Basic Statistical Methods2004PDF

Basic Statistical Basic Statistical MethodsMethods

SHC790SHC790J vd WesthuizenJ vd Westhuizen

ContextContext

Data DescriptionData Description ProbabilityProbability InferenceInference

Data DescriptionData Description


How can we summarise small How can we summarise small amounts of data?amounts of data? Remember that the following is Remember that the following is

required before we continue:required before we continue: Data needs to be a random sampleData needs to be a random sample How accurately were the How accurately were the

measurements mademeasurements made Notice the difference between Notice the difference between

measurements and countsmeasurements and counts Measurements Measurements More accurateMore accurate Counts Counts Always whole numbersAlways whole numbers


Example TableExample Table Bridge Column Bridge Column

Support HeightSupport Height Most important Most important

thing about thing about numbers in this listnumbers in this list How low a How low a

typical bridge is.typical bridge is. How variable How variable

the numbers the numbers are.are. 8.898.8999

8.278.2788

8.528.5277

7.997.9966

9.029.0255

8.988.9844

8.158.1533

8.078.0722

8.788.7811

Height (m)Height (m)NumberNumber


Mean HeightMean Height

Symbol for Symbol for mean if mean if observations is observations is referred to as referred to as xsxs

8.898.8999

8.278.2788

8.528.5277

7.997.9966

9.029.0255

8.988.9844

8.158.1533

8.078.0722

8.788.7811


518.89

89.827.852.899.702.998.815.807.878.8

=

++++++++=HeightMeanBridge

nxx

ni i =

=1


Median HeightMedian Height

Bridge Median Height = 8.52Bridge Median Height = 8.52

If there is an even If there is an even number of observations number of observations we take the average of we take the average of the two middle ones the two middle ones

9.029.0255

8.988.9844

8.898.8999

8.788.7811

8.528.5277

8.278.2788

8.158.1533

8.078.0722

7.997.9966



The ModeThe Mode The mode is the observation that The mode is the observation that

occurs most frequently. In our occurs most frequently. In our previous example there are no previous example there are no mode.mode.

This is more useful for counted This is more useful for counted data. For example number of data. For example number of vehicles per hour over a 24 hour vehicles per hour over a 24 hour period.period.


Guidelines for mean, median and Guidelines for mean, median and modesmodes In case of no outliers use mean.In case of no outliers use mean. In case of outliers In case of outliers

Median Mode

Meas

urem

ents Counts

Vehicles per hour - peak hours have a substantial

influence


Other measures of location (See Other measures of location (See 10)10) Root mean squareRoot mean square

Geometric mean of n positive Geometric mean of n positive numbers = nth root of their numbers = nth root of their productproduct

Harmonic meanHarmonic mean


Measure of variationMeasure of variation RangeRange

The range of a set of numbers is the The range of a set of numbers is the largest minus the smallestlargest minus the smallest

The range of bridge height is 9.02 The range of bridge height is 9.02 7.99 = 1.03 m7.99 = 1.03 m

Its very sensitive Its very sensitive To unusually big To unusually big or small observationsor small observations


IQR and SIQR (InterIQR and SIQR (Inter--quartile quartile Range and SemiRange and Semi--InterInter--quartile quartile range) range) Distance between Distance between quartilesquartiles QuartileQuartile

25% 25% 25% 25%

Separate ordered data into 4 equal groups

1st quartile 2nd quartile 3rd quartile(median)(Lower) (Upper)


To determine the lower and To determine the lower and upper quartiles we recommend upper quartiles we recommend the following:the following: Lower quartileLower quartile

0.25n+0.5 (n 0.25n+0.5 (n number of number of observation)observation)

Upper quartileUpper quartile 0.75n+0.50.75n+0.5


ExampleExample n=9n=9 LQ = 0.25(9)+0.5 = 2.75LQ = 0.25(9)+0.5 = 2.75 UQ = 0.75(9)+0.5 = 7.25UQ = 0.75(9)+0.5 = 7.25

9.029.0255

8.988.9844

8.898.8999

8.788.7811

8.528.5277

8.278.2788

8.158.1533

8.078.0722

7.997.9966


LQ = 8.13

UQ = 8.91 IQR = 8.91 8.13 = 0.78

Data DescriptionData Description Mean Absolute Mean Absolute

Deviation(MAD)Deviation(MAD)

354.0919.3..

=

=DAM xx xx

3.190.00Sum

0.370.378.898.89

0.25-0.258.278.27

0.000.008.528.52

0.53-0.537.997.99

0.500.509.029.02

0.460.468.988.98

0.37-0.378.158.15

0.45-0.458.078.07

0.260.268.788.78

Height (m) Height (m) --xx

Absolute deviation from mean

Deviation from meanObservation

nxx

DAM

=..


Standard DeviationStandard Deviation SD is approximately the average of the SD is approximately the average of the

difference between each value and the difference between each value and the mean.mean.

SD is what is most commonly used as a SD is what is most commonly used as a measure of variationmeasure of variation

The square of the SD is called varianceThe square of the SD is called variance

( )1

2

=

nxxS

2var S=


Standard Deviation Standard Deviation -- ExampleExample

( )1

2

=

nxxS

1.34890Sum

0.13690.378.898.89

0.0625-0.258.278.27

008.528.52

0.2809-0.537.997.99

0.250.59.029.02

0.21160.468.988.98

0.1369-0.378.158.15

0.2025-0.458.078.07

0.06760.268.788.78

Height (m) Height (m) --xx

Squared deviation from mean

Deviation from meanObservation

S =0.411

xx ( )2xx


Standard Deviation Standard Deviation Typical Typical ObservationsObservations

Large dataset Large dataset (1000 observations)(1000 observations)

Few observations futher Few observations futher than 3 SD from the meanthan 3 SD from the mean

MediumMedium--sized dataset sized dataset (100 observations)(100 observations)


Small datasetSmall dataset(10 observations)(10 observations)


Dataset sizeDataset sizeDescriptionDescription


The boxThe box--andand--whisker plotwhisker plot

Med

ian

20

10 20 30 40 50 60 70

LQ UQ

55

Whiskers


Cumulative frequency plotCumulative frequency plot Vertical axis represents the Vertical axis represents the

number/percentage of observations number/percentage of observations that are less than or equal to the x that are less than or equal to the x value on the horizontal axisvalue on the horizontal axis


Flood ExampleFlood Example

SIZE OF FLOOD

0 12 13 16 17 20 26 33 38 50 61 70

FREQ

UEN

CY

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


Flood Example Flood Example IQRIQR The boxThe box--andand--whisker plotwhisker plot

SIZE OF FLOOD

0 12 13 16 17 20 26 33 38 50 61 70

FREQ

UEN

CY

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

16 23 3812 61


Large DatasetsLarge Datasets See arm length measurements See arm length measurements

below.below.

723800725784708795758757791761

737777780763765860812754776788

787835777808761727811765751759

771830767811767774802706796801

817829805834802801811831817801

803776796806786757833811791796

801809727804781763774747783764

816720802789830854830795785759

809740798775806754754758694765

810786767782833846845846817741


Tabulate this dataTabulate this data Choosing classesChoosing classes

Use convenient round Use convenient round numbersnumbers

Sufficient many classes Sufficient many classes not wide groupingnot wide grouping

Sufficient few classes Sufficient few classes between 5 between 5 15 classes15 classes

Each observation must Each observation must go into only one classgo into only one class

1000870 and over

1002850-869

9811830-849

8711810-829

7616800-809

607790-799

5311780-789

428770-779

3412760-769

2213740-759

96720-739

32700-719

11Up to 699

CumulativeCount ObservationsRange

Example 1

Data DescriptionData Description StemStem--andand--leaf plotsleaf plots

69 470 6 67172 0 7 7 5 373 774 2 0 775 8 4 4 9 7 9 1 4 7 876 7 5 4 3 9 7 7 5 1 5 3 177 5 4 6 4 1 7 6 778 2 6 5 9 3 1 6 7 8 0 379 8 5 6 1 6 1 580 6 9 2 4 9 1 6 3 1 1 2 5 1 2 8 081 7 0 6 1 7 1 7 1 1 282 983 3 0 0 3 1 4 0 584 6 5 685 486 0


Histogram presentationHistogram presentation

Arm Length (mm)

650 700 750 800 850 900

Prop

ortio

n of

obs

erva

tion

per c

lass

wid

th

0

2

4

6

8

10

12

14

16

Note the change in range size and the frequency half at range doublethe size of the selected standard

range.Rather keep range size constant


Cumulative frequency plotCumulative frequency plot

Arm Length (mm)

680 700 720 740 760 780 800 820 840 860 880

% O

bser

vatio

ns

=

=

==

=


Temperature Rainfall exampleTemperature Rainfall example Class ExampleClass Example

2.22.23.33.32.92.93.23.24.74.73.13.13.93.93.23.23.43.43.03.03.63.63.13.1

RainfallRainfallTemperatureTemperature

Temperature (Thousands of degrees)

2.9 3.0 3.1 3.2 3.3

Rai

nfal

l (H

unde

rds

mm

)

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0





64.9575.0759.5920.718.9

7.264.8410.892.23.3

9.288.4110.242.93.2

14.5722.099.614.73.1

12.4815.2110.243.93.2

10.211.5693.43

11.1612.969.613.63.1

Temperature-Rainfall(Rainfall)2(Temperature)2RainfallTemperature



57.0655.3055.0

255.0

=

=

=

r

r

SSS

ryyxx

xy

Sxx = 59.59 1/6(18.9)2 = 0.055

Syy = 75.07 1/6(20.7)2 = 3.655

Sxy = 64.95 1/6(18.9)(20.7) = -0.255

( ) 15.415.064.445.364.4055.0255.0

==

=

=

=

ab

b

SS

bxx

xy

ProbabilityProbability

What is meant by probability?What is meant by probability? The probability of an event is the The probability of an event is the

chance that it will occur.chance that it will occur. A priori approach.A priori approach.

Sometimes, the experimental setSometimes, the experimental set--up is so clear, we know the up is so clear, we know the probabilities in advance of probabilities in advance of collecting any data collecting any data


A priori approach.A priori approach. ExamplesExamples

CoinCoin P[x = head] = P[x = tail] = 0.5P[x = head] = P[x = tail] = 0.5

DiceDice P[x = 1] = P[x = 2] = P[x = 3] = P[x = 4] P[x = 1] = P[x = 2] = P[x = 3] = P[x = 4]

=P[x = 5] = P[x = 6] = 1/6=P[x = 5] = P[x = 6] = 1/6

CardsCards P[x = 4 of harts] = 1/52P[x = 4 of harts] = 1/52 P[x = Ace] = 4/52 = 1/13P[x = Ace] = 4/52 = 1/13


EmpiricalEmpirical By having enough experimental By having enough experimental

datadata ExamplesExamples

Break 100 concrete cubes. 30 cubes Break 100 concrete cubes. 30 cubes strength is more than 50MPastrength is more than 50MPa

P[x > 50] = 30/100 = 0.3P[x > 50] = 30/100 = 0.3


Subjective assessmentsSubjective assessments What is the probability that it will What is the probability that it will

rain on the 27rain on the 27thth of July in Pretoria.of July in Pretoria. Argument: During winter iArgument: During winter itt does does

not rain regularly in Pretoria.not rain regularly in Pretoria. No experimental informationNo experimental information Intuition:Intuition:

P[x = rain] = 0.05P[x = rain] = 0.05


Probability ScaleProbability Scale

Prob

abilit

y

Swim through the Atlantic ocean

Dice

Coin

Pass Statistics

Dying 1.00.9

0.5

0.167

0


MutuallyMutually--exclusive eventsexclusive events This means they cannot occur This means they cannot occur

together.together. Consequently, the Consequently, the probabilityprobability of of

one one oror the other of two mutuallythe other of two mutually--exclusive events occurring is the exclusive events occurring is the sum of their individual sum of their individual probabilitiesprobabilities..


Example Example MutuallyMutually--exclusive eventsexclusive events Number of students on campus is Number of students on campus is

25000.25000. B.Eng = 5000B.Eng = 5000 B.Com = 8000B.Com = 8000 Choosing a studentChoosing a student

P[B.Eng] = 5000/25000 = 0.2P[B.Eng] = 5000/25000 = 0.2 P[B.Com] = 8000/25000 = 0.32P[B.Com] = 8000/25000 = 0.32

0][52.032.02.0][

=

=+=

BComBIngPBComBIngP


Venn Diagram Venn Diagram MutuallyMutually--exclusiveexclusive

B.Com

B.Eng

Other

P[B.Eng]0.2

P[B.Com]0.32

P[Other]0.48Sum of Ps = 1


Independent probabilityIndependent probability Where two experiments is not influencing Where two experiments is not influencing

each other.each other. Probability of both happening is the Probability of both happening is the

productproduct of their of their individual probabilitiesindividual probabilities ExampleExample

What sex is a person? Man What sex is a person? Man A, Women A, Women BB Does the person own a vehicle? Yes Does the person own a vehicle? Yes C, No C, No

DD P[A] = 0.45, P[B] = 0.55, P[C] = 0.8, P[D] = 0.2P[A] = 0.45, P[B] = 0.55, P[C] = 0.8, P[D] = 0.2


ExampleExample

No

No

Ve

hicl

e Ve

hicl

e -- Y

esYes

Man Sex Women

0.09 0.11

0.36 0.44

[ ][ ][ ][ ] 11.02.055.0

44.08.055.009.02.045.036.08.045.0

==

==

==

==

DBPCBPDAPCAP


General addition RuleGeneral addition Rule

A C

A is made up of {A but not C} plus {A and C}C is made up of {C but not A} plus {A and C}Therefore A or C is {A} plus {C}, minus {A and C}


ExampleExample

No

No

Ve

hicl

e Ve

hicl

e -- Y

esYes

Man Sex Women

0.09 0.11

0.36 0.44

[ ][ ][ ][ ]

91.044.08.055.0][][][][89.036.08.045.0][][][][

11.02.055.044.08.055.009.02.045.036.08.045.0

=+=+=

=+=+=

==

==

==

==

CBPCPBPCBPCAPCPAPCAP

DBPCBPDAPCAP


Same Example Same Example Probability TreeProbability Tree

Sex

Vehicle Vehicle

ManP=0.45

WomanP=0.55

YesP=0.8

YesP=0.8

NoP=0.2

NoP=0.2

0.36 0.09 0.44 0.11


General Multiplication RuleGeneral Multiplication Rule

A C

yprobabilitlconditionaaastoreferredisACPthatgivenmeanslineverticaltheWhere

CAPCPCandAPor

ACPAPCandAP

}\{""\

}\{}{}{

}\{}{}{

=

=


ExampleExample 12 people12 people 9 9 Native bornNative born 3 3 Foreign bornForeign born If we select 2 people, what is the If we select 2 people, what is the

probability that both are foreign born?probability that both are foreign born? P[F1] = 3/12 = 0.25P[F1] = 3/12 = 0.25 Once F occurred we know that there are 11 Once F occurred we know that there are 11

remaining of whom 2 are foreign born.remaining of whom 2 are foreign born. P[F2P[F2\\F1] = 2/11 = 0.1818F1] = 2/11 = 0.1818 P[F1 and F2] = 0.25 x 0.1818 = 0.04545 = P[F1 and F2] = 0.25 x 0.1818 = 0.04545 =

1/221/22


ExampleExample 200 Students200 Students 77 Accounting77 Accounting 64 Law64 Law 92 Study neither92 Study neither Other 3 numbers to Other 3 numbers to

total 200 total 200 92 = 10892 = 108 Only Accounting =108 Only Accounting =108

64 = 4464 = 44 Only Law = 108 Only Law = 108 77 = 77 =

3131 Both = 77 Both = 77 44 or 64 44 or 64

31 = 3331 = 33 74

7744]\[

165.020033]&[

==

==

AAOnlyP

LAP

44 31

92

33

A L


Class ExampleClass Example A concrete beam will fail if the A concrete beam will fail if the

concrete is to weak or the load concrete is to weak or the load is to high.is to high.

P[Weak] = 0.2P[Weak] = 0.2 P[High] = 0.3P[High] = 0.3 P[Failure] = ?P[Failure] = ?


PP[[Week] = 0.2Week] = 0.2 P[High] = 0.3P[High] = 0.3 P[Failure] = ?P[Failure] = ?

44.006.05.0][)3.0)(2.0(3.02.0][

][][][][][][][][][

==

+=

+=

+=

FailurePFailureP

HighPWeakPHighPWeakPFailurePHighWeakPHighPWeakPFailureP

0.2 0,30.06

W H

Yes

-Loa

d to

hig

h

No

0.3

0.7

0.2 0.80.06

0.14 0.56

0.24

Yes Concrete weak No

P[Failure]

ProbabilityProbability P[Week] = 0.2P[Week] = 0.2 P[High] = 0.3P[High] = 0.3 P[Failure] = ?P[Failure] = ? 44.006.05.0][

)3.0)(2.0(3.02.0][][][][][][][][][][

==

+=

+=

+=

FailurePFailureP

HighPWeakPHighPWeakPFailurePHighWeakPHighPWeakPFailureP

Concrete

WeakP = 0.2

OKP = 0.8

Load Load

HighP = 0.3

OKP = 0.7

HighP = 0.3

OKP = 0.7

0.06 0.14 0.24 0.56

P[Failure] = 0.44


Class ExampleClass ExampleA lowA low--water bridge is designed to allow for water bridge is designed to allow for flood occurring once every 10 years. flood occurring once every 10 years. Damage occurs during each flood. The Damage occurs during each flood. The bridge is also located in an active seismic bridge is also located in an active seismic region and the probability of a destructive region and the probability of a destructive earthquake occurring in a year is 30%. earthquake occurring in a year is 30%. Determine the probability of damage during Determine the probability of damage during any given year assuming that floods and any given year assuming that floods and earthquakes are statistically independent.earthquakes are statistically independent.


Class example Class example SolutionSolution P[Flood] = 0.1P[Flood] = 0.1 P[No flood] = 0.9P[No flood] = 0.9 P[Quake] = 0.3P[Quake] = 0.3 P[No quake] = 0.7P[No quake] = 0.7

37.003.01.03.0][][][][][

03.01.03.0][

=+=

+=

==

FloodQuakePFloodQuakePFloodPQuakePFloodQuakeP

FloodQuakeP


Class exampleClass exampleVehicles are classified into light (10 ton). Vehicle counts show that 60% is light, 30% Vehicle counts show that 60% is light, 30% medium and 10% heavy vehicles. Vehicles medium and 10% heavy vehicles. Vehicles are weighed regularly. If it is known that the are weighed regularly. If it is known that the probability that a vehicle is light and probability that a vehicle is light and overloaded is 0.12. Calculate the overloaded is 0.12. Calculate the probability that the next vehicle is heavy or probability that the next vehicle is heavy or that the next vehicle is overloaded. that the next vehicle is overloaded. Assume statically independence.Assume statically independence.


Class example solutionClass example solution P[light] =0.6P[light] =0.6 P[light and overloaded] = 0.12P[light and overloaded] = 0.12 P[light and overloaded] = P[light and overloaded] =

P[light]P[overloaded]P[light]P[overloaded] 0.12 = 0.6 x P[overloaded]0.12 = 0.6 x P[overloaded] P[overloaded] = 0.12/0.6 = 0.2P[overloaded] = 0.12/0.6 = 0.2


Example continueExample continue

Overloaded Overloaded Overloaded

Vehicle

Light0.6

Medium0.3

Heavy0.1

Yes0.2

Yes0.2

Yes0.2

No0.8

No0.8

No0.8

0.12 0.48 0.06 0.24 0.02 0.08

28.002.01.02.0][][][][][

=+=

+=

HeavyOverloadedPHeavyOverloadedPheavyPoverloadedPHeavyOverloadedP


Exclusive and IndependentExclusive and Independent Exclusive:Exclusive:

Events are ones that never occur together.Events are ones that never occur together. IndependentIndependent

The proportion of times A occurs is the same The proportion of times A occurs is the same whether or not B occurswhether or not B occurs

Using a Venn diagramUsing a Venn diagram

w yx

A B

z

A & B mutually exclusiveX = 0

A & B independentw/x = z/yw & x refer to events in Ay & z refer to events outside ASee paragraph 66 NB!


Example Example Independent featuresIndependent featuresTotal of 200 students, 80 study Total of 200 students, 80 study biology, 90 study neither biology nor biology, 90 study neither biology nor geography, and the choice of geography, and the choice of whether a student does or does not whether a student does or does not study biology is independent of their study biology is independent of their choice of studying geography.choice of studying geography.Let us use the Venn diagram to Let us use the Venn diagram to determine all required numbers in determine all required numbers in the Venn diagram. the Venn diagram.


Example solutionExample solution

w yx

B G

z

z = 90 given

w+x = 80

Total outside B = 200-80Total outside B = 120

Number inside G but outside B = 120 90 = 30y=30

w/x = z/y90/30 = w/xw=3xw+x = 4x =80x = 20w = 60


Conditional ProbabilityConditional ProbabilityN amount of tests consist of:N amount of tests consist of:N = nN = n1111+ n+ n2222+ n+ n1212+ n+ n2121

nn1212 nn2121nn1111

A B

nn2222][

][]/[

][][]/[

][

][

][

2111

11

2111

11

1121

1112

11

APBAPABP

BPBAP

nnN

Nn

nnnBAP

NnnBP

NnnAP

NnBAP

=

=

+

=

+=

+=

+=

=


Class Example: van As & Class Example: van As & JoubertJoubert

Area 1

Area 2

400 vehicles/hour

600 vehicles/hour

1% busses

3% busses

Calculate the probability that the next vehicle will be a bus


Example Example SolutionSolution

978.0022.01][022.0018.0004.0][018.06.003.0][][]/[][004.04.001.0][][]/[][

][][]/[

][][]/[

03.0]/[01.0]/[

6.0][4.0][

2

222

1

111

2

2

2

1

1

1

2

1

2

1

==

=+=

==

=

==

=

=

=

=

=

=

=

MPBP

ABPAPABPABP

ABPAPABPABP

APABPABP

APABPABP

ABPABP

AfromVPAfromVP

Motor Vehicle Bus

2

Are

a

1

0.004

0.018


BayesBayes TheoryTheory

][][]/[]/[

][][]/[

][][]/[

][][]/[

BPAPABPBAP

APABPABP

BPABPBAP

BPBAPBAP

=

=

=

=

lll


Class ExampleClass Example

ReadyReady--mix concrete is delivered at a site. This mix concrete is delivered at a site. This concrete is used to cast columns.concrete is used to cast columns.BB00 Weak concreteWeak concreteBB11 Good concreteGood concreteKK00 Column failColumn failKK11 Column not failingColumn not failingP[BP[B00] = 0.1] = 0.1P[BP[B11] = 0.9] = 0.9P[KP[K00/B/B00] = 0.6] = 0.6P[KP[K11/B/B11] = 0.98] = 0.98P[KP[K00] = ?] = ?P[BP[B00/K/K00] = ?] = ?


Example SolutionExample Solution

77.0078.0

1.06.0][

][]/[]/[

:078.0060.0018.0][

882.09.098.0][]/[][][

][]/[

06.01.06.0][]/[][][

][]/[

0

00000

0

111111

1111

000000

0000

=

==

=+=

===

=

===

=

KPBPBKPKBP

BayeKP

BPBKPBKPBPBKPBKP

BPBKPBKPBP

BKPBKP

P[B1]0.9

P[B0]0.1

P[K1] P[K0]

0.882

0.06

0.01

8

0.04


Class Example Class Example -- van As & van As & JoubertJoubert

Gravel from a quarry. 70% of the material Gravel from a quarry. 70% of the material are classified as good material. 20% of the are classified as good material. 20% of the good material fail during testing and 10% of good material fail during testing and 10% of weak material does not fail during testing. weak material does not fail during testing. When testing the material calculate the When testing the material calculate the probability that the material will pass the probability that the material will pass the test. Then calculate the probability the test. Then calculate the probability the material will be good material given that the material will be good material given that the material pass the test.material pass the test.


Example Example SolutionSolution

Material

Good0.7

Weak0.3

Test Test

Fail0.2

Pass0.8

Fail0.9

Pass0.1

0.14 0.56 0.27 0.03

P[Pass] = 0.56 +0.03 = 0.59

949.0]/[03.056.07.08.0]/[

][][]/[]/[

=

+

=

=

PassGoodP

PassGoodP

PassPGoodPGoodPassPPassGoodP


Combinations and PermutationsCombinations and Permutations Lets start with an example: How many 6 letter Lets start with an example: How many 6 letter

words can me make with the word GROUND.words can me make with the word GROUND. 11stst letter letter -- 6 choices6 choices 22ndnd letter letter -- 5 choices5 choices 33rdrd letter letter -- 4 choices4 choices 44thth letter letter -- 3 choices3 choices 55thth letter letter -- 2 choices2 choices 66thth letter letter 1 choice1 choice Total = 6 x 5 x 4 x 3 x 2 x 1 = 6!Total = 6 x 5 x 4 x 3 x 2 x 1 = 6!

How many 3 letter words can we make?How many 3 letter words can we make? 11stst letter letter -- 6 choices6 choices 22ndnd letter letter -- 5 choices5 choices 33rdrd letter letter -- 4 choices4 choices Total = 6 x 5 x 4 = 6!/3!Total = 6 x 5 x 4 = 6!/3!


PermutationsPermutations nnPPxx is the amount permutations, is the amount permutations,

consisting of x elements from a consisting of x elements from a set of n elementsset of n elements

nnPPxx = n!/(n= n!/(n--x)!x)! For permutations the order is For permutations the order is

important. Therefore GROimportant. Therefore GROORGORG For combinations the order is not For combinations the order is not

important. Therefore GRO=ORGimportant. Therefore GRO=ORG


CombinationsCombinations nnCCxx amount of combinations that amount of combinations that

consist of x elements from a set of n consist of x elements from a set of n elementselements

If we can make If we can make nnCCxx combinations, for combinations, for each combination we can find x! each combination we can find x! permutations.permutations.

((nnCCxx) x! = ) x! = nnPPxx = n!/(n= n!/(n--x)!x)! Therefore (Therefore (nnCCxx) = n!/[(n) = n!/[(n--x)!x!]x)!x!]

!)!(! Cxn xxnnn

x =

=


ExampleExample National LotteryNational Lottery 49 balls choose 6 balls the order not 49 balls choose 6 balls the order not

important. (Number of combinations)important. (Number of combinations)

If you buy 1 ticket every week you will If you buy 1 ticket every week you will win the lotto once in 269 000 yearswin the lotto once in 269 000 years

649

61014

!6)!649(!49

!)!(!

=

=

=

xxnnn

x


Binominal DistributionBinominal Distribution To introduce it, suppose we have To introduce it, suppose we have

a sixa six--sided dice, with four sides sided dice, with four sides painted blue and two sides painted blue and two sides painted red. At any roll, there is a painted red. At any roll, there is a oneone--third chance of getting a red third chance of getting a red face and a twoface and a two--thirds chance of thirds chance of getting a blue face.getting a blue face.

Determine the probability for 3 red Determine the probability for 3 red faces consecutively.faces consecutively.

ProbabilityProbability Binominal DistributionBinominal Distribution

Enables us to Enables us to calculate the calculate the probability of probability of obtaining exactly x obtaining exactly x successes, when the successes, when the probability of success probability of success at any one trail is pat any one trail is p

( )( )( )( )( )( ) 037.0

31

321

311]3[

222.031

323

311]2[

444.031

323

311]1[

296.0321

311]0[

1][

30

3

333

3

21

2

233

2

12

1

133

1

3

0

033

0

=

=

==

=

=

==

=

=

==

=

=

==

==

pXP

pXP

pXP

pXP

ppxXP xxnn

x

Dice

R1/3

R1/3

R1/3

B2/3

B2/3

B2/3

RRR = 0.037

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 1 2 3

Amount Red

Prob

abili

ty


Class ExampleClass ExampleA company manufacture water pipes that A company manufacture water pipes that can resist 500kPa water pressure. It is can resist 500kPa water pressure. It is known that the probability that a pipe will known that the probability that a pipe will burst at 500kPa is 0.05. If 4 water pipes burst at 500kPa is 0.05. If 4 water pipes are selected out of a group of pipes, what are selected out of a group of pipes, what is the probability that:is the probability that:

1. Exactly 1 pipe will burst1. Exactly 1 pipe will burst2. More than 2 pipes burst2. More than 2 pipes burst3. All 4 pipes will burst3. All 4 pipes will burst4. Any pipe burst4. Any pipe burst

ProbabilityProbability Example SolutionExample Solution

( )( )( )( ) ( ) ( )( )( ) ( ) ( )[ ]( )( ) ( ) ( )[ ]( )( ) ( ) ( )[ ]( )( ) ( ) ( )[ ] 00001.005.095.0105.005.01]4[

00048.005.095.0405.005.01]3[

0135.005.095.0605.005.01]2[

1715.005.095.0405.005.01]1[

8145.005.095.0105.005.01]0[

1][

404444

4

313344

3

222244

2

131144

1

040044

0

====

====

====

====

====

==

XP

XP

XP

XP

XP

ppxXP xxnn

x

1. P[X = 1] = 0.17152. P[X > 2] = 0.00048 + 0.00001 = 0.000493. P[X = 4] = 0.000014. Any pipe burst = 1 P[X = 0] = 1 0.8145 = 0.1855

Note the sum of P[X = x] = 1

ProbabilityProbability Poisson DistributionPoisson Distribution

( )

( )

( ) ( ) ( )

( ) ( )

!][

1

11

111

1

1

!111

)!(!!

1)!(!

!1)!(!

!

)(

1)!(!

!

xexXP

en

n

nxnnn

largenForn

nxn

xnnnppxnx

n

nnxnxnpp

xnxn

valuectedexpenpLetstantconnpwheresmallPandgearlnFor

ppxnx

nondistributiBinominal

x

n

x

x

x

n

x

xxxn

xnxxxn

xxn

==

+

+=

=

==

=


Class ExampleClass ExamplePretoria experience yearly an Pretoria experience yearly an average 2.8 large rain storms average 2.8 large rain storms (> 50mm/hour). Calculate the (> 50mm/hour). Calculate the probability that next year there probability that next year there will be:will be:

1.1. 112.2. 223.3. 334.4. More than 2, large rain storms. More than 2, large rain storms.


Class Example Class Example -- SolutionSolution

( )

531.0467.01]2[

238.017.0!0

8.21]2[

]2[]1[]0[1]2[

222.0!3

8.2]3[

238.0!2

8.2]2[

17.0!1

8.2]1[

!][

8.20

8.23

8.22

8.21

==>

++

=>

=+=+==>

===

===

===

2.8=

==

XP

eXP

xPxPxPXP

eXP

eXP

eXP

xexXP

x


Poisson Distribution vs. BinominalPoisson Distribution vs. Binominal

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

n

P[X=

x]Poisson Binominal


Class ExampleClass Example Vehicle counts at a shopping centre Vehicle counts at a shopping centre

shows there is an average of 120 shows there is an average of 120 vehicles arriving per hour.vehicles arriving per hour. Using first Using first the the Poisson distribution Poisson distribution

calculate the probability that in the next calculate the probability that in the next minute;minute;

TThenhen using theusing the Binominal DistributionBinominal Distributiontaking intervals of 2 secondstaking intervals of 2 seconds calculatecalculate the the probability of;probability of;

1.1. No vehicles arriveNo vehicles arrive2.2. 2 vehicles arrive2 vehicles arrive3.3. At least 2 vehicles arriveAt least 2 vehicles arrive


Class Example Class Example PoissonPoisson

594.0]2[])1[]0[(1]2[

271.0]2[135.0]0[!

][

min/2

==+==

==

==

==

=

60120

=

XPXPXPXP

XPXP

xexXP

vehiclesx


Class Example Class Example BinominalBinominal Take 2 sec intervals. Therefore n = 30Take 2 sec intervals. Therefore n = 30

034.0]5[090.0]4[1870.0]3[280.0]2[271.0]1[126.0]0[

)1()!(!

!][

//

302

0667.036001202

==

==

==

==

==

==

==

===

==

XPXPXPXPXPXP

ppxnx

nxXP

minerimentsexpminvehicles

np

p

xnx


Normal DistributionNormal Distribution

Y-Ax

is

X-Axis

ProbabilityProbability Normal DistributionNormal Distribution

=

2

21

21)(

x

exy

Y-A

xis

X-Axis

ProbabilityProbability Standard Normal DistributionStandard Normal Distribution

( )

=

2

21

21)(

z

ezy

Y-A

xis

Z-Axis

z )(zy

-1 1

)( =

xz

21

0


Properties of the normal distributionProperties of the normal distribution Continues functionContinues function y(x) is the probability densityy(x) is the probability density Describe completely through 2 parametersDescribe completely through 2 parameters Probability density 0 where x Probability density 0 where x dydy//dxdx = 0 Maximum at = 0 Maximum at y(y()=)=

Area under function = 1Area under function = 1 Standard Normal DistributionStandard Normal Distribution

= 0= 0 SD = 1SD = 1 Function symmetricFunction symmetric

ythenythen

121


Normalising/Transform the Normal Normalising/Transform the Normal DistributionDistribution Concrete Normal DistributionConcrete Normal Distribution

Y-A

xis

Z-Axis-1 1

016.02

1=

=

2550

1

xz

Y-A

xis

X-Axis

2550

=

=

0 50 100


Normalising/Transform the Normal Normalising/Transform the Normal DistributionDistribution Steel Normal DistributionSteel Normal Distribution

Y-Ax

is

X-Axis

10200

=

=

180 200 250

Y-Ax

is

X-Axis-1 1

040.02

1=

=

10200

1

xz


Class ExampleClass Example Make use of the standard normal distribution tableMake use of the standard normal distribution table Calculate the probability for z:Calculate the probability for z:1.1. Smaller than 0Smaller than 02.2. Smaller than 0.8Smaller than 0.83.3. Greater than 0.8Greater than 0.84.4. Greater than 1.2Greater than 1.25.5. Between 0.8 and 1.2Between 0.8 and 1.26.6. Between Between --0.3 and 1.20.3 and 1.2 For what value of z will the probability be to get a value For what value of z will the probability be to get a value

smaller as zsmaller as z1.1. 0.90.92.2. 0.20.2 What is the probability that a value betweenWhat is the probability that a value between1.1. --1 and 11 and 12.2. --2 and 22 and 23.3. --3 and 3 SD of the mean3 and 3 SD of the mean


Example SolutionExample Solution ProbabilitiesProbabilities1.1. y(0) = 0.5y(0) = 0.52.2. y(0.8) = 0.78814y(0.8) = 0.788143.3. 11--y(0.8) = 1 y(0.8) = 1 0.78814 = 0.211860.78814 = 0.211864.4. 11--y(1.2) = 1 y(1.2) = 1 0.88493 = 0.115070.88493 = 0.115075.5. y(1.2)y(1.2)--y(0.8) = 0.09679y(0.8) = 0.096796.6. y(1.2)y(1.2)--(1(1--y(0.3)) =0.50284y(0.3)) =0.50284 Z ValuesZ Values1.1. y(z) = 0.9 therefore z = 1.28y(z) = 0.9 therefore z = 1.282.2. y(z) = 0.8 ; z =0.85 therefore z = y(z) = 0.8 ; z =0.85 therefore z = --0.850.85 ProbabilitiesProbabilities1.1. y(0


Class exampleClass example A large amount of concrete cubA large amount of concrete cubees ws wereere tested. The tested. The

data set have a normal distribution with a data set have a normal distribution with a =50 =50 MPaMPa and and = 6 = 6 MPaMPa..

Calculate the probability that a cubCalculate the probability that a cubees strength:s strength:1.1. Larger than 65 Larger than 65 MPaMPa2.2. Less than 38 Less than 38 MPaMPa3.3. Between 45 and 60 Between 45 and 60 MPaMPa4.4. Between 59 and 60 Between 59 and 60 MPaMPa (without table)(without table)5.5. Between 1 standard deviation more than the Between 1 standard deviation more than the

average and the 3average and the 3rdrd quartilequartile6.6. Calculate the strength to ensure only 5% of all Calculate the strength to ensure only 5% of all

cubcubees will be less than this strength.s will be less than this strength.7.7. The borders where 80% of the strength will fall into. The borders where 80% of the strength will fall into.



Y-A

xis

X-Axis

MPaMPa

650

=

=

0 50 10044 56

6

Y-Ax

is

X-Axis-1 1

066.02

1=

=

650

1

xz

Normal Distribution

Standard Normal Distribution


Class ExampleClass Example

99379.0)

5.26

5065.1

=(=

=

z

z

Y-Ax

is

Z-Axis

=

65065

1z

Required Area

00621.0]65[)5.21]65[

99379.0)5.2.1

=>

(=>=(

xPxP

cont



74927.020327.095254.0]67.1833.0[]6045[20327.0833.01833.0

833.06

5045]45[

95254.0667.1

667.16

5060]60[.3

02275.097725.011

26

5038.2

====)(=)(=

=

=)(=

=

==(2)=(2)=

=

zxP

zxP

zxP

z

( )

( )

MPaxx

zxxz

xz

zthanlessatlookingButzz

P

zzparpageSee

xPxy

AvgMPax

exyx

1.40)6(65.150

65.1%5

65.195.0).609134.075.084134.0

84134.0)167.075.0).5

9620018984.0018984.0)5960(]6059[

018984.0)59&60(5.59

1.42

21

=

=

+=

=

=

=

==(==

=(==(

===

=

2=


Y-Ax

is

Z-Axis

10%

( )

68.5732.4268.750

28.1:68.57

50)6(28.1

28.1.0

=

+=

+=

=

=

90=

x

Bordersxx

zx

xz

zforz


Using the cumulative frequency plot Using the cumulative frequency plot to graphically judge whether data is to graphically judge whether data is normally distributednormally distributed

Arm Length (mm)

690 700 710 720 730 740 750 760 770 780 790 800 810 820 830 840 850Z co

rres

pond

ing

to p

ropo

rtion

of o

bser

vatio

n le

ss th

an x

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

Straight line indicate a normal distribution


Expectations of discrete random Expectations of discrete random variables.variables. Let the probability that a random Let the probability that a random

variable X takes the value x be P[X=x].variable X takes the value x be P[X=x]. The Expectation of X is:The Expectation of X is:

The summation taking place over all the The summation taking place over all the possible values of X. That is, the possible values of X. That is, the expectation is in effect the mean. expectation is in effect the mean.

( ) ( ) == ][ xXPxXE


Expectation Expectation -- ExampleExample

0.20.2

11

0.10.10.20.20.40.40.10.1P[X=x]P[X=x]

44332200xx

( ) ( ) ( ) ( ) ( )

2)(4.06.08.02.00)(

1.042.034.022.011.00)(

=

++++=

++++=

xExExE


Expectations of continuous random Expectations of continuous random variables. Thenvariables. Then

ExampleExample= dxxxfxE )()(

32)(

61

21)(

211)(

0

20211)(

2

0

32

2

0

=

=

=

=

=

=

xE

xxxE

dxxxxE

isitxotherfor

andbetweenxxf

x

x


Variances and standard deviations of Variances and standard deviations of discrete random variables.discrete random variables. The variance of a random variable X, is The variance of a random variable X, is

the expectation of [Xthe expectation of [X--E(X)]E(X)]22, that is., that is.

( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ]

1.1)(

2.1)(4.02.002.04.0)(

1.022.014.002.011.02)(

][)]([)(

22222

2

==

=

++++=

++++=

==

XV

XVXVXV

datapreviousfromExamplexXPXExXV


Variances and standard deviations of Variances and standard deviations of continuous random variables.continuous random variables.ThenThen

[ ]

47.0)(92)(

94

97

95

81)(

211

32)(

211)(

)()()(

2

0

234

2

0

2

2

==

=

++=

=

=

=

=

=

XV

XV

xxxxXV

dxxxXV

xxfwhereexampleourFor

dxxfXExXV

x

x

ProbabilityProbability Example Example IQIQ The following table provide information The following table provide information

presenting a group of students IQ levels presenting a group of students IQ levels related to the different IQ levels related to the different IQ levels probabilitiesprobabilities

0.010.011401400.020.021301300.080.081201200.220.221101100.330.331001000.220.2290900.090.0980800.020.0270700.010.016060

P[X=x]P[X=x]IQ LevelIQ Level


IQ Test IQ Test -- ExampleExample

IQ Level P[X=x] E(X=x)-P[X=x]x V(X) Probability Cumulative Probability z60 0.01 0.6 15.8404 0.0004 0.004 0.004 -2.6570 0.02 1.4 17.7608 0.0025 0.025 0.029 -1.980 0.09 7.2 35.2836 0.0100 0.100 0.129 -1.1390 0.22 19.8 21.1288 0.0228 0.228 0.357 -0.37

100 0.33 33 0.0132 0.0297 0.298 0.654 0.4110 0.22 24.2 22.8888 0.0223 0.223 0.877 1.16120 0.08 9.6 32.6432 0.0096 0.096 0.973 1.93130 0.02 2.6 18.2408 0.0024 0.024 0.997 2.65140 0.01 1.4 16.1604 0.0003 0.003 1.000

Total 99.8 179.96 0.0999

E(X)=mean 99.80V(X) 179.96Std Deviation 13.41

( )2

211

2=

x

exy


IQ

40 60 80 100 120 140 160

P[X

=x]

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35


IQ

60 70 80 90 100 110 120 130 140

P[X=

x]

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35


IQ

40 60 80 100 120 140 160

P[X

Documents

Basic Statistical Methods2004PDF