For n trials, this is the binomial distribution with p = .005.
Thus the standards deviation is sqrt(n*p*q) where q = 1-p.
However, do not try to use the normal approximation to the binimial unless N is very large (since is p is so small here).
Here's a link with more:
http://www.stat.wvu.edu/SRS/Modules/.../binomial.html