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  • ...lure occur during the test <math>f=0\,\!</math>. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</math>. We want to determine th ...math> equation. The equation for the <math>MTTF\,\!</math> for the Weibull distribution is:
    2 KB (360 words) - 18:55, 18 September 2023
  • ...% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</math>. ...priate values into the cumulative binomial equation, which for the Weibull distribution appears as:
    3 KB (470 words) - 18:30, 18 September 2023
  • This example validates the results for a parametric binomial test design in Weibull++. ...these samples be tested? The failure times are assumed to follow a Weibull distribution with beta = 2.
    787 bytes (109 words) - 16:25, 28 September 2015
  • ...tion is different from the time of the required reliability. An underlying distribution should be assumed. **'''Non-Parametric Binomial:''' No distribution assumption is needed for this test design method. It can be used for one sh
    22 KB (3,517 words) - 21:22, 10 December 2015
  • {{template:LDABOOK|11|The Mixed Weibull Distribution}} ...ized that the times-to-failure for each mode may follow a distinct Weibull distribution, thus requiring individual mathematical treatment. Another reason is that e
    16 KB (2,352 words) - 18:12, 22 December 2015
  • ...analyze the results, we will obtain slightly different parameters for the distribution each time, and thus slightly different reliability results. However, by emp ...gures, we see that <math>X\,\!</math> marks the spot below which 5% of the distribution's population lies. Similarly, <math>Y\,\!</math> represents the point above
    43 KB (6,535 words) - 21:49, 18 September 2023
  • ...math>G(\widehat{\theta }).</math> For example, the mean of the exponential distribution is a function of the parameter <math>\lambda </math>: <math>G(\lambda )=1/\ ...he mean and standard deviation, respectively. Using the same one-parameter distribution, the variance of the function <math>G\left( \widehat{\theta } \right)</math
    45 KB (6,841 words) - 18:59, 30 June 2011
  • ...y function based on any assumed life distribution. In Weibull++ this life distribution can be either the 2-parameter Weibull, lognormal, normal or the 1-parameter ...wing analysis shows the data set for the voltage spikes. Using the Weibull distribution and the MLE analysis method (recommended due to the number of suspensions i
    19 KB (2,734 words) - 21:33, 18 September 2023
  • If the cumulative binomial probability <math>B(k;P,N)\,\!</math> of up to <math>{{N}_{2}}\,\!</math> f The cumulative binomial distribution probability is given by:
    30 KB (4,764 words) - 22:36, 3 June 2014
  • ...mes-to-failure or success data) to estimate the parameters of the selected distribution. Several parameter estimation methods are available. This section presents ...sections illustrate the steps in this method using the 2-parameter Weibull distribution as an example. This includes:
    52 KB (7,778 words) - 01:14, 10 March 2023
  • ...system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, the first equation above can now be expressed in terms ...distribution and the times-to-failure of the third component with a normal distribution. Then the first equation above can be written as:
    40 KB (6,296 words) - 20:09, 8 February 2017
  • ...a component affects the failure rates of other components (i.e., the life distribution characteristics of the other components change when one component fails), t ...f the system with such a configuration can be evaluated using the binomial distribution, or:
    66 KB (9,879 words) - 18:52, 5 January 2016