Example: Lognormal Distribution Probability Plot - Revision history

Lisa Hacker at 21:45, 18 September 2023

2023-09-18T21:45:19Z

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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the [~~[The_Lognormal_Distribution#Probability_Plotting\|~~Life ~~Data Analysis Reference book]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''.

	</noinclude>		</noinclude>

Richard House at 22:22, 25 September 2012

2012-09-25T22:22:54Z

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	\text{3000} & \text{79}\text{.89 }\!\!%\!\!\text{ } \\		\text{3000} & \text{79}\text{.89 }\!\!%\!\!\text{ } \\
	\text{9000} & \text{91}\text{.70 }\!\!%\!\!\text{ } \\		\text{9000} & \text{91}\text{.70 }\!\!%\!\!\text{ } \\
	\end{matrix}</math></center>		\end{matrix}\,\!</math></center>


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	[[Image:WB.10 lpp3.png\|center\|420px\| ]]		[[Image:WB.10 lpp3.png\|center\|420px\| ]]

	The natural logarithm of the time where the fitted line intersects is equivalent to <math>{\mu }'\,\!</math> . In this case, <math>{\mu }'=6.45\,\!</math> . The value for <math>{{\sigma }_{{{T}'}}}\,\!</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%\,\!</math> and <math>Q(t)=15.85%.\,\!</math> At <math>Q(t)=15.85%\,\!</math> , ln <math>(t)=4.55\,\!</math> . Therefore, <math>{\sigma'}=6.45-4.55=1.9\,\!</math> .		The natural logarithm of the time where the fitted line intersects is equivalent to <math>{\mu }'\,\!</math>. In this case, <math>{\mu }'=6.45\,\!</math>. The value for <math>{{\sigma }_{{{T}'}}}\,\!</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%\,\!</math> and <math>Q(t)=15.85%.\,\!</math> At <math>Q(t)=15.85%\,\!</math>, ln <math>(t)=4.55\,\!</math>. Therefore, <math>{\sigma'}=6.45-4.55=1.9\,\!</math>.

Richard House at 03:29, 28 August 2012

2012-08-28T03:29:55Z

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	[[Image:WB.10 lpp3.png\|center\|420px\| ]]		[[Image:WB.10 lpp3.png\|center\|420px\| ]]

	The natural logarithm of the time where the fitted line intersects is equivalent to <math>{\mu }'</math> . In this case, <math>{\mu }'=6.45</math> . The value for <math>{{\sigma }_{{{T}'}}}</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%</math> and <math>Q(t)=15.85%.</math> At <math>Q(t)=15.85%</math> , ln <math>(t)=4.55</math> . Therefore, <math>{\sigma'}=6.45-4.55=1.9</math> .		The natural logarithm of the time where the fitted line intersects is equivalent to <math>{\mu }'\,\!</math> . In this case, <math>{\mu }'=6.45\,\!</math> . The value for <math>{{\sigma }_{{{T}'}}}\,\!</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%\,\!</math> and <math>Q(t)=15.85%.\,\!</math> At <math>Q(t)=15.85%\,\!</math> , ln <math>(t)=4.55\,\!</math> . Therefore, <math>{\sigma'}=6.45-4.55=1.9\,\!</math> .

Lisa Hacker at 00:51, 16 August 2012

2012-08-16T00:51:44Z

← Older revision		Revision as of 00:51, 16 August 2012
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	</noinclude>		</noinclude>
	~~Eight~~ units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.		8 units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.

	'''Solution'''		'''Solution'''

Richard House at 07:49, 7 August 2012

2012-08-07T07:49:40Z

← Older revision		Revision as of 07:49, 7 August 2012
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	Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.		Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.


	'''Solution'''		'''Solution'''

Kate Racaza: proofread

2012-07-23T08:58:24Z

proofread

← Older revision		Revision as of 08:58, 23 July 2012
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	{~~\| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;"~~		<noinclude>{{Banner Weibull Examples}}
	\|-		''This example appears in the [[The_Lognormal_Distribution#Probability_Plotting\|Life Data Analysis Reference book]]''.
	~~\| valign="middle" align="left" bgcolor=EEEDF7\|~~[[~~Image:Weibull-Examples-banner.png\|400px~~\|~~center~~]]
	\|}
	~~<br>~~

			</noinclude>
	Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.		Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.


	'''Solution'''		'''Solution'''

	In order to plot the points for the probability plot, the appropriate unreliability estimate values must be obtained. These will be estimated through the use of median ranks, which can be obtained from statistical tables or the Quick Statistical Reference in Weibull++. The following table shows the times-to-failure and the appropriate median rank values for this example:		In order to plot the points for the probability plot, the appropriate unreliability estimate values must be obtained. These will be estimated through the use of median ranks, which can be obtained from statistical tables or the Quick Statistical Reference in Weibull++. The following table shows the times-to-failure and the appropriate median rank values for this example:

	<center><math>\begin{matrix}		<center><math>\begin{matrix}
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	[[Image:WB.10 lpp3.png\|center\|420px\| ]]		[[Image:WB.10 lpp3.png\|center\|420px\| ]]

	The natural logarithm of the time where the fitted line intersects .. is equivalent to <math>{\mu }'</math> . In this case, <math>{\mu }'=6.45</math> . The value for <math>{{\sigma }_{{{T}'}}}</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%</math> and <math>Q(t)=15.85%.</math> At <math>Q(t)=15.85%</math> , ln <math>(t)=4.55</math> . Therefore, <math>{\sigma'}=6.45-4.55=1.9</math> .		The natural logarithm of the time where the fitted line intersects is equivalent to <math>{\mu }'</math> . In this case, <math>{\mu }'=6.45</math> . The value for <math>{{\sigma }_{{{T}'}}}</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%</math> and <math>Q(t)=15.85%.</math> At <math>Q(t)=15.85%</math> , ln <math>(t)=4.55</math> . Therefore, <math>{\sigma'}=6.45-4.55=1.9</math> .

Chuck Smith at 23:33, 21 March 2012

2012-03-21T23:33:49Z

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	\|}		\|}
	<br>~~'''Lognormal Distribution Probability Plotting Example'''~~		<br>

	Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.		Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.

Chuck Smith: Created page with '{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;" |- | valign="middle" align="left" bgcolor=EEEDF7|[[Image:Weibull-E…'

2012-03-21T23:33:28Z

Created page with '{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;" |- | valign="middle" align="left" bgcolor=EEEDF7|[[Image:Weibull-E…'

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<br>'''Lognormal Distribution Probability Plotting Example'''

Eight units are put on a life test and tested to failure. The failures occurred at 45, 140, 260, 500, 850, 1400, 3000, and 9000 hours. Estimate the parameters for the lognormal distribution using probability plotting.

'''Solution'''

In order to plot the points for the probability plot, the appropriate unreliability estimate values must be obtained. These will be estimated through the use of median ranks, which can be obtained from statistical tables or the Quick Statistical Reference in Weibull++. The following table shows the times-to-failure and the appropriate median rank values for this example:

<center><math>\begin{matrix}
\text{Time-to-} & \text{Median} \\
\text{Failure (hr}\text{.)} & \text{Rank ( }\!\!%\!\!\text{ )} \\
\text{ 45} & \text{ 8}\text{.30 }\!\!%\!\!\text{ } \\
\text{ 140} & \text{20}\text{.11 }\!\!%\!\!\text{ } \\
\text{ 260} & \text{32}\text{.05 }\!\!%\!\!\text{ } \\
\text{ 500} & \text{44}\text{.02 }\!\!%\!\!\text{ } \\
\text{ 850} & \text{55}\text{.98 }\!\!%\!\!\text{ } \\
\text{1400} & \text{67}\text{.95 }\!\!%\!\!\text{ } \\
\text{3000} & \text{79}\text{.89 }\!\!%\!\!\text{ } \\
\text{9000} & \text{91}\text{.70 }\!\!%\!\!\text{ } \\
\end{matrix}</math></center>

These points may now be plotted on normal probability plotting paper as shown in the next figure.

[[Image:WB.10 lpp2.png|center|500px| ]]

Draw the best possible line through the plot points. The time values where this line intersects the 15.85% and 50% unreliability values should be projected up to the logarithmic scale, as shown in the following plot.

[[Image:WB.10 lpp3.png|center|420px| ]]

The natural logarithm of the time where the fitted line intersects .. is equivalent to <math>{\mu }'</math> . In this case, <math>{\mu }'=6.45</math> . The value for <math>{{\sigma }_{{{T}'}}}</math> is equal to the difference between the natural logarithms of the times where the fitted line crosses <math>Q(t)=50%</math> and <math>Q(t)=15.85%.</math> At <math>Q(t)=15.85%</math> , ln <math>(t)=4.55</math> . Therefore, <math>{\sigma'}=6.45-4.55=1.9</math> .