l berkley davis copyright 2009 mer301: engineering reliability lecture 4 1 mer301: engineering...

L Berkley DavisCopyright 2009

MER301: Engineering ReliabilityLecture 4

1

MER301: Engineering Reliability

LECTURE 4:

Chapter 3:Lognormal Distribution, The Weibull Distribution, and Discrete Variables



2

Summary of Topics

Lognormal Distribution

Weibull Distribution

Probability Density and Cumulative Distribution Functions of Discrete Variables

Mean and Variance of Discrete Variables



3

Normal Distribution Many Physical Phenomena are

characterized by normally distributed variables

Engineering Examples include variation in such areas as: Dimensions of parts Experimental measurements Power output of turbines



4


Special case of the normal distribution where and the variable w is normally distributed Chemical processes and material properties are

often characterized by lognormal distributions

Parameters and are the mean and variance of W, respectively

wex

2



5


)1()()(22 eXEXV



6




7

Lognormal Example 4.1

Gas Turbine CO Emissions

is a normally distributed function of combustor fuel/air ratio

Mean value of CO will need to be 9ppm or less

flameT

)2750(07324.05.4 flameTeCO

)9( ppmCOP



8

Lognormal Example 4.1(cont)CO as a Function of Temperature

0

10

20

30

40

50

60

2700 2720 2740 2760 2780 2800

Flame Temperature

CO

Em

iss

ion

s

Series1



9

Lognormal Example 4.1

excel spreadsheet for the CO example

Tflame CO(ppm)2745 6.492731 18.12754 3.362769 1.122768 1.22776 0.672717 50.452746 6.032766 1.392734 14.532740 9.362725 28.082722 34.98

2735 13.52738 10.842718 46.892741 8.72744 6.982752 3.892745 6.492745 6.492744 6.982770 1.042749 4.842747 5.612742 8.082780 0.52763 1.742786 0.32

2740 9.362775 0.722726 26.12758 2.52764 1.612779 0.542749 4.842742 8.082760 2.162744 6.982761 2.012728 22.542737 11.662727 24.252745 6.492750 4.5

2750 4.5

2745 6.492783 0.4

2724 30.212739 10.07

279027802770276027502740273027202710

10

5

0

Tflame

Fre

qu

en

cy

50454035302520151050

15

10

5

0

CO(ppm)

Fre

qu

en

cy


Skewness and Kurtosis: Tflame Example

Skewness Skewness characterizes the degree of

asymmetry of a distribution around its mean. Positive skewness indicates a distribution with an asymmetric tail extending towards more positive values. Negative skewness indicates a distribution with an asymmetric tail extending towards more negative values" (Microsoft, 1996). Samples from Normal distributions produce a skewness statistic of about zero.

ses can be estimated roughly using a formula from Tabachnick & Fidell,1996

Kurtosis kurtosis characterizes the relative

peakedness or flatness of a distribution compared to the normal distribution. Positive kurtosis indicates a relatively peaked distribution. Negative kurtosis indicates a relatively flat distribution. Samples from Normal distributions produce a kurtosis statistic of about zero

sek can be estimated roughly using a formula from Tabachnick & Fidell, 1996

279027802770276027502740273027202710

10

5

0

Tflame

Fre

qu

en

cy

zerocontainsN

skewness ......6

2

zerocontainsN

kurtosis .......24

2

50N

Mean 2748.36Standard Error 2.477286204Median 2745Mode 2745Standard Deviation 17.51705874Sample Variance 306.8473469Kurtosis -0.492454159Skewness 0.351282489Range 69Minimum 2717Maximum 2786Sum 137418Count 50


Lognormal Example 4.1(cont) Tflame

Mean 2748.36Standard Error 2.477286204Median 2745Mode 2745Standard Deviation 17.51705874Sample Variance 306.8473469Kurtosis -0.492454159Skewness 0.351282489Range 69Minimum 2717Maximum 2786Sum 137418Count 50

CO(ppm)

Mean 9.8932Standard Error 1.638305582Median 6.49Mode 6.49Standard Deviation 11.58456986Sample Variance 134.2022589Kurtosis 3.843417031Skewness 1.995733832Range 50.13Minimum 0.32Maximum 50.45Sum 494.66Count 50

0 5 10 15 20 25 30 35 40 45 50

0

5

10

15

CO

Fre

quen

cy

Histogram of CO, with Normal Curve

2710 2720 2730 2740 2750 2760 2770 2780 2790

0

5

10

Temperature

Fre

quen

cy

Histogram of Temperature, with Normal Curve



12

Lognormal Example 4.1(cont) CO is given by the equation

Let and

So that or

)2750(07324.05.4 flTeCO5.4/COX )2750(07324.0 flTW

WeX WCOX )5.4/ln()ln(

0 5 10 15 20 25 30 35 40 45 50

0

5

10

15

CO

Fre

qu

en

cy

Histogram of CO, with Normal Curve

2.52.01.51.00.5-0.0-0.5-1.0-1.5-2.0-2.5

15

10

5

0

ln(CO/4.5)

Fre

quen

cy

Histogram of ln(CO/4.5), with Normal Curve

279027802770276027502740273027202710

10

5

0

Tflame

Fre

qu

en

cy

Histogram of Tflame, with Normal Curve

2.52.01.51.00.5-0.0-0.5-1.0-1.5-2.0-2.5

15

10

5

0

W=-0.07324(Tfl-2750)

Fre

qu

en

cy

Histogram of W=-0.07324(Tfl-2750), with Normal Curve


Lognormal Example 4.1(cont)


13

0 10 20 30 40 50

95% Confidence Interval for Mu

4 9 14

95% Confidence Interval for Median

Variable: CO

A-Squared:P-Value:

MeanStDevVarianceSkewnessKurtosisN

Minimum1st QuartileMedian3rd QuartileMaximum

6.6009

9.6770

4.5000

4.3330.000

9.893211.5846134.2021.995733.84342

50

0.3200 1.9425 6.490011.045050.4500

13.1855

14.4359

8.0800

Anderson-Darling Normality Test


95% Confidence Interval for Sigma


Descriptive Statistics

-2.5 -1.5 -0.5 0.5 1.5 2.5


-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6


Variable: ln(CO/4.5)

A-Squared:P-Value:



-0.24504

1.07213

0.00000

0.5990.114

0.119721.283481.64732-3.5E-01-4.9E-01

50

-2.64351-0.84201 0.36619 0.89740 2.41691

0.48449

1.59939

0.58531






0 10 20 30 40 50

Rounded CO

Boxplot of Rounded CO

-3 -2 -1 0 1 2 3

X=ln(CO/4.5)

Boxplot of X=ln(CO/4.5)



14


CO is given by the equation

Test Ln(CO/4.5) and CO for normality …. Ln(CO/4.5) is normally distributed and CO is not

Average: 0.119725StDev: 1.28348N: 50

Anderson-Darling Normality TestA-Squared: 0.599P-Value: 0.114

-2 0 2

.001

.01

.05

.20

.50

.80

.95

.99

.999

Pro

babi

lity

ln(CO/4.5)

Normality Test ln(CO/4.5)

Average: 9.8932StDev: 11.5846N: 50

Anderson-Darling Normality TestA-Squared: 4.333P-Value: 0.000

0 10 20 30 40 50

.001

.01

.05

.20

.50

.80

.95

.99

.999

Pro

babi

lity

CO

Normality Test CO

)2750(07325.05.4 flTeCO



15

Lognormal Example 4.1(cont) CO is given by the equation

Let and

So that or

Now we want

)2750(07324.05.4 flTeCO5.4/COX )2750(07324.0 flTW

WeX WX )ln(

))2ln(

())2ln(

()()9(

))2ln(())2ln()(ln(

)2()25.4/()9(

WZPW

PxXPppmCOP

WPXP

XPCOPppmCOP



16


From the analysis of the data for flame temperature and W

Then

2720 2735 2750 2765 2780


2744 2749 2754


Variable: Temperature

A-Squared:P-Value:



2743.38

14.63

2742.00

0.5980.114

2748.36 17.52

306.8470.351282-4.9E-01

50

2717.002737.752745.002761.502786.00

2753.34

21.83

2750.00






-2.5 -1.5 -0.5 0.5 1.5 2.5


-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6


Variable: W=.07324(Tfl

A-Squared:P-Value:



-0.24450

1.07169

0.00000

0.5980.114

0.120111.282951.64596-3.5E-01-4.9E-01

50

-2.63664-0.84226 0.36620 0.89719 2.41692

0.48472

1.59873

0.58592






12011.0 28295.1

6736.0)4459.0()28295.1

12011.0)2ln((

WW ZPZP

)2750(07324.0 flTW


Lognormal Example 4.1(cont) Summary of the CO Lognormal Distribution


17

ppm

eeXEXV

ppmCOE

eCOEXE

ZPZPxXP

CO

CO

WW

67.23

666.27)1()569.2()1()()(

6.11559.11569.25.4)(

569.2)5.4/()(

6736.0)4459.0()28295.1

12011.0)2ln(()(

647.122

2/

2

2

Mean CO

Standard Deviation of CO

System Does not meet CO Requirements-combustor needs a factor of 5 improvement

in CO performance



18


Widely used to analyze and predict failure for physical systems failure may be a function of time, cycles,

starts, landings, etc

Can provide reasonably accurate failure predictions with small samples Important in safety critical systems



19

The Weibull Distribution



20

The Weibull Failure Function….

2/1)2/1(

1!0)1(

)!1()(

!

nn

Note



21

Two parameters define the Weibull distribution:

, the shape parameter, is a measure of the time

dependency of the probability of failure. Completely random failures(random errors, external shocks) have a = 1. Failures which increase in probability over time(wearout, old age) have > 1, and failures whose probability decreases over time(manufacturing errors) have 0 < < 1.

,the scale parameter, is the time at which a cumulative 63.2% of the population is expected to have failed



Infant Mortality “Useful” Life Wear Out

Twear-outTburn-in

ManufacturingDefects

ExternalStress Failures

Wear OutFailures

Haz

ard

Rate

, h(t) Bath Tub Curve

(composite)

Service Age (Hours or Starts, etc...)

Infant Mortality “Useful” Life Wear Out

Twear-outTburn-in

ManufacturingDefects

ExternalStress Failures

Wear OutFailures

Haz

ard

Rate

, h(t) Bath Tub Curve

(composite)

Service Age (Hours or Starts, etc...)

Unio n Col l eg eMec ha nic al E ngi ne eri ng

ME R30 1: E ngi ne erin g R el iabi l i tyLe ct ur e 4

22


dependency of the probability of failure


1 11


beta= 0.50delta= 1000X Ln(x) Weibull0 010 2.3026 0.09550 3.912 0.2

100 4.6052 0.271200 5.2983 0.361500 6.2146 0.5071000 6.9078 0.6322000 7.6009 0.7575000 8.5172 0.893

10000 9.2103 0.958


23

beta= 1.00delta= 1000

X Ln(X) Weibull0 010 2.3026 0.0150 3.912 0.049

100 4.6052 0.095200 5.2983 0.181500 6.2146 0.3931000 6.9078 0.6322000 7.6009 0.8655000 8.5172 0.993

10000 9.2103 1

beta= 2.00delta= 100010 2.3026 050 3.912 0.002

100 4.6052 0.01200 5.2983 0.039500 6.2146 0.2211000 6.9078 0.6322000 7.6009 0.9825000 8.5172 1

10000 9.2103 1

functionfailuretruecumulative

factorscale

factorshape

valuex

cumulativexweibull

WorksheetExcelWeibull

.""

.

.

)",,,("

...."



22





Effect of Beta on Weibull CDFDelta=1000

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10

ln(x)

Wel

bu

ll C

DF

beta=0.5

beta =1

beta=2

24



22




Infant Mortality

Old Age

Useful Life

63.2% Failure Rateat x=delta =1000

X=1000



25

Many kinds of failure data plot as a straight line with slope The x- axis is time and the y-axis is the cumulative failure density function F(t),

Weibull plots are used to predict cumulative failures at any time. For instance, with = 1.66 and

= 177051, after 30000 time units 5% of the population will have failed.

ln(ln(1/(1-63.2%)) = 0. So, is the y-intercept of the straight line plot.

Weibull Plots-Cumulative Density Function

ReliaSoft Weibull++ 7 - www.ReliaSoft.comProbability - Weibull

Time, (t)

Un

reli

ab

ilit

y,

F(

t)

10.0 100000.0100.0 10000.01000.00.1

0.5

1.0

5.0

50.0

90.0

99.0

0.1

10.0

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.4

1.6

2.0

3.0

4.0

6.0

Probability-WeibullCB@90% 2-Sided [R]

Data 1Weibull-2PMLE SRM MED FMF=9/S=447

Data PointsSusp PointsProbability LineTop CB-IIBottom CB-II

Target Rel

Jagmeet SinghGE5/13/200810:47:44 AM

t=30000


"--------- Weibull CDF "---------"X ln(X) beta=0.5 beta =1 beta=210 2.3026 0.095 0.01 050 3.912 0.2 0.049 0.002

100 4.6052 0.271 0.095 0.01200 5.2983 0.361 0.181 0.039500 6.2146 0.507 0.393 0.221

1000 6.9078 0.632 0.632 0.6322000 7.6009 0.757 0.865 0.9825000 8.5172 0.893 0.993 1

10000 9.2103 0.958 1 1

Weibull Plots-Cumulative Density Function

12

5.0

(Two Cycle Weibull Paper)

X=100 X=1000 X=10000

1000 1000



27

Discrete Distribution Probability Mass Function

Describes how the total probability of 1 is distributed among various possible values of the variable X


The Sum of Two Dice…

Define the following probabilities Let A= probability of 2=1/36 Let B= probability of 3=2/36 Let C= probability of 4=3/36 Let D= probability of 5=4/36 Let E= probability of 6=5/36 Let F= probability of 7=6/36

Let H= probability of 9=4/36 Let I= probability of 10=3/36 Let J= probability of 11=2/36 Let K= probability of 12=1/36


For a Probability Mass Function

The Sum of Two Dice…The Probability Mass Function

Define the following probabilities Let A= probability of 2=1/36 Let B= probability of 3=2/36 Let C= probability of 4=3/36 Let D= probability of 5=4/36 Let E= probability of 6=5/36 Let F= probability of 7=6/36

Let H= probability of 9=4/36 Let I= probability of 10=3/36 Let J= probability of 11=2/36 Let K= probability of 12=1/36

1)()()()()()()()()()()( KPJPIPHPGPFPEPDPCPBPAP

n

i

n

iii xXPxf

1 1

1)()(


The Sum of Two Dice… The Cumulative Distribution

Define the following probabilities Let A= probability of 2=1/36 Let B= probability of 3=2/36 Let C= probability of 4=3/36 Let D= probability of 5=4/36 Let E= probability of 6=5/36 Let F= probability of 7=6/36 Let G= probability of 8=5/36 Let H= probability of 9=4/36 Let I= probability of 10=3/36 Let J= probability of 11=2/36 Let K= probability of 12=1/36


30

Dice 11 2 3 4 5 6

1 1,1 2,1 3,1 4,1 5,1 6,12 1,2 2,2 3,2 4,2 5,2 6,2

Dice 2 3 1,3 2,3 3,3 4,3 5,3 6,34 1,4 2,4 3,4 4,4 5,4 6,45 1,5 2,5 3,5 4,5 5,5 6,56 1,6 2,6 3,6 4,6 5,6 6,6

Case Dice Sum I ProbabilityA 2 1 0.0278B 3 2 0.0556C 4 3 0.0833D 5 4 0.1111E 6 5 0.1389F 7 6 0.1667G 8 7 0.1389H 9 8 0.1111I 10 9 0.0833J 11 10 0.0556K 12 11 0.0278

iP(X<=xi)

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10 12 14

Sum of Two Dice

Cu

mu

lati

ve

Dis

trib

uti

on

P(X<=xi)


The Sum of Two Dice… The Mean….



31

Dice 11 2 3 4 5 6

1 1,1 2,1 3,1 4,1 5,1 6,12 1,2 2,2 3,2 4,2 5,2 6,2

Dice 2 3 1,3 2,3 3,3 4,3 5,3 6,34 1,4 2,4 3,4 4,4 5,4 6,45 1,5 2,5 3,5 4,5 5,5 6,56 1,6 2,6 3,6 4,6 5,6 6,6

Dice Sum P(X<=xi) SumXP(X<=xi)2 0.0278 0.05563 0.0834 0.16684 0.1667 0.33325 0.2778 0.55556 0.4167 0.83347 0.5834 1.16698 0.7223 1.11129 0.8334 0.9999

10 0.9167 0.83311 0.9723 0.611612 1 0.3336

Mean=7


Dice Sum P(X<=xi) Px(xi-mean)^22 0.0278 0.6951946143 0.0834 0.8899113874 0.1667 0.7500499015 0.2778 0.4447111346 0.4167 0.1390945287 0.5834 8.1683E-088 0.7223 0.1387056089 0.8334 0.444088974

10 0.9167 0.74935018111 0.9723 0.88928866712 1 0.694805414

Variance=5.835 Std Dev=2.4156

The Sum of Two Dice… The Variance….



32

Dice 11 2 3 4 5 6

1 1,1 2,1 3,1 4,1 5,1 6,12 1,2 2,2 3,2 4,2 5,2 6,2

Dice 2 3 1,3 2,3 3,3 4,3 5,3 6,34 1,4 2,4 3,4 4,4 5,4 6,45 1,5 2,5 3,5 4,5 5,5 6,56 1,6 2,6 3,6 4,6 5,6 6,6



33

Probability Mass Function

3-29



34

Example 4.2 Consider a group of five potential blood

donors – A, B, C, D, and E – of whom only A and B have type O+ blood. Five blood samples, one from each individual, will be typed in random order until an O+ individual is identified. Let X=the number of typings necessary to

identify an O+ individual Determine the probability mass function of X



35

Cumulative Distribution Function

3-31



36

Example 4.3

For the previous example (4.2), determine F(x) for each value of x in the set of possible values x=1 to x=4



37

Expected Value or Mean of theDiscrete Distribution



38

Example 4.4 Consider a university having 15,000

students X= number of courses for which a randomly

selected student is registered. The probability mass function can be found by

knowing how many students signed up for any specific number of classes

Determine the probability mass function f(x). Calculate the mean/expected number of courses

per student.



39

Variance of Discrete Distributions



40

Example 4.5 In example 4.4 the

density function is given as shown Determine the

variance and the standard deviation.

x n f(x)1 150 0.012 450 0.033 1950 0.134 3750 0.255 5850 0.396 2550 0.177 300 0.02

N P15000 1



41

Expected Value of a Function If the random variable X has a set of

possible values x1,x2,…,xn and a probability mass function f(x), the the expected value of a function h(X) can be estimated as

where )()()(

1)( ii

n

iXh xfxhXhE

1)(

....,

1

21

n

ii

n

xf

xxxX



42

Example 4.6 Let X be the number of cylinders in the

engine of the next car to be tuned up at a certain facility. The cost of the tune up

h(x)=20+3x+0.5x2

Assume 50%,30%,and 20% of cars have four, six, and eight cylinders, respectively

Since x is a random variable, so is h(x) Write the density function f(y) for y=h(x) Determine the expected value for Y=h(X)



43

Summary of Topics



Probability Density and Cumulative Distribution Functions of Discrete Variables

Mean and Variance of Discrete Variables

l berkley davis copyright 2009 mer301: engineering reliability lecture 4 1 mer301: engineering...

Documents

berkley davis copyright

lognormal distribution

weibull distribution

cont slide

lognormal example

probability of failure

cumulative failures

variance of discrete