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$\begingroup$ I didn't know about EstimatedDistribution, thanks for that! By the way, given that this is a probability distribution, the scaling factor ought to just be the total number of observations times the spacing between bins, i.e. c = 0.1 Total@data[[All,2]]. That gives $264.9$ compared to $266.675$ from your method; I don't have a good explanation for the difference. $\endgroup$

user484
– user484

2015-07-28 01:24:23 +00:00
Commented Jul 28, 2015 at 1:24
$\begingroup$ @Rahul they make different assumptions about what the data represent. In your case we assume that the data points are left endpoints on the bins of a histogram (a perfectly reasonable assumption in this case). In the case of least squares there is a presumption that the data contain noise. $\endgroup$

Andy Ross
– Andy Ross

2015-07-28 01:56:21 +00:00
Commented Jul 28, 2015 at 1:56

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