Re: [Scilab-users] Problems with cdfbet

[Heinz Nabielek]

On 04.11.2022, at 09:46, Dang Ngoc Chan, Christophe <christophe.d...@sidel.com> wrote: Hello Heinz, De : users <users-boun...@lists.scilab.org> De la part de Heinz Nabielek Envoyé : mardi 1 novembre 2022 22:02 I need to compute the inverse of the cumulative beta function, but I just cannot handle cdfbet. From what I undestand in https://help.scilab.org/docs/6.1.1/en_US/cdfbet.html the notation for the beta function in Scilab is : P = \int_{0}^{X} t^{A-1} * (1-t)^{B-1} dt What you want is X knowing [P, A, B] right ? The syntax seems to be [X,Y]=cdfbet("XY",A,B,P,Q) With Q = 1 - P. With your examples: --> [X,Y]=cdfbet("XY",1 ,10 ,0.95 ,0.05) X  = 0.2588656 Y  = 0.7411344 --> [X,Y]=cdfbet("XY",1 ,100 ,0.95 ,0.05) X  = 0.0295130 Y  = 0.9704870 Regards Christophe Dang Ngoc Chan Dear Christophe et Stéphane, great many thanks for the patient explanations. This way I was able to compute acceptance limits in the large-scale manufacture of small objects that are difficult to find in textbooks. Greetings Heinz   ... I am certain that you can generate a much more efficient Scilab code. N=10^(3:9); def=[300 100 30 10 3 1 0]; for i=1:7; NN=N(i); for j=1:7; defects=def(j); A(i,j)=cdfbet("XY", defects+1, NN+1-defects, .95,.05); end; plot2d(1e9,1,logflag='ll');xgrid(); plot(N', A,'--','LineWidth',3); a=gca();a.font_size=3; legend ('n=','300 defects','100 defects','30 defects','10 defects','3 defects','1 defect','0 defects'); xlabel('N = number of particles investiged','fontsize',3); ylabel('one-sided upper 95% defect fraction','fontsize',3); title ('Acceptance limit of defect fraction n/N at 95% confidence','fontsize',4);

Acceptance limit of defect fraction n over N at 95% confidence.pdf
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______________ Dr Heinz Nabielek Schüttelstrasse 77A/11 A-1020 Wien, Österreich Tel +43 1 276 56 13 cell +43 677 616 349 22 heinznabie...@me.com

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