225 iPhones go on sale on black Friday, and 100 customers are in line to buy them. If the random number of iPhones that each customer wishes to buy is distributed Poisson with mean 2, approximate the probability that all 100 customers get their desired number of iPhones?
I have tried ${225-200}/200/underroot(100)$=1.25 but I know probablity cannot be greater than one, so this is wrong. Can someone correct me?
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The individual distributions have mean and variance $2$, hence (assuming independence) we can approximate the sum as a normal distribution with mean and variance $200$ (so standard deviation $\sqrt{200}\approx 14$) the sum of them (assuming above what is mentioned. As $225$ is the mean plus just a bit less than two standard deviations. So without looking up any tables, $0.95$ might be a good estimate for the desired probability.
$Y=X_1+X_2+\cdots+X_{100}$ is Poisson with $\lambda=100\cdot 2=200$ You must calculate $P(Y\le 225)$ where Y is Poisson(200)