A cruise ship has $200$ passengers. There are two large lifeboats, one on each side of the ship, each with capacity $C$ people. Suppose that, in an emergency, each passenger chooses a lifeboat at random (independent of the other passengers) and runs to this randomly chosen lifeboat. What is the smallest capacity $C$ which guarantees the probability of at least $.90$ that every passenger will find room in their chosen lifeboat? (Use a normal approximation)
I can't seem to figure out what type of distribution this is. Also, if I did have the distribution and we are going to use a normal approximation I don't see how we will be able to find such a $C$. Any suggestions are greatly appreciated.
The number $X$ of people that go to the first (one) boat is binomially distributed with parameters $p=\frac12$ and $n=200$. Hence, you are looking for $C$ such that $$P(X\le C)\ge 0.9$$ Since $p=1/2$ (optimal case for approximation, perfect symmetry) you can approximate $X$ with a normal random variable $Y$ with parameters $$Y\sim \mathcal N(μ=np, σ^2=np(1-p))$$ so your equation becomes $P(Y\le C)\ge 0.9$. Plugging in $n=200, p=1/2$ and making $Y$ standard normal $$Z=\frac{Y-100}{\sqrt{50}}\sim \mathcal N(0,1)$$ you get $$0.9\le P(Y\le C)=P\left(Z\le \frac{C-100}{\sqrt{50}}\right)=\Phi\left(\frac{C-100}{\sqrt{50}}\right)$$ which implies that $$\frac{C-100}{\sqrt{50}}\ge 1.282\implies C\ge 109.065 $$ which means that $C=110$ will do (since it must be an integer).