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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 sh22 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 e16 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 above43 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)</math45 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 i19 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