Commenter Mugabe (aka chocolate babies?) has claimed to have scored a perfect 800 on all three sections of the old GRE.  Some commenters, such as “Deal with it!” believe him, others are deeply skeptical.

In one study, the quantitative GRE correlated 0.71 with WAIS-R full-scale IQ, though the sample was small and restricted. So the question becomes, if the GRE is considered equivalent to an IQ test, what IQ did a perfect 2400 GRE equate to?

For decades people have been given GRE subscale scores (verbal, math, analytical) and the percentiles that go with them, but had no idea how to interpret their COMBINED GRE scores.  In order to interpret any standardized test score, the first thing you want to know is the Z score, and in order to know the Z score, you must know the mean and the Standard Deviation (SD).  According to one source, the means and SDs (in parentheses)  for 1.1 million people taking the GRE from 1994 to 1997 were 474 (114) for the verbal, 558 (139) for the Quantitative, and 547 (130) for the analytical.

However knowing your Z score tells you little, if you don’t know anything about the population from which your Z score was derived.  A low Z score from brilliant population could be more impressive than a high Z score from a low IQ population.  Thus instead of using the data on 1.1 million GRE takers of unknown intelligence, I am going to use GRE data from a smaller sample of GRE takers with known SAT scores: The above table was poorly formatted, but what it shows is that in a sample of people who took both the GRE and the SAT (circa 1990), the mean GRE verbal was 510.1 (SD = 107.7), GRE Quantitative was 573.4 (SD = 125.6) and the GRE analytical was 579.7 (SD = 117.6).  Moreover, in this sample, the correlation between V & Q scores was 0.56, V & A scores 0.65, and Q & A scores 0.73.

Armed with these means, SDs, and correlations, we can estimate the mean and SD of the COMBINED GRE score (V + Q + A)

According to commenter Mugabe:

var(X+Y+Z) = var(X) + var(Y) + var(Z)+2*(sigma(X)sigma(Y)rho(X,Y) + sigma(X)sigma(Z)rho(X,Z) + sigma(Y)sigma(Z)rho(Y,Z))

Mugabe advises us not to use this formula because the math subscale has a low ceiling, but let’s be rebellious and use it anyway, substituting V, Q, and A for generic X, Y, and Z(note variance equals squared standard deviation):

var(V+Q+A) = var(V) + var(Q) + var(A)+2*(sigma(V)sigma(Q)rho(V,Q) + sigma(V)sigma(A)rho(V,A) + sigma(Q)sigma(A)rho(Q,A))

var(V+Q+A) = 11,599.29 + 15,775.36 + 13,829.76 + 2*(107.7)(125.6)(0.56) + (107.7)(117.6)(0.65) + (125.6)(117.6)(0.73)

var(V+Q+A) = 41,204.41 + 15,150.37 + 8,232.59 + 10,782.51

var(V+Q+A) = 75,369.88

Taking the square root of the variance, tells us that the SD of the combined composite (in our sample) should be 274.53, and of course the mean of the composite is simply a matter summing the sample’s mean on each of the three sub-scales (1,663.2).

Thus Mugabe claims to have scored 2.68 SD above our sample GRE takers.

But in order to assign that an IQ, we need to know how talented the sample was.  Their pre-1995 verbal SATs have a mean of 518 and an SD of 104.7 (which on the IQ scale equates to a mean of 120 (SD 14.14)(U.S. norms)).  If we assume they were as self-selected for combined GRE talent as they were SAT verbal talent, then being 2.68 SD above them on the combined GRE equates to an IQ of:

2.68(14.14) + 120 = 158 (U.S. norms) or 157 (U.S. white norms)

However because of ceiling bumping, this is a very conservative estimate, but the g loading of the GRE, especially at high levels, has not been well researched, and some researchers question the validity of g as a measure of intelligence (but that’s a whole other post).