# 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?

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?