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 1% of flights are over-booked?
Age
16 to 18
Challenge level
Image
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?
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.
In order to avoid lengthy calculations in the discrete case we 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 $N(0,1)$ we have to put $$Y = {X-\mu\over \sigma}$$ then $Y$ 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 $\text{Prob}[X\leq 400.5$] and look up the probability for the corresponding value of $Y$ in the Normal table.
If you use a Normal distribution table you need to check to see if it gives the area $\Phi(Y)$ under the Normal curve to the left of $Y$, that is the probability that the variable is less than $Y$, or to the right of $Y$.
$T$ 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.
In order to avoid lengthy calculations in the discrete case we 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 $N(0,1)$ we have to put $$Y = {X-\mu\over \sigma}$$ then $Y$ 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 $\text{Prob}[X\leq 400.5$] and look up the probability for the corresponding value of $Y$ in the Normal table.
If you use a Normal distribution table you need to check to see if it gives the area $\Phi(Y)$ under the Normal curve to the left of $Y$, that is the probability that the variable is less than $Y$, or to the right of $Y$.
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 $411$.
This example involves using the Normal probability distribution as an approximation to the Binomial distribution.
You can look up the Standard Normal Probability table, for example at http://www.statsoft.com/textbook/sttable.html where you will also find some of the theory about the distribution, or you can use an online calculator which will give you readings automatically, for example the one at http://davidmlane.com/hyperstat/z_table.html .
In this example you are not given the number of tickets sold and asked to find the probability that too many passengers will turn up for the flight, but rather the inverse problem, and solving this requires some trial and improvement method.