1. 1 Special Continuous Probability Distributions -Exponential Distribution -Weibull Distribution Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Systems Engineering Program Department of Engineering Management, Information and Systems Stracener_EMIS 7370/STAT 5340_Fall 08_09.25.08

2. 2 Exponential Distribution

3. 3 A random variable X is said to have the Exponential Distribution with parameters , where  > 0, if the probability density function of X is:,for  0,elsewhere The Exponential Model - Definition

4. 4 Probability Distribution Function for < 0 for  0 *Note: the Exponential Distribution is said to be without memory, i.e. P(X > x 1 + x 2 | X > x 1 ) = P(X > x 2 ) Properties of the Exponential Model

5. 5 Mean or Expected Value Standard Deviation Properties of the Exponential Model

6. 6 Suppose the response time X at a certain on-line computer terminal (the elapsed time between the end of a user’s inquiry and the beginning of the system’s response to that inquiry) has an exponential distribution with expected response time equal to 5 sec. (a) What is the probability that the response time is at most 10 seconds? (b) What is the probability that the response time is between 5 and 10 seconds? (c) What is the value of x for which the probability of exceeding that value is 1%? Exponential Model - Example

7. 7 The E(X) = 5=θ, so λ = 0.2. The probability that the response time is at most 10 sec is: The probability that the response time is between 5 and 10 sec is: Exponential Model - Example or P (X>10) = 0.135

8. 8 The value of x for which the probability of exceeding x is 1%: Exponential Model - Example

9. 9 Weibull Distribution

10. 10 Definition - 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. The Weibull Probability Distribution Function

11. 11 Probability Density Function f(t) t t is in multiples of  1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 β=0.5 β=5.0 β=3.44 β=2.5 β=1.0 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 The Weibull Probability Distribution Function

12. 12 for x  0     F(x) The Weibull Probability Distribution Function

13. 13 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 ln t. Weibull Probability Paper (WPP)

14. 14 Weibull Probability Paper links http://perso.easynet.fr/~philimar/graphpapeng.htm http://www.weibull.com/GPaper/index.htm Weibull Probability Paper (WPP)

15. 15 F(x) in %   x                   Cumulative probability in percent 1.8 in.  1 in. Use of Weibull Probability Paper

16. 16 100pth Percentile and, in particular Mean or Expected Value Note: See the Gamma Function Table to obtain values of (a) Properties of the Weibull Distribution

17. 17 Standard Deviation of X where Properties of the Weibull Distribution

18. 18 Values of the Gamma Function The Gamma Function 

19. 19 Mode - The value of x for which the probability density function is maximum i.e., x mode 0 f(x) x Max f(x)=f(x mode ) Properties of the Weibull Distribution

20. 20 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 Weibull Distribution - Example

21. 21 (a) P( X  105) = F(105; 20, 100) (b) P(98  X  102) = F(102; 20, 100) - F(98; 20, 100) Weibull Distribution - Example Solution

22. 22 (c) P( X  x) = 0.10 P( X  x) Then Weibull Distribution - Example Solution

23. 23 The random variable X can modeled by a Weibull distribution with  = ½ and  = 1000. The spec time limit is set at x = 4000. What is the proportion of items not meeting spec? Weibull Distribution - Example

24. 24 The fraction of items not meeting spec is That is, all but about 13.53% of the items will not meet spec. Weibull Distribution - Example