Question

On any given flight, the goal of an airline is to fill the plane as much as possible, without exceeding the capacity of the plane. In order to achieve this, the airlines routinely overbook their flights in consideration of last minute cancellations. We assume that a customer cancels his/her ticket in the last minute with probability 0.06, independent of the other customers. We also assume that the airline is not able to sell more tickets in order to replace the canceled ones. What is the probability that a particular flight will be over capacity if the airline sells 309 tickets, for a plane that has a maximum capacity of 293 seats? In solving this problem, use the Central Limit Theorem, and in particular, use the De Moivre-Laplace normal approximation to the binomial distribution (with 1/2 correction) and be very careful when you choose the boundaries for probability computation. You will also need to use the standard normal CDF table that is in the summary notes that was made available to you for use during the exams. Use this table precisely as follows: In using the standard normal CDF table, first compute the input argument for the standard normal CDF with your calculator, then round this input argument value to two decimal digits after the decimal point, and finally locate the entry in the table which corresponds to the rounded input value. If you need the value of the standard normal CDF for arguments larger than 3.49 (not available in the table), you can use 1.0000. Your final answer for the problem should have four decimal digits after the decimal point.

(PROBABILITY ABD STATISTICS QUESTION)

Answer #1

Let X be a Binomial random variable which denotes the number of customers who cancel their ticket in last minute.

The flight with a maximum capacity of 293 will be over capacity if more than 293 passengers turns up out of the 309 bookings. Or if less than 16 passengers cancel their ticket in last minute. i.e if X < 16

Here, n = 309 and p = 0.06

np = 18.54

np(1 - p) = 17.43

Mean = 18.54

Standard deviation, = = 4.175

Since np(1 - p) > 10, X can be approximated to Normal distribution.

X ~ N(18.54, 4.175)

Using correction of continuity, the required probability

= P(X < 15.5)

= P{Z < (15.5 - 18.54)/4.175}

= P(Z < -0.728)

= **0.2333**

The probability that a particular flight will be over capacity
if the airline sells 309 tickets = **0.2333**

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