Well, we also know that as the sample size is increased (this is where that theorem I mentioned comes in), the mean becomes more and more likely, and the variance ( = s.d.²) is reduced.
It turns out for a random variable, the s.d. is reduced by 1/√n. when the mean is normalized.
Ignoring the math, that means that the expected sample mean gets closer to the population mean as you include more and more of the population into the sample. (It also means that the size of the sample increases at the square of the improvement of the s.d. → diminishing returns).
So if I buy 3 dozen eggs, the expected value is 3 x mean or 2.4 eggs, this does not change. On the other hand, the s.d. does change. Now its 0.5/√3 or about 0.29. An IMPORTANT note: the s.d. I am referring to is NOT the population s.d. (which does not change), it is the sample s.d..
I made the following assumptions: continuous random variable (not correct, but that is your problem, its an approximation that I am comfortable with), no correlation between location of broken eggs (also not correct, chances are the carton that is crushed will have more than one broken egg, same thing with driving a truck into the boxes or dropping the pallet they ar on) - the cartons are independent.
I leave it to you to switch this discussion of a continuous variable over to the discrete case.