Over-booking
The probability that a passenger books a flight and does not turn up is 0.05. For an aeroplane with 400 seats how many tickets can be sold so that only 2% of flights are over-booked?
Problem
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?
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?
Getting Started
Let $S$ be the number of seats in the aircraft;
$T$ be the number of tickets sold;
$p$ be the probability that any given passenger arrives for the flight.
Let $X$ be the number of passengers that arrrive for a given flight.
Then $X$ has the Binomial distribution for $T$ trials with the probability of success $0.95$. You can calculate the mean $\mu$ and variance $\sigma^2$ for this distribution.
Most modern advanced scientific calculators can handle Binomial distributions with 400+ trials, so an answer can be reached fairly quickly using trial and improvement.
Alternatively, we can approximate the Binomial distribution by a Normal distribution with the same mean and variance, i.e. by $$N(\mu,\sigma^2)$$.
Thus we now assume that $X$ has distribution $N(\mu,\sigma^2)$. In order to use the standard Normal probability tables $N(0,1)$ (or the standard Normal distribution our calculator), we have to put $$Z = {X-\mu\over \sigma}$$ then $Z$ has distribution $N(0,1)$.
So as to allow for the approximation to the discrete data by the continuous Normal distribution, we want to find $\text{Prob}[X\leq 400.5$] using either the inverse Normal function on a calculator or by looking up the appropriate value of $Z$ in the Normal table.
If you use a Normal distribution table you need to check to see if it gives the area $\Phi(Z)$ under the Normal curve to the left of $Z$, that is the probability that the variable is less than $Z$, or to the right of $Z$.
Student Solutions
Andrei fromTudor Vianu National College, Romania, gives a very clear account of the use of the binomial and normal distributions to solve this problem.
The passengers who have bought tickets either turn up for the flight or do not turn up. Taking $X$ as the random variable for the number of passengers who turn up for the flight, then $X$ is binomially distributed with parameters $p$, the probability of arriving for the flight, and $n$ as the number of tickets sold. The probability distribution is: $$P(x;n,p)= {n \choose x}p^x(1-p)^{n-x},\ x=0,1,...,n.$$ The mean of the distribution is $E(x)=np$ and the variance $\sigma ^2 ={np(1-p)}.$
In this problem $n=400$ and $p=0.95$.
So, $E(X)=380$ and $\sigma = 4.36$ and the expected number of empty seats is $20$.
It is known that, if the value of $n$ is large, the variable $X$ could be considered to have a probability distribution that approximates to the standard normal distribution, with the same mean and variance. \par To verify that the normal distribution could be, in the conditions of the problem, a good approximation for the binomial distribution, I have to verify that both the mean $\mu =np$ and the variance $\sigma^2 = n(1-p)$ are greater than 5. Here $np=380$ and $n(1- p)=20$. So, the use of the normal distribution is acceptable.
Using the applet at http://davidmlane.com/hyperstat/z_table.html , I tried to find the number of tickets, $x$, that the airline should sell to satisfy the conditions of the problem.
Let $x$ be the number of tickets sold, which, as explained before, could be considered to have a normal distribution $N(\mu,\sigma^2)$. The mean of the distribution is $x\times 0.95$, and the standard deviation is $\sqrt{x\times 0.95\times 0.05}$. The area under the curve and below $400$ is $98$ per cent or $0.98$ and the area above $400$ is $2$ per cent or $0.02$ (the probability that too many passengers will turn up for the flight).
Trying for some values of $x$ I obtained the number of tickets that the airline must sell. Put $$y = {x-np\over \sqrt{np(1-p)}};$$ then $y$ has distribution $N(0,1)$. The probability that all passengers who arrive for the flight can actually get a seat is ${\rm Prob}\{x \leq 400.5\}$ (because $x=400$ is fine, but $x=401$ is not). Thus $${\rm Prob}\{x \leq 400.5\} = {\rm Prob}\left\{y\leq {400.5-np\over \sqrt{np(1-p)}}\right\}$$ and this can now be found from tables of the normal distribution.
We find that if $411$ tickets are sold then the probability of too many passengers arriving is less than $2$ percent but for $412$ it is more than $2$ percent so the ideal number of tickets to be sold is $411$.
Teachers' Resources
Why do this problem?
This is a real-life example of the Binomial distribution in practice, so as well as providing a motivating context, it gives students the opportunity to consider underlying assumptions and their validity.
Possible approach
This problem involves a Binomial distribution with a large number of trials. This can be approached directly via the Binomial distribution on an advanced scientific calculator using trial and improvement. Alternatively, students can use the Normal probability distribution as an approximation to the Binomial distribution, standardise and solve directly for the number of trials required. You may wish to guide students down one of these two paths to reinforce specific content or you may wish to allow students to choose an approach and compare results and efficiency.
The tools which students are offered may influence their choice such as Standard Normal Probability tables in formula booklets or online, online calculators like the one at http://davidmlane.com/hyperstat/z_table.html (or in Geogebra) and/or advanced scientific calculators or calculator emulator.
Having decided on the approach you wish students to take or to allow them to choose more freely, you may want to introduce the topic through personal experience. Has anyone ever been denied boarding through over-booking or been offered incentives to take a later flight? Why might airlines deliberately over-book? What are the risks to the airline if they do over-book and if they don't?
The problem helps students to see the mathematical reasons for over-booking and also where limits on over-booking may be reasonably set.
Key questions
How might we model the number of passengers who arrive for a flight?
What assumptions are we making if we accept this model?
Which parameters are known and which are we looking for? How might we go about this?
Possible extension
Students could consider the problem in terms of costs. What "losses" would airlines suffer if they didn't over-book? What costs might they have to pay for passengers who are denied boarding? What number of tickets sold maximises their profit if our Binomial model holds? Students might use a spreadsheet to investigate expected outcomes for different numbers of ticket sales.
Possible support
The problem could be reduced to a more manageable size of plane, such as one which holds just $10$ passengers. A simple spreadsheet simulation could be used to investigate over-booking in the reduced case.