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

Exponential Distribution

The Exponential Model - Definition 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

Properties of the Exponential Model Probability Distribution Function for < 0 for  0 *Note: the Exponential Distribution is said to be without memory, i.e. P(X > x1 + x2 | X > x1) = P(X > x2)

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

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

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

Weibull Distribution

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

The Weibull Probability Distribution Function for x  0 b = 5 b = 3 b = 1 F(x) b = 0.5

Weibull Probability Paper (WPP) 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)

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

Use of Weibull Probability Paper 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. 10 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 1000 q x

Properties of the Weibull Distribution 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

Properties of the Weibull Distribution Standard Deviation of X where

The Gamma Function  Values of the Gamma Function

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

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

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)

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

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