asian institute of technology january...

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Economic Risk and Decision Analysis Economic Risk and Decision Analysis for Oil and Gas Industryfor Oil and Gas Industry

CE81.9008CE81.9008School of Engineering and TechnologySchool of Engineering and Technology

Asian Institute of TechnologyAsian Institute of Technology

January SemesterJanuary Semester

Presented byPresented byDr. Thitisak BoonpramoteDr. Thitisak Boonpramote

Department of Mining and Petroleum Engineering, Department of Mining and Petroleum Engineering, ChulalongkornChulalongkorn UniversityUniversity

Descriptive StatisticsDescriptive Statistics

Part B:Working With Grouped Data

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Grouping DataGrouping Data

Condensing large data sets into groups simplifies calculations of parametersSteps in grouping

1. Define classes for data set to be analyzed2. Determine frequency of data element appearances in

each class3. Calculate absolute and relative frequency4. Calculate class mark (CM), or midpoint, for each class5. Proceed with calculation of parameters

Guidelines for Defining ClassesGuidelines for Defining Classes

Number of classes should be between 5 and 20Define classes so that every element in data set falls into one and only one classNo class should be empty

3

Guidelines for Defining ClassesGuidelines for Defining Classes

Approximate number of classes (Nc) for data set with N elements

NN c log322.31+=

Class interval (CI) should be same for entire data set

cNXX minmaxCI −

=

Determining MeanDetermining Mean

( )

∑

∑

=

== n

ii

n

iii

f

fX

1

1CM

Class mark(mid value of individual class)

Frequency of individual class

4

Class interval

Determining Median, ModeDetermining Median, Mode

⎟⎠⎞

⎜⎝⎛ −

+=b

anL 2CIMedian

Lower boundary of median class

Number of elements in sample

Cumulative frequency of class preceding median class

Number of elements in median class

Mode taken as CM of class with highest frequency of data elements

Determining Geometric, Harmonic MeanDetermining Geometric, Harmonic Mean

( )

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

∑

∑

=

=n

ii

n

iii

m

f

f

1

1CMln

expG

Geometric mean

Harmonic mean

( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

∑

∑

=

=n

iii

n

ii

m

f

f

1

1

CMexpH

5

Determining Standard Deviation, VarianceDetermining Standard Deviation, Variance

( ) ( )

n

nffs

n

iii

n

iii

2

1

2

1CMCM ⎥

⎦

⎤⎢⎣

⎡−

=∑∑==

Standard deviation

Variance = s 2

Calculate Parameters From Grouped DataCalculate Parameters From Grouped Data

Calculate measures of central tendency for grouped drill-bit data

536972768089…

139

123456…20

Ft. DrilledBit numberGroup data

Calculate class interval

5322.5)20log(322.31

log322.31

≅=+=+= NN c

172.175

53139

CI minmax

≅=−

=

−=

NXX

6



Σf = 20

23

b = 663

Freq. (f)

60.578.596.5

114.5132.5

Class Mark (CM)

0.208591.8150213,011.002,020.0

0.03310.03820.06220.05240.0226

8.205312.089327.417328.443414.6597

7.320.5018,486.7555,873.5078,661.5052,668.75

121.0235.5579.0697.0397.5

52-6970-87

88-105106-123124-141

f /(CM)f ln(CM)f (CM)2f (CM)Class



Σf = 20

fCM

0.208591.8150213,011.002,020.0


Mean ( )ft 101

200.020,2

1

1 ===

∑

∑

=

=n

ii

i

n

ii

f

CMfX

7



Σf = 20

fCM

0.208591.8150213,011.002,020.0


Median

ft 17.1026

522017882

=

⎟⎠⎞

⎜⎝⎛ −

+⎟⎠⎞

⎜⎝⎛ −

+=b

anCLLM



Σf = 20

fCM

0.208591.8150213,011.002,020.0


Geometric mean

( )ft 57.98

208150.91exp

lnexp 5980.4

1

1 ==⎟⎠⎞

⎜⎝⎛=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

∑

∑

=

= ef

CMfG n

ii

i

n

ii

m

8



Σf = 20

fCM

0.208591.8150213,011.002,020.0


Harmonic mean

( )ft 92.95

2085.020

1

1 ==

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

∑

∑

=

=n

iiii

n

ii

m

CMff

fH



Σf = 20

fCM

0.208591.8150213,011.002,020.0


Standard deviation

( ) ( )

ft 20.2155.44920

020,204011,21320

20020,2011,213CMCM

22

11

2

==−

=

−=

⎥⎦

⎤⎢⎣

⎡−

=∑∑==

n

nffs

n

iii

n

iii

9



Σf = 20

fCM

0.208591.8150213,011.002,020.0


Variance (s 2)

55.44920.21 22 ==s

Frequency Distribution (Histogram)Frequency Distribution (Histogram)

Presents distribution of frequencies of values of variablesSometimes used for ungrouped data, usually discrete variablesMore commonly used for grouped data, either discrete or continuous variables

Absolute frequency distribution shows actual number of data elements in each classRelative frequency distribution shows proportion of data items in each class

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Kinds of DistributionsKinds of Distributions

Cumulative include number or proportion of data elements in and below each class Decumulative include number or proportion of data elements in and above each class

Frequency CurveFrequency Curve

Frequency polygon in which data points are fitted to smooth curveTo fit normal curve to histogram

( )2

5.0

2sCI ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−×

= sXX

enxfπ

Class interval used to draw histogram

Number of observations(n = 1 if drawn on

proportional frequency)

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OgiveOgive

Presents cumulative frequencies (relative or absolute) vs. classboundaries

Cumulative (frequencies equal to or less than) ogive has frequencies plotted at upper boundary of each classDecumulative (frequencies equal to or greater than) ogive has frequencies plotted at lower boundary of each class

Plot Bit RecordsPlot Bit Records

Construct histogram and ogives (both kinds)

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20/20=1.0018/20=0.9015/20=0.759.20=0.453/20=0.15

2/20=0.105/20=0.25

11/20=0.5517/20=0.8520/20=0.00

23+2=5

6+5=116+11=173+17=20

23663

60.578.596.5

114.5132.5

52-6970-87

88-105106-123124-141

Decum. Rel. Freq.

Cum. Rel. Freq.

Cum. Freq (CF)

Freq. (f )

Class Mark (CM)Class

12


Construct histogram and ogives (both kinds)Histogram, frequency curve

Freq. (Number

of Bits)

7.5

6.0

4.5

3.0

1.5

0.052 69 87 105 123 141

Footage Drilled by Bits


Construct histogram and ogives (both kinds)Ogives (cumulative)

0 69 87 105 123 141

Freq. (Number

of Bits)

100

80

60

40

20

0


Less than or equal to…

13


Construct histogram and ogives (both kinds)Ogives (decumulative)

0 69 87 105 123 141

Freq. (Number

of Bits)

100

80

60

40

20

0


Greater than or equal to…


Fit a normal curve to the histogram

0.17020.49921.2070

…1.33470.5661

0.02850.08370.2024

…0.22380.0949

-3.5567-2.4805-1.5978

…-1.4972-2.3549

405060…

140150

0.17020.49921.2070

…1.33470.5661

0.02850.08370.2024

…0.22380.0949

-3.5567-2.4805-1.5978

…-1.4972-2.3549

405060…

140150

2

5.0 ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

sXxp pe ( )

π220s

xCIexf p=xs

sXxp ⎟⎟⎠

⎞⎜⎜⎝

⎛ −−= 5.0 ( )

π2CI20

sexf p=

14


Fit a normal curve to the histogram

Freq. (Number

of Bits)

7

6

5

4

3

2

1

040 60 80 100 120 140 160


QuartilesQuartiles

Quartiles (symbol Q) divide data set into four equal parts

Interquartilerange

Median

Range

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Calculating QuartilesCalculating Quartiles

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

1

111

4CIb

anLQ

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

2

222

4CIb

anLQ

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

3

333

4CIb

anLQ

Lower limits of classescontaining quartiles

Analogous to ain equation for computational median

Analogous to bin equation for computational median

Other Subdivisions of Data SetsOther Subdivisions of Data Sets

Deciles (symbol D) divide data set into 10 equal partsPercentiles (symbol P) divide data set into 100 equal parts

Median, second quartile (Q2), fifth decile (D5), and 50th

percentile (P50) are identical.

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Position Indicator for Position Indicator for FractileFractile

( )1+= nkiF

Fractile indicator (Pi, Di, or QI)at which fractile is read from ungrouped data

Magnitude of desired fractile

If D7 is desired, i = 7

Maximum number of divisions

Determine Footage Drilled at Determine Footage Drilled at FractilesFractiles

Use ungrouped drilling recordsDetermine footage drilled for P 50, D7, Q3, and D5

536972…

139

123…20

Ft. DrilledBit number

( ) ( ) 5.1012010050

50 =+=PF

10th value = 102; 11th value = 105∴ P50 = (102+105)/2 = 103.5

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Determined Footage Drilled at Determined Footage Drilled at FractilesFractiles


536972…

139

123…20


( ) ( ) 7.14120107

7 =+=DF

14th value = 110; 15th value = 115∴ D7 = 110+0.7(115-110)=113.5



536972…

139

123…20


( ) ( ) 75.1512043

3 =+=QF

15th value = 115; 16th value = 116∴ Q3 = 115+0.75(116-115)=115.75

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536972…

139

123…20


( ) ( ) 5.10120105

5 =+=DF

Since D5 = P50,D5 = 103.5

Coefficient of Coefficient of PeakednessPeakedness, , aa44

Dimensionless value

( )

4

1

4

4

444

moment centralfourth

sn

XXf

sm

sa

n

iii∑

=

−

=

==

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Coefficient of Coefficient of SkewnessSkewness, , aa33

When a3 = 0, curve is symmetric or bell shapedWhen a3 < 0, curve is skewed to rightWhen a3 > 0, curve is skewed to left

( )

3

1

3

3

333

moment central third

sn

XXf

sm

sa

n

iii∑

=

−

=

==

Using Spreadsheets (Excel)Using Spreadsheets (Excel)

May need to install plug-in(s)May enter data in random order

No need to sortMay need to enter data in single column or row

Excel uses ‘kurtosis’ for ‘peakedness’

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Find Means, Median, ModeFind Means, Median, Mode

Enter data in single columnUse Tools > Data Analysis > Descriptive Analysis

Enter range of dataClick ‘Grouped by columns’Click ‘Summary Statistics’

Statistics From ExcelStatistics From Excel

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22


Histogram From ExcelHistogram From Excel

Upper boundary of each class interval

23



24

01234567

69 87 105 123 More

Class Interval

Frequency

0%

20%

40%

60%

80%

100%

Cum


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