Commenter pumpkinhead asks:
What would be the general formula for working the composite IQ from 3 or more sub-tests?
Let’s say your verbal IQ is in the top 1% (135), your spatial IQ is in the top 1% (135) and your working memory IQ is in the top 1% (135). One might think your composite IQ is also in the top 1% (IQ 135) since that’s the average of all three, however this ignores the fact that it’s much harder to score 135 in three completely different tests than it is to score 135 on just one, so someone who is in the top 1% in all three tests is perhaps in the top 0.3% overall (IQ 141).
Because IQ is simply the number of standard deviations you score above the mean, multiplied by 15 and added to 100, in order to calculate your composite score on multiple tests, you need to know the mean and standard deviation of the composite score. Knowing the mean is fairly straightforward. Since the mean IQ by definition is 100, the mean of a composite of three tests is just triple the mean of each sub-test, which would be 300.
The standard deviation of the composite is trickier, because unless the correlation between subtests is perfect, you can not simply triple it.
Instead you use the following formula:
Composite σ = SQUARE ROOT OF: σ12 + σ22 + σ32 + 2r12σ1σ2 + 2r13σ1σ3 + 2r23σ2σ3
σ refers to standard deviation, r refers to the correlation (between any pair of the three subtests in its subscript) and subscripts refer to tests 1, 2 and 3 respectively, so for example r12 refers to the correlation between subtests 1 and 2 and σ3 refersto the standard deviation of subtest 3. This formula can be extrapolated to any number of tests.
So applying this formula to verbal + spatial + working memory IQ and assuming a correlation of 0.67 between verbal and spatial , 0.57 between verbal and working memory, and 0.5 between spatial and working memory:
Composite σ = SQUARE ROOT OF: 152 + 152 + 152 + 2(0.67)(15)(15) + 2(0.57)(15)(15) + 2(0.5)(15)(15)
Composite σ = SQUARE ROOT OF: 225 + 225 + 225 + 301.5 + 256.5 + 225
Composite σ = SQUARE ROOT OF: 1458
Composite σ = 38
So someone who score IQ 135 on all three subtests has a total score of 405 (135 * 3 = 405) which is 105 points higher than the average person who has a total score of 300 (IQ 100 * 3 = 300) and with a composite SD of 38, they are 2.76 standard deviations above average (105/38 = 2.76), which equates to an IQ of 141 (because 2.76 * 15 = 41 and 41 + 100 = 141)
[update Janary 2. 2019: this article was revised to remove a needless concern I had about the formulas]
Pumpkin Person 
verbal, spatial, working memory
Composite σ = SQUARE ROOT OF: 152 + 152 + 152 + 2.13(0.67)(15)(15) + 1.4(0.57)(15)(15) + -0.33(0.5)(15)(15)
Composite σ = SQUARE ROOT OF: 225 + 225 + 225 + 321.0975 + 179.55 – 37.125
Composite σ = SQUARE ROOT OF: 1138.5225
Composite σ = 33.74
First problem is if correlation is non linear with g wich would explain that it’s easier to have to 155 than one 165 (maybe 10 times easier if the formula has 0 predictive value at this level of G).
On the other hand, for very low correlation, it may be good, because high g people may invest a lot of it in one specific area, so you don’t want to deflate the scores too much.
The Pb is that it’s useless for very high g. For Mensa, it could help them ask three scores instead of one without pushing the level at 136 instead of 132.
Currently, as they accept any one score above 132, they probably admit some 120 people, with external tests, and 125 with their own test.
Fascinating
Interesting.
Perhaps this would also be a way to combine the results of the two Cattell IQ tests that Mensa sometimes uses (e.g. in the UK), namely Cattell III B and Cattell III A (the first being verbal in nature, the second being visual). The correlation between the two has not been published to my knowledge, but an assumed correlation between 0.6 and 0.8 doesn’t seem unreasonable.
A possible difficulty I would raise concerns the implicit assumption of normality of the combined score. It was my understanding that in general the sum of normally distributed variables need not be normal unless the distributions are statistically independent, whereas the above argument stipulates positive correlations.
Regards