# 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...

• @Henry i am sorry, i am asking about (ii). I do not understand how it is different to (iv), maybe it is not. I would like an independent confirmation. I was meant to say 1,3,4 are easy... Commented Oct 27, 2014 at 23:55
• The only thing I can think of is that you cannot have more than $100$ defects per batch, so might adjust the Poisson probabilities accordingly. But since the probability that you get a value over $100$ for a Poisson distribution with $\lambda=5$ is less than $10^{-91}$, it does not seem worth worrying about. Commented Oct 28, 2014 at 0:02

(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.