April 2007 - Poisson and Normal Approximation

Q7 - Abridged version pasted here:

A charity issues a large number of certificates each costing £10 and each being repayable one year after issue. Of these certificates, 1% are randomly selected to receive a prize of £10 such that they are repaid as £20. The remaining 99% are repaid at their face value of £10.

Consider a person who purchases 200 of these certificates.
Use a Poisson approximation to this binomial distribution to approximate the
probability that this person is repaid more than £2,040.

The solution says P(S>2040) = P (N>4) where N is a Poisson (2).

I understand how the parameter for Poisson was calculated. I also understand P(S>2040) but I don't get the P(N>4) expression.

Can anyone help me understand how they got the expression P(N>4)?

 

N is the random variable which counts the number of winning certificates out of 200 he has bought.

S > 2040
i.e. 20 * N + 10 * (200 - N) > 2040
i.e. 2000 + 10 * N > 2040
i.e. N > 4

In case you are wondering how they computed P(N > 4) or equivalently P(N <= 4), you can do it by hand or refer to the back pages of the tables. I need to check but I think the back pages contains CDF of Poisson for a set of given values of lambda.