Let ${X_i}$ be independent, identically distributed, random variables each with mean $M$ and variance $\sigma^2$. Let $Y(z)$ be the number of these random variables we need to add together to exceed z, that is, the smallest integer such that
$X_1 + X_2 + ... + X_Y > z.$
It seems clear that as $z$ approaches infinity, $Y(z))$ should be about $z/M$, so that $E(Y(z))M/z$ should approach $1$. But what is the variance of $Y(z)$ for large $z$? Intuitively, when we add about $z/M$ copies of $X_i$ together, the variance of the resulting sum is about $\sigma^2 z/M$. So typical values will differ from $z$ by about $\sigma \sqrt{z/M}$, which we can fix by adding or subtracting about $\sigma\sqrt{z/M}/M$ copies of $X_i$. So I would conjecture that if we denote the variance of $Y(z)$ by $V(z)$, then
$$lim_{z \to \infty} V(z) M^3/(\sigma^2 z)$$
should exist and be positive. Is this true? What is the limit?
joriki's comments are correct. $Y(z)/z$ converges in distribution to a normal distribution with mean $M$ and variance $\sigma^2/M^3$. See (http://www.columbia.edu/~ks20/stochastic-I/stochastic-I-RT-II.pdf) for a proof. Thanks to did for giving me the term "renewal process" which is what enabled me to google for this result.