   SEARCH HOME Math Central Quandaries & Queries  Question from Nikita, a student: An airline company knows that 8% of it's passengers will not show up for their scheduled flights. A plane has 175 seats. a) What is the probability that 10 passengers or fewer will not show up? b) What is the probability that 10 to15 passengers will not show up? c)What is the probability that exactly 10 passengers will not show up? d) What is the probability that more than 19 passengers will not show up? First note that any solution to this problem either needs more data or some (not necessarily realistic) assumptions. In particular, you are presumably meant to assume that no-shows are completely independent [a dangerous assumption, as many people fly in groups] and that the airline does not use this knowledge to overbook planes [in which case the number of passengers would exceed 175.]

With these assumptions, the number of no-shows is a binomial RV. What is N? What is p?

This gives the easiest answer for (c). For the other questions you probably want to use a normal approximation, using one or two table lookups instead of a long sum. Approximate the distribution by a normal RV with the same mean and standard deviation (your text will have formulae for the mean and SD of a binomial RV.) And break halfway between the extreme legal values and the first illegal ones.

So if your mean was 20 and your SD was 5, you would find P(13 ≤ X ≤ 23) by breaking at 12.5 (=20 - 1.5 × 5) and 23.5 (= 20 + 0.7 × 5) and finding as the probability that a standard normal RV is between -1.5 and 0.7.

Good Hunting!
RD     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.