Solved – If the mean and median underestimate the true central tendency, why use them

descriptive statisticsmeanmedianmode

In skewed distributions, both the mean and the median can easily underestimate or overestimate the true central tendency. For example, have a look at this violin plot:

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The median is shown here in red. It seems to overestimate the true central tendency. This is the same for the mean. The best central tendency description would be the mode of that distribution (which would get the peak).

This is a basic question: but why use mean and median for skewed distributions? Do people even use them? Wouldn't it make sense to get the maximum point of the Kernel density estimate?

Best Answer

In skewed distributions, both the mean and the median can easily underestimate or overestimate the true central tendency. [...] Wouldn't it make sense to get the maximum point of the Kernel density estimate?

No it wouldn't, at least not always.

Take as an example the exponential distribution: it is parametrized by $\lambda$, its expected value is $\lambda^{-1}$, the same as its mean, its median is $\lambda^{-1} \ln(2)$ and its mode is always $0$, regardless of parametrization. So all the values from this distribution are greater than or equal to the mode -- what does that tell us? Not much...

It is the opposite with the mean which marks the probability mass center; the median also provides us with similar information. Yes, $0$ is the most likely value, but we are more interested here in the tail of the distribution (since it's a tail-only distribution).

Finally, the mode would be a useless summary statistic if you would like to compare different exponential variables.

Check out also: If mean is so sensitive, why use it in the first place?

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