Special Continuous Probability Distributions Weibull Distribution

1. PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Special Continuous Probability DistributionsWeibull Distribution

2. Weibull Distribution – Probability Density Function A random variable X is said to have the Weibull Probability Distribution with parameters  and , where  > 0 and  > 0, if the probability density function of is: , for ≥ 0 , elsewhere where,  is the Shape Parameter,  is the Scale Parameter. Note: If  = 1, the Weibull reduces to the Exponential Distribution.

3. Weibull Probability Density Function Shape Probability Density Function t is in multiples of  f(t) 1.8 β=5.0 1.6 β=0.5 1.4 β=3.44 1.2 β=1.0 β=2.5 1.0 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 t

4. Weibull Distribution - Probability Distribution Function , for x ≥ 0 F(X) for various β and =100 F(x) b = 5 b = 3 b = 1 b = 0.5

5. Weibull Probability Paper Derived from double logarithmic transformation of the Weibull Distribution Function. Of the form where Any straight line on Weibull Probability paper is a Weibull Probability Distribution Function with slope, and intercept, - ln, where the ordinate is ln{ln(1/[1-F(t)])} the abscissa is lnt.

6. Weibull Probability Paper Links http://perso.easynet.fr/~philimar/graphpapeng.htm http://www.weibull.com/GPaper/index.htm

7. Use of Weibull Probability Paper b 8 4 3 2 1.5 1.0 0.8 0.7 0.5 99.0 95.0 90.0 80.0 70.0 50.0 40.0 30.0 20.0 10.0 5.0 4.0 3.0 2.0 1.0 0.5 Cumulative probability in percent F(x)in % 1.8 in. = b 1 in. X 10 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 1000 q

8. Weibull Distribution – Mean and Pecentile Mean or Expected Value 100pth Percentile and, in particular Note: See the Gamma Function Table to obtain values of (a)

9. Weibull Distribution – Standard Deviation Standard Deviation of X where

10. The Gamma Function  Values of the Gamma Function

11. Example If X~W(2, 100), then

12. Example If X~W(1/5, 100),

13. f(x) Max f(x)=f(xmode) 0 x xmode Weibull Distribution - Mode Mode - The value of x for which the probability density function is maximum i.e., 0

14. Weibull Distribution - Example Let X = the ultimate tensile strength (ksi) at -200 degrees F of a type of steel that exhibits ‘cold brittleness’ at low temperatures. Suppose X has a Weibull distribution with parameters  = 20, and  = 100. Find: (a) P( X  105) (b) P(98  X  102) (c) the value of x such that P( X  x) = 0.10

15. Weibull Distribution - Example Solution (a) P( X  105) = F(105; 20, 100) (b) P(98  X  102) = F(102; 20, 100) - F(98; 20, 100)

16. Weibull Distribution - Example Solution Continued (c) F(x) = P( X  x) = 0.10 F(x) Then

17. Weibull Distribution – Example Continued The random variable X can modeled by a Weibull distribution with  = ½ and  = 1000. The spec limit is set at x = 4000. What is the proportion of items not meeting spec?

18. Weibull Distribution – Example Solution The fraction of items not meeting spec is That is, about 13.53% of the items will not meet spec.