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
1
Stracener_EMIS 7370/STAT 5340_Fall 08_09.25.08
Exponential Distribution
2
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:
1
f ( x) 

0
e

x

,
for x 0
,
elsewhere
3
Properties of the Exponential Model
• Probability Distribution Function
F (x)  P(X  x) 
for x< 0
0
1- e

x

for x  0
*Note: the Exponential Distribution is said to be
without memory, i.e.
• P(X > x1 + x2 | X > x1) = P(X > x2)
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Properties of the Exponential Model
• Mean or Expected Value
  E (X )  
• Standard Deviation
 
5
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%?
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Exponential Model - Example
The E(X) = 5=θ, so λ = 0.2.
The probability that the response time is at most 10 sec is:
P ( X  10)
 F (10,0.2)
 1  e (.2 )(10)
 1  0.135
 0.865
or P (X>10) = 0.135
The probability that the response time is between 5 and 10 sec is:
P(5  X  10)
 F (10;0.2)  F (5;0.2)
 (1  e  2 )  (1  e 1 )
 0.233
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Exponential Model - Example
The value of x for which the probability of exceeding x is 1%:
P( X  x)  1  e  x  0.99
e x  0.01
 λx  ln( 0.01)
4.605
x
0 .2
x  23.025 sec
F (10)  0.99
 1 e
  N (10)
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Weibull Distribution
9
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 x

is:
x
f ( x) 
  1   
x
e


,
0
,
forx  0
elsewhere
Where,  is the Shape Parameter,  is the Scale
Parameter. Note: If  = 1, the Weibull reduces to
the Exponential Distribution.
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The Weibull Probability Distribution Function
Probability Density Function
f(t)
1.8
β=5.0
1.6
1.4
1.2
β=0.5
β=3.44
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
t is in multiples of 
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The Weibull Probability Distribution Function
for x  0
F(x)  PX  x   1 - e
x
 
θ
β
F(t) for various  and  = 100
F(x)
probability, p
1
5
0.8
3
1
0.6
  0.5
0.4
0.2
0
0
50
100
x
150
200
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Weibull Probability Paper (WPP)
• Derived from double logarithmic transformation of
the Weibull Distribution Function.
 ( t /  )
• Of the form
where
F(t )  1  e
y  ax  b




1
y  ln ln 


1

F
(
t
)




a   b   ln  x  ln t
•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.
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Weibull Probability Paper (WPP)
Weibull Probability Paper links
http://perso.easynet.fr/~philimar/graphpapeng.htm
http://www.weibull.com/GPaper/index.htm
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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
F(x)
in %
Cumulative probability in percent
50.0
40.0
30.0
1.8 in.  
20.0
1 in.
10.0
5.0
4.0
3.0
2.0
1.0
0.5
10
2
3
4 5 6 7 8 9 10
2

3
4 5 6 7 8 9 1000
15
x
Properties of the Weibull Distribution
• 100pth Percentile
x p   - ln(1 - p)
1

and, in particular
x0.632  
• Mean or Expected Value
1 
  E(X)     1
 
Note: See the Gamma Function Table to
obtain values of (a)
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Properties of the Weibull Distribution
• Standard Deviation of X
  2  2  1 
     1     1
  
  
1
2
where
 (a)  (a)
2
2
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The Gamma Function 

(a )   e x dx
x
a 1
0
(a  1)  a(a )
Values of the
Gamma Function
y=a
1
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
1.1
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.2
1.21
1.22
1.23
1.24
 (a)
1
0.9943
0.9888
0.9836
0.9784
0.9735
0.9687
0.9642
0.9597
0.9555
0.9514
0.9474
0.9436
0.9399
0.9364
0.933
0.9298
0.9267
0.9237
0.9209
0.9182
0.9156
0.9131
0.9108
0.9085
a
1.25
1.26
1.27
1.28
1.29
1.3
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.4
1.41
1.42
1.43
1.44
1.45
1.46
1.47
1.48
1.49
 (a)
0.9064
0.9044
0.9025
0.9007
0.899
0.8975
0.896
0.8946
0.8934
0.8922
0.8912
0.8902
0.8893
0.8885
0.8879
0.8873
0.8868
0.8864
0.886
0.8858
0.8857
0.8856
0.8856
0.8858
0.886
a
1.5
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.6
1.61
1.62
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.7
1.71
1.72
1.73
1.74
 (a)
0.8862
0.8866
0.887
0.8876
0.8882
0.8889
0.8896
0.8905
0.8914
0.8924
0.8935
0.8947
0.8959
0.8972
0.8986
0.9001
0.9017
0.9033
0.905
0.9068
0.9086
0.9106
0.9126
0.9147
0.9168
a
1.75
1.76
1.77
1.78
1.79
1.8
1.81
1.82
1.83
1.84
1.85
1.86
1.87
1.88
1.89
1.9
1.91
1.92
1.93
1.94
1.95
1.96
1.97
1.98
1.99
2
 (a)
0.9191
0.9214
0.9238
0.9262
0.9288
0.9314
0.9341
0.9369
0.9397
0.9426
0.9456
0.9487
0.9518
0.9551
0.9584
0.9618
0.9652
0.9688
0.9724
0.9761
0.9799
0.9837
0.9877
0.9917
0.9958
1
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Properties of the Weibull Distribution
• Mode - The value of x for which the probability
density function is maximum
i.e.,
f x mode   max f ( x)
1
x mode   1  1  
f(x)
Max f(x)=f(xmode)
0
xmode
x
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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
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Weibull Distribution - Example Solution
(a)
P( X  105) = F(105; 20, 100)
 1 e
(b)
 (105/100) 20
 1  0.070  0.930
P(98  X  102)
= F(102; 20, 100) - F(98; 20, 100)
e
 ( 0.98) 20
e
 (1.02) 20
 0.513  0.226  0.287
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Weibull Distribution - Example Solution
(c)
P( X  x) = 0.10
P( X  x)
Then
e
 ( x / 100) 20
 1 e
 ( x /100) 20
 0.10
 0.90
( x / 100) 20   ln 0.90
( x / 100) 20   ln 0.90
x / 100   ln 0.90
1/ 20
x  100 ln 0.90
1/ 20
x  89.36
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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?
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Weibull Distribution - Example
The fraction of items not meeting spec is
PX  4000  1  P( X  4000)
 1  F(4000)
1/2
e
 4000 


1000


 e 2
 0.1353
That is, all but about 13.53% of the items will not meet
spec.
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