#Weibull distribution examples pdf

The first 5 points fit a 2 parameter Weibull and gave an R^2 values of 0.99. I experimented with using only the first 5 data points from a sample set of 26 to do a Weibull analysis. One of the advantages of the Weibull is you can form a distribution with a much smaller sample size than say a histogram. What is the minimum sample size to generate a Weibull distribution? The confidence intervals on a sample of 5 were quite wide.

I then look at what type of failure mode each is typically used to model and select based on that.Д«TW, I split your data set and performed a test of whether the shape and scale parameters were statistically different, and the p values were > 0.4.

I usually end up with 2, sometimes 3, distributions remaining that are essentially equivalent.

You will typically be able to eliminate 1 or 2 more because these values are obviously very poor, usually extremely high).

Review the remaining distributions with highest equivalent fits, looking at the means and standard errors.

I usually eliminate the higher level distributions based on this.

Review remaining distributions to identify the higher level distributions (e.g., 3-parameter Weibull), and determine both whether there this a physical reason for a location parameter (as I discussed previously) AND whether the additional parameter is significantly better than the base distribution.

Use the probability plots to rule out obviously poor fits (i.e., curves and doglegs that may indicate mixtures of different failure modes, which must be modeled separately).

I use the following approach (in Minitab): There is a lot of engineering judgment that goes with it. The data is closest to the 5 point CDF is the beta cumulative probability density function but I intuitively think the regression method makes the most sense since. Beta cumulative probability density function (first step in calculating the MR for any data set) and the second using the regression values for the set of 21 data (used to calculate the MR). The median ranking has a huge effect on this, so what is the correct way to do this? I have used 2 methods. Should I use a proportions method? The sample size is still large.Дў) I would like to plot the remaining 21 data points over the CDF calculated from my first 5 points. This equals huuuuuuge sample sizes using this method. This seems to work well when your standard deviation is small but with failure data the standard deviation is typically large and on occasion not normally distributed.

So my questions areДЎ) How does one determine the minimum sample size to use for a Weibull? There are equations for calculations the sample size for a mean e.g n = Z^2 * sigma^2 / sampling_error^2. The second data set (remaining 21 points) changes to a 3 parameter Weibull with an R^2 value somewhere in the neighborhood of 0.976. The objective was to see how well these first 5 points predicted the distribution of the entire set. ylabel ( 'Fraction of samples, normalized' ) plt. legend ( loc = 'best', frameon = False ) plt.

hist ( r_weibull, density = True, histtype = 'stepfilled', alpha = 1.8 ) ax. rvs ( c, size = 1000 ) print ( 'Random Weibull variates, first', N, ': ', r_weibull ) ax.

#Weibull distribution examples pdf

pdf ( x ), 'k-', lw = 2, label = 'Frozen PDF' ) # Samples of Weibull variates (empirical) r_weibull = weibull_min. pdf ( x, c ), 'r-', lw = 5, alpha = 0.6, label = 'Weibull_min PDF' ) # Frozen PDF rv = weibull_min ( c ) ax. ppf ( 0.99, c ), 100 ) # ppf - Percent point function (inverse of cdf - percentiles). stats ( c, moments = 'mvsk' ) # var -> variance = Пѓ^2 (sigma square) print ( 'Weibull mean:', mean, ' Weibull variance:', var, ' ' ) x = np. subplots ( 1, 1 ) c = 1 from scipy.stats import weibull_min mean, var, skew, kurt = weibull_min.