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I know that the IQ statistic is designed to give it a mean of 100 and that you'll certainly never find someone with an IQ below 1 or above 300, but that tells us very little about the variance or general shape of the distribution. So why is it that every graph of IQ scores that I've seen appears to be a truncated normal distribution? Is it some property of the test design, some property of the test subjects, or some deep theorem in statistics that I've overlooked?

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  • $\begingroup$ Can you provide examples? $\endgroup$ – user17122 Mar 28 at 22:27
  • $\begingroup$ @baca Every Google Images result for "IQ graph" that I've ever seen. $\endgroup$ – J. Mini Mar 29 at 1:26
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    $\begingroup$ I don't understand. The first pictures for me when I go to google "IQ curves" are these ones: iqtestforfree.net/images/iq_bell_curve.gif iqcomparisonsite.com/Images/NormalCurveSmall.gif external-preview.redd.it/… Those are perfectly normal curves (pun intended). $\endgroup$ – user17122 Mar 29 at 4:12
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    $\begingroup$ @baca They're obviously not a normal normal distribution. Nobody has a negative IQ score and nobody has an IQ over 10,000. My question is why they're normal distributions in the first place. Why they're truncated is obvious. $\endgroup$ – J. Mini Mar 29 at 14:20
  • $\begingroup$ Although not a duplicate, the answer to this question probably also answers yours. $\endgroup$ – Arnon Weinberg Mar 30 at 16:49
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IQ isn't normal, it's normalized to have mean 100 and standard deviation 15, usually via a percentile method.

The reason IQ looks roughly normal is because intelligence (however it is defined) is a complex trait. Complex traits are predicted to have a roughly normal distribution based on the central limit theorem: a sum of many individual factors (including genetic and environmental ones) will tend to be distributed normally in a population, even if the underlying factors themselves are not normal.

There is no real concrete measure "IQ": it isn't measuring a real-world physical property the way you measure mass or length. Instead, you use tests intended to get some measure of that abstract trait, and then normalize individuals based on the group statistics. Actual tests administered to measure IQ will have a minimum and maximum score: you can at worst get every question wrong, at best get every question right.

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  • $\begingroup$ "Complex traits are predicted to have a roughly normal distribution based on the central limit theorem" - which version? The classical one that I know and love seems insufficient. Have I missed a lemma? $\endgroup$ – J. Mini Mar 30 at 17:05
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    $\begingroup$ I'm unconvinced. If the raw test score were the average of some independent factors, then we might have a shot at using the CLT, but sums don't behave as nicely. You could normalise, but I doubt that we know the population parameters. $\endgroup$ – J. Mini Mar 30 at 17:27
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    $\begingroup$ So we assume that the raw test score is a sum of independent random variables that we do not know the distributions of. But as it's a sum of independent random variables, it must have some normal distribution with unknown parameters. We then fiddle around with that a bit to get the nice IQ graphs that we all know and love. Is that right? $\endgroup$ – J. Mini Mar 30 at 17:51
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    $\begingroup$ @J.Mini Not the raw test score, the underlying thing that is being measured. But yes, IQ distributions get fiddled with: that's the normalization procedure. When you design an IQ test, you don't yet know how the scores on the test will map to IQ scores. You design the test, give it to lots of people, and use the distribution you observe to decide what score is "100" (the median or mean score) and what score is "85" (one standard deviation below the mean, or an equivalent percentile). $\endgroup$ – Bryan Krause Mar 30 at 18:00
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    $\begingroup$ I'd also say that it isn't that it must be this way. If there was some single dominant factor in IQ, the distribution might not be normal at all: it could be bimodal, for example. But we do observe that it is approximately normal, and we should not be surprised to find this. $\endgroup$ – Bryan Krause Mar 30 at 18:02

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