3-Parameter Weibull Example: Difference between revisions
Chris Kahn (talk | contribs) |
Chris Kahn (talk | contribs) No edit summary |
||
| Line 4: | Line 4: | ||
|} | |} | ||
Six identical units are reliability tested under the same stresses and conditions. All units are tested to failure and the following times-to-failure are recorded: 48, 66, 85, 107, 125 and 152 hours. Find the parameters of the | Six identical units are reliability tested under the same stresses and conditions. All units are tested to failure and the following times-to-failure are recorded: 48, 66, 85, 107, 125 and 152 hours. Find the parameters of the 3-parameter Weibull distribution using probability plotting. | ||
Revision as of 04:51, 23 July 2012
 |
Six identical units are reliability tested under the same stresses and conditions. All units are tested to failure and the following times-to-failure are recorded: 48, 66, 85, 107, 125 and 152 hours. Find the parameters of the 3-parameter Weibull distribution using probability plotting.
Solution
The following figure shows the results. Note that since the original data set was concave down, 17.26 was subtracted from all the times-to-failure and replotted, resulting in a straight line, thus γ = 17.26. (We used Weibull++ to get the results. To perform this by hand, one would attempt different values of γ, using a trial and error methodology, until an acceptable straight line is found. When performed manually, you do not expect decimal accuracy.)
