The number of new cars sold by a dealer each week is distributed as a Poisson variable with mean 10. Suppose that 80% of the customers need a loan to pay for the car, and of these 62.5% are likely to take the loan from the dealer. Suppose that the loan amounts are on average 22,000, with a standard deviation of 5,000. Find the approximate probability that the total new car loan issued by the dealer in one week will exceed 600,000.
Here is where I am. Let $N$ be the number of cars sold and $N~P(10)$. Let $L$ be a random variable representing the loan amount. We want to compute $P(0.8\times0.625\times N\times L>600,000)$. $N$ and $L$ times together. How to resolve this? Please help!!!
The number of loan takers N will be distributed $N\sim\text{Poisson}(10\cdot .8 \cdot .625)=\text{Poisson}(5)$.
The cost of the loan (assuming a normal distribution) will be $L|N\sim\text{Normal}(22000N, (5000N)^2)$.
The probability that the loans will exceed 600000 is
$$\begin{split}P(L>600000)&=\sum_{N=0}^\infty P(L>600000|N)P(N)\\ &=\sum_{N=0}^\infty P(Z>\frac{600000-22000N}{5000N})P(N)\\ &=\sum_{N=0}^\infty \Phi(\frac{22000N-600000}{5000N})\frac{e^{-5}5^N}{N!}\end{split}$$
The prob stabilizies after about N=25 to be $2.369295e-07$