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.


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, $\mathbb{P}(\text{overbook}) = 0.886$

  • $\begingroup$ 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

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 $\mathrm{Pr}[S>120]$.


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