1. -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_

2. Exponential Distribution

3. 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

4. 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)

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

6. 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%?

7. 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:

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

9. Weibull Distribution

10. 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.

11. 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 t
t is in multiples of 

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

13. 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)

14. Weibull Probability Paper (WPP)
Weibull Probability Paper links

15. Use of Weibull Probability Paper
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. q x

16. 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

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

18. The Gamma Function 
Values of the
Gamma Function

19. 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)

20. 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

21. 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)

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

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

24. 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.