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I read in a publication that the average salary for lawyers in America is $\bar x=\$163,595$. Of these salaries, the average for men is $\bar x_m=\$183,687$, and for women, it is $\bar x_w=\$163,595$. I'm thinking, how is it possible for the average of women's salary be equal the average of the entire set?

Note that $\bar x_w = \bar x < \bar x_m$, and $\bar x_w,\, \bar x,\, \bar x_m > 0$.

Can we prove that this is possible/impossible? Can we find 1 simple example where $\bar x_w = \bar x$ (given $\bar x_w,\, \bar x,\, \bar x_m > 0$, and $\bar x_m > \bar x$) is possible?

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    $\begingroup$ No it's not possible unless there are $0$ men. The overall average will be a positive weighted sum of the two gender averages ( weighted by the portion of population of lawyers in each gender) which will be properly between the two subset averages. That is (overall average) = (men ave.)(men % of pop.) + (woman ave.)(woman % of pop.). I'm assuming these are actual means, not sample means or some other sort of estimate. $\endgroup$ – Ned May 25 '20 at 22:56
  • $\begingroup$ @twosigma This would not work if the remaining elements have an average that is higher. $\endgroup$ – Gilead May 25 '20 at 22:56
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    $\begingroup$ @ThomasWinckelman: the subsets would need to be mutually exclusive. In your example, the first subset is (1,3,5) with a mean of 3, but the second subset is (1,5) also with a mean of 3. In my example, the second (higher) subset needs to be strictly greater than the mean of the entire set, which it's not in this case. $\endgroup$ – Gilead May 25 '20 at 23:02
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    $\begingroup$ @ThomasWinckelman The men and women subsets are disjoint and exhaustive (i.e. a partition) so that sort of example doesn't apply here. $\endgroup$ – Ned May 25 '20 at 23:03
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    $\begingroup$ I was able to find those numbers at datausa.io/profile/soc/… but with the women's average at 132,637. This seems to be from some governmental source but I can't figure out which. I suspect your source comes from the same ultimate source as this one but with an error. $\endgroup$ – Michael Lugo May 26 '20 at 15:23
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This is the calculation behind Ned's comment.

Let $m$ be the number of men, $w$ the number of women and $a_m$, $a_w$ respective averages. Then, the overall average is

$$a = \frac{ma_m + wa_w}{m+w}.$$ If $a = a_w$, from the above we get $ma_m + wa_w = (m+w)a_w$, i.e. $m(a_m-a_w) = 0$. Thus, either $m = 0$ or $a_m = a_w$.

Moreover, if $a_m \geq a_w$, then

$$a_ w = \frac{ma_w + wa_w}{m + w} \leq \frac{ma_m + wa_w}{m + w} \leq \frac{ma_m + wa_m}{m + w} = a_m.$$

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