Whynot just apply the arithmetic mean if you have no outliers?
Dear all,
I am facing an interesting problem: I have a number of datasets containing observations. Each dataset has observations, whose distribution is skewed. Some datasets are left-skewed, for example:
.1, .1, .6, .76,.8,.98,.99,.99,.99,1
while other are right-skewed:
.01,.01,.05,.07,.08,.2,.2,.4,.4,1
Is there a measure of central tendency that works as follows: if the sample distribution is left-skewed, points at the left side receive more weight than points at the right side in the final "average". On the other hand, when the distribution is right-skewed, points at the right side receive more weight than points at the left side. More weight means "contribute more" in the "overall mean"
I have tested the geometric, harmonic, arithmetic and exponential means, but none of them seem to work well in my case.
Any tips are appreciated!
Alonso
P.S. Median is not appropriate in my case (biological reasons).