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An airline has a plane with 400 seats. The probability that a passenger fails to turn up for their flight is 0.04.

The airline has an overbooking policy. Find the largest number of passengers this airline can book and still be at least 85% sure that everyone who shows up has a seat.

Working:

Need to find a value $x$ when $1 - \text{cumulative probability}$ is at least 0.85 or more, but as close to 0.85 as possible. If we call this $x$-value $Z$ we need:

$[1 – P(X = 0) + P(X = 1) +…+ P(X = Z)] $

So $n=400 p=0.04 q=0.96 x=0,1,2,...$

Can't seem to add a picture of the answers but I got it as $411$.

Would anyone be able to verify that this method is correct ?

Community
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user321652
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1 Answers1

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Hopefully it helps noticing that when the number of trials is large and the probability is small within a binomial distribution you can approximate the distribution to the poisson distribution using

$$ X: Binomial(n,p) \rightarrow Y: Poisson(\lambda = np) $$

Goshawk
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  • In which case would the method I used still be correct? – user321652 Mar 13 '16 at 16:39
  • Unfortunately i'm unsure. I would say $X: Binomial(400,0.04) \rightarrow Y : Poisson(\lambda = 16)$ and as $P(Y \le 20) = 0.8681.. \gt 0.85$ we have that we can be 86% sure (more than 85% certain) that 20 passengers won't turn up. Allowing 420 passengers to be booked with 86% certainty that only 400 will turn up. Hopefully someone can verify your method though. – Goshawk Mar 13 '16 at 17:11