what premium should the company charge each policy holder to assure that the premium income will cover the cost of the claims? 
A car insurance company has $2,500$ policy holders. The expected claim
  paid to a policy holder during a year is $1,000$ with a standard
  deviation of $900$. What premium should the company charge each policy
  holder to assure that with probability $0.999$ the premium income will
  cover the cost of the claims?

I'm having trouble with interpreting this problem. I want to correctly denote the random variables to use CLT or Chebyshev's inequality.
So I suppose $X$ to be the cost of the claims, then we have $E[X]=1000$, $Var(X) = 900^2$.
 Let $y$ be the cost of the premium, so we need to find $ P(Y-X\geq 0)$ right?
 A: No; you need to look at the aggregate claims random variable.  The idea is that the insurer needs to collect sufficient premiums to cover the aggregate claims with at least $0.999$ probability.
Assuming each claim is independent, then the aggregate claims $S = \sum_{i=1}^{2500} X_i$ where $X_i$ is the annual claim size for policyholder $i$, for the book of business is approximately normally distributed with mean $$\operatorname{E}[S] = 2500 \operatorname{E}[X_i] = 2500000,$$ and variance $$\operatorname{Var}[S] = 2500 \operatorname{Var}[X_i] = 2250000.$$  Consequently, $$Z = \frac{S - \operatorname{E}[S]}{\sqrt{\operatorname{Var}[S]}}$$ is approximately standard normal, and we want to find some number $P$ such that $$\Pr[S \le P] = 0.999;$$ equivalently, $$\Pr\left[ Z \le \frac{P - \operatorname{E}[S]}{\sqrt{\operatorname{Var}[S]}}\right] = 0.999.$$  Since the $99.9^{\rm th}$ percentile of the standard normal is approximately $3.09023$, it follows that $$\frac{P - 2500000}{1500} = 3.09023,$$ or $$P = 2504635.35.$$  This represents the total premiums the insurer needs to collect; dividing by $2500$ gives the per-policyholder premium, assuming that all policyholders are rated equally.  What is the excess above the expected loss per policyholder?
