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

4. 4 Example Data: 2.9, 3.5, 4, 2.5, 3.1 N=5 m 12.51/6=0.1667 22.92/6=0.3333 33.13/6=0.5000 43.54/6=0.6667 545/6=0.8333

5. 5 Normal Probability paper 50% 84% 43214321 3.2 3.95

6. 6 Shear strength of concrete m Shear strength s m/N+1 lnS m Shear strength s m/N+1 lnS 10.350.0714-1.05 80.580.5714-0.54 20.400.1429-0.92 90.680.6429-0.39 30.410.2143-0.89 100.70.7143-0.36 40.420.2857-0.87 110.750.7857-0.29 50.430.3571-0.84 120.870.8571-0.14 60.480.4286-0.73 130.960.9286-0.04 70.490.5000-0.71

7. 7 Normal Probability Paper 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 S

8. 8 0 -0.2 -0.4 -0.6 -0.8 -1.2 Normal Probability Paper -0.58 -0.22 50% 84% lnS

9. 9 Lognormal Probability Paper S 1.0 0.7 0.5 0.3

10. 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. 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. 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. 13 Validity of method rely on (combine some intervals if necessary) 8.Check: If  probability model substantiated with confidence level Otherwise  Model not substantiated

14. 14

15. 15 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)

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

17. 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

18. 18