The Statement of the Problem
A certain scientific theory supposes that mistakes in cell division occur according to a Poisson process with rate $2.5$ per year, and that an individual dies when $196$ such mistakes have occurred. Assuming this theory, find the probability that an individual dies before age $67.2$.
What I've Figured Out
Ok, so, an individual's lifetime is given by the random variable $S_{196}$, i.e. the waiting time until the $196$th cell division. It follows that this random variable has a gamma distribution with parameters $196$ and $2.5$. So, all I have to do is calculate the cdf of this gamma distribution below $67.2$, i.e.
$$ \mathbb P(S_{196} < 67.2) $$
where
$$S_{196} \sim \Gamma(196, 2.5).$$
Naturally, I'm using software for this computation, and I'm getting some strange numbers back. I believe the answer should be $0.02275$, which makes sense, but I'm not getting very close to that. I can't figure out what I'm doing wrong here, so if anyone can help, it'd be appreciated. Thanks.
Best Answer
A little crude, perhaps, but let's use the normal approximation. Our "life" random variable $W$ is the sum of $196$ exponentially distributed random variables, each with mean $\frac{1}{2.5}$ and variance $\frac{1}{(2.5)^2}$. We are invited to assume these are independent.
Thus $W$ has mean $\frac{196}{2.5}$, that is, $78.4$.
The standard deviation of $W$ is $\sqrt{196/(2.5)^2}$, which is $5.6$.
Note that $67.2$ is exactly $2$ standard deviation units below the mean. For a normal, the probability of being $2$ or more standard deviation units below the mean is $\approx 0.02275$.