Download Lecture4

MER301: Engineering
Reliability
LECTURE 4:
Chapter 3:
Lognormal Distribution, The Weibull
Distribution, and Discrete Variables
L Berkley Davis
Copyright 2009
MER301: Engineering Reliability
Lecture 4
1
Summary of Topics
 Lognormal Distribution
 Weibull Distribution
 Probability Density and Cumulative
Distribution Functions of Discrete
Variables
 Mean and Variance of Discrete
Variables
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
2
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
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
3
Lognormal Distribution
 Special case of the normal
distribution where x  e w and the
variable w is normally distributed

 Chemical processes and material properties
are often characterized by lognormal
distributions
Parameters
and 2 are the mean


and variance of W, respectively
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
4
Lognormal Distribution
V ( X )  E( X )  (e 1)
2
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
2
5
Lognormal Distribution
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
6
Lognormal Example 4.1
 Gas Turbine CO Emissions
CO  4.5e
0.07324(T flame 2750)
 T flame is a normally distributed
function of combustor fuel/air ratio
 Mean value of CO will need to be
9ppm or less
P(CO  9 ppm)
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
7
Lognormal Example 4.1(cont)
CO as a Function of Temperature
CO Emissions
60
50
40
30
Series1
20
10
0
2700
2720
2740
2760
2780
2800
Flame Temperature
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
8
Tflame
2745
2731
2754
2769
2768
2776
2717
2746
2766
2734
2740
2725
2722
CO(ppm)
6.49
18.1
3.36
1.12
1.2
0.67
50.45
6.03
1.39
14.53
9.36
28.08
34.98
2735
2738
2718
2741
2744
2752
2745
2745
2744
2770
2749
2747
2742
2780
2763
2786
13.5
10.84
46.89
8.7
6.98
3.89
6.49
6.49
6.98
1.04
4.84
5.61
8.08
0.5
1.74
0.32
2740
2775
2726
2758
2764
2779
2749
2742
2760
2744
2761
2728
2737
2727
2745
2750
2750
9.36
0.72
26.1
2.5
1.61
0.54
4.84
8.08
2.16
6.98
2.01
22.54
11.66
24.25
6.49
4.5
4.5
2745
2783
6.49
0.4
2724
2739
Lognormal Example 4.1
30.21
L Berkley
Davis
10.07
Copyright 20009
10
5
15
Frequency
Frequency
 excel spreadsheet for the CO example
0
2710 2720 2730 2740 2750 2760 2770 2780 2790
10
5
Tflame
0
0
5
10 15 20 25 30 35 40 45 50
CO(ppm)
MER301: Engineering Reliability
Lecture 4
9
Skewness and Kurtosis: Tflame Example
N  50
2748.36
2.477286204
2745
2745
17.51705874
306.8473469
-0.492454159
0.351282489
69
2717
2786
137418
50
10
Frequency
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count

Skewness

5
0
2710 2720 2730 2740 2750 2760 2770 2780 2790
Tflame
distributions produce a skewness
statistic of about zero.


6
skewness  2
...contains...zero
N
L Berkley Davis
Copyright 20009
ses can be estimated roughly using a
formula from Tabachnick & Fidell,1996
Kurtosis

24
kurtosis  2
....contains...zero
N
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
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
Lognormal Example 4.1(cont)
Tflame
CO(ppm)
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
2748.36
2.477286204
2745
2745
17.51705874
306.8473469
-0.492454159
0.351282489
69
2717
2786
137418
50
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
9.8932
1.638305582
6.49
6.49
11.58456986
134.2022589
3.843417031
1.995733832
50.13
0.32
50.45
494.66
50
Histogram of Temperature, with Normal Curve
Histogram of CO, with Normal Curve
10
Frequency
Frequency
15
5
10
5
0
0
2710
2720
2730
2740
2750
2760
Temperature
2770
2780
2790
0
5
10
15
20
25
CO
L Berkley Davis
Copyright 20009
30
35
40
45
50
Lognormal Example 4.1(cont)
CO is given by the equation

CO  4.5e 0.07324(T fl 2750)
X  CO / 4.5 and W  0.07324(T fl  2750)
Let

So that
X  eW or
ln( X )  ln( CO / 4.5)  W
Histogram of CO, with Normal Curve
Histogram of Tflame, with Normal Curve
10
15
Frequency
15
10
Histogram of W=-0.07324(Tfl-2750), with Normal Curve
5
15
5
0
5
10
15
20
25
10
0
30
35
405
45
2710
50
2720
2730
2740
2750
Tflame
CO
Frequency
0
Frequency
Frequency
Histogram of ln(CO/4.5), with Normal Curve
2760
10
2770
2780
2790
5
0
0
-2.5
-2.0
-1.5
-1.0
-0.5
-0.0
0.5
1.0
1.5
2.0
2.5
ln(CO/4.5)
L Berkley Davis
Copyright 20009
-2.5
-2.0
-1.5
-1.0
-0.5
-0.0
0.5
1.0
1.5
2.0
W=-0.07324(Tfl-2750)
MER301: Engineering Reliability
Lecture 4
12
2.5
Lognormal Example 4.1(cont)
Descriptive Statistics
Boxplot of Rounded CO
Variable: CO
Anderson-Darling Normality Test
A-Squared:
P-Value:
0
10
20
30
40
50
95% Confidence Interval for Mu
4.333
0.000
Mean
StDev
Variance
Skewness
Kurtosis
N
9.8932
11.5846
134.202
1.99573
3.84342
50
Minimum
1st Quartile
Median
3rd Quartile
Maximum
0.3200
1.9425
6.4900
11.0450
50.4500
95% Confidence Interval for Mu
6.6009
4
9
14
13.1855
95% Confidence Interval for Sigma
9.6770
14.4359
0
10
20
4.5000
30
40
50
Rounded CO
95% Confidence Interval for Median
95% Confidence Interval for Median
8.0800
Descriptive Statistics
Boxplot of X=ln(CO/4.5)
Variable: ln(CO/4.5)
Anderson-Darling Normality Test
A-Squared:
P-Value:
-2.5
-1.5
-0.5
0.5
1.5
2.5
95% Confidence Interval for Mu
0.599
0.114
Mean
StDev
Variance
Skewness
Kurtosis
N
0.11972
1.28348
1.64732
-3.5E-01
-4.9E-01
50
Minimum
1st Quartile
Median
3rd Quartile
Maximum
-2.64351
-0.84201
0.36619
0.89740
2.41691
95% Confidence Interval for Mu
-0.24504
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.48449
95% Confidence Interval for Sigma
1.07213
95% Confidence Interval for Median
L Berkley Davis
Copyright 20009
0.00000
-3
-2
-1
0
1
2
3
1.59939
95% Confidence Interval for Median
X=ln(CO/4.5)
0.58531
MER301: Engineering Reliability
Lecture 4
13
Lognormal Example 4.1(cont)
Normality Test ln(CO/4.5)
.999
.999
.99
.99
.95
.95
Probability
Probability
Normality Test CO
.80
.50
.20
.05
.80
.50
.20
.05
.01
.01
.001
.001
0
10
20
30
40
50
-2
CO
Average: 9.8932
StDev: 11.5846
N: 50

0
2
ln(CO/4.5)
Anderson-Darling Normality Test
A-Squared: 4.333
P-Value: 0.000
Average: 0.119725
StDev: 1.28348
N: 50
Anderson-Darling Normality Test
A-Squared: 0.599
P-Value: 0.114
CO is given by the equation
CO  4.5e 0.07325(T fl 2750)

Test Ln(CO/4.5) and CO for normality ….

Ln(CO/4.5) is normally distributed and CO is not
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
14
Lognormal Example 4.1(cont)


CO is given by the equation
Let
CO  4.5e 0.07324(T fl 2750)
X  CO / 4.5 and W  0.07324(T fl  2750)
So that
X  eW or
ln( X )  W
Now we want
P(CO  9 ppm)  P(CO / 4.5  2)  P( X  2)
P(ln( X )  ln( 2))  P(W  ln( 2))
P(CO  9 ppm)  P( X  x)  P(
L Berkley Davis
Copyright 20009
W 


ln( 2)  
MER301: Engineering Reliability
Lecture 4

)  P ( ZW 
ln( 2)  

15
)
Lognormal Example 4.1(cont)
W  0.07324(T fl  2750)
Descriptive Statistics
Descriptive Statistics
Variable: Temperature
Variable: W=.07324(Tfl
Anderson-Darling Normality Test
A-Squared:
P-Value:
2720
2735
2750
2765
2780
95% Confidence Interval for Mu
Anderson-Darling Normality Test
0.598
0.114
Mean
StDev
Variance
Skewness
Kurtosis
N
2748.36
17.52
306.847
0.351282
-4.9E-01
50
Minimum
1st Quartile
Median
3rd Quartile
Maximum
2717.00
2737.75
2745.00
2761.50
2786.00
A-Squared:
P-Value:
-2.5
-1.5
-0.5
0.5
1.5
2.5
95% Confidence Interval for Mu
95% Confidence Interval for Mu
2743.38
2744
2749
2754
14.63

2742.00
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
21.83
Minimum
1st Quartile
Median
3rd Quartile
Maximum
-2.63664
-0.84226
0.36620
0.89719
2.41692
0.5
0.6
0.48472
95% Confidence Interval for Sigma
1.07169
1.59873
95% Confidence Interval for Median
95% Confidence Interval for Median
0.00000
0.58592
From the analysis of the data for flame temperature and W
  0.12011

2750.00
0.12011
1.28295
1.64596
-3.5E-01
-4.9E-01
50
-0.24450
-0.3
95% Confidence Interval for Median
95% Confidence Interval for Median
Mean
StDev
Variance
Skewness
Kurtosis
N
95% Confidence Interval for Mu
2753.34
95% Confidence Interval for Sigma
0.598
0.114
Then
L Berkley Davis
Copyright 20009
P( Z W
  1.28295
ln( 2)  0.12011

)  P( Z W  0.4459)  0.6736
1.28295
MER301: Engineering Reliability
Lecture 4
16
Lognormal Example 4.1(cont)
 Summary of the CO Lognormal Distribution
ln( 2)  0.12011
P ( X  x )  P ( ZW 
)  P( ZW  0.4459)  0.6736
1.28295
  2 / 2
E ( X )  E (CO / 4.5)  e
 2.569
E (CO)  CO  4.5  2.569  11.559  11.6 ppm
V ( X )  E ( X )  (e
 CO  23.67 ppm
2
2
Mean CO
 1)  (2.569) 2  (e1.647  1)  27.666
Standard Deviation of CO
System Does not meet CO Requirements
-combustor needs a factor of 5 improvement
L Berkley Davis
MER301: Engineering Reliability
Copyright 20009
4
in CO Lecture
performance
17
Weibull Distribution
 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
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
18
The Weibull Distribution
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
19
The Weibull Failure Function….
Note!
(n)  (n  1)!
(1)  0! 1
(1 / 2)   1 / 2
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
20
The Weibull Distribution
Two parameters define the Weibull distribution:
b, the shape parameter, is a measure of the time
dependency of the probability of failure. Completely random
failures(random errors, external shocks) have a b = 1. Failures which
increase in probability over time(wearout, old age) have b > 1, and
failures whose probability decreases over time(manufacturing errors)
have 0 < b < 1.
 ,the scale parameter, is the time at which a cumulative
63.2% of the population is expected to have failed
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
21
b, the shape parameter, is a measure of the time
dependency of the probability of failure
“Useful” Life
Hazard Rate, h(t)
Infant Mortality
Bath Tub Curve
(composite)
MER30 1:
Engi ne erin g R eliability
b 1
b 1
Unio n Coll eg e
Mec ha nic al Engi ne eri ng
Le ct ur e 4
External
Stress Failures
Wear Out
b 1
Wear Out
Failures
Manufacturing
Defects
Tburn-in
Service Age (Hours or Starts, etc...)
L Berkley Davis
Copyright 20009
22
Twear-out
b, the shape parameter, is a measure of the time
dependency of the probability of failure
beta= 0.50
delta= 1000
X
Ln(x)
Weibull
0
0
10
2.3026
0.095
beta=
50
3.912
0.2
delta=
100
4.6052
0.271
Unio n Coll eg e
X
Mec ha nic al Engi ne eri ng
200
5.2983
0.361
0
500
6.2146
0.507
10
1000
6.9078
0.632
50
2000
7.6009
0.757
100
5000
8.5172
0.893
200
10000
9.2103
0.958
500
1000
2000
5000
10000
L Berkley Davis
Copyright 20009
"Weibull ..Excel..Worksheet
"  weibull ( x, b ,  , cumulative)"
x  value
b  shape. factor
  scale. factor
Weibull cumulative " true"  failure. function
1.00
1000
MER30 1: Engi ne erin g R eliability
Ln(X)Le ct ur e 4
0
2.3026
0.01
3.912
0.049
4.6052
0.095
5.2983
0.181
6.2146
0.393
6.9078
0.632
7.6009
0.865
8.5172
0.993
9.2103
1
MER301: Engineering Reliability
Lecture 4
22
beta= 2.00
delta= 1000
10
2.3026
50
3.912
100
4.6052
200
5.2983
500
6.2146
1000
6.9078
2000
7.6009
5000
8.5172
10000
9.2103
0
0.002
0.01
0.039
0.221
0.632
0.982
1
1
23
b, the shape parameter, is a measure of the time
dependency of the probability of failure
Effect of Beta on Weibull CDF
Delta=1000
1.2
1
Welbull CDF
Unio n Coll eg e
63.2%
Mec ha nic al Engi ne eri ng
Failure Rate
at x=delta =1000
0.8
MER30 1: Engi ne erin g R eliability
Le ct ur e 4
22
beta=0.5
0.6
beta =1
Infant Mortality
beta=2
0.4
Old Age
Useful Life
0.2
0
0
L Berkley Davis
Copyright 20009
2
4
6
ln(x)
X=1000
8
10
24
Weibull Plots-Cumulative Density Function
Many kinds of failure data plot as a straight line with slope b . The x- axis is
time and the y-axis is the cumulative failure density function F(t),
ReliaSoft Weibull++ 7 - www.ReliaSoft.com
0 .5
0 .7
0 .6
1 .0
0 .9
0 .8
.0
.6
.4
.2
2
1
1
1
99.0
6 .0
4 .0
3 .0
Probability - Weibull
b
90.0

U n re lia b ility , F ( t)
50.0
h
10.0
Data 1
Weibull-2P
MLE SRM MED FM
F=9/S=447

Data Points
Susp Points
Probability Line
Top CB-II
Bottom CB-II
Target Rel
5.0
1.0
0.5
0.1
10.0
Weibull plots are used to predict
cumulative failures at any time.
For instance, with b = 1.66 and
= 177051, after 30000
time units 5% of the population
will have failed.
Probability-Weibull
CB@90% 2-Sided [R]
100.0
1000.0
Time, (t)
10000.0
100000.0
t=30000
ln(ln(1/(1-63.2%)) = 0. So, h is
the y-intercept of the straight line
plot.
Jagmeet Singh
GE
5/13/2008
10:47:44 AM
b. h.
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
25
Weibull Plots-Cumulative Density Function
(Two Cycle Weibull Paper)
b  0. 5
b 1
  1000
X=100
L Berkley Davis
Copyright 20009
X=1000
b 2
X
10
50
100
200
500
1000
2000
5000
10000
X=10000
ln(X)
2.3026
3.912
4.6052
5.2983
6.2146
6.9078
7.6009
8.5172
9.2103
  1000
"--------- Weibull CDF "---------"
beta=0.5
beta =1
beta=2
0.095
0.01
0
0.2
0.049
0.002
0.271
0.095
0.01
0.361
0.181
0.039
0.507
0.393
0.221
0.632
0.632
0.632
0.757
0.865
0.982
0.893
0.993
1
0.958
1
1
Discrete Distribution Probability
Mass Function
 Describes how the total probability
of 1 is distributed among various
possible values of the variable X
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
27
The Sum of Two Dice…
 Define the following probabilities






Let
Let
Let
Let
Let
Let
Dice 2
L Berkley Davis
Copyright 20009
A= probability of 2=1/36
B= probability of 3=2/36
C= probability of 4=3/36
D= probability of 5=4/36
E= probability of 6=5/36
F= probability of 7=6/36
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6




Dice 1
3
3,1
3,2
3,3
3,4
3,5
3,6
Let
Let
Let
Let
H= probability of 9=4/36
I= probability of 10=3/36
J= probability of 11=2/36
K= probability of 12=1/36
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
The Sum of Two Dice…
The Probability Mass Function

Dice 2
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
Dice 1
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
Define the following probabilities






Let
Let
Let
Let
Let
Let
A= probability of 2=1/36
B= probability of 3=2/36
C= probability of 4=3/36
D= probability of 5=4/36
E= probability of 6=5/36
F= probability of 7=6/36




n
For a Probability Mass Function
Let
Let
Let
Let
H= probability of 9=4/36
I= probability of 10=3/36
J= probability of 11=2/36
K= probability of 12=1/36
n
 f ( x )   P( X  x )  1
i 1
i
i 1
i
P( A)  P( B)  P(C )  P( D)  P( E )  P( F )  P(G)  P( H )  P( I )  P( J )  P( K )  1
L Berkley Davis
Copyright 20009
The Sum of Two Dice…
The Cumulative Distribution
Define the following probabilities











Let
Let
Let
Let
Let
Let
Let
Let
Let
Let
Let
A= probability of 2=1/36
B= probability of 3=2/36
C= probability of 4=3/36
D= probability of 5=4/36
E= probability of 6=5/36
F= probability of 7=6/36
G= probability of 8=5/36
H= probability of 9=4/36
I= probability of 10=3/36
J= probability of 11=2/36
K= probability of 12=1/36
L Berkley Davis
Copyright 20009
Case
A
B
C
D
E
F
G
H
I
J
K
Dice Sum
2
3
4
5
6
7
8
9
10
11
12
iI
1
2
3
4
5
6
7
8
9
10
11
1
2
3
4
5
6
Dice 1
3
3,1
3,2
3,3
3,4
3,5
3,6
6
6,1
6,2
6,3
6,4
6,5
6,6
5
5,1
5,2
5,3
5,4
5,5
5,6
4
4,1
4,2
4,3
4,4
4,5
4,6
P(X<=xi)
Probability
0.0278
0.0556
0.0833
0.1111
0.1389
0.1667
0.1389
0.1111
0.0833
0.0556
0.0278
MER301: Engineering Reliability
Lecture 1
1.2
1
Cumulative Distribution

Dice 2
2
2,1
2,2
2,3
2,4
2,5
2,6
1
1,1
1,2
1,3
1,4
1,5
1,6
0.8
0.6
P(X<=xi)
0.4
0.2
0
0
2
4
6
8
10
12
Sum of Two Dice
30
14
The Sum of Two Dice…
The Mean….

Dice 2
Define the following
probabilities











Let
Let
Let
Let
Let
Let
Let
Let
Let
Let
Let
Dice Sum
2
3
4
5
6
7
8
9
10
11
12
A= probability of 2=1/36
B= probability of 3=2/36
C= probability of 4=3/36
D= probability of 5=4/36
E= probability of 6=5/36
F= probability of 7=6/36
G= probability of 8=5/36
H= probability of 9=4/36
I= probability of 10=3/36
J= probability of 11=2/36
K= probability of 12=1/36
L Berkley Davis
Copyright 20009
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
P(X<=xi)
0.0278
0.0834
0.1667
0.2778
0.4167
0.5834
0.7223
0.8334
0.9167
0.9723
1
2
2,1
2,2
2,3
2,4
2,5
2,6
Dice 1
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
SumXP(X<=xi)
0.0556
0.1668
0.3332
0.5555
0.8334
1.1669
1.1112
0.9999
0.833
0.6116
0.3336
Mean=7
MER301: Engineering Reliability
Lecture 1
31
6
6,1
6,2
6,3
6,4
6,5
6,6
The Sum of Two Dice…
The Variance….

Define the following
probabilities











Let
Let
Let
Let
Let
Let
Let
Let
Let
Let
Let
2
2,1
2,2
2,3
2,4
2,5
2,6
Dice Sum P(X<=xi)
2
0.0278
3
0.0834
4
0.1667
5
0.2778
6
0.4167
7
0.5834
8
0.7223
9
0.8334
10
0.9167
11
0.9723
12
1
A= probability of 2=1/36
B= probability of 3=2/36
C= probability of 4=3/36
D= probability of 5=4/36
E= probability of 6=5/36
F= probability of 7=6/36
G= probability of 8=5/36
H= probability of 9=4/36
I= probability of 10=3/36
J= probability of 11=2/36
K= probability of 12=1/36
L Berkley Davis
Copyright 20009
Dice 2
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
Dice 1
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
Px(xi-mean)^2
0.695194614
0.889911387
0.750049901
0.444711134
0.139094528
8.1683E-08
0.138705608
0.444088974
0.749350181
0.889288667
0.694805414
Variance=5.835
Std Dev=2.4156
MER301: Engineering Reliability
Lecture 1
32
6
6,1
6,2
6,3
6,4
6,5
6,6
Probability Mass Function
3-29
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
33
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
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
34
Cumulative Distribution Function
3-31
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
35
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
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
36
Expected Value or Mean of the
Discrete Distribution
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
37
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.
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
38
Variance of Discrete
Distributions
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
39
Example 4.5
 In example 4.4
the density
function is given
as shown
 Determine the
variance and the
standard
deviation.
L Berkley Davis
Copyright 20009
x
1
2
3
4
5
6
7
MER301: Engineering Reliability
Lecture 4
n
150
450
1950
3750
5850
2550
300
N
15000
f(x)
0.01
0.03
0.13
0.25
0.39
0.17
0.02
P
1
40
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
n
 h ( X )  Eh( X )   h( xi )  f ( xi )
where
i 1
X  x1 , x 2 ....x n 
n
 f (x )  1
i 1
L Berkley Davis
Copyright 20009
i
MER301: Engineering Reliability
Lecture 4
41
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)
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
42
Summary of Topics
 Lognormal Distribution
 Weibull Distribution
 Probability Density and Cumulative
Distribution Functions of Discrete
Variables
 Mean and Variance of Discrete
Variables
L Berkley Davis
Copyright 20009
MER301: Engineering Reliability
Lecture 4
43