> We can further assume, arbitrarily, that P(B) = 0.02 (one in every fifty people make good engineers). So the likelihood that a person is a good engineer given that they are female is P(B|A).
> What we learn from this is that sex is not a good screening criterion for finding good engineers. You realize an advantage of only 0.8%, not 60%.
if you multiply two numbers (here 0.4 and 0.6) by the same constant (here 0.02 / 0.5) you can't then say "You realize an advantage of only 0.8%, not 60%"... 0.016/0.024 == 40/60.
Re: OP's point ("When you pick out an individual there's still a 60% chance than one is better than the other given no other information than their sex"), I agree that's wrong, but not for the reason given. He's not reversing a conditional probability, but rather asking a question about the unconditional probability (the base rate), so the 60/40 conditional information is not relevant.
If you pick two random individuals, one male and one female, and the IQ average of the two groups is equal (roughly true, from the article), the chance that one has a higher IQ than the other is 50%.
julia> sum([randn() > (randn() * 10) for i=1:1_000_000])
499884
IE assuming that men have 10x the IQ variance, the variance makes literally no difference when comparing two random people.
Your post looks like a non-sequitur to me. But for the sake of math...
For an arbitrary variance multiplier of 10 (std. dev. sqrt(10)), there are no non-imaginary solutions for a high-pass threshold where you can achieve a 60:40 split between two populations that have the same median.
You can get about a 58:42 split, putting your threshold a bit below the common median. But to achieve a 60:40 split, you need one population to have 25 times the variance, or for the populations to have different medians. Above that, you have two possible real solutions for the threshold value.
You can try it yourself, and hopefully my math is correct for this one:
60/40 * erfc( x / sqrt(2))/2 = erfc( sqrt( variance_ratio ) * x / sqrt(2))/2
Since the threshold value equates to a different standard deviation for each population, you can say with some certainty that given the knowledge of which population a person is in, there is a different probability that person is above the threshold value, which works out to be 60% and 40%, respectively. But that's just begging the question, since those are the same numbers you used to work out the threshold value, and you still had to assume values for the medians and variances of both populations.
I'm super confused by this reply. The article says that the stddev of the distribution of male IQ is larger than the stddev of the distribution for female IQ, and also seems to state the mean and median are equal. GGP threw out 60/40 as their believed split of men to women being good engineers. I took this to mean that GGP was saying he believed engineering talent to be independent of gender when IQ was present, and that engineering required a higher than average IQ. More formally:
In that model the 60/40 split comes from a conditional distribution of gender given an IQ above some threshold. So something like:
julia> women = sum([randn() > 1 for i=1:10_000_000])
julia> men = sum([(randn() * 2) > 1 for i=1:10_000_000])
julia> women / (women + men)
0.3395449535000696
Which isn't exactly 60/40, but is fairly close. It says that [iff the 60/40 split is true], the stddev difference needed is less than 2x.
> We can further assume, arbitrarily, that P(B) = 0.02 (one in every fifty people make good engineers). So the likelihood that a person is a good engineer given that they are female is P(B|A).
> What we learn from this is that sex is not a good screening criterion for finding good engineers. You realize an advantage of only 0.8%, not 60%.if you multiply two numbers (here 0.4 and 0.6) by the same constant (here 0.02 / 0.5) you can't then say "You realize an advantage of only 0.8%, not 60%"... 0.016/0.024 == 40/60.
Re: OP's point ("When you pick out an individual there's still a 60% chance than one is better than the other given no other information than their sex"), I agree that's wrong, but not for the reason given. He's not reversing a conditional probability, but rather asking a question about the unconditional probability (the base rate), so the 60/40 conditional information is not relevant.
If you pick two random individuals, one male and one female, and the IQ average of the two groups is equal (roughly true, from the article), the chance that one has a higher IQ than the other is 50%.
IE assuming that men have 10x the IQ variance, the variance makes literally no difference when comparing two random people.