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(Sorry if talk in layman terms)

Consider this you have 2 groups of populations. Let's assume we measure intelligence here. Let's assume the distributions are perfectly normal.

One group is centered around 100 iq. Second group is centered around 120 iq.

Now suppose first group is way larger than second group. To a point where 120 iq guys are not less common (in absolute numbers) in first group than they are common in second group.

Now the question is, in which group you are more likely to find anomalies like say, 185 iqs?

My question is basically what's more significant for anomalies to occur quality or quantity?. Is quality group more likely to have anomalies than a quantity group? And at what point quantity group beats the quality group? Is it enough that in absolute terms quantity group has more of the centered items than does the quality group?

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    $\begingroup$ If the variances of the two normal distributions are the same, then you are more likely to find your "anomalies" in the second group. This is because a normal distribution has a lighter tail than an exponential distribution. $\endgroup$
    – Henry
    Commented Oct 15, 2022 at 10:13
  • $\begingroup$ Thx. But when should the quantity beat it? It cant be that the second group always beats the first. Since I am talking about absolute numbers of 185 iq anomalies. At some point the quantity of first group should be so large that you can have 100X times the numbers of 120 iqs that exist in the second group. Surely at some point it should beat it. When is it that point? $\endgroup$
    – bilanush
    Commented Oct 15, 2022 at 10:45
  • $\begingroup$ It depends on the variances or standard deviations. It also depends on how to read "To a point where 120 iq guys are not less common (in absolute numbers) in first group than they are common in second group": if that means $n_1\phi\left(\frac{120-100}{\sigma_1}\right) = n_2\phi\left(\frac{120-120}{\sigma_2}\right)$ and $\sigma_1=\sigma_2=20$ then $\frac{n_1}{n_2} \approx 1.65$ rather than @JimB's $\approx 53.99$. My point was that the ratio of the normal densities decreases to the right (while it would be constant with an exponential distribution) $\endgroup$
    – Henry
    Commented Oct 16, 2022 at 0:06

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The following assumes that there are two infinite populations and one takes simple random samples of size $n_1$ and $n_2$, respectively, from those populations. (If there the two populations are just finite populations, then the assumption of a normal distribution doesn't apply. If so, the question should be made with more specifics.)

If "when should the quantity beat it" means when the expected number of folks over 185 IQ points in the first group exceeds the expected number of folks with over 185 IQ points in the second group, then the associated equation for that is the following:

$$n_1\left(1-\Phi\left(\frac{185-100}{\sigma_1}\right)\right)> n_2\left(1-\Phi\left(\frac{185-120}{\sigma_2}\right)\right)$$

where $\Phi(.)$ is the unit normal cumulative distribution function. I suppose you might phrase the question as "what is the probability that a sample of $n_1$ from group 1 will have more folks with higher than 185 IQ points than a sample of $n_2$ from group 2?". But that is a different question from solving the above equation so you need to be more specific.

But suppose the above equation is what you want to solve for $n_2=100$ and $\sigma_1=\sigma_2=20$.

$$n_1>100 \frac{\left(1-\Phi\left(\frac{185-120}{20}\right)\right)}{\left(1-\Phi\left(\frac{185-100}{20}\right)\right)}\approx 5399$$

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  • $\begingroup$ In your final result n1 and n2 are the sizes of the groups? So it means group1 should be 53x times tre size of the second group? $\endgroup$
    – bilanush
    Commented Oct 15, 2022 at 18:15
  • $\begingroup$ No. Those are the sizes of samples for the corresponding normal distributions. I'll edit my answer to make that clearer. If the groups consist of a finite number of individuals, then the "normal distributions" are just approximations and a different approach is needed to get an exact answer. I think you are mixing up concepts of finite populations and normal distributions. $\endgroup$
    – JimB
    Commented Oct 15, 2022 at 19:39

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