Binomial/Poisson distribution question Someone asked me to help with assignment, but I am confused about the question, so basically, it is really easy:
Batches of 100 components have a mean number of 5 defects per batch. What is the probability at least 9 defective component in a batch? Calculate using
(i) Binomial distribution
(ii) Poisson distribution
(iii) Normal approximation to binomial
(iv) Poisson approximation to binomial
(v) Briefly discuss how good the approximations to the binomial distribution are, in reference to (ii), (iii), (iv)
(i), (iii), (iv) are easy and so is (v) once I got the number.
I do not understand what it means by calculating using Poisson distribution? Clearly this is not Poisson distributed by the way it is described.... 
If I see batches of 100 as a kind of 'time' of arrival (like in a Poisson arrival), then part (ii) gives Poisson(5), which is the same as part (iv) anyway, so I do not follow...
 A: (1) Let $X \sim \mathrm{Binomial}(n = 100, p = 0.05)$.  Then $$\Pr[X \ge 9] = 1 - \sum_{x=0}^8 \binom{n}{x}p^x (1-p)^{n-x}.$$
(2) Let $\lambda = 5$ be the rate parameter for the number of defects per batch; then $N \sim \mathrm{Poisson}(\lambda = 5)$ models the number of defects per batch, and we wish to calculate $$\Pr[N \ge 9] = 1 - \sum_{k=0}^8 e^{-\lambda} \frac{\lambda^k}{k!}.$$
(3) With $np = 5$, the normal approximation to the binomial is not particularly good, but we will use it anyway:  We then have $X \dot\sim \mathrm{Normal}(\mu = np = 5, \sigma^2 = np(1-p) = 4.75)$, and the desired probability with continuity correction is $$\Pr[X \ge 9] \approx \Pr\left[Z \ge \frac{8.5 - 5}{\sqrt{4.75}}\right],$$ where $Z$ is standard normal.
(4) I'm not sure how this is substantively different than (2).  To me, the exact distribution of defects per batch of 100 is binomial, not Poisson, so it can only be approximated with a Poisson distribution.  Thus I don't see how (2) and (4) should be different--both are necessarily approximations to the binomial.
