Template:Example: Lognormal Distribution Probability Plot - Revision history

Lisa Hacker: Redirected page to Example: Lognormal Distribution Probability Plot

2012-08-15T06:41:57Z

Redirected page to Example: Lognormal Distribution Probability Plot

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	[[~~Category~~: ~~For Deletion~~]]		#REDIRECT [[Example:_Lognormal_Distribution_Probability_Plot]]

Kate Racaza: Replaced content with 'Category: For Deletion'

2012-07-23T09:22:14Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 09:22, 23 July 2012
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	~~'''Lognormal Distribution Probability Plotting Example'''~~		[[Category: For Deletion]]
	[[~~Example~~: ~~Lognormal Distribution Probability Plot~~]]
	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\|thumb\|250px\| ]]~~

	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\|thumb\|250px\| ]]~~

	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> .

Nicolette Young at 21:19, 25 April 2012

2012-04-25T21:19:19Z

← Older revision		Revision as of 21:19, 25 April 2012
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	These points may now be plotted on normal probability plotting paper as shown in the next figure.		These points may now be plotted on normal probability plotting paper as shown in the next figure.

	[[Image:WB.10 lpp2.png\|center\|~~500px~~\| ]]		[[Image:WB.10 lpp2.png\|center\|thumb\|250px\| ]]

	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.		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~~\| ]]		[[Image:WB.10 lpp3.png\|center\|thumb\|250px\| ]]

	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 23:32, 21 March 2012

2012-03-21T23:32:46Z

← Older revision		Revision as of 23:32, 21 March 2012
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	'''Lognormal Distribution Probability Plotting Example'''		'''Lognormal Distribution Probability Plotting Example'''
			[[Example: Lognormal Distribution Probability Plot]]
	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.

Nicolette Young at 19:10, 14 March 2012

2012-03-14T19:10:21Z

← Older revision		Revision as of 19:10, 14 March 2012
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	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.		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\|~~400px~~\| ]]		[[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> .

Nicolette Young at 19:10, 14 March 2012

2012-03-14T19:10:05Z

← Older revision		Revision as of 19:10, 14 March 2012
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	These points may now be plotted on normal probability plotting paper as shown in the next figure.		These points may now be plotted on normal probability plotting paper as shown in the next figure.

	[[Image:WB.10 lpp2.png\|center\|~~400px~~\| ]]		[[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.		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.

Nicolette Young at 19:09, 14 March 2012

2012-03-14T19:09:12Z

← Older revision		Revision as of 19:09, 14 March 2012
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	These points may now be plotted on normal probability plotting paper as shown in the next figure.		These points may now be plotted on normal probability plotting paper as shown in the next figure.

	[[Image:~~ldachp9ex1~~.~~gif\|thumb~~\|center\|400px\| ]]		[[Image:WB.10 lpp2.png\|center\|400px\| ]]

	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.		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:~~ldachp9ex1~~.2.~~gif\|thumb~~\|center\|400px\| ]]		[[Image:WB.10 lpp3.png\|center\|400px\| ]]

	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> .

Harry Guo at 21:03, 13 February 2012

2012-02-13T21:03:46Z

← Older revision		Revision as of 21:03, 13 February 2012
Line 29:		Line 29:
	[[Image:ldachp9ex1.2.gif\|thumb\|center\|400px\| ]]		[[Image:ldachp9ex1.2.gif\|thumb\|center\|400px\| ]]

	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 ~~}_{{{T}~~'}}}=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> .

Harry Guo at 18:12, 13 February 2012

2012-02-13T18:12:59Z

← Older revision		Revision as of 18:12, 13 February 2012
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	'''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:

Harry Guo: moved Template:Example: Lognormal Distribution to Template:Example: Lognormal Distribution Probability Plot

2012-02-13T18:11:10Z

moved Template:Example: Lognormal Distribution to Template:Example: Lognormal Distribution Probability Plot

← Older revision	Revision as of 18:11, 13 February 2012
(No difference)