How to tell quantitatively whether 1D data is clustered around 1 or 3 values? I've got some data on the time between heart beats of a human. One indication of ectopic (extra) beats is that these intervals are clustered around three values instead of one. How can I obtain a quantitative measure of this?
I'm looking to compare multiple data sets, and these two 100-bin histograms are representative of all of them.

I could compare the variances, but I want my algorithm to be able to detect whether there is one or three clusters in each case without comparing to the other cases.
This is for offline processing, so there's a lot of computation power available, if that's needed.
 A: Fit a mixture distribution to the data, something like a mixture of 3 normal distributions, then compare the likelihood of that fit to a fit of a single normal distribution (using likelihood ratio test, or AIC/BIC).  The flexmix package for R may be of help. 
A: If you want to use K-means clustering, then you need a way to compare the $K=1$ and $K=3$ cases. One approach would be to use the gap statistic from Tibshirani et al. and choose the $K$ that provides the better value. There's an R implementation available in SLmisc, though that particular function will try $K=1,2,3$, so you will need to take care to ensure that only $K=1$ or $K=3$ can be returned as the optimal value.
A: Use a K-means clustering algorithm to identify the various means
Look for function KNN in R-seek to find the appropriate function
A: I advise strongly against using k-means here. The results for different values of k aren't very well comparable. The method is just a crude heuristic. If you really want to use clustering, use EM clustering, since your data seems to contain normal distributions. And validate your results!
Instead, the obvious approach is to try fitting a single Gaussian function and (for example using the Levenberg-Marquard method) fit three Gaussian functions, maybe constrained to the same height (to avoid degeneration).
Then test, which of the two distributions fits better.
