Template:Example: 3P Weibull Distribution: Difference between revisions

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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 three-parameter Weibull distribution using probability plotting.
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 three-parameter Weibull distribution using probability plotting.


==== Solution ====
'''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 <span class="texhtml">γ = 17.26</span>. (We used Weibull++ to get the results. To perform this by hand, one would attempt different values of <span class="texhtml">γ,</span> using a trial and error methodology, until an acceptable straight line is found. When performed manually, you do not expect decimal accuracy.)  
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 <span class="texhtml">γ = 17.26</span>. (We used Weibull++ to get the results. To perform this by hand, one would attempt different values of <span class="texhtml">γ,</span> using a trial and error methodology, until an acceptable straight line is found. When performed manually, you do not expect decimal accuracy.)  
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[[Image:lda6.9.gif|thumb|center|500px|]]

Revision as of 17:15, 9 February 2012

Three-parameter Weibull distribution example

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 three-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.)

Lda6.9.gif