I think that you are probably okay with the second approach. However, it may be necessary to scale the random numbers in which case I think you are alright if the transformation is linear: a*(random_number)+b from the theorem that if X is N(mu,sigma) then a*X+b is N(a*mu+b, a^2*sigma).

Consider the following situation. Suppose you go to the social security administration and get 10,000 random social security numbers. Then you divide these 10,000 social security numbers into 10 different sets of 1,000 numbers each. For each social security number you find the person and measure their height (heights are normally distributed within the American population(of course this could be time consuming to do in reality

)). So the first set has records 1-1,000 the second set has records 1,001-2,000, etc. Now let's say you do a histogram of the first group (1-1,000). This will be a nice bell shaped histogram. The second group (1,001-2,000) will also be a nice bell shaped histogram.

If you do the entire group (1-10,000) you will also get the bell shaped curve.

Certainly you would also get the same result if instead of taking the first 1-1,000 you considered records (1-9,999) and took every record for which the last digit is a 1 (this will be one thousand records), and made a histogram of it. Or if you took every record for which the last digit is a 2 and made a histogram.

So all of these subsets are still numbers from a normal distribution.

You will also get the bell shaped curve if instead of plotting the heights you plotted some linear combination (1/2)*height + 10. This comes from our theorem that if X is N(mu,sigma) then a*X+b is N(a*mu+b, a^2*sigma).

But if the scaling is not linear than you may not have a normal distribution anymore. Certainly it is not the case that X is N(mu,sigma) then any function f(X) is also normal distributed. For example the function f(x) may be f(x) = 0 * x which is definitley not normally distributed. One could also imagine a function which is stepwise which is not normally distributed.

I think also that if f(x) = x^2 then f(X) is not normally distributed.

David