 # Why is an average of an average usually incorrect? explain why taking Karen Simpson 2021-12-14 Answered
Why is an average of an average usually incorrect? 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, lets
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If there are ${n}_{1},{n}_{2},\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{3}$ 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:"
$\frac{{n}_{1}}{{n}_{1}+{n}_{2}+{n}_{3}}{a}_{1}+\frac{{n}_{2}}{{n}_{1}+{n}_{2}+{n}_{3}}{a}_{2}+\frac{{n}_{3}}{{n}_{1}+{n}_{2}+{n}_{3}}{a}_{3}$
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 couldnt
###### Not exactly what you’re looking for? Kayla Kline
The average of averages is only equal to the average of all values in two cases:
if the number of elements of all groups is the same; orthe trivial case when all the group averages are zero
Heres