3
$\begingroup$

The problem said:

An airplane has 120 seats. The probability that a ticketed passenger will show up for a flight is 0.95. Assume that all passengers act independently and that the airline has sold 130 tickets for a particular flight. Using the Normal approximation to the Binomial (with appropriate continuity correction), compute the approximate probability that the flight will be overbooked.

I know I need to use the central limit theorem, and I start by defining indicator variables:

Xi= 1 if passanger i show up for a fligth, 0 otherwise.

Then the approximate probability that the flight will overbooked, defined by even OB is:

P(OB)=P(sum from 1 to 130 of Xi > 120)  

I know also that P(xi)=0.95

But I'm missing something to finish this problem, and found the correct solution wich the book said is: 0.886 or 88.6%. Thanks for your help.

$\endgroup$
1
$\begingroup$

Let $X_i \sim Ber(p)$, then $Y = \sum_{i=1}^n X_i \sim Bin(n,p)$.

From the CLT we have $$\frac{Y - E[Y]}{\sqrt{Var(Y)}} = \frac{Y - np} {\sqrt{np(1-p)}} \sim N(0,1),$$

Plugging the values you have: $$\frac{120 - 130 \times 0,95} {\sqrt{130 \times 0,95 \times (1-0,95)}} =-1.40848,$$ So, $$\mathbb{P}(\text{overbook})=\mathbb{P}(Y>120) = \mathbb{P}(Z>1.40848),$$ where $Z \sim N(0,1)$. Hence, $\mathbb{P}(\text{overbook}) = 0.886$

$\endgroup$
2
  • $\begingroup$ Would you mind sharing why did the minus sign ($-1.40848$) go away in the last part? $\endgroup$ – Никита Васильев Feb 26 '20 at 18:24
  • $\begingroup$ The bell curve is symmetrical. Therefore, $\malthcal{P} (Z > a) = \malthcal{P} (Z < - a)$, for any real $a$. $\endgroup$ – Guilherme Thompson Feb 26 '20 at 18:36
1
$\begingroup$

First let's compute the variance of $X_i$:$$\sigma^2 = \mathrm{Var}[X_i] = 0.95-0.95^2 = \frac{14}{400}.$$

We now the define the random variable $S$ as the number of passengers that want to board the plane: $$S = \sum_{i=1}^n X_i.$$

The mean of $S$ is obvious: $$\mathrm{E}[S] = pn = 0.95 \times 130 = 123.5.$$ (This is bad already: we expect there to be more passengers than seats.)

The variance of $S$ can be estimated using the CLT:$$\mathrm{Var}[S]\rightarrow n\sigma^2 = 4.55.$$

Enter both values into a normal distribution calculator and you find the probability $\mathrm{Pr}[S>120]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.