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..
>