Problem about insurance company

Question

Probability that a person makes an accident is $0.05$. Suppose that the amount of money that a insurance company needs to give him is distributed as $\operatorname{Exp}(\lambda)$, and on average the company needs to give him $800\$$. How we can estimate the probability that sum of money that company gives to 1000 customers is higher that $50000\$$?

What I think. So the probability that a person need money from the insurance company is simulated by a random variable $X \sim \operatorname{Bernoulli(0.05)}$. And the sum of money that a person needs from insurance company is simulated by a random variable $Y \sim \operatorname{Exp}(\frac{1}{800})$. So I need to find the probability $\mathbb{P}(\sum\limits_{i=1}^{1000}Y_i \ge 50000 \$)$ where $Y_1, Y_2, \dots, Y_{1000}$ are i.i.d. But I don't know if my interpretation is good, and I don't know how to continue.

Answer 1

What you need to compute is $$ P(\sum_{i=1}^{1000} X_i Y_i>50000) $$ where $X_i\sim Bern(0.05)$, $Y_i\sim Exp(1/800)$.

Note that $$ E[XY]=E[X]E[Y]=0.05*800=40, $$ and $$ E[X^2Y^2]=E[X^2]E[Y^2]=(0.05)\cdot(800^2\cdot 2)=64000. $$ $$ Var[XY]=E[X^2Y^2]-(E[XY])^2=62400. $$ Therefore $$ E[\sum_{i=1}^{1000} X_i Y_i]=40000 $$ $$ Var[\sum_{i=1}^{1000} X_i Y_i]=62400000 $$ By CLT, $\sum_{i=1}^{1000} X_i Y_i$ approximately admits a Gaussian distribution. $$ P(\sum_{i=1}^{1000} X_i Y_i>50000)=P(\frac{(\sum_{i=1}^{1000} X_i Y_i-40000)}{\sqrt{62400000}}>\frac{10000}{\sqrt{62400000}})=P(U>1.26)=10.4\% $$