Normal Distribution Probability Plotting Example - Revision history

Lisa Hacker at 18:51, 18 September 2023

2023-09-18T18:51:53Z

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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the [~~[The Normal Distribution#Estimation of the Parameters\|~~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 19:45, 31 January 2014

2014-01-31T19:45:21Z

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	[[Image:~~chp8PP2~~.~~gif~~\|center\|500px\| ]]		[[Image:Chp8PP2_new.png\|center\|500px\| ]]

Richard House at 19:06, 31 January 2014

2014-01-31T19:06:22Z

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	[[Image:~~chp8PP~~.~~gif~~\|center\|500px\| ]]		[[Image:Chp8PP_new.png\|center\|500px\| ]]

Richard House at 20:46, 25 September 2012

2012-09-25T20:46:05Z

← Older revision		Revision as of 20:46, 25 September 2012
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	\text{110} & \text{77}\text{.15 }\!\!%\!\!\text{ } \\		\text{110} & \text{77}\text{.15 }\!\!%\!\!\text{ } \\
	\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\		\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\
	\end{matrix}</math></center>		\end{matrix}\,\!</math></center>


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	The estimate of <math>\mu </math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma </math> is determined as follows:		The estimate of <math>\mu \,\!</math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma \,\!</math> is determined as follows:


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	\widehat{\sigma }= & \frac{112-88}{2}=\frac{24}{2} \\		\widehat{\sigma }= & \frac{112-88}{2}=\frac{24}{2} \\
	\widehat{\sigma }= & 12\text{ hours}		\widehat{\sigma }= & 12\text{ hours}
	\end{align}</math>		\end{align}\,\!</math>


	Alternately, <math>\widehat{\sigma }</math> could be determined by measuring the distance from <math>t(Q=15.85%)</math> to <math>t(Q=50%)</math> , or <math>t(Q=50%)</math> to <math>t(Q=84.15%)</math> , as either of these two distances is equal to the value of one standard deviation.		Alternately, <math>\widehat{\sigma }\,\!</math> could be determined by measuring the distance from <math>t(Q=15.85%)\,\!</math> to <math>t(Q=50%)\,\!</math>, or <math>t(Q=50%)\,\!</math> to <math>t(Q=84.15%)\,\!</math>, as either of these two distances is equal to the value of one standard deviation.

Lisa Hacker at 00:51, 16 August 2012

2012-08-16T00:51:15Z

← Older revision		Revision as of 00:51, 16 August 2012
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	</noinclude>		</noinclude>
	~~Seven~~ units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters using probability plotting.		7 units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters using probability plotting.

Kate Racaza: updated banner and fixed typos

2012-08-15T06:52:41Z

updated banner and fixed typos

← Older revision		Revision as of 06:52, 15 August 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 Normal Distribution#Estimation of the Parameters\|Life Data Analysis Reference book]]''.
	~~\| valign="middle" align="left" bgcolor=EEEDF7\|~~[[~~Image:Weibull_Examples_Banner.png~~\|~~400px\|center~~]]
	\|}		</noinclude>
	<br>
	Seven units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters using probability plotting.		Seven units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters 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:
			In order to plot the points for the probability plot, the appropriate estimates for the unreliability values must be obtained. These values will be estimated through the use of median ranks, which can be obtained from statistical tables or from the Quick Statistical Reference (QSR) tool in Weibull++. The following table shows the times-to-failure and the appropriate median rank values for this example:


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

	~~[[Image:chp8PP.gif\|thumb\|center\|300px\| ]]~~

			[[Image:chp8PP.gif\|center\|500px\| ]]

	Draw the best possible line through the plot points. The time values where this line intersects the 15.85%, 50%, and 84.15% unreliability values should be projected down to the abscissa, as shown in the following plot.

	~~[[Image:chp8PP2~~.~~gif\|thumb\|center\|300px\| ]]~~		Draw the best possible line through the plot points. The time values where the line intersects the 15.85%, 50%, and 84.15% unreliability values should be projected down to the abscissa, as shown in the following plot.


	The estimate of <math>\mu </math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma </math> is determined ~~by Eqn. (sigplot)~~:		[[Image:chp8PP2.gif\|center\|500px\| ]]


			The estimate of <math>\mu </math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma </math> is determined as follows:


	::<math>\begin{align}		::<math>\begin{align}
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	\widehat{\sigma }= & 12\text{ hours}		\widehat{\sigma }= & 12\text{ hours}
	\end{align}</math>		\end{align}</math>


	Alternately, <math>\widehat{\sigma }</math> could be determined by measuring the distance from <math>t(Q=15.85%)</math> to <math>t(Q=50%)</math> , or <math>t(Q=50%)</math> to <math>t(Q=84.15%)</math> , as either of these two distances is equal to the value of one standard deviation.		Alternately, <math>\widehat{\sigma }</math> could be determined by measuring the distance from <math>t(Q=15.85%)</math> to <math>t(Q=50%)</math> , or <math>t(Q=50%)</math> to <math>t(Q=84.15%)</math> , as either of these two distances is equal to the value of one standard deviation.

Chris Kahn at 09:43, 26 July 2012

2012-07-26T09:43:25Z

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	<br>		<br>

Nicolette Young: 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…'

2012-03-22T00:08:42Z

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>
Seven units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters 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{Rank ( }\!\!%\!\!\text{ )} \\
\text{85} & \text{ 9}\text{.43 }\!\!%\!\!\text{ } \\
\text{90} & \text{22}\text{.85 }\!\!%\!\!\text{ } \\
\text{95} & \text{36}\text{.41 }\!\!%\!\!\text{ } \\
\text{100} & \text{50}\text{.00 }\!\!%\!\!\text{ } \\
\text{105} & \text{63}\text{.59 }\!\!%\!\!\text{ } \\
\text{110} & \text{77}\text{.15 }\!\!%\!\!\text{ } \\
\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\
\end{matrix}</math></center>

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

[[Image:chp8PP.gif|thumb|center|300px| ]]

Draw the best possible line through the plot points. The time values where this line intersects the 15.85%, 50%, and 84.15% unreliability values should be projected down to the abscissa, as shown in the following plot.

[[Image:chp8PP2.gif|thumb|center|300px| ]]

The estimate of <math>\mu </math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma </math> is determined by Eqn. (sigplot):

::<math>\begin{align}
\widehat{\sigma }= & \frac{t(Q=84.15%)-t(Q=15.85%)}{2} \\
\widehat{\sigma }= & \frac{112-88}{2}=\frac{24}{2} \\
\widehat{\sigma }= & 12\text{ hours}
\end{align}</math>

Alternately, <math>\widehat{\sigma }</math> could be determined by measuring the distance from <math>t(Q=15.85%)</math> to <math>t(Q=50%)</math> , or <math>t(Q=50%)</math> to <math>t(Q=84.15%)</math> , as either of these two distances is equal to the value of one standard deviation.