 day period. Construct a histogram and hypothesize a distribution.

Inventory Demand Data2 7 1 3 6 1 32 0 1 5 11 5 32 8 1 7 4 8 44 0 2 20 0 2 51 6 12 7 0 5 118 6 2 0 4 2 48 10 6 6 5 2 63 6 5 0 1 3 10 2 1 8 5 6 10 1 9 4 1 4 2

Relative HistogramRelative HistogramClass Freq Rel. 0.0 - 10.0 15 0.37510.1 - 20.0 11 0.27520.1 - 30.0 6 0.15030.1 - 40.0 3 0.07540.1 - 50.0 2 0.05050.1 - 60.0 2 0.05060.1 - 70.0 0 0.00070.1 - 80.0 1 0.025

Relative Frequency Histogram

0

0.1

0.2

0.3

0.4

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Failure Times

Re

lati

ve

Fre

qu

en

cy

Relative HistogramRelative HistogramHistogram vs Relative Histogram

0

2

4

6

8

10

12

14

16

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Failure Times

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

Freq.

Rel. Freq.

HistogramHistogram

Class Excel Exercise



Data Analysis Data Analysis



Empirical DistributionsEmpirical Distributions


Empirical CumulativeEmpirical Cumulative Rank Order the data smallest to largest

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

ixF i

0.5)(

Empirical CumulativeEmpirical Cumulative

RankObsn

ixF i

0.5)(

1 2.4 0.052 2.7 0.153 2.8 0.254 2.9 0.355 3.0 0.456 3.0 0.557 3.1 0.658 3.3 0.759 3.5 0.8510 3.9 0.95

Empirical CumulativeEmpirical Cumulative

xF i )(

2.4 0.052.7 0.152.8 0.252.9 0.353.0 0.453.0 0.553.1 0.653.3 0.753.5 0.853.9 0.95

ObsEmpirical Cumulative

0.0

0.2

0.4

0.6

0.8

1.0

2.0 2.5 3.0 3.5 4.0

Gpa

F(x

)

Time to FailureTime to Failure

Time to Fail

0

0.20.4

0.60.8

11.2

0.0 20.0 40.0 60.0 80.0

Time

Cu

mu

lati

ve

Exponential Exponential DistributionDistribution

f x e x( ) Density

Cumulative

Mean 1/

Variance 1/2

F x e x( ) 1

, x > 0

Exponential Distribution

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 1 2 3 4 5

x

F(x

)

Inventory DataInventory Data

Inventory Demand

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 5 10 15 20 25

Demand Size

Cu

mu

lati

ve

Diameter ErrorsDiameter Errors

Diameter Errors

0.000

0.200

0.400

0.600

0.800

1.000

1.200

-2.00 -1.00 0.00 1.00 2.00 3.00

Error

Cu

mu

lati

ve

Normal DistributionNormal DistributionNormal Distribution

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-4.00 -2.00 0.00 2.00 4.00

Z

F(Z

)

Scatter Plots (Paired Scatter Plots (Paired Data)Data)

Shows the relationship between paired data

Example: Suppose for example we wish to look at state per student expenditures versus achievement results on the Stanford Achievement Test

Scatter Plots (Paired Scatter Plots (Paired Data)Data)

SAT vs Student Expenditure

30

40

50

60

0 2,000 4,000 6,000 8,000 10,000 12,000

$ per student

Av

era

ge

SA

T S

co

re


Box Plots Box Plots


Box PlotsBox Plots

Problem with empirical is we may simply not have enough data

For small data sets, analysts often like to provide a rough graphical measure of how data is dispersed

Consider our student data2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

Box PlotsBox Plots

Ranked student Gpa data2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

Min = 2.4 Max = 3.9

2.4 3.9

Box PlotsBox Plots


Median = (3.0+3.0)/2 = 3.0

2.4 3.93.0

Box PlotsBox Plots


Median Bottom= (2.7+2.8)/2 = 2.75

2.4 3.93.02.75

Box PlotsBox Plots


Median Top = (3.3+3.5)/2 = 3.4

2.4 3.93.02.75 3.4

Box PlotsBox Plots


2.4 3.93.02.75 3.4

Fail Time DataFail Time Data

Min = 0.5 Max = 73.8

Lower Quartile = 6.1

Median = 14.3

Upper Quartile = 26.7

0.5 6.1 14.3 26.7 73.8

Class ProblemClass Problem The following data represents sorted observations

on deviations from desired diameters of ball bearings. Compute a box plot.

-1.27 0.24 0.66 1.49 2.20-0.54 0.27 0.77 1.55 2.28-0.51 0.30 0.89 1.55 2.29-0.16 0.31 0.94 1.59 2.29-0.12 0.38 0.95 1.63 2.300.00 0.44 1.00 1.70 2.310.01 0.45 1.01 1.71 2.360.19 0.48 1.03 1.71 2.440.19 0.56 1.15 1.90 2.480.21 0.59 1.18 1.97 2.580.22 0.60 1.30 2.10 2.66


Statistical Measures Statistical Measures


Aside: Mean, Aside: Mean, VarianceVariance

Mean:

Variance:

xp x xdiscretex

( ) ,

2 2 ( ) ( )x p xx

ExampleExample

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

ExampleExample

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

Binomial MeanBinomial Mean

= 1p(1) + 2p(2) + 3p(3) + . . . + np(n)

xp xx

( )

xnxn

x

ppxnx

nx

)1()!(!

!

0

Binomial MeanBinomial Mean

= 1p(1) + 2p(2) + 3p(3) + . . . + np(n)

xp xx

( )

xnxn

x

ppxnx

nx

)1()!(!

!

0

Miracle 1 occurs

= np

Binomial MeasuresBinomial Measures

Mean:

Variance:

xp xx

( )

2 2 ( ) ( )x p xx

= np

= np(1-p)

Binomial DistributionBinomial Distribution

0.0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5

x

P(x

)

0.0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5

x

P(x

)

n=5, p=.3 n=8, p=.5

x

0.0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6 7 8

P(x

)

n=4, p=.8

0.0

0.1

0.2

0.3

0.4

0.5

0 2 4

x

P(x

)

n=20, p=.5

Measures of Measures of CentralityCentrality

Mean

Median

Mode


Mean

xp x xdiscretex

( ) ,

xf x dx xcontinuous( ) ,

Sample Mean

n

i

i

n

xX

1


Exercise: Compute the sample mean for the student Gpa data2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9


Failure Data

2.7 25.8 19.6 4.5 0.56.4 18.3 41.6 5.8 73.813.9 32.2 27.7 5.1 12.034.9 21.0 10.2 46.1 37.914.9 24.1 1.0 29.8 3.37.1 59.9 9.4 12.9 7.911.1 2.1 16.0 22.5 8.63.8 51.8 1.6 17.1 14.7

X 1.19


MedianCompute the median for the student Gpa data2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

0.32

0.30.3

X


ModeClass mark of most frequently occurring interval

For Failure data, mode = class mark first interval

0.5X


Measure Student Gpa Failure DataMean 3.00 19.10Median 3.04 14.40Mode --- 5.00

Sample mean X is a blue estimator of true mean

X E[ X ] = u.b.

Measures of Measures of DispersionDispersion

Range

Sample Variance


RangeCompute the range for the student Gpa data2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

Min = 2.4 Max = 3.9

Range = 3.9 - 2.4 = 1.5


Variance

2 2 ( ) ( )x p xx

2 2

( ) ( )x f x dx

Sample variance

x

11

22

2

n

nxs

n

ii


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

I Xi Xi

2

1 2.4 5.762 2.7 7.293 2.8 7.844 2.9 8.415 3.0 9.006 3.0 9.007 3.1 9.618 3.3 10.899 3.5 12.2510 3.9 15.21Sum = 30.6 95.3Avg = 3.1 9.5

x

11

22

2

n

nxs

n

ii


Exercise: Compute the variance for failure time data

s2 = 302.76

2.7 25.8 19.6 4.5 0.56.4 18.3 41.6 5.8 73.813.9 32.2 27.7 5.1 12.034.9 21.0 10.2 46.1 37.914.9 24.1 1.0 29.8 3.37.1 59.9 9.4 12.9 7.911.1 2.1 16.0 22.5 8.63.8 51.8 1.6 17.1 14.7

An AsideAn Aside

For Failure Time data, we now have three measures for the data

Expontial ??

s2 = 302.76

X 1.19

Power Supply Failure Times

0

5

10

15

20

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Time Class

Fre

qu

en

cy

An AsideAn Aside

Recall that for the exponential distribution = 1/

s2 = 1/2

If E[ X ] = and E [s2 ] = s2, then

1/ = 19.1or

1/2 = 302.76

s2 = 302.76X 1.19

0575.ˆ

0524.ˆ

south dakota school of mines & technology introduction to probability & statistics...

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