Template:One parameter exponential distribution example Probability Plot

Example 1
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 $$100%-MR$$, since the y-axis represents the reliability and the $$MR$$ values represent unreliability estimates.

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

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 $$\lambda $$. This is because at $$t=m=\tfrac{1}{\lambda }$$:


 * $$\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}$$

These steps are shown graphically in the next pages.



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



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