|
|
(One intermediate revision by one other user not shown) |
Line 1: |
Line 1: |
| '''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|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.
| |
| | |
| [[Image:weibull EPP2.png|center|400px|]]
| |
| | |
| Since <math>\tfrac{1}{\lambda }=33</math> hours, <math>\lambda =0.0303</math> failures/hour.
| |