Probability Plotting Example - Revision history

Lisa Hacker at 18:33, 18 September 2023

2023-09-18T18:33:30Z

← Older revision		Revision as of 18:33, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	~~''This example appears in the~~ ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference] and in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]''.		''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference] and in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]''.
	</noinclude>		</noinclude>
	Let's assume six identical units are reliability tested at the same application and operation		Let's assume six identical units are reliability tested at the same application and operation

Lisa Hacker at 18:33, 18 September 2023

2023-09-18T18:33:07Z

← Older revision		Revision as of 18:33, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference] and ~~in the This example appears~~ in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]''.		''This example appears in the ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference] and in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]''.
	</noinclude>		</noinclude>
	Let's assume six identical units are reliability tested at the same application and operation		Let's assume six identical units are reliability tested at the same application and operation

Lisa Hacker at 18:32, 18 September 2023

2023-09-18T18:32:40Z

← Older revision		Revision as of 18:32, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]'' and in the ''This example appears in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]~~''.~~''.		''This example appears in the ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference] and in the This example appears in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]''.
	</noinclude>		</noinclude>
	Let's assume six identical units are reliability tested at the same application and operation		Let's assume six identical units are reliability tested at the same application and operation

Lisa Hacker at 18:32, 18 September 2023

2023-09-18T18:32:10Z

← Older revision		Revision as of 18:32, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the [~~[Parameter Estimation\|~~Life ~~Data Analysis Reference book]~~] and in the [~~[Distributions_Used_in_Accelerated_Testing\|~~Accelerated ~~Life Testing Reference book]~~]''.		''This example appears in the ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]'' and in the ''This example appears in the [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference]''.''.
	</noinclude>		</noinclude>
	Let's assume six identical units are reliability tested at the same application and operation		Let's assume six identical units are reliability tested at the same application and operation

Nicolette Young at 20:27, 21 December 2015

2015-12-21T20:27:33Z

← Older revision		Revision as of 20:27, 21 December 2015
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	paper is simply a log-linear paper.		paper is simply a log-linear paper.

	[[Image:ALTA4.1.png\|center\|~~400px~~]]		[[Image:ALTA4.1.png\|center\|500px]]

	Draw the best possible straight line that goes through the <math>t=0\,\!</math> and <math>		Draw the best possible straight line that goes through the <math>t=0\,\!</math> and <math>
Line 49:		Line 49:
	straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833\,\!</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012\,\!</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)\,\!</math>.		straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833\,\!</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012\,\!</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)\,\!</math>.

	[[Image:ALTA4.2.png\|center\|~~400px~~]]		[[Image:ALTA4.2.png\|center\|500px]]

	Now any reliability value for any mission time <math>t\,\!</math> can be obtained. For example, the		Now any reliability value for any mission time <math>t\,\!</math> can be obtained. For example, the

Richard House at 22:13, 24 September 2012

2012-09-24T22:13:43Z

← Older revision		Revision as of 22:13, 24 September 2012
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	\text{1,051} & \text{5} \\		\text{1,051} & \text{5} \\
	\text{1,744} & \text{6} \\		\text{1,744} & \text{6} \\
	\end{matrix}</math></center>		\end{matrix}\,\!</math></center>

	Obtain their median rank plotting positions. Median rank positions are used instead of other ranking methods because median ranks are at a		Obtain their median rank plotting positions. Median rank positions are used instead of other ranking methods because median ranks are at a
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	\text{1,051} & \text{73}\text{.56} \\		\text{1,051} & \text{73}\text{.56} \\
	\text{1,744} & \text{89}\text{.10} \\		\text{1,744} & \text{89}\text{.10} \\
	\end{matrix}</math></center>		\end{matrix}\,\!</math></center>

	On an exponential probability paper, plot the times on the x-axis and their corresponding		On an exponential probability paper, plot the times on the x-axis and their corresponding
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	[[Image:ALTA4.1.png\|center\|400px]]		[[Image:ALTA4.1.png\|center\|400px]]

	Draw the best possible straight line that goes through the <math>t=0</math> and <math>		Draw the best possible straight line that goes through the <math>t=0\,\!</math> and <math>
	(t)=100%</math> point and through the plotted points (as shown in the plot below).		(t)=100%\,\!</math> point and through the plotted points (as shown in the plot below).

	At the <math>Q(t)=63.2%</math> or <math>R(t)=36.8%</math> ordinate point, draw a		At the <math>Q(t)=63.2%\,\!</math> or <math>R(t)=36.8%\,\!</math> ordinate point, draw a
	straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)</math>.		straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833\,\!</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012\,\!</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)\,\!</math>.

	[[Image:ALTA4.2.png\|center\|400px]]		[[Image:ALTA4.2.png\|center\|400px]]

	Now any reliability value for any mission time <math>t</math> can be obtained. For example, the		Now any reliability value for any mission time <math>t\,\!</math> can be obtained. For example, the
	reliability for a mission of 15 hours, or any other time, can now be obtained either from the plot or analytically.		reliability for a mission of 15 hours, or any other time, can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <math>t=15</math>		To obtain the value from the plot, draw a vertical line from the abscissa, at <math>t=15\,\!</math>
	hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read		hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read
	<math>R(t)</math> . In this case, <math>R(t=15)=98.15%</math> . This can also be obtained		<math>R(t)\,\!</math>. In this case, <math>R(t=15)=98.15%\,\!</math>. This can also be obtained
	analytically, from the exponential reliability function.		analytically, from the exponential reliability function.

Richard House at 16:55, 5 September 2012

2012-09-05T16:55:48Z

← Older revision		Revision as of 16:55, 5 September 2012
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	stress levels. All of these units fail during the test after operating for the following times (in hours): 96, 257, 498, 763, 1051 and 1744.		stress levels. All of these units fail during the test after operating for the following times (in hours): 96, 257, 498, 763, 1051 and 1744.

	The steps for using the probability plotting method to determine the parameters of the exponential ~~<math>~~pdf~~</math>~~ representing the		The steps for using the probability plotting method to determine the parameters of the exponential ''pdf'' representing the
	data are as follows:		data are as follows:

Melinda Caroline at 03:18, 17 August 2012

2012-08-17T03:18:28Z

← Older revision		Revision as of 03:18, 17 August 2012
Line 47:		Line 47:

	At the <math>Q(t)=63.2%</math> or <math>R(t)=36.8%</math> ordinate point, draw a		At the <math>Q(t)=63.2%</math> or <math>R(t)=36.8%</math> ordinate point, draw a
	straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)</math>.		straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)</math>.

	[[Image:ALTA4.2.png\|center\|400px]]		[[Image:ALTA4.2.png\|center\|400px]]

Melinda Caroline at 03:10, 17 August 2012

2012-08-17T03:10:32Z

← Older revision		Revision as of 03:10, 17 August 2012
Line 52:		Line 52:

	Now any reliability value for any mission time <math>t</math> can be obtained. For example, the		Now any reliability value for any mission time <math>t</math> can be obtained. For example, the
	reliability for a mission of 15 hr, or any other time, can now be obtained either from the plot or analytically.		reliability for a mission of 15 hours, or any other time, can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <math>t=15</math>		To obtain the value from the plot, draw a vertical line from the abscissa, at <math>t=15</math>
	hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read		hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read
	<math>R(t)</math> . In this case, <math>R(t=15)=98.15%</math> . This can also be obtained		<math>R(t)</math> . In this case, <math>R(t=15)=98.15%</math> . This can also be obtained
	analytically, from the exponential reliability function.		analytically, from the exponential reliability function.

Richard House at 07:46, 7 August 2012

2012-08-07T07:46:48Z

← Older revision		Revision as of 07:46, 7 August 2012
Line 41:		Line 41:
	paper is simply a log-linear paper.		paper is simply a log-linear paper.

	~~<br>~~ [[Image:ALTA4.1.png\|center\|400px]]		[[Image:ALTA4.1.png\|center\|400px]]

	~~<br>~~

	Draw the best possible straight line that goes through the <math>t=0</math> and <math>		Draw the best possible straight line that goes through the <math>t=0</math> and <math>
Line 51:		Line 49:
	straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)</math>.		straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> (This is always at 63.2% because <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%)</math>.

	~~<br>~~ [[Image:ALTA4.2.png\|center\|400px]] ~~<br>~~		[[Image:ALTA4.2.png\|center\|400px]]

	Now any reliability value for any mission time <math>t</math> can be obtained. For example, the		Now any reliability value for any mission time <math>t</math> can be obtained. For example, the