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Re: Re: reliability function drawing

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73709] Re: [mg73711] Re: reliability function drawing
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Mon, 26 Feb 2007 06:09:25 -0500 (EST)
  • Reply-to: hanlonr at cox.net

Needs["Statistics`"];
Needs["Graphics`"];

dist=WeibullDistribution[a, b];

PDF[dist,t]

(a*t^(a - 1))/(b^a*E^(t/b)^a)

CDF[dist,t]

1 - E^(-(t/b)^a)

Reliability Function:

1-CDF[dist,t]

E^(-(t/b)^a)

DisplayTogetherArray[Partition[Table[
        Plot3D[1-CDF[dist,t],
          {t,0,5},{a,0.25,1.5},
          AxesLabel->{"\nt","  a",None},
          PlotLabel->"b = " <> ToString[b]],
        {b,4}],2],ImageSize->600];

DisplayTogetherArray[Partition[Table[
        Plot[Evaluate[Table[1-CDF[dist,t],
              {a,0.25,1.5,0.25}]],{t,0,5},
          PlotStyle->
            Table[Hue[(8a+13)/25],{a,1/4,3/2,1/4}],
          Frame->True,Axes->False,
          FrameLabel->{"t",None},
          PlotLabel->"b = " <> ToString[b]],
        {b,4}],2],ImageSize->600];


Bob Hanlon

---- "=C3=96m=C3=BCr TOSUN" <omurtosun at akdeniz.edu.tr> wrote:
>
> sorry about message..
>
> i'm using weibull distribution with 2 parameters(shape and scale
> parameters which i have calculated) and trying to draw it's reliability
> function graphic versus time which has an exponential distribution
>
> i search the help files but i think i missed and couldn't see smt. about
> this..
>



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