I am not sure if i should be asking this on the Mathematics of Physics Stack Exchange site.

I am taking data for a physics research application and have a question about using median vs. mean averaging. I am concerned with the amount it takes to acquire each data point (location). For each point I currently take multiple measurements, average them, and use the standard deviation of the mean as my error calculation for the point.

My problem is that it is only reasonable for me to take a relatively small number of measurement(under 10) for each data point. It seems to me; if any of the measurement vary drastically from the actual value, the median of the measurements will converge to a value close to the actual value in fewer data points than the mean. What do you think of this assumption?

If I use the median of my measurements what would be the correct way to calculate the error for each data point?

  • $\begingroup$ The statistics site is likely to be even better (and more responsive) to this question. $\endgroup$
    – cardinal
    Commented Jan 15, 2013 at 2:59
  • 1
    $\begingroup$ Is there a way to move the question without reposting it? $\endgroup$
    – James
    Commented Jan 15, 2013 at 3:11

1 Answer 1


It is true that the median is more robust (subject to outliers) than the mean.

My understanding is that the reason statistics tends to use the mean (and squared errors for that matter) is that in the long run, on average, assuming symmetrical distributions, they get closer to the true answer than medians and absolute deviations.

However, if you are not interested in being correct on average over the long run and are more interested in being close to right on any given point estimate, the median is probably a reasonable statistic for you to select.

It is unclear from your question what exactly you want to do with your "error estimate". Do you want to use it to do a statistical test? If all you need is another summary statistic to describe the central tendency of dispersion around your observed measure of central tendency AND you want to continue to leverage whatever 'advantage' the median is giving you... then I would recommend calculating the median absolute deviation. That is the median of |X-Mdn|.


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