Am Dienstag, 28. Mai 2002 02:42 schrieb Joel Hammer: > I ran the data both on Excel and with my own bash script. They get close > results. > I have attached my data.txt and a better ps file, with the labels better > spaced out.
Sorry, Joel, my statistics used to be quite good 25 years ago, but now they got a bit rusty. I fed your data into OpenOffice Calc and got results similar to yours (only my variance is a bit different, var=0.49772 ...). Now Alan Jackson's posting gave me the clue: the discrepancy in your graph between the plotted bar graph and the normal curve is a problem of scaling. One idea behind the normal curve is that probabilities or relative frequencies are represented as areas under the curve; that yields an area of 1 under the total curve. In a bar graph, relative frequencies often are not represented as AREAS, but as HIGHTS of the respective rectangles. All these hights (in your graph you have 21 of them) then add up to 1. Now your data are grouped into groups of width 0.2, which can be seen as the width of each the rectangles. Thinking of areas, that means that the rectangles have a total area of 0.2. So, for bringing together the bar graph and the curve, you can either divide all the bar heights by 0.2, or multiply the curve by that factor. In gnuplot, plotting 0.2*f(x) instead of your f(x) should give you a reasonable result. Klaus _______________________________________________ Linux-users mailing list - http://linux-sxs.org/mailman/listinfo/linux-users Subscribe/Unsubscribe info, Archives,and Digests are located at the above URL.
