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


I guess that the answer consists in solving the inequality:

$$\frac{\operatorname{Var}[S]}{(s-\operatorname{E}[S])^2} \leq 0.001,$$

which solution is:

$$s \geq \sqrt{\frac{\operatorname{Var}[S]}{0.001}} + \operatorname{E}[S]$$ 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]$.

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