Charles. Use this distribution in reliability analysis, such as calculating a device’s mean … Best regards, Thanks alot best regards Thanks for catching this error. You want P(B|A), which by Bayes’ theorem is equal to P(A and B) / P(A). Ed, So, am suggesting a revision of the current schedule to do a maintenance at every 500 hours. We now solve these equations for α and β. Something is still off. See section 4.1.1. Hi Miguel, Hello Charles: If you specify only the mean and variance you can calculate the shape and scale parameters as shown in Example 2; in this case the function would Weibull1(mean, var). Is this correct? The probability P(x) of less than x can be calculated by the formula =WEIBULL.DIST(0.2,25.07,.55,TRUE), which yields the value 9.6812E-12. Yes, this is true. I would like to run a Weibull analysis to determine how many cycles to run a selected sample set to. How many cycles do I have to run a sample set of ‘X’ to have a confidence level of ‘Y’ with a β of ‘Z’? Year 7 = 88% (versus 100%). For some values of alpha and beta, few values will be ignored, but for others many (or perhaps even all) values will be ignored. Because of that, the time between failures is 0 and ln(t) is (- INF). I just checked and it seems that everything is correct in Example 1. This sounds like a homework assignment, and I have a rule that I won’t answer such questions. Yes, but the Excel seems to reverse the roles of alpha and beta, which is why I corrected this on the website. Confidence Level(95.0%) 4351.970951, Thank you for your consideration. The 1st moment is denoted by . The pdf should be calculated by =WEIBULL.DIST(0.2,25.07,.55,FALSE), yielding the value 1.21354E-09. In other words, if a machine runs for 100 hours, then breaks, then runs for another 50 hours and breaks again, are the data points 100 and 50 or 100 and 150? You also need to decide what happens with machines that fail in the first year. Now suppose you have 1000 screens assume that when a screen fails it is simply taken out of service. ), You can calculate the mean and standard deviation of the data that you have and then use the approach described in Example 2 of the referenced webpage to estimate the values of the shape and scaling factors. Sorry, but I have not implemented these techniques for the Weibull distribution. ; The shape parameter, k. is the Weibull shape factor.It specifies the shape of a Weibull distribution and takes on a value of between 1 and 3. What has me turned around is when I calculated MRL I would expect the values to decrease but instead the opposite is happening which I assume is driven by Beta<1 value. 2. Jain, Given that all widgets have survived 200 hours so far, need the probability that one of the three widgets fails in the next 3 hours, another of the remaining two widgets fails within 15 minutes of the preceding widget failure, and the third widget survives the entire 3 hours and 15 minutes? The 3rd moment is denoted by . I used the following: Your hep will be appreciated. How to Calculate the Weibull Distribution Mean and Variance. Miguel, i was wandering if you can help me. Do you believe this data follows a specific distribution? Your email address will not be published. How to fit the parameters to the data is explained on the following webpage. Maintain 95% of production … so if there is 1 failure for Option A, we must repair immediately. The result seems unattractive compared to no failures at all which is modeled as (Prob(no failure in next 3 hours and 15 minutes given survived 200 hours))^3. Which one is more accurate and why ? Yes, Excel got alpha and beta mixed up. Just by trial and error and eyeballing, I came up with a shape factor of 4.1 and a scale factor of 5.8. I have wind data for the year 2013, how can I calculate the shape and scale factors of the data or do I just estimate my own values? You have a great site with lots of good information. Since the standard deviation is 400, the variance is 400 squared, which is 160,000. Real Statistics Function: Since Excel doesn’t provide an inverse function, you can use the following function provided by the Real Statistics Resource Pack instead. Charles. 3. If I want to estimate budget for next year maintenance/repair, I would estimate the hours of operations of all machines to determine which ones will reach or pass the MTTF. What scaling factor and shape factor would come closest to those values? Charles. it may not be the right place but i hope you could help me. If they get repaired does their MTTF change? If you have a discrete distribution, then you can sum pdf values as you have described. 95%) for the calculated Beta and Alpha so I can change them in the Monte Carlo as well? Any wind speed values which fall within these ranges (bins) are grouped together and counted. Charles. The WEIBULL.DIST function uses the following arguments: The Weibull Probability Density Function is given by the equation: Where x is the independent variable, α is the shape parameter, and β is the scale parameter. The scale parameter, c, is the Weibull scale factor in m/s; a measure for the characteristic wind speed of the distribution. Then I took the log of each value to obtain a population with an SEV distribution. Charles. Now the challenge is to map out which Option to choose given the following: The life of an X-ray machine in hours follows the Weibull distribution, with β = 3 and η=1000 hours. We begin by tackling the simpler problem that “Given widget 1 has survived 200 hours so far, what is the probability that it fails in the next 3 hours?” Let A = widget 1 has survived 200 hours so far and B = widget 1 fails in the next 3 hours. I don’t know whether this is the same thing as the CDF and 1-CDF charts that you are referring to. Charles. It is also quite likely that you will get more accurate results when using Solver instead of Goal Seek. http://www.real-statistics.com/regression/confidence-and-prediction-intervals/ Click here for more information about this version. 35% are discarded in 5 years I don’t understand what sort of values X takes; can you explain this better? I have now corrected the referenced webpage to be consistent with this choice of symbols, and so the Excel Weibull function becomes WEIBULL(x, beta, alpha, cum). Charles, I utilized 3 different methods to model the Weibull (two from your site) and they give me relatively similar information MTTF of 20072, 21062, 23400 hours(MLE, MOM, MedRR with R2=0.92) with different parameters. Hello, Once we obtain the value for β, we can calculate α using the equation. The scale or characteristic life value is close to the mean value of the distribution. Any insights on how I should go about this? Charles, hey charles, How to solve these problems is described on the referenced webpage. My question, I want to determine which one has the smallest MSE or RMSE with respect to my data. I hope I got it correct this time. #VALUE! https://www.jstor.org/stable/3009409?seq=1#page_scan_tab_contents When the given value of alpha or beta argument is less than or equal to 0. Charles. I have time to failure for each of the subcomponents, and it’s independent of each other, I can’t calculate the MTTF because I don’t have the total no of failures for each component, but I have the total no of failures for the machine, Charles. 2)What is the probability that a randomly selected item from a population having a Weibull distribution with a shape parameter of 3 and a scale parameter of 109.3 has a value between 85.6 and 97.5? Thus, ln(1,160,000)-2*ln(1,000) = ln(1,160,000/1,000^2) = ln(1,160,000/1,000,000) = ln(1.16). Returns the Weibull distribution for a supplied set of parameters, List of the most important Excel functions for financial analysts. This example shows how to calculate the shape and scale parameters from the mean and standard deviation. In variance equation (example 2) why you use 12/beta, instead of 2/beta, as it is told at “key statistical properties of the Weibull distribution” section?