Problem
There are two restaurants on the campus of a university. Each can feed 120 students. We know that there are 200 students attending the university who will want to eat lunch in one of the restaurants. The restaurant is chosen randomly by the student, for example by tossing a fair coin.
What is the probability that there will not be enough meals in one of the restaurants?
My Answer
Each students choice is a Bernoulli trial with even probabilities on each of the two outcomes that is:
$X_i = 1 $ with $ p=1/2$ or $X_i = 0 $ with $ p=1/2$ where $X_i$ is a random variable which denotes student i's choice.
So we want $P(\Sigma_1^{200} X_i \gt 120) \bigcup P(\Sigma_1^{200} X_i \lt 80)$
But from here I don't really know where to go with it?
Any help would be great.
Take a look only on one of these restaurants, hence the number of students that enters it distributed Binomial with $n=200$ and $p=1/2$. Denote it by $Y$. Then use the continuous correction and the Normal approximation to compute the probability of interest: $$ P(Y> 120) = P(Y\ge 120.5) \approx 1 - \phi \left(\frac{120.5 - np}{\sqrt{npq}} \right), $$ finally by multiplying it by the number of restaurants you should receive the answer.