The problem is as follows.
An insurance company sells 10,000 similar car insurance policies. They estimate that the amount paid out in claims on a typical policy has mean £240 and standard deviation £800. Estimate how much they need to put aside in reserves to be 99% sure that the reserve will exceed the total amount claimed.
Let $S_n$ be the total amount claimed. The solution given is:
$$ P\left(\frac{S_n - nμ}{σ\sqrt{n}}< Φ^{-1}(0.99)\right) = 0.99. $$
Then you rearrange the inequality for $S_n$ and evaluate the RHS. I do not understand why there is $Φ^{-1}(0.99)$ involved. Below is my thought process.
Let $y$ be the amount in reserves. I want to find $y$ such that $P(y - S_n > 0) = 0.99$. By CLT, the random variable $X_n := \frac{S_n - nμ}{σ\sqrt{n}} \sim N(0,1) $ for large $n$. I can write $y - S_n$ in terms of $X_n$ and then it is:
$$ P( y - (σ\sqrt n X_n + nμ)) > 0 ) = P\left(X_n < \frac{y - nμ}{σ\sqrt n} \right) = 0.99 $$
Then using the fact that for the standard normal distribution $P(X < 2.326) \approx 0.99$ allows me to solve for $y$. Both of these give the same answer but I would like to know what is the thought process for getting the first expression?
Let $R$ be the reserves, which have to be put aside and $S_n$ be the the total amount of claims. We need to find $R$, such that $S_n$, doesn‘t exceed it with probability $0.99$, so:
$$P(S_n<R)=P(\frac{S_n-n\mu}{\sqrt{n}\sigma}<\frac{R-n\mu}{\sqrt{n}\sigma})\sim \Phi(\frac{R-n\mu}{\sqrt{n}\sigma})$$
Furthermore: $$\Phi(\frac{R-n\mu}{\sqrt{n}\sigma})=0.99$$ which is equivalent to: $$\frac{R-n\mu}{\sqrt{n}\sigma}=\Phi^{-1}(0.99)$$ So we see: $$P(S_n<R) \sim P(\frac{S_n-n\mu}{\sqrt{n}\sigma}<\Phi^{-1}(0.99))=0.99$$