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.. >