Template:Example: Normal Distribution Probability Plot - Revision history

Lisa Hacker: Redirected page to The Normal Distribution

2012-08-13T01:51:02Z

Redirected page to The Normal Distribution

← Older revision		Revision as of 01:51, 13 August 2012
Line 1:		Line 1:
	~~'''Normal Distribution Probability Plotting Example'''~~		#REDIRECT [[The Normal Distribution]]

	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:WB.9 chp8pp.png\|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:WB.9 chp8PP2.png\|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:

	~~::<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.

Richard House at 02:27, 8 August 2012

2012-08-08T02:27:20Z

← Older revision		Revision as of 02:27, 8 August 2012
Line 19:		Line 19:
	\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\		\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\
	\end{matrix}</math></center>		\end{matrix}</math></center>



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

	[[Image:WB.9 chp8pp.png\|center\|~~250px~~\| ]]		[[Image:WB.9 chp8pp.png\|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.		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:WB.9 chp8PP2.png\|center\|~~250px~~\| ]]		[[Image:WB.9 chp8PP2.png\|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:		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:
Line 39:		Line 35:
	\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.

Nicolette Young at 18:58, 25 April 2012

2012-04-25T18:58:09Z

← Older revision		Revision as of 18:58, 25 April 2012
Line 24:		Line 24:
	These points can now be plotted on normal probability plotting paper as shown in the next figure.		These points can now be plotted on normal probability plotting paper as shown in the next figure.

	[[Image:WB.9 chp8pp.png\|center\|~~400px~~\| ]]		[[Image:WB.9 chp8pp.png\|center\|250px\| ]]


	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.		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:WB.9 chp8PP2.png\|center\|~~400px~~\| ]]		[[Image:WB.9 chp8PP2.png\|center\|250px\| ]]

Nicolette Young at 19:21, 14 March 2012

2012-03-14T19:21:22Z

← Older revision		Revision as of 19:21, 14 March 2012
Line 24:		Line 24:
	These points can now be plotted on normal probability plotting paper as shown in the next figure.		These points can now be plotted on normal probability plotting paper as shown in the next figure.

	[[Image:~~chp8PP~~.~~gif\|thumb~~\|center\|~~300px~~\| ]]		[[Image:WB.9 chp8pp.png\|center\|400px\| ]]


	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.		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~~\| ]]		[[Image:WB.9 chp8PP2.png\|center\|400px\| ]]

Harry Guo at 18:23, 10 February 2012

2012-02-10T18:23:27Z

← Older revision		Revision as of 18:23, 10 February 2012
Line 39:		Line 39:
	\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.

Harry Guo at 18:23, 10 February 2012

2012-02-10T18:23:01Z

← Older revision		Revision as of 18:23, 10 February 2012
Line 32:		Line 32:


	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)~~:		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:

	::<math>\begin{align}		::<math>\begin{align}

Harry Guo: moved Template:Example: Weibull Normal Distribution to Template:Example: Normal Distribution Probability Plot

2012-02-10T18:17:24Z

moved Template:Example: Weibull Normal Distribution to Template:Example: Normal Distribution Probability Plot

← Older revision	Revision as of 18:17, 10 February 2012
(No difference)

Harry Guo: /* Normal Distribution Example */

2012-02-10T18:17:05Z

Normal Distribution Example

← Older revision		Revision as of 18:17, 10 February 2012
Line 1:		Line 1:
	~~===~~ Normal Distribution Example~~===~~		'''Normal Distribution Probability Plotting Example'''

	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~~====~~		'''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:

Nicolette Young: Created page with '=== Normal Distribution Example=== 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 d…'

2012-02-08T20:32:37Z

Created page with '=== Normal Distribution Example=== 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 d…'

New page

=== Normal Distribution Example===
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.