Let X = number of people who are pescatarian.

Then X ~ Bin (250, 0.0275).

Then since n q = 250 0.9725 = 243.125 > 10 we will use the normal approximation.

Then mean () = p = 0.0275 and standard deviation () = {pq/n} = {(0.0275 0.9725)/250 = 0.01034.

Then z-score is given by, z = (p - )/.

For p = 0.04 the z-score is

z = (p - )/ = (0.04 - 0.0275)/0.01034 = 1.209.

Then P (p > 0.04) = P (Z > 1.209) = 0.1131.

Thus, the probability that there will be more than 4% pescatarian among the guests and the event planner will not have enough pescatarian meals is 0.113.