Thanks Richard!

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On Mon, May 23, 2011 at 5:56 PM, <Richard.Cotton@hsl.gov.uk> wrote:

>
> > In order to do this I can use the relation between count and density, but
> I
> > would like to know if there is a way for me to predict it upfront.
>
> In the code for hist.default, you'll see the line
>
> dens <- counts/(n * diff(breaks))
>
> > Here is an example:
> >
> > set.seed(242)
> > z = rnorm(30)
> > hist_z <- hist(z)
> > hist_z$counts / hist_z$density # the relation is 15
>
> In your example, n is 30, and the breaks are evenly spaced every 0.5.
>
> Regards,
> Richie.
>
> Mathematical Sciences Unit*
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