Template:One parameter exponential distribution example Probability Plot - Revision history

Lisa Hacker: Redirected page to 1P Exponential Example

2012-08-06T08:32:24Z

Redirected page to 1P Exponential Example

← Older revision		Revision as of 08:32, 6 August 2012
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	~~'''One Parameter Exponential Probability Plot Example'''~~		#REDIRECT [[1P_Exponential_Example]]

	Six units are put on a life test and tested to failure. The failure times are 7, 12, 19, 29, 41, and 67 hours. Estimate the failure rate parameter for a one-parameter exponential distribution using the probability plotting method.

	~~'''Solution'''~~

	In order to plot the points for the probability plot, the appropriate reliability estimate values must be obtained. These will be equivalent to <math>100%-MR</math>, since the y-axis represents the reliability and the <math>MR</math> values represent unreliability estimates.

	~~<center><math>\begin{matrix}~~
	~~\text{Time-to-} & \text{Reliability} \\~~
	~~\text{failure, hr} & \text{Estimate, }% \\~~
	~~7 & 100-10.91=89.09% \\~~
	~~12 & 100-26.44=73.56% \\~~
	~~19 & 100-42.14=57.86% \\~~
	~~29 & 100-57.86=42.14% \\~~
	~~41 & 100-73.56=26.44% \\~~
	~~67 & 100-89.09=10.91% \\~~
	~~\end{matrix}</math></center>~~


	Next, these points are plotted on exponential probability plotting paper. A sample of this type of plotting paper is shown next, with the sample points in place. Notice how these points describe a line with a negative slope.
	Once the points are plotted, draw the best possible straight line through these points. The time value at which this line intersects with a horizontal line drawn at the 36.8% reliability mark is the mean life, and the reciprocal of this is the failure rate <math>\lambda </math>.
	~~This is because at <math>t=m=\tfrac{1}{\lambda }</math>:~~

	~~::<math>\begin{align}~~
	~~R(t)= & {{e}^{-\lambda \cdot t}} \\~~
	~~R(t)= & {{e}^{-\lambda \cdot \tfrac{1}{\lambda }}} \\~~
	~~R(t)= & {{e}^{-1}}=0.368=36.8%.~~
	~~\end{align}</math>~~

	~~These steps are shown graphically in the next pages.~~

	[[~~Image:weibull EPP.png\|center\|250px\|~~]]

	~~As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.~~

	~~[[Image:weibull EPP2.png\|center\|250px\|]]~~

	~~Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.~~

Nicolette Young at 17:53, 25 April 2012

2012-04-25T17:53:27Z

← Older revision		Revision as of 17:53, 25 April 2012
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	These steps are shown graphically in the next pages.		These steps are shown graphically in the next pages.

	[[Image:weibull EPP.png\|center\|~~400px~~\|]]		[[Image:weibull EPP.png\|center\|250px\|]]

	As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.		As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.

	[[Image:weibull EPP2.png\|center\|~~400px~~\|]]		[[Image:weibull EPP2.png\|center\|250px\|]]

	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

Nicolette Young at 18:46, 2 April 2012

2012-04-02T18:46:52Z

← Older revision		Revision as of 18:46, 2 April 2012
Line 38:		Line 38:

	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

	~~[[1P Exponential Example]]~~

Nicolette Young at 18:46, 2 April 2012

2012-04-02T18:46:22Z

← Older revision		Revision as of 18:46, 2 April 2012
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	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

	[[1P ~~Weibull~~ Example]]		[[1P Exponential Example]]

Nicolette Young at 18:46, 2 April 2012

2012-04-02T18:46:06Z

← Older revision		Revision as of 18:46, 2 April 2012
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	~~{\| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;"~~
	\|-
	~~\| valign="middle" align="left" bgcolor=EEEDF7\|[[Image:Weibull-Examples-banner.png\|400px\|center]]~~
	~~\|}<br><br>~~
	'''One Parameter Exponential Probability Plot Example'''		'''One Parameter Exponential Probability Plot Example'''

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	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

			[[1P Weibull Example]]

Nicolette Young at 23:22, 21 March 2012

2012-03-21T23:22:38Z

← Older revision		Revision as of 23:22, 21 March 2012
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			{\| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;"
			\|-
			\| valign="middle" align="left" bgcolor=EEEDF7\|[[Image:Weibull-Examples-banner.png\|400px\|center]]
			\|}<br><br>
	'''One Parameter Exponential Probability Plot Example'''		'''One Parameter Exponential Probability Plot Example'''

Nicolette Young at 16:40, 13 March 2012

2012-03-13T16:40:11Z

← Older revision		Revision as of 16:40, 13 March 2012
Line 35:		Line 35:
	As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.		As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.

	[[Image:weibull EPP2.png\|center\|~~300px~~\|]]		[[Image:weibull EPP2.png\|center\|400px\|]]

	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

Nicolette Young at 16:39, 13 March 2012

2012-03-13T16:39:58Z

← Older revision		Revision as of 16:39, 13 March 2012
Line 35:		Line 35:
	As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.		As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.

	[[Image:weibull EPP2.png~~\|thumb~~\|center\|300px\|]]		[[Image:weibull EPP2.png\|center\|300px\|]]

	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

Nicolette Young at 16:38, 13 March 2012

2012-03-13T16:38:57Z

← Older revision		Revision as of 16:38, 13 March 2012
Line 31:		Line 31:
	These steps are shown graphically in the next pages.		These steps are shown graphically in the next pages.

	[[Image:weibull EPP.png~~\|thumb~~\|center\|~~300px~~\|]]		[[Image:weibull EPP.png\|center\|400px\|]]

	As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.		As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.

Nicolette Young at 16:38, 13 March 2012

2012-03-13T16:38:19Z

← Older revision		Revision as of 16:38, 13 March 2012
Line 31:		Line 31:
	These steps are shown graphically in the next pages.		These steps are shown graphically in the next pages.

	[[Image:~~weibullEPP~~.png\|thumb\|center\|300px\|]]		[[Image:weibull EPP.png\|thumb\|center\|300px\|]]

	As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.		As can be seen in the plot below, the best-fit line through the data points crosses the <math>R=36.8%</math> line at <math>t=33</math> hours.

	[[Image:~~weibullEPP2~~.png\|thumb\|center\|300px\|]]		[[Image:weibull EPP2.png\|thumb\|center\|300px\|]]

	Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.		Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.