Romain Francois schrieb:
>Do you want that : > >h<-hist(x,breaks=10,freq = TRUE) >xfit<-seq(min(x),max(x),length=40) >yfit<-dnorm(xfit,mean=mean(x),sd=sd(x)) > >lines(xfit,yfit * 150 * (h$breaks[2]-h$breaks[1])) > > > > Right thats what I want ... but does it make sense to fit the line with a try and error multipier (150) Is there no way to compute the frequency and the distribution line with standardised function? I used SPSS with the data from x<-5+rnorm(150) Hit the graph-histogramm menue choosed: display normal curve and got the result: http://biostatistic.de/temp/spss1.jpg - just as easy as possible So, they have any standardised function. maybe it does not make sence, but i am not able to see why there is such a esay to use function in SPSS but not in R Maybe anybody is able to explane whiy and my intention: As a previous computer scientist I am trying to find a way to eleminate SPSS an use R Sure there is a big lack of my statistic knowledge - but the often the SPSS user have similar lack but it is easy to click and view instead to try the similar steps in R But if I am not able to find the steps in R for the common .. "SPSS-clicks" , I will never be able to suggest R in the institute to the people with mor statistical knowledge but no knowledge about computer science ... and command line interpreter Regards Knut -- Viele Grüße Knut Krüger -- Reitpark Einthal Leitung: 1 Tierarzt, 1 Berufsreiter Homepage http://www.einthal.de Eine fachgerechte Betreuung rund um die Uhr. ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
