In shipment A, there are 990 correct and 10 faulty units. In shipment B, there are 1940 correct and 60 faulty units. 100 units out of each shipment is inspected. Calculate with an APPROPRIATE approximation the probability of finding five or more faulty units. Emphasis from book, not me. My solution was A = Bin(100,0.01) ≈ N(1,0.1) B = Bin(100,0.03) ≈ N(3,1.71) A+B ≈ N(4,1.71) P(A+B > 4.5) = 1-P(A+B < 4.5) = /normal table/ = 0.386 Which produced a fairly good answer, but the book also made it clear that they were looking for a poisson approximation. I am not entirely clear on why that is. Anyone?