1 Choice of Distribution 1.Theoretical Basis e.g. CLT, Extreme value 2.Simplify calculations e.g. Normal or Log Normal 3.Based on data: - Histogram - Probability.

11

Choice of DistributionChoice of Distribution

1. Theoretical Basise.g. CLT, Extreme value

2. Simplify calculationse.g. Normal or Log Normal

3. Based on data:- Histogram- Probability paper

22

(a)Arrange the data in ascending order

(b)Let

(c)Plot VS

(d)See if follows a straight line

Nm xxxx ,...,...,, 21

1

N

mPm

mx mP

Use of Probability PaperUse of Probability Paper

33

Arithmetic Plot Probability Paper

1

0 x

)(xFx

Pro

bab

ility

Sca

le

x

)(xFx

Concept of Probability PaperConcept of Probability Paper

44

ExampleExample

Data: 2.9, 3.5, 4, 2.5, 3.1 N=5

　m 　 　 　

1 2.5 1/6= 0.1667

2 2.9 2/6= 0.3333

3 3.1 3/6= 0.5000

4 3.5 4/6= 0.6667

5 4 5/6= 0.8333

mx 1/ NmPm

55

Normal Probability paper

75.02.395.3

ˆ

2.3ˆ

5.084.0

5.0

xx

x

50% 84%

4

3

2

1

3.2

3.95

66

Shear strength of concreteShear strength of concrete

m Shear strength s

m/N+1 lnS m

Shear strength s

m/N+1

lnS

1 0.35 0.0714 -1.05 8 0.58 0.5714 -0.54

2 0.40 0.1429 -0.92 9 0.68 0.6429 -0.39

3 0.41 0.2143 -0.89 10 0.7 0.7143 -0.36

4 0.42 0.2857 -0.87 11 0.75 0.7857 -0.29

5 0.43 0.3571 -0.84 12 0.87 0.8571 -0.14

6 0.48 0.4286 -0.73 13 0.96 0.9286 -0.04

7 0.49 0.5000 -0.71

77

Normal Probability Paper

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

S

88

0

-0.2

-0.4

-0.6

-0.8

-1.0

-1.2

Normal Probability Paper

-0.58

-0.22

50% 84%

0.36 -0.58;

withlognormal is so

S

S

S

36.0

)58.0(22.0

58.0

ln

ln

lnS

99

Lognormal Probability Paper

S

1.0

0.7

0.5

0.3

58.056.0lnln mx

1010

• Even though the data points appear to fall on a straight line, but how good is it?

• Would it be accepted or rejected at a prescribed confidence level?

• If it appears to fit several probability models, which one is better?

Goodness of fit test of distribution• Chi-square test ( )

• Kolmogorov-Smirnov test (K-S)

2

1111

Procedures of Chi-Square test ( )Procedures of Chi-Square test ( )

1. Draw histogram2. Draw proposed distribution (frequency

diagram) normalized by no. of occurrence same area as histogram

3. Select appropriate intervals4. Determine = observed incidences per interval

= predicted incidences per interval

based on model

2

in

ie

1212

Procedures of Chi-Square test ( )Procedures of Chi-Square test ( )

5. Determine for each interval

6. Determine for all intervals

Note: Larger Z less fit

7. Compare Z with the standardized value

2

i

ii

e

en 2)(

k

i i

ii

e

enZ

2)(

fC ,1

level of confidence No. of parameters in

proposed distribution, estimated from data

f = k – 1 – m

1313

• Validity of method rely on (combine some intervals if necessary)

55 iek ,

8.Check:

If probability model substantiated with confidence level

Otherwise Model not substantiated

fCZ ,1 1

1414

1515

Example 6.7 –Cracking strength of concrete

= 10.73

= 7.97

f = 8 – 3 = 5

= 11.1

As both & <

Both models substantiated

(LN is better than N)

NZ

LNZ

5 ,95.0C

NZLNZ 5 ,95.0C

1616

Kolmogorov-Smirnov (K-S) TestKolmogorov-Smirnov (K-S) Test

)(xSn

n

kk

xx

xxxn

k

xx

1

0

1

1{Arrange the data in ascending order:

Nk xxxx ,...,...,, 21

Sample CDF

1717

• Compare of sample with CDF, of proposed model.

• Identify the largest discrepancy between the two curves.

• Compare with a standardized value

reject model

model substantiated

)(xSn)(xFX

maxD

maxD

DD

DD

max

max

1818

265.005.026 D