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 paper

2 (a)Arrange the data in ascending order  (b)Let (c)Plot VS (d)See if follows a straight line Use of Probability Paper

3 Arithmetic Plot Probability Paper 1 0x Probability Scale x Concept of Probability Paper

4 Example Data: 2.9, 3.5, 4, 2.5, 3.1 N=5 m 12.51/6= /6= /6= /6= /6=0.8333

5 Normal Probability paper 50% 84%

6 Shear strength of concrete m Shear strength s m/N+1 lnS m Shear strength s m/N+1 lnS

7 Normal Probability Paper S

Normal Probability Paper % 84% lnS

9 Lognormal Probability Paper S

10 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)

11 Procedures of Chi-Square test ( ) 1.Draw histogram 2.Draw proposed distribution (frequency diagram) normalized by no. of occurrence  same area as histogram 3.Select appropriate intervals 4.Determine = observed incidences per interval = predicted incidences per interval based on model

12 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 level of confidence No. of parameters in proposed distribution, estimated from data f = k – 1 – m

13 Validity of method rely on (combine some intervals if necessary) 8.Check: If  probability model substantiated with confidence level Otherwise  Model not substantiated

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15 Example 6.7 –Cracking strength of concrete = = 7.97 f = 8 – 3 = 5 = 11.1 As both & <  Both models substantiated (LN is better than N)

16 Kolmogorov-Smirnov (K-S) Test Arrange the data in ascending order: Sample CDF

17 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

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