Template:Example: Lognormal Distribution RRY - Revision history

Lisa Hacker: Redirected page to The Lognormal Distribution

2012-08-13T03:45:08Z

Redirected page to The Lognormal Distribution

← Older revision		Revision as of 03:45, 13 August 2012
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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:~~

	~~{\|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"~~
	\|-
	~~\|colspan="2" style="text-align:center"\| 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>\rho </math> , using rank regression on Y.~~

	~~'''Solution'''~~

	~~Construct Table 2, as shown next.~~

	~~<center><math>\overset{{}}{\mathop{\text{Table }\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}=\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>\widehat{\rho }=0.9754</math>~~

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

	~~[[Image:~~Lognormal Distribution ~~Example 2 Data and Result.png\|center\|550px\|~~ ]]

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

Richard House at 04:58, 8 August 2012

2012-08-08T04:58:08Z

← Older revision		Revision as of 04:58, 8 August 2012
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	Fourteen units were reliability tested and the following life test data were obtained:		Fourteen units were reliability tested and the following life test data were obtained:

	{\|align="center" border=1 cellspacing=0		{\|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
	\|-		\|-
	\|colspan="2" style="text-align:center"\| Table 1 - Life Test Data		\|colspan="2" style="text-align:center"\| Table 1 - Life Test Data
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	\end{matrix}</math></center>		\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 median rank values ( <math>F({{t}_{i}})</math> ) can be found in rank tables or by using the Quick Statistical Reference in Weibull++ .
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	Given the values in the table above, calculate <math>\widehat{a}</math> and <math>\widehat{b}</math>:		Given the values in the table above, calculate <math>\widehat{a}</math> and <math>\widehat{b}</math>:


	::<math>\begin{align}		::<math>\begin{align}
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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\|~~250px~~\| ]]		[[Image:Lognormal Distribution Example 2 Data and Result.png\|center\|550px\| ]]

	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.

Nicolette Young at 21:20, 25 April 2012

2012-04-25T21:20:23Z

← Older revision		Revision as of 21:20, 25 April 2012
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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.

Nicolette Young at 19:11, 14 March 2012

2012-03-14T19:11:34Z

← Older revision		Revision as of 19:11, 14 March 2012
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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\|~~400px~~\| ]]		[[Image:Lognormal Distribution Example 2 Data and Result.png\|thumb\|center\|500px\| ]]

	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.

Harry Guo at 16:57, 2 March 2012

2012-03-02T16:57:53Z

← Older revision		Revision as of 16:57, 2 March 2012
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	{\|align="center" border=1 cellspacing=0		{\|align="center" border=1 cellspacing=0
	\|-		\|-
	\|colspan="2" style="text-align:center"\| Table - Life Test Data		\|colspan="2" style="text-align:center"\| Table 1 - Life Test Data
	\|-		\|-
	!Data point index		!Data point index
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	'''Solution'''		'''Solution'''

	Construct Table 9.2, as shown next.		Construct Table 2, as shown next.

	<center><math>\overset{{}}{\mathop{\text{Table 9}\text{.2 - Least Squares Analysis}}}\,</math></center>		<center><math>\overset{{}}{\mathop{\text{Table }\text{2 - Least Squares Analysis}}}\,</math></center>

	<center><math>\begin{matrix}		<center><math>\begin{matrix}

Harry Guo at 16:55, 2 March 2012

2012-03-02T16:55:51Z

← Older revision		Revision as of 16:55, 2 March 2012
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	{\|align="center" border=1 cellspacing=0		{\|align="center" border=1 cellspacing=0
	\|-		\|-
	\|colspan="2" style="text-align:center"\| Table ~~9.1~~ - Life Test Data		\|colspan="2" style="text-align:center"\| Table - Life Test Data
	\|-		\|-
	!Data point index		!Data point index
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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:~~5folio~~.png\|thumb\|center\|400px\| ]]		[[Image:Lognormal Distribution Example 2 Data and Result.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.		The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.

Harry Guo at 21:17, 13 February 2012

2012-02-13T21:17:26Z

← Older revision		Revision as of 21:17, 13 February 2012
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	and:		and:

	::<math>{{\sigma ~~}_{T}~~}=\sqrt{({{e}^{2\cdot 3.516+{{0.9663}^{2}}}})({{e}^{{{0.9663}^{2}}}}-1)}=66.69\text{ hours}</math>		::<math>{\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 ~~using Eqn. (RHOln)~~:		The correlation coefficient can be estimated as:

	::<math>\widehat{\rho }=0.9754</math>		::<math>\widehat{\rho }=0.9754</math>

Harry Guo at 21:11, 13 February 2012

2012-02-13T21:11:58Z

← Older revision		Revision as of 21:11, 13 February 2012
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	{\|align="center" border=1 cellspacing=0		{\|align="center" border=1 cellspacing=0
	\|-		\|-
	\|colspan="2" style="text-align:center"\| Table 9.1 - Life Test Data ~~for Example 2~~		\|colspan="2" style="text-align:center"\| Table 9.1 - Life Test Data
	\|-		\|-
	!Data point index		!Data point index
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	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	N & T_{i} & F(T_{i}) & {T_{i}}'& y_{i} & {{T_{i}}'}^{2} & y_{i}^{2} & T_{i} y_{i} \\		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{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{2} & \text{10} & \text{0}\text{.1170} & \text{2.3026}& \text{-1.1901} & \text{5.3019} & \text{1.4163} & \text{-2.7403} \\
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	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 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> ).		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> ~~using Eqns. (aaln) and (bbln)~~:		Given the values in the table above, calculate <math>\widehat{a}</math> and <math>\widehat{b}</math>:


	::<math>\begin{align}		::<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{\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}		& \widehat{b}= & \frac{10.4473-(49.2220)(0)/14}{183.1530-{{(49.2220)}^{2}}/14}
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	and:		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>		::<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:		or:
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	::<math>\widehat{a}=\frac{0}{14}-(1.0349)\frac{49.2220}{14}=-3.6386</math>		::<math>\widehat{a}=\frac{0}{14}-(1.0349)\frac{49.2220}{14}=-3.6386</math>

	:Therefore~~, from Eqn. (bln)~~:		:Therefore:

	::<math>{{\sigma ~~}_{{{T}~~'}}}=\frac{1}{\widehat{b}}=\frac{1}{1.0349}=0.9663</math>		::<math>{\sigma'}=\frac{1}{\widehat{b}}=\frac{1}{1.0349}=0.9663</math>

	:and ~~from Eqn. (aln)~~:		and:

	::<math>{\mu }'=-\widehat{a}\cdot {{\sigma ~~}_{{{T}~~'}}}=-(-3.6386)\cdot 0.9663</math>		::<math>{\mu }'=-\widehat{a}\cdot {\sigma'}=-(-3.6386)\cdot 0.9663</math>

	or:		or:
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	::<math>{\mu }'=3.516</math>		::<math>{\mu }'=3.516</math>

	The mean and the standard deviation of the lognormal distribution are obtained using ~~Eqns. (mean) and (sdv)~~:		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>		::<math>\overline{T}=\mu ={{e}^{3.516+\tfrac{1}{2}{{0.9663}^{2}}}}=53.6707\text{ hours}</math>

Harry Guo at 18:15, 13 February 2012

2012-02-13T18:15:53Z

← Older revision		Revision as of 18:15, 13 February 2012
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	'''Solution'''		'''Solution'''

	Construct Table 9.2, as shown next.		Construct Table 9.2, as shown next.

Harry Guo at 18:15, 13 February 2012

2012-02-13T18:15:44Z

← Older revision		Revision as of 18:15, 13 February 2012
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	'''Lognormal Distribution RRY Example'''		'''Lognormal Distribution RRY Example'''

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