Finite-population radioactive decay ($N(t)\sim \mathrm{Binomial}(N_0,e^{-\lambda t})$)

Take $N_0$ independently decaying atomic nuclei, each with lifetime $T_i$ drawn from an exponential distribution of rate $\lambda$. With no replenishment, the survivor count isEach indicator variable has success probability $p(t)$, so the binomial distribution givesThe smooth population law is the expected value $\overline{N}(t)=\mathbb{E}[N(t)]$, satisfying ${\overline{N}}^{\prime }=-\lambda \overline{N}$. One realization instead loses whole atomic nuclei at discrete times. An observed count above the smooth curve need not contradict the mathematical model.

For an original chosen example, let $N_0=1000$ and half-life $3\mathrm{h}$. Then $\lambda =\log 2/(3\mathrm{h})$. After $9\mathrm{h}$, the survival probability is $1/8$, givingThe standard deviation describes variation between repeated populations governed by these assumptions. It is not an uncertainty interval for a noiseless deterministic curve.

The half-life and the mean lifetime answer different questions:The first is the median of one lifetime, or the time when half the population survives on average. The second is the lifetime's expected value. In this example they are $3\mathrm{h}$ and approximately $4.33\mathrm{h}$.