Hello, Part (iii) (b) states the expected frequencies, of which we are asked to complete the table but I can't replicate the results given. Using the method used in the solutions, I would say that the expected frequency of y=0 is: [(1-0.18728) x [(1.1505^0 / 0!) x e^-1.1505]] x 200 = 51.44... Could anyone who's not queueing up outside Northern Rock currently point out my silly mistake?
Hi The formula for P(Y=0) is different to the formula for P(Y=r) for r=1,2,3,... as it says at the top of the questions. So P(X=0) = e^(-1.1505) P(Y=0) = a + (1-a)P(X=0) = 0.1873 + 0.8127 e^(-1.1505) = 0.4445 Expected frequency = 0.4445 x 200 = 88.9 (I've used a instead of alpha)
