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?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @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
  • 2
    $\begingroup$ 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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.