probability plot - weibull - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: probability plot - weibull (/thread-10301.html) probability plot - weibull - Marcus_W - 03-09-2018 08:26 PM Hello, I own my HP Prime for some day and I am fascinated about the possibilities. I have now one question. I would like to calculate failure probabilities with "Weibull probability plots" based on measured data. So my question is, is there any probability to generate a Weibull probability plot on a HP Prime? I have seen the weibull and weibull_cdf function but I don't know what the parameters (k,n,t,x,x2) are and how I could calculate these parameters with the HP Prime based on my measured data. Thank you. & Best regards, Marcus RE: probability plot - weibull - parisse - 03-10-2018 07:11 AM The syntax is : weibull_cdf(Real(k),Real(lambda),[Real(theta)],Real(x1),[Real(x2)]) Cf. https://en.wikipedia.org/wiki/Weibull_distribution for parameter definitions, theta is an additional optionnal parameter, a shift in x. RE: probability plot - weibull - Marcus_W - 03-10-2018 01:07 PM Hey Parisse, thank you for you answer regarding weibull_cdf. I think this function is useful when I know the parameters of my weibull distribution. My problem is, that I have measured data and would like to build/fit a weibull function based on this data. Has anyone an idea how to realize this with my HP Prime? Best regards, Marcus RE: probability plot - weibull - Marcus_W - 03-10-2018 11:44 PM Hey, I think I found a solution, based on this: http://www.real-statistics.com/distribution-fitting/fitting-weibull-regression/ I am not so good in coding, but maybe there is someone how can use this: Code: ``` LOCAL len :=0; LOCAL n :=0; LOCAL param := 0; LOCAL beta := 0; LOCAL lambda := 0; LOCAL value := 0; EXPORT weibull_ifitr(data,percent) BEGIN C1:=data; len:=length(C1); C2:=sort(C1); C3:=ln(C2); FOR n FROM 1 TO len DO C4(n):=(n-0.5)/len; END; C5:=ln(-ln(1-C4)); param:=linear_regression(C3,C5); beta:=param(1); lambda:=e^(-param(2)/param(1)); value:=weibull_icdf(beta,lambda,percent); RETURN value; END;``` Comments are welcome. bye, Marcus