In which normal distribution you are more likely to find anomalies? (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?

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 $\sigma_1=\sigma_2=20$.
$$n_1>100 \frac{\left(1-\Phi\left(\frac{185-120}{20}\right)\right)}{\left(1-\Phi\left(\frac{185-100}{20}\right)\right)}\approx 5399$$
