225 phones go on sale, and 100 customers are in line to buy them. If the random number of phones that each customer wishes to buy is distributed Poisson with mean 2, approximate the probability that all customers get their desired number of phones? Use Central limit theorem to solve the problem.
I understand that problem states to model the number of phones each customer buys as a Poisson distribution with $\lambda=2 $.
Let X be the number of phones a customer buys:
$ X \sim P_\lambda\;\;where\; \lambda = 2\;\; E[X] = 2\;\;and\;\;V[X] = 2 $
Then we expect that each customer buys 2 phones.
And Expected total number of sales will be - by linearity of expectation:
$ E[X_1 + X_2+...X_n] = E[X_i] + E[X_2] + ...+ E[X_n] = 200 $
I don't understand how to apply all this information to compute probability that all customers get their desired number of phones. Maybe one customer wants to buy 1 phone and the other customer wants to buy 3 phones? Do we assume that all customers buy the expected number of phones? And how can I apply CLT to solve this problem?
Solution:
Let Y be $ X_1+...+X_{100} $ Then E[Y] = 200, and Var(Y) = 200
Invoke CLT, then $ Y \sim N(200,200) $
The question asks P(Y <= 225) - probability that total demand <= 225
$ P(Y<= 225)\approx \Phi (\frac{225-220}{\sqrt{200}}) \approx \Phi (1.77) = 0.9616 $