Over-booking

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 $411$.