I have a bimodal distribution, and if plotted with Mathematica it looks like this:
Now, the lowest value from the actual data is 8196 and 690720, but as seen in the plot, Mathematica lets the data range go from 0 to 744572. Is Mathematica choosing a bad histogram data range?
What is in general a good choice for defining the center of the bins and data range, so that I can fit a distribution through the histogram?
My approach would be:
(1) Calculate (bin width) = (Max-Min)/(number of bins) [I'm aware that there are different rules how to choose the optimal number of bins depending on the underlying distribution, let's just assume this is 12] (2) then I have 12 equal bins, starting from 8196 and ending at 690720, each having a width of 56877 (3) The first bin goes then from 8196 to 65073=8196+56877 and so on (4) As the center of the bin I define the middle between 8196 and 65073 which is 36624.5 and I position my first bin there. (5) Then I get 12 data pairs of bin center position and number of observations and I can fit a bimodal distribution through it
Am I making a mistake if I do that, or what is the reasoning behind Mathematica's choice of the histogram range exceeding the actual data range?
Edit: I've uploaded the raw data here: raw data
Edit2: For clarity explaining the mysterious frequency of 17 which was pointed out by Stephan: The data is a confocal photoluminescence map where a laser scans an emitter and it looks like this: The laser scans row by row so in the middle of each row the emitter lights up which explains the frequency of 17 when the raw data is plotted as it originates from a single list.