Summation of binomial/poisson question A company has 800 customers, each with a probability 0.02 of dying during the next year (independent of all other policy holders). Find the probability that between 8 and 16 customers will die within the next year.
One could look at this as a Poisson problem and work out all of the individual probabilities but there clearly is a more efficient method. The binomial and poisson tables also dont help for this.
So it seems that some technique such as mgf or transformation is needed.
Does anyone have a good alternative approach to tackling this?
 A: First, @Andre Nicholas is absolutely correct that modern software allows an exact solution with very little trouble. This question allows a simple
comparison of results obtained by various approximation methods commonly suggested in textbooks. Second, because you do not say what you have tried, I'm answering a slightly different question, that uses the same methods required for the problem stated.
You have a binomial random variable $X$ with distribution $Bin(n=800,p=.02)$
I'm solving the problem for 'between 8 and 12 inclusive'. Then we seek 
$$P\{8 \le X \le 12\} = P\{X \le 12\} - P\{X \le 7\}.$$
Exact. Using R statistical software, the statement 'pbinom(12, 800, .02) - pbinom(7, 800, .02)' returns the answer  0.1809859, accurate to as many decimal places as given. Another method is to use the statement 'sum(dbinom(8:12, 800, .02))', which returns exactly the same result. The latter statement adds five probabilities together:
$$P\{8 \le X \le 12\} = P\{X=8\} + P\{X=9\} +P\{X=10\} +P\{X=11\} +P\{X=12\}.$$
Exact computation by hand would indeed be tedious; for example 
$$P\{X = 8\} = C(800, 8)(.02)^6(.98)^{792},$$
according to the formula for the binomial point mass function, where the binomial coefficient $C(~,~)$ is equal to
$C(800, 8) = 800!/[8! (792!)].$ And that's just for the first of five terms.
Poisson approximation. In some circumstances, the Poisson distribution provides a reasonably good approximation to the binomial distribution. The best-fitting Poisson distribution has mean $\lambda = 800(.02) = 16.$ You might be able to find a table of the Poisson cumulative distribution function to evaluate the approximation as 0.193 - 0.010 = 0.183 (which is not too far from the exact result 0.181 rounded to three places). Otherwise, it's back to software again. And if you're using software, you might as well get the exact binomial result. If binomial $n$ is 'large' and $p$ is 'small', binomial probabilities can often be well-approximated using the Poisson distribution. But in today's computational environment, that fact is more of theoretical interest than of practical importance.
Normal approximation. The binomial random variable $X$ has $\mu = np = 16$ and $\sigma^2 = np(1-p) = 15.68.$ When $np$ and $np(1-p)$ both exceed 5, binomial probabilities can sometimes be satisfactorily approximated by a normal distribution. Especially when $p$ is far from 1/2 (as here), result may be inaccurate in the second decimal place. Here $X$ is approximately $Norm(\mu, \sigma)$. The approximation becomes
$$P\{8 \le X \le 12\} = P\{7.5 < X < 12.5\} =
P\{\frac{7.5 - 16}{3.96} < \frac{X - \mu}{\sigma} < \frac{12.5 - 16}{3.96}\}
\approx P\{-2.15 < Z < -0.88\} \approx 0.1725,$$
where $Z$ is a standard normal random variable, and the final result can be obtained by subtracting two values obtained from a printed normal table. This approximation is popular because it involves only simple arithmetic and 
use of a normal table. But it does not work very well in this particular problem.
Bottom line. We live in a time when exact probability approximations of most distributions used in practice can be easily obtained from software. Approximations based on limiting approximations to various distributions may be useful as theoretical exercises, but exact computations should be used in applied problems wherever possible.
Addendum:  As in the comment by @Did answers to the original question are:
exact binomial 0.55650, Poisson approximation 0.55596. Also, normal approximation 0.534.
