weibull and exponential distributions
TRANSCRIPT
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RELIABILITY ENGINEERING UNIT
ASST4403
Lecture 8: WEIBULL AND EXPONENTIAL DISTRIBUTIONS
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earn ng outcomes
Familiarity with the properties of Weibull and
exponential distributions
Ability to find characteristics of a random variablewith Weibull distribution
Ability to find characteristics of a random variablewith ex onential distribution
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: shape parameter1 t
t
: characteristic life
(scale parameter)
e
(Hyper-exponential)
f(t)
=3.5 xponen a
(Normal)
(Rayleigh)
3t
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What is behind
Used first to model fatigue/s reng
An extreme value distribution
(the smallest extreme value)
Most flexible
(Waloddi Weibull:1887-1979, Swedishengineer)
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: shape parameter1 tt
: c arac er s c et e
t
1( ) 1 0.632F t e
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Weibull Distribution
0.8
0.9
, c arac er s c e: me a w c
cumulative probability = 63.2%
0.6
0.7
0.3
0.4
.
63.2%
0.1
0.2
t0
Note: when =1 =
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: shape parameter1
1
The mean value
2
2 2 2 1 The standard
deviation
The median (the random variable has 50% chance of being
lower and 50% greater than this value) /15.0 )5.0ln( mediantt
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Gamma function
,
Otherwise the value (z) can be found in a table
o e a so or a z
Example: (4)=(4-1)!=3!=321=6 2.6 =1.42962
(3.7)=(2.7+1) =2.7 (2.7)= 2.7 1.54469=4.1707 8
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0.
0.
ame growing
0.
0.Same &growing
0.
0.
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Effect of and on the shape of the Weibull pdf
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3-Parameter
We u str ut on
t1
etf )(
1f(t)
f(t)
0
63.2%
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Weibull distribution : examples
A weibull distributed random variable has shapeparame er an sca e parame er , r. n
The mean an ar eva on The median The robabilit the random variable takes a value
between 800 and 1,200 hours The time at which the random variable has 90%
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The mean)
11(1000)
11(
hr23.88688623.01000)5.1(1000
)2
11()11(1000)
11()
21(
22
The median hr25.46388623.011000 2
hrtt median 55.832)5.0ln(1000)5.0ln(2/1/1
5.0
e pro a y e ran om var a e a es a va uebetween 8,000 and 24,000 hours
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e me a w c e ran om vara e as
chance to be greater than
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The probability the random variable takes a value,
)800()1200()1200800Pr(
228001200
FFT
2904.011 10001000
ee
e me a w c e ran om vara e aschance to be greater than
tTtT
t
1.0)Pr(9.0)Pr(
1000
9.09.0
2
9.0
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... 9.09.0
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Exponential Distribution
A special case when =1 in Weibull distribution Very important in reliability engineering
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Exponential Distribution
1
f(t)
0.8
0.9
0.5
0.6
.
f(t)
0.3
0.4
0.1
0.2
0
t, time
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t
etf
)(
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x
t 0, 0( ) ,t
tf t e
( ) { }F t P X t
01
tx te dx e
63.2% = t17
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t Q: Why 63.2%?1( ) 1 1 0.632t
tF t e e
1 mean
Standard1
63.2%
= t18
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Exponential distribution : examples
Redo the example for Weibull distributed randomvara e w s ape parame er an sca eparameter 1,000 hr.
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Exponential distribution :
examp es The mean)
11(1000)
11(
hr100011000)2(1000
)1
11()21(1000)
11()
21(
22
The median hr1000!1!21000
2
hrtt median 15.693)5.0ln(1000)5.0ln(1/1/1
5.0
e pro a y e ran om var a e a es a va uebetween 8,000 and 24,000 hours
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e me a w c e ran om vara e as
chance to be greater than
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Exponential distribution :
examp es The probability the random variable takes a value,
)800()1200()1200800Pr(
8001200
FFT
1481.011 10001000
ee
e me a w c e ran om vara e aschance to be greater than
tTtT
t
1.0)Pr(9.0)Pr(
1000
9.09.0
9.0
... ..
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