Central Limit Theorem for functions of random variables

Question

"An insurance company wishes to collect a yearly premium on life insurance policies issued to two groups. There are 1000 people in $X_1$ and they have a 0.01 chance of dying in the year, and there are 500 people in $X_2$ and they have a 0.03 chance of dying in the year. Benefitors in $X_1$ get 20 000 dollars and benefitors in $X_2$ get 10 000 dollars. What premium should be collected such that it is equal to the 90th percentile of the distribution of the payout?"

I set it up: $$X_1\sim bin(1000,0.01)$$ $$X_2\sim bin(500,0.03)$$ $$Y = 20X_1+10X_2$$ where $Y$ is thus (in thousands of dollars) the total payout associated with each collection.

and we are looking for an $\alpha$ such that $$P(Y \leq \alpha) = 0.9$$

We are to use the central limit theorem to approximate the solution. This is where I get slightly lost. I planned to use the normalization formula

$$W = \frac{\sum_{i=1}^{n}X_i - n\mu}{\sqrt{n}\sigma}$$ since $$Y = \sum_{i=1}^{n}a_iX_i$$.

We can calculate $\mu_y$ as $$20E(X_1)+ 10E(X_2) = 20*10 = 10*15 = 350$$ and use $$\sigma_y^{2} = \sum_i a_{i}^{2} \sigma_{Xi}^{2} = 20^2(1000*0.01(1-0.01)) + 10^2(500*0.03(1-0.03)) = 5415$$

Now we have $$P \left( \frac{Y - n\mu}{\sqrt{n}\sigma} \leq \frac{\alpha - n\mu}{\sqrt{n}\sigma} \right) =$$ $$\phi \left( \frac{\alpha - n\mu}{\sqrt{n}\sigma} \right) = $$ $$ \phi \left( \frac{\alpha - 2*350}{\sqrt{2*5415}} \right) = $$ $$ 0.9 = \phi(1.29) $$

Which gives us $\alpha = 6.75...$ which is obviously quite wrong. I think my mistake is in my original assumption, that we can apply CLT to $Y$ "as is", and that I am missing something in its implementation. Any thoughts would be greatly appreciated!

Answer 1

IF you wanted to use a normal approximation, then:

1. No, you could not have used the CLT since it specifies that each $X_i$ be independent and identically distributed. In your scenario, the random variables are not identically distributed.

2. That's ok! Even if the random variables are not $iid$, for "large" $n$, it can still provide a "reasonable" approximation. In your work, you gave $$ \Phi \left( \frac{\alpha - n\mu}{\sqrt{n}\sigma} \right) = \Phi \left( \frac{\alpha - 2\cdot 350}{\sqrt{2\cdot 5415}} \right) = 0.9.$$ I did not check that each value was correct, but it follows from what you gave that $$\alpha = \Phi^{-1}(.9)\cdot \sqrt{2\cdot 5415}+2\cdot 350 = 833.3676.$$ I think you made a small mistake somewhere.