Consider the binomial distribution with n trials and P(S) = p. a Show that p(y) p(y 1) =

Chapter 3, Problem 3.63

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Consider the binomial distribution with n trials and P(S) = p. a Show that p(y) p(y 1) = (n y + 1)p yq for y = 1, 2,..., n. Equivalently, for y = 1, 2,..., n, the equation p(y) = (n y + 1)p yq p(y 1) gives a recursive relationship between the probabilities associated with successive values of Y . b If n = 90 and p = .04, use the above relationship to find P(Y < 3). c Show that p(y) p(y 1) = (n y + 1)p yq > 1 if y < (n + 1)p, that p(y) p(y 1) < 1 if y > (n+1)p, and that p(y) p(y 1) = 1 if(n+1)p is an integer and y = (n+1)p. This establishes that p(y) > p(y 1) if y is small (y < (n + 1)p) and p(y) < p(y 1) if y is large (y > (n + 1)p). Thus, successive binomial probabilities increase for a while and decrease from then on. d Show that the value of y assigned the largest probability is equal to the greatest integer less than or equal to (n + 1)p. If (n + 1)p = m for some integer m, then p(m) = p(m 1).