# Probability and Statistics: Over-booking flights

I am finding difficulty trying to answer the second part of this question. The answer is one this website but I do now get how they worked it out. https://nrich.maths.org/4932
I also know that binomial and normal distribution plays a big part in this.

An airline flies a plane with 400 seats. Each passenger who buys a ticket arrives for the flight (that is, does not miss the flight) with probability 0.95. If the airline sells 400 tickets what is the expected number of empty seats? (I found the mean to be 380 and the Sd to be 20)

The airline regularly books more than 400 passengers for its flights. How many tickets can the airline sell if it wants to have to refuse passengers who arrive for the flight with tickets in no more than about two per cent of the flights? (I do not know how to reach the answer of 411. Can someone explain with step by step working?)

• Surely there is no way the answer can be $211$ as then the airline would never have to refuse passengers. Oct 10, 2018 at 3:07
• The website says 411, not 211. Oct 10, 2018 at 3:27
• Are you familiar with the technique for using a normal distribution as an approximation for a binomial distribution? The web site assumes you know this. Oct 10, 2018 at 3:29
• @DavidK I mean 411, that was by accident. I know the basics and I know the normal approximation to the binomial distribution which includes the continuity correction but when I do all of them I do not get the answer. Oct 10, 2018 at 3:32
• Then show us in the question exactly how you did it and what answer you got. Maybe then someone can figure out where the discrepancy in answers is. To format the formulas, start at math.stackexchange.com/help/notation and follow further links as needed. Oct 10, 2018 at 3:35

Let the number of tickets sold be $$n$$. The number of arriving passengers is distributed as $$\operatorname{binom}(n,0.95)$$, which we approximate as $$\mathcal N(0.95n,0.0475n)$$ because we do not know what $$n$$ is yet.
For a standard normal variable $$Z$$, $$P(Z>x)=0.02$$ gives $$x=2.055$$. Thus, for the airline to have to refuse passengers less than 2% of the time, 400 has to be greater than 2.055 standard deviations above the mean: $$0.95n+2.055\sqrt{0.0475n}<400$$ This is a quadratic in $$\sqrt n$$ and we solve this as $$\sqrt n<20.285$$, i.e. $$n<411.489$$. Thus the largest number of tickets the airline can sell is 411.
• @ChaseRyanTaylor because computing the binomial probability for large $n$ is hard to do by hand. Oct 10, 2018 at 4:01
• @ParclyTaxel You might add, even harder in this case since $n$ is unknown. Oct 10, 2018 at 4:18