Why the average of averages can be wrong—and when it works

A mean is usually written as one number, but it comes from a total divided by a count. If a group of $n$ values has mean $m$, then the values in that group add up to $nm$. Knowing the mean without the count loses part of the information needed to combine groups.

That is why two displayed averages should not automatically receive equal influence. An average based on eight observations represents four times as much data as an average based on two observations. Combining them fairly means reconstructing the group totals first.

Suppose group A contains $2$ values with mean $90$, while group B contains $8$ values with mean $70$. The tempting shortcut is $(90+70)/2=80.$ But this treats the two groups as if they contained the same number of observations.

Recover the totals instead. Group A contributes $2\times90=180$ and group B contributes $8\times70=560$. Across all $10$ observations, the total is $740$, so the combined mean is $\frac{180+560}{2+8}=\frac{740}{10}=74.$ This result is correctly closer to $70$, because eight of the ten observations belong to group B.

For groups with sizes $n_1,n_2,\ldots,n_k$ and means $m_1,m_2,\ldots,m_k$, the combined mean is $\bar{x}=\frac{n_1m_1+n_2m_2+\cdots+n_km_k}{n_1+n_2+\cdots+n_k}.$ Each product $n_im_i$ reconstructs a group total. The numerator then adds all values, while the denominator counts all observations.

The weights can also be written as proportions. In the example, group A has weight $2/10$ and group B has weight $8/10$, giving $(2/10)90+(8/10)70=18+56=74.$ Both forms express the same structure; choose the one that makes the data easiest to check.

If two groups have the same size $n$, the combined mean becomes $\frac{nm_1+nm_2}{2n}=\frac{m_1+m_2}{2},$ so the shortcut is valid. It can also happen to give the right result when unequal groups have identical means, because weighting identical numbers changes nothing.

More generally, equal weighting is safe only when each displayed average represents an equal share of the underlying observations, or when additional information proves that the unequal weights do not affect the result. “There are two averages” is not enough evidence for weights of one half each.

A combined mean must lie between the smallest and largest group means, provided every group size is positive. It should also move toward the mean of the larger group. In the example, the answer must lie between $70$ and $90$ and should be much closer to $70$ because group B contains $80\%$ of the observations. The value $74$ passes both checks.

These checks do not replace the calculation, but they expose many mistakes quickly. An answer above $90$, below $70$, or closer to the two-value group would contradict the structure before any detailed recalculation is needed.

The deeper lesson is not a new formula to memorise. It is a question to ask whenever a summary statistic appears: what count, total, or population does this number represent? The same habit matters when comparing class results, experiment batches, response rates, prices, and other aggregated data.

A useful tutor should therefore ask for the group sizes before combining averages and should make the missing assumption visible. If you want to practise the reasoning with your own numbers, open Euler’s tutor; the worked guides in Euler Learn use the same pattern of derivation followed by an independent check.