0
$\begingroup$

I was reading an article the other day about how 90% of people think of themselves as "above average" in terms of intelligence. Assuming that by average they are referring to arithmetic mean, what would be the maximum possible percentage of a sample or population that could actually have some measure of intelligence, say IQ, that was above the mean of the group? Is there some kind of mathematical proof of this?

$\endgroup$

1 Answer 1

1
$\begingroup$

In a sample there must be at least one record with a value less than the mean of the sample (unless all records are equal). Hence the greatest proportion of records greater than the mean is 1-1/n.

$\endgroup$
1
  • $\begingroup$ +1 For completeness, you ought also to demonstrate this maximum is attainable. But your argument makes that obvious. For instance, a sample with values $(m-(n-1)\epsilon, m+\epsilon,\ldots, m+\epsilon)$ with any $\epsilon\gt 0$ attains the maximum proportion of $(n-1)/n$. $\endgroup$
    – whuber
    Sep 17, 2017 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.