# Chebyshev's inequality to compute the premium income

This is a problem 9.7 from Anderson "Introduction to probability"

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? Compute the answer with both Chebyshev's inequality and CLT.

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 $$i$$, then $$\operatorname{E}[S] = 2500 * \operatorname{E}[X_i] = 2500 * 1000 = 2500000,$$ and variance $$\operatorname{Var}[S] = 2500 \operatorname{Var}[X_i] =2500 * 900^2.$$
We want to fins s (the premium income), such that $$P(S\leq s)=0.999.$$ This is equivalent to $$P(S\geq s)=0.001.$$ By Chebyshev's inequality $$P(S\geq s) = P(S-\operatorname{E}[S]\geq s- \operatorname{E}[S])\leq P(|S-\operatorname{E}[S]|\geq s-\operatorname{E}[S])\leq \frac{\operatorname{Var}[S]}{(s-\operatorname{E}[S])^2}$$

My question is how do we get the desired equality from this inequality? Thanks

$$\frac{\operatorname{Var}[S]}{(s-\operatorname{E}[S])^2} \leq 0.001,$$
$$s \geq \sqrt{\frac{\operatorname{Var}[S]}{0.001}} + \operatorname{E}[S]$$ or $$s \leq \sqrt{\frac{\operatorname{Var}[S]}{0.001}} - \operatorname{E}[S].$$
So the solution is $$\sqrt{\frac{\operatorname{Var}[S]}{0.001}} + \operatorname{E}[S]$$.