Can someone explain why taking an average of an average usually results in a wrong answer? Is there ever a case where the average of the average can be used correctly?

As an example, let's say that an assessment is given to three schools and I want to find out the average score for all three schools combined and the average score per school. When I attempt to add the three individual scores and divide by three I get a number that is very close (+/- 1 percent) to the actual overall average.

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    $\begingroup$ It's not wrong, it needs interpretation. If school A has 10000 students, with average marks in chemistry, say 2.56, and school B has one pupil (ok, bit extreme, but why not), with average mark in chemistry 1, then you may either want to know the average marks w.r.t. the number of schools (2.56 + 1)/2 or w.r.t to pupils (10000*2.56 +1)/10001. Both number make sense but have, of course, completely different implications and interpretations. $\endgroup$
    – user20266
    Commented Jan 2, 2012 at 21:20
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    $\begingroup$ Say two of the schools are small and very good, and the third school is huge, and pretty bad. If you average over students, the good results from the small schools hardly make a dent. If you first find the average per school, and average the averages, this will tend to hide the poor results from the huge school. In the real world, disparities of size or performance are often not very dramatic, so the two procedures can end up giving fairly similar numbers. One thing we can be sure of: whoever is trying to make a point will choose the procedure that supports his/her point. $\endgroup$ Commented Jan 2, 2012 at 21:22
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    $\begingroup$ An interesting real-world case. When a university reports the "average class size", the result is quite different if they report the class size for the average student, or the class size for the average instructor. $\endgroup$
    – GEdgar
    Commented Jan 2, 2012 at 21:38
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    $\begingroup$ Here is an off-the-cuff heuristic for you: their units of measurement are not the same. You begin with scores, produce scores/students for each school, then you finally compute (scores/students)/schools. That would be completely different than taking (all scores) / (all students) collectively. $\endgroup$
    – rschwieb
    Commented Jun 7, 2018 at 15:28
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    $\begingroup$ This distortion of averages is a common trick of journalists and politicians to tilt statistical evidence to support their positions on various issues. $\endgroup$ Commented Jun 29, 2018 at 10:38

3 Answers 3


If there are $n_1$, $n_2$, and $n_3$ students in the three schools, and the average test score for each school is $a_1$,$a_2$,$a_3$, respectively, the correct average is a "weighted average:"


The average of the averages is:

$$\frac{1}{3}a_1 + \frac{1}{3}a_2 + \frac{1}{3}a_3$$

These two values will be exactly the same if each school has exactly the same number of students, and will tend to be "close" if the schools are relatively close in size and/or the scores for the three schools are close.

If a school system created a small school consisting of all the smartest students, they could bump up the second value - the "average of averages" - but they couldn't do that if they take the correct weighted average.

  • $\begingroup$ I changed some $n_1$'s to $n_2$ and $n_3$... $\endgroup$ Commented Jan 3, 2012 at 0:52
  • $\begingroup$ I don't know if i have to open new question. How would you weight average potions of impresions & clicks keywords? $\endgroup$
    – jrosell
    Commented Mar 4, 2016 at 7:03
  • $\begingroup$ If s1 is number of students in school 1 and n1 is the score of school 1, then a1 is average for school 1, which is a1 = (n1)/(s1). Similarly, a2= (n2)/(s2) and a3 = (n3)/(s3). I would like to know what does this value represent (n1 + n2 + n3) / (s1 + s2 + s3) Does this value represents the average of all the schools as we are dividing the total score of all schools and dividing it by total number of students? How this is different with the given "weighted average" value? $\endgroup$ Commented Jul 26, 2017 at 7:36
  • $\begingroup$ $n_1$ is not the score for the school, it is the sum of the scores of the students in the school. That is not a "score for the school" in any useful meaning. And it is the weighted average. $\endgroup$ Commented Jul 26, 2017 at 15:49
  • $\begingroup$ @ThomasAndrews: Thanks for pointing out. I meant the same. n1 is the sum of scores for all the students in the school 1. Similarly the case for n2 and n3. If wa = (n1 + n2 + n3)/(s1 + s2 + s3) , does wa represent the weighted average. But substituting the values for a1, a2, and a3 in the formula given in answer we get different value than wa. $\endgroup$ Commented Jul 27, 2017 at 5:47

For example: the average of $ \{ 2,2,2,2,2,2,2,2,2,2,2,2,2\} $ is $2$, ($N=13$) and the average of $\{4\}$ is $4$, ($N=1$). The average of the averages is $3$. But the average of all numbers is $30/14$ ≃ $2.14$.

I hope this is enough to explain what goes wrong (you're giving equal weights to the "first averages" when you take their average, which isn't the correct thing to do if you want the average of all the numbers).


Thomas Andrews already answered the question, but I'd like to present a more analytical solution to the problem.

The average of averages is only equal to the average of all values in two cases:

  1. if the number of elements of all groups is the same; or
  2. the trivial case when all the group averages are zero

Here's why this is so.

Consider two sets $X = \{x_1, x_2, ..., x_n\}$ and $Y = \{y_1, y_2, ..., y_m\}$ and their averages:

$$ \bar{x} = \frac{\sum_{i=1}^{n}{x_i}}{n} \,,\, \bar{y} = \frac{\sum_{i=1}^{m}{y_i}}{m} $$

The average of the averages is:

$$ average(\bar{x}, \bar{y}) = \frac{\frac{\sum_{i=1}^{n}{x_i}}{n} + \frac{\sum_{i=1}^{m}{y_i}}{m}}{2} = \frac{\sum_{i=1}^{n}{x_i}}{2n} + \frac{\sum_{i=1}^{m}{y_i}}{2m} $$

Now consider the whole group $Z = \{x_1, x_2, ..., x_n, y_1, y_2, ..., y_m\}$ and its average:

$$ \bar{z} = \frac{\sum_{i=1}^{n}{x_i} + \sum_{i=1}^{m}{y_i}}{n + m}$$

For the general case, we can see that these averages are different:

$$ \frac{\sum_{i=1}^{n}{x_i}}{2n} + \frac{\sum_{i=1}^{m}{y_i}}{2m} \ne \frac{\sum_{i=1}^{n}{x_i} + \sum_{i=1}^{m}{y_i}}{n + m} $$

This answers the first OP question, as to why the average of averages usually gives the wrong answer.

However, if we make $n = m$, we have:

$$ \frac{\sum_{i=1}^{n}{x_i}}{2n} + \frac{\sum_{i=1}^{m}{y_i}}{2n} = \frac{\sum_{i=1}^{n}{x_i} + \sum_{i=1}^{n}{y_i}}{2n} $$

This is why the average of averages is equal to the average of the whole group when the groups have the same size.

The second case is trivial: $\bar{x} = \bar{y} = average(\bar{x}, \bar{y}) = 0$.

Note that the above reasoning can be extended for any number of groups.

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    $\begingroup$ The simplicity of the two rules make this the most useful answer for me. $\endgroup$
    – Rob
    Commented Apr 14, 2017 at 18:57
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    $\begingroup$ This is a great answer and both the summary of the two cases and the proof are elegant. I am not sure that it answers why the average of averages usually gives the wrong answer, it only shows how the two calculations are different. The application and reasoning is more interesting. What is it exactly that the calculation is measuring, how will it influence behaviour? If it is the average of all students, then why is that the choice? Is it to compare all students with each other, and is that fair? If it is the average between schools then why? What are the dangers of misinterpretation? $\endgroup$
    – Davos
    Commented Sep 21, 2018 at 12:30
  • $\begingroup$ Here is a group of groups of values where the group average is not 0 nor are the group sizes the same, yet the overall average is the same as the average of averges (i.e., a counter-example to your assertion of only two cases): [[2,2,2,2,2,2,2], [-37, 41], [1, 1, 4], [2]] $\endgroup$ Commented Aug 4, 2023 at 0:44
  • $\begingroup$ I guess. The 3rd case, as you described, could be when the averages are the same by any other reason (by construction or by chance maybe). Well spotted! Thanks. $\endgroup$
    – Ricardo
    Commented Aug 23, 2023 at 20:51

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