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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

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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. –  Chris Godsil Jan 25 '13 at 12:58

1 Answer 1

up vote 2 down vote accepted

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

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And the theory behind it is assuming the error distribution to be normal, which is symmetric. –  mez Jan 25 '13 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. –  Rahul Jan 25 '13 at 12:53
    
Normal distribution is a common model for the error of physical measurements. en.wikipedia.org/wiki/Normal_distribution –  mez Jan 25 '13 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. –  Rahul Jan 25 '13 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 '13 at 13:19

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