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''' | |||
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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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.)
