IQ scores are supposedly put on a normal distribution with a standard deviation of 15.

The population of the world is about 7,674,000,000.

So, if I know my statistics and made no numerical errors, the smartest person in the world (even given the unrealistic assumption that we could measure everybody's intelligence), would only have an IQ of 194.8083.

screenshot from Wolfram Alpha

Double the population of the world and the max measurable IQ only goes to 196.4.

Given this, why do I see occasional reports of measured IQs above 200? A 200 IQ, statistically, would require a population of at least 76 billion people, ten times more than there are humans now. A 220 IQ would require 10,000 times as many people as that.

Does the math and statistics just get ignored for those high scores, or is the scale different somehow?


2 Answers 2


Most "professional" IQ tests are capped at around 160 points (Mensa: 162, Stanford Binet: 160, WAIS-IV: 160), so scores above 200 are not relevant in analysis using data from those.

There's also a logical flaw in your claim that an IQ above 195 should not be measurable "statistically": even if it's extremely unlikely, it's possible by the definition of a continuous normal distribution.

Score outliers above 195 are typically measured by non-standard IQ tests or subjectively evaluated. They should not be considered as data points in the context of statistical analysis.


Your math appears to be correct. If one understands the relation between reliability and validity, one understands that an IQ upwards of 160 would be practically impossible to measure. This is because validity is limited by reliability. We cannot create a measure of IQ that could identify someone over 160 because we can't locate enough people to reliably create such a test. We cannot obtain a normative sample of a large enough size to get a handful of people with an IQ of 175, 5 SDs above the mean. Just remember, a measure can't be valid if it isn't reliable.

  • 1
    $\begingroup$ Please add sources to your claims. $\endgroup$
    – AliceD
    Commented Dec 5, 2022 at 8:25

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