Question

An airline estimates that 80% of passengers who reserve the tickets actually show up for the flights. Based on this information, it has to decide how many tickets it will sell for each flight, which is typically more than the number of seats actually available. In the economy section of a particular aircraft, 200 seats are available. The airline sells 225 seats. What is the probability that more passengers will show up than there are seats for?

Answer 1

The same can be modeled as a Binomial distribution as we have passengers either turning up or not turning up

Please note

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Here n = 225, p = 80% = 0.8, q = 1 – p = 0.2.

P(more passengers show up than there are seats for) = P(X > 200) = P(X > 201) + P(X > 202) + .....+P(X > 225)

The best way to do this would be by using software (excel or calculators) to find the cumulative value, as manual calculations would be time consuming.

Using Excel, us BINOMDIST(200, 225, 0.8, True). This gives us P(x 200 ).= 0.9999

Therefore P(X > 200) = 1 - 0.9999 = 0.0001

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Method 2: If calculators are not available and nor is excel.

Since the numbers are very large, we can also use the normal approximation to the binomial.

Mean = n * p = 225 * 0.8 = 180

SD = Sqrt(n * p * (1 - p)) = Sqrt( 225 * 0.8 * 0.2) = 6

To Find P(X > 200) we use the continuity correction factor, which for P(X > 200) = P(X > 200.5)

P(X > 200.5) = 1 - P(X < 200.5)

Z = (200.5 - 180) / 6 = 3.42

The p value for P(X < 200) = 0.9997

Therefore P(X > 200) = 1 - 0.9997 = 0.0003 (which is very close to the actual value of 0.0001)

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Continuity Correction Factor Table

1) If P(X = n)       Use P(n – 0.5 < X < n+0.5)

2) If P(X > n)       Use P(X > n + 0.5)

3) If P(X <= n)    Use P(X < n + 0.5)

4) If P(X < n)       Use P(X < n - 0.5)

5) If P(X => n)    Use P(X > n - 0.5)