Consider the following data points: $1, 1, 2, 3, 4$

I understand that... the average is the total of the numbers divided by the count of numbers in the set, the median is the central value based on location in the set, the mode is the value occurring most often in the set, and the midrange is highest and lowest values divided by $2$.

Is there a name for the calculation of the average based on the unique values found in the set? So... $\frac{1 + 2 + 3 + 4}{4} = 2.5$? And is there a use for it?

Pardon a possibly elementary question, I'm a programmer but I'm not exactly a math-oriented person, and this has been bugging me recently.

  • $\begingroup$ Thanks for taking the time to answer. I selected the answer that I understood and generated some interesting comments. This topic came up when I was basically trying to sort what is a "typical" book for the publishing company I work at: page count, chapter count, art count, budget, and the like. But I also thought of it for a music application I'm working on doing just some odd "what happens if we do this?" kind of math (and I wonder if there is a name for that as well). I just didn't know if there was a name or purpose to something like this type of average. Thanks again for taking the time $\endgroup$ – Philip Regan Sep 1 '10 at 0:04

From a practical viewpoint, I wouldn't think so. For that example $ (1, 1, 2, 3, 4) $ your procedure gives $ 2.5 $. If the data points had instead been $ (1, 1.01, 2, 3, 4) $, however, the result would jump to about $ 2.2 $. The most common useful properties would seem to depend continuously upon the data, however.

Since I don't know the application for which you're using that kind of average, though, it may well turn out to be helpful. In this case, I'd recommend simply spelling it out as you did here.

  • $\begingroup$ indeed such an average should at least be defined for discrete data, I fear. $\endgroup$ – mau Aug 31 '10 at 8:21
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    $\begingroup$ One amusing possibility is that the dataset represents a random sample (with replacement) from an urn known to contain finitely many distinct values. An average like the one proposed here is likely to be a better estimator of the population average than the usual average. (It's definitely a better estimator in some cases where the number of distinct values in the population is known.) $\endgroup$ – whuber Aug 31 '10 at 17:37

You could conceive of your result as being a weighted average. In a batch of data consisting of $n_i \ge 1$ instances of $x_i$, $1 \le i \le n$, your average is

$$ \frac{ \sum_{i=1}^{n} x_i}{n} = \frac{ \sum_{i=1}^{n} \left(\frac {1}{n_i} n_i x_i \right) }{n} = \frac{ \sum_{i=1}^{n} \sum_{j=1}^{n_j} \left( \frac {1}{n_i} x_i \right) }{ \sum_{i=1}^{n} \sum_{j=1}^{n_j} \frac {1}{n_i}}$$

exhibiting the weights as $1/n_i$.


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