Hi Sam, ?hist gives:
"density values f^(x[i]), as estimated density values. If all(diff(breaks) == 1), they are the relative frequencies counts/n and in general satisfy sum[i; f^(x[i]) (b[i+1]-b[i])] = 1, where b[i] = breaks[i]." 1st case, density != frequency because all(diff(breaks) == 1) is FALSE (diff = 0.5); sum(h$density*diff) = 1 2nd case, density == frequency because all(diff(breaks) == 1) is TRUE (diff = 1); sum(h$density*diff) = 1 Regards, Pascal -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://www.memritv.org http://openvotingconsortium.org http://thereligionofpeace.com http://mideasttruth.com http://palestinefacts.org ((lambda (x) `(,x ',x)) '(lambda (x) `(,x ',x))) ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
