I am a bit stuck on the following question:
Feller discusses the statistics of flying bomb hits in an area in the south of London during the Second World War. The area in question was divided into 24 × 24 = 576 small areas. The total number of hits was 537. There were 229 squares with 0 hits, 211 with 1 hit, 93 with 2 hits, 35 with 3 hits, 7 with 4 hits, and 1 with 5 or more. Assuming the hits were purely random, use the Poisson approximation to find the probability that a particular square would have exactly k hits. Compute the expected number of squares that would have 0, 1, 2, 3, 4, and 5 or more hits and compare this with the observed results.
The solutions for this problem are as follows: $\text{E[number of squares with 0 hits]} = 227$
$\text{E[number of squares with 1 hits]} = 211$
$\text{E[number of squares with 2 hits]} = 99$
$\text{E[number of squares with 3 hits]} = 31$
$\text{E[number of squares with 4 hits]} = 9$
$\text{E[number of squares with 5 or more hits]} = 1$
I generally seem to have troubles figuring values of $\lambda$ out but I do know that it is some form of probability which I cant seem to find that matches such results. I also know that $E[X] = \lambda$ for poisson variables so it would be nice if someone could explain how these numbers were derived or how to figure out $\lambda$.
There are $576$ areas and $537$ hits. That's $n=537$, $p=\tfrac 1{576}$ for a Binomial Distribution (of the count of successful hits in a square).
The Poisson approximation to a Binomial uses $\lambda = np$ as its rate parameter. The probability of a square having $k$ hits is thus approximated:
$$\mathsf P(X=k) \approx \frac{(np)^k~\mathsf e^{-np}}{k!} = \frac{537^k~\mathsf e^{-537/576}}{576^k~k!}$$
The expected estimated number of squares with exactly $k$ hits is then:
$$\mathsf E(N_{X=k}) \approx n~\mathsf P(X=k) = \frac{537^k~\mathsf e^{-537/576}}{576^{k-1}~k!}$$
(Due to the linearity of expectation and that $\mathsf P(A) = \mathsf E(\mathbf 1_A)$.)
Hence:
$$\mathsf E(N_{X=0}) \approx n~\mathsf P(X=0) = \frac{537^0~\mathsf e^{-537/576}}{576^{-1}~0!} \approx 226{\small .74} $$