Template:Example: Lognormal Distribution RRY: Difference between revisions

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The above example can be repeated using Weibull++ , using RRY.
The above example can be repeated using Weibull++ , using RRY.


[[Image:Lognormal Distribution Example 2 Data and Result.png|thumb|center|500px| ]]  
[[Image:Lognormal Distribution Example 2 Data and Result.png|thumb|center|250px| ]]  


The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.
The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.

Revision as of 21:20, 25 April 2012

Lognormal Distribution RRY Example

Fourteen units were reliability tested and the following life test data were obtained:

Table 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]\displaystyle{ \rho }[/math] , using rank regression on Y.

Solution

Construct Table 2, as shown next.

[math]\displaystyle{ \overset{{}}{\mathop{\text{Table }\text{2 - Least Squares Analysis}}}\, }[/math]
[math]\displaystyle{ \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]


The median rank values ( [math]\displaystyle{ F({{t}_{i}}) }[/math] ) can be found in rank tables or by using the Quick Statistical Reference in Weibull++ .

The [math]\displaystyle{ {{y}_{i}} }[/math] values were obtained from the standardized normal distribution's area tables by entering for [math]\displaystyle{ F(z) }[/math] and getting the corresponding [math]\displaystyle{ z }[/math] value ( [math]\displaystyle{ {{y}_{i}} }[/math] ).

Given the values in the table above, calculate [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math]:


[math]\displaystyle{ \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]\displaystyle{ \widehat{b}=1.0349 }[/math]

and:

[math]\displaystyle{ \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]\displaystyle{ \widehat{a}=\frac{0}{14}-(1.0349)\frac{49.2220}{14}=-3.6386 }[/math]
Therefore:
[math]\displaystyle{ {\sigma'}=\frac{1}{\widehat{b}}=\frac{1}{1.0349}=0.9663 }[/math]

and:

[math]\displaystyle{ {\mu }'=-\widehat{a}\cdot {\sigma'}=-(-3.6386)\cdot 0.9663 }[/math]

or:

[math]\displaystyle{ {\mu }'=3.516 }[/math]

The mean and the standard deviation of the lognormal distribution are obtained using equations in section Lognormal Statistical Properties:

[math]\displaystyle{ \overline{T}=\mu ={{e}^{3.516+\tfrac{1}{2}{{0.9663}^{2}}}}=53.6707\text{ hours} }[/math]

and:

[math]\displaystyle{ {\sigma}=\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 as:

[math]\displaystyle{ \widehat{\rho }=0.9754 }[/math]

The above example can be repeated using Weibull++ , using RRY.

Lognormal Distribution Example 2 Data and Result.png

The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.