Download HW1_Prob3.rtf

Mark Andersen
Reliability Engineering
HW 1
Februrary 4, 2008
Problem 3
a)
Descriptive Statistics: Fail Time (Hz)
Variable
Fail Time (Hz)
N
22
N*
0
Variable
Fail Time (Hz)
Maximum
173.40
Mean
72.39
SE Mean
8.18
StDev
38.36
Minimum
17.88
Q1
44.73
Median
61.68
b)
Histogram of Hertz
7
6
Frequency
5
4
3
2
1
0
40
80
120
Fail Time (Hertz)
c)
Distribution ID Plot for Fail Time (Hz)
160
Q3
100.26
Probability Plot for Fail Time (Hz)
LSXY Estimates-Complete Data
Weibull
C orrelation C oefficient
Weibull
0.984
Lognormal
0.990
Exponential
*
Loglogistic
0.991
Lognormal
99
90
50
P er cent
P er cent
90
10
50
10
1
10
1
10
100
Fail T ime ( H z)
E xponential
100
Fail T ime (H z)
Loglogistic
99
90
50
P er cent
P er cent
90
10
50
10
1
1
10
100
Fail T ime ( H z)
1
10
1000
100
Fail T ime (H z)
Out of these graphs, the exponential is not a good fit, and the Lognormal and Loglogistic
are the best fits.
Distribution ID Plot for Fail Time (Hz)
Probability Plot for Fail Time (Hz)
LSXY Estimates-Complete Data
3-Parameter Weibull
C orrelation C oefficient
3-P arameter Weibull
0.989
3-P arameter Lognormal
0.993
2-P arameter E xponential
*
3-P arameter Loglogistic
0.992
3-Parameter Lognormal
99
90
90
Percent
Percent
50
10
50
10
1
1
10
100
Fail Time (Hz) - Threshold
2-Parameter Exponential
50
100
Fail Time (Hz) - Threshold
3-Parameter Loglogistic
99
90
50
Percent
Percent
90
10
50
10
1
01
0. 0
10
0.0
00
0.1
00
1. 0
00
.0
10
00
00
.0
0.0
00
10
10
Fail Time (Hz) - Threshold
1
10
100
Fail Time (Hz) - Threshold
200
2-parameter exponential is not a good fit; 3-parameter Lognormal provides the best fit.
Distribution ID Plot for Fail Time (Hz)
Probability Plot for Fail Time (Hz)
LSXY Estimates-Complete Data
C orrelation C oefficient
S mallest E xtreme V alue
0.898
N ormal
0.961
Logistic
0.962
Normal
Smallest Extreme Value
99
90
90
Percent
Percent
50
10
50
10
1
0
200
100
Fail Time (Hz)
1
0
100
50
Fail Time (Hz)
150
Logistic
99
Percent
90
50
10
1
0
100
50
Fail Time (Hz)
150
Normal provides best fit.
**However, from Histogram we can see that this does not follow a normal distribution.
3-Parameter Loglogistic
Goodness-of-Fit
Distribution
Weibull
Lognormal
Exponential
Loglogistic
3-Parameter Weibull
3-Parameter Lognormal
2-Parameter Exponential
3-Parameter Loglogistic
Smallest Extreme Value
Normal
Logistic
Anderson-Darling
(adj)
1.021
0.744
4.616
0.755
0.846
0.747
2.192
0.763
3.311
1.160
1.118
Correlation
Coefficient
0.984
0.990
*
0.991
0.989
0.993
*
0.992
0.898
0.961
0.962
Table of MTTF
Distribution
Weibull
Lognormal
Mean
71.8534
74.3210
Standard
Error
7.4325
9.8527
95% Normal CI
Lower
Upper
58.6677 88.0027
57.3150 96.3729
Exponential
Loglogistic
3-Parameter Weibull
3-Parameter Lognormal
2-Parameter Exponential
3-Parameter Loglogistic
Smallest Extreme Value
Normal
Logistic
60.5627
75.7660
72.4721
73.5634
68.2119
74.6895
71.4949
72.3873
72.3873
11.8104
10.4729
7.9522
9.0339
10.3120
9.6469
8.0717
8.2714
8.4251
41.3248
57.7851
58.4481
55.8572
50.7200
55.7819
55.6746
56.1755
55.8744
88.7563
99.3421
89.8611
91.2696
91.7362
93.5972
87.3152
88.5990
88.9002
e)
The Weibull distribution provides a good fit and the Histogram from the data appears to
follow a Weibull distribution
Distribution Overview Plot for Fail Time (Hz)
Distribution Overview Plot for Fail Time (Hz)
LSXY Estimates-Complete Data
P robability Density F unction
Table of S tatistics
S hape
2.19377
S cale
81.1337
M ean
71.8534
S tDev
34.5652
M edian
68.6505
IQ R
48.1811
F ailure
22
C ensor
0
A D*
1.021
C orrelation
0.984
Weibull
0.012
90
P DF
P er cent
0.008
0.004
0.000
0
50
100
Fail T ime ( H z)
50
10
1
150
10
S urv iv al F unction
Hazard F unction
100
0.06
Rate
P er cent
100
Fail T ime ( H z)
50
0.04
0.02
0
0.00
0
50
100
Fail T ime ( H z)
150
0
50
100
Fail T ime ( H z)
150
NOTE: The parameter values for the Weibull distribution are:
Shape = 2.19377 a.k.a. ά
Scale = 81.1337 a.k.a. λ
f (t )  (t ) 1 e ( t ) so f (t )  2.19 * 81.13(81.13t ) 2.191 e (81.13t ) 2.19 ,
This becomes f (t )  177.99 * (81.13t ) 2.191 e (177.99*t )
F (t )  1  e  ( t ) so F (t )  1  e (81.13t ) 2.19
This becomes
F (t )  1  e (177.99*t )
R(t )  e ( t ) so R(t )  e (81.13t ) 2.19
This becomes
R(t )  e  (177.99*t )
z (t )  (t ) 1 so z (t )  2.19 * 81.13(81.13t ) 2.191 ,
This becomes
z (t )  177.99 * (81.13t ) 2.191
MTTF  ( 1 )( 1  1) so MTTF  ( 1
) ( 1
 1) ,


81.13
2.19
This becomes MTTF
 0.0123 *  *1.455  0.017944 * 