# How to average numbers with high standard deviation

I have a set of n numbers that I need to review and come up with the closet average. The set of numbers may or may not have a high standard deviation. Below is an example...

Set of numbers..

• $0.6618 •$0.6509

• $0.6835 •$0.9561

• $15.4250 (should not be averaged, out of bounds) •$15.4400 (should not be averaged, out of bounds)

• $4.7500 (should not be averaged, out of bounds) •$0.5948

• $0.6485 •$0.6856

A simple average of these numbers is \$4.0496 however my needs require me to remove the values that are way out of bounds. Ideally my average would be around$0.6973

• – user856
Jan 25, 2013 at 12:50
• I do not think you want the "closet average". But even if I replace this by "closest average", it not clear what you are asking for. Jan 25, 2013 at 12:58

One solution is to use the median instead, which is resistant to outliers. For your data, the median is \$0.6846.

• And the theory behind it is assuming the error distribution to be normal, which is symmetric.
– mez
Jan 25, 2013 at 12:50
• @mezhang: I did not know that. Do you have a reference? It's usually the arithmetic mean that turns up when you assume normally distributed errors.
– user856
Jan 25, 2013 at 12:53
• Normal distribution is a common model for the error of physical measurements. en.wikipedia.org/wiki/Normal_distribution
– mez
Jan 25, 2013 at 12:56
• @mezhang: That's not an explanation for "the theory behind [the median] is assuming the error distribution to be normal". If you assume the error distribution to be normal, you get the mean, not the median, as the maximum-likelihood estimate. I don't know of a theory that starts from normally distributed errors and arrives at the median, which is what your original comment seems to be implying.
– user856
Jan 25, 2013 at 13:05
• try to search central limit theorem for median. I have not found good reference yet, but this book only mentioned that it is more complicated than central limit theorem on the mean. here
– mez
Jan 25, 2013 at 13:19