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'''Lognormal Distribution RRY Example'''
#REDIRECT [[The Lognormal Distribution]]
 
Fourteen units were reliability tested and the following life test data were obtained:
 
{|align="center" border=1 cellspacing=0
|-
|colspan="2" style="text-align:center"| Table 9.1 - Life Test Data
|-
!Data point index
!Time-to-failure
|-
|1 ||5
|-
|2 ||10
|-
|3 ||15
|-
|4 ||20
|-
|5 ||25
|-
|6 ||30
|-
|7 ||35
|-
|8 ||40
|-
|9 ||50
|-
|10 ||60
|-
|11 ||70
|-
|12 ||80
|-
|13 ||90
|-
|14 ||100
|}
 
Assuming the data follow a lognormal distribution, estimate the parameters and the correlation coefficient,  <math>\rho </math> , using rank regression on Y.
 
'''Solution'''
 
Construct Table 9.2, as shown next.
 
<center><math>\overset{{}}{\mathop{\text{Table 9}\text{.2 - Least Squares Analysis}}}\,</math></center>
 
<center><math>\begin{matrix}
  N & t_{i} & F(t_{i}) & {t_{i}}'& y_{i} & {{t_{i}}'}^{2} & y_{i}^{2} & t_{i} y_{i}  \\
  \text{1} & \text{5} & \text{0}\text{.0483} & \text{1}\text{.6094}& \text{-1}\text{.6619} & \text{2}\text{.5903} & \text{2}\text{.7619} & \text{-2}\text{.6747}  \\
  \text{2} & \text{10} & \text{0}\text{.1170} & \text{2.3026}& \text{-1.1901} & \text{5.3019} & \text{1.4163} & \text{-2.7403}  \\
  \text{3} & \text{15} & \text{0}\text{.1865} & \text{2.7080}&\text{-0.8908} & \text{7.3335} & \text{0.7935} & \text{-2.4123}  \\
  \text{4} & \text{20} & \text{0}\text{.2561} & \text{2.9957} &\text{-0.6552} & \text{8.9744} & \text{0.4292} & \text{-1.9627}  \\
  \text{5} & \text{25} & \text{0}\text{.3258} & \text{3.2189}& \text{-0.4512} & \text{10.3612} & \text{0.2036} & \text{-1.4524}  \\
  \text{6} & \text{30} & \text{0}\text{.3954} & \text{3.4012}& \text{-0.2647} & \text{11.5681} & \text{0.0701} & \text{-0.9004}  \\
  \text{7} & \text{35} & \text{0}\text{.4651} & \text{3.5553} & \text{-0.0873} & \text{12.6405} & \text{-0.0076}& \text{-0.3102}  \\
  \text{8} & \text{40} & \text{0}\text{.5349} & \text{3.6889}& \text{0.0873} & \text{13.6078} & \text{0.0076} & \text{0.3219}  \\
  \text{9} & \text{50} & \text{0}\text{.6046} & \text{3.912} & \text{0.2647} & \text{15.3039} & \text{0.0701} &\text{1.0357}  \\
  \text{10} & \text{60} & \text{0}\text{.6742} & \text{4.0943} & \text{0.4512} & \text{16.7637} & \text{0.2036}&\text{1.8474}  \\
  \text{11} & \text{70} & \text{0}\text{.7439} & \text{4.2485} & \text{0.6552} & \text{18.0497}& \text{0.4292} & \text{2.7834} \\
  \text{12} & \text{80} & \text{0}\text{.8135} & \text{4.382} & \text{0.8908} & \text{19.2022} & \text{0.7935} & \text{3.9035}  \\
  \text{13} & \text{90} & \text{0}\text{.8830} & \text{4.4998} & \text{1.1901} & \text{20.2483}&\text{1.4163} & \text{5.3552}  \\
    \text{14} & \text{100}& \text{1.9517} & \text{4.6052} & \text{1.6619} & \text{21.2076} &\text{2.7619} & \text{7.6533}  \\
  \sum_{}^{} & \text{ } & \text{ } & \text{49.222} & \text{0} & \text{183.1531} & \text{11.3646} & \text{10.4473}  \\
 
\end{matrix}</math></center>
 
 
The median rank values ( <math>F({{t}_{i}})</math> ) can be found in rank tables or by using the Quick Statistical Reference in Weibull++ .
 
The  <math>{{y}_{i}}</math>  values were obtained from the standardized normal distribution's area tables by entering for  <math>F(z)</math>  and getting the corresponding  <math>z</math>  value ( <math>{{y}_{i}}</math> ).
 
Given the values in the table above, calculate  <math>\widehat{a}</math>  and  <math>\widehat{b}</math>:
 
 
::<math>\begin{align}
  & \widehat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime }{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime })(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime 2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{\prime })}^{2}}/14} \\
&  &  \\
& \widehat{b}= & \frac{10.4473-(49.2220)(0)/14}{183.1530-{{(49.2220)}^{2}}/14} 
\end{align}</math>
 
or:
 
::<math>\widehat{b}=1.0349</math>
 
and:
 
::<math>\widehat{a}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\widehat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,t_{i}^{\prime }}{N}</math>
 
or:
 
::<math>\widehat{a}=\frac{0}{14}-(1.0349)\frac{49.2220}{14}=-3.6386</math>
 
:Therefore:
 
::<math>{\sigma'}=\frac{1}{\widehat{b}}=\frac{1}{1.0349}=0.9663</math>
 
and:
 
::<math>{\mu }'=-\widehat{a}\cdot {\sigma'}=-(-3.6386)\cdot 0.9663</math>
 
or:
 
::<math>{\mu }'=3.516</math>
 
The mean and the standard deviation of the lognormal distribution are obtained using equations in section [[Lognormal Statistical Properties]]:
 
::<math>\overline{T}=\mu ={{e}^{3.516+\tfrac{1}{2}{{0.9663}^{2}}}}=53.6707\text{ hours}</math>
 
and:
 
::<math>{{\sigma }_{T}}=\sqrt{({{e}^{2\cdot 3.516+{{0.9663}^{2}}}})({{e}^{{{0.9663}^{2}}}}-1)}=66.69\text{ hours}</math>
 
The correlation coefficient can be estimated using Eqn. (RHOln):
 
::<math>\widehat{\rho }=0.9754</math>
 
The above example can be repeated using Weibull++ , using RRY.
 
[[Image:5folio.png|thumb|center|400px| ]]
 
The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.

Latest revision as of 03:45, 13 August 2012

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