Template:One parameter exponential distribution example Probability Plot: Difference between revisions

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====Example 1: One Parameter Exponential 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 to Example 1=====
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:weibullEPP.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.
 
[[Image:weibullEPP2.png|thumb|center|300px|]]
 
Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.

Latest revision as of 08:32, 6 August 2012