Question

Airlines often sell more tickets for a flight than there are seats because some ticket holders do not show up for the flight. Assume that an airplane has 170 seats for passengers and that the probability that a person holding a ticket appears for the flight is 0.93. If the airline sells 176 tickets, what is the probability that everyone who appears for the flight will get a seat?

a: 0.9625 B: 0.9515 C. 09778 D. 09608

Answer #1

Option C. **0.9778** is correct.

Let X =No. of ticket holders who show up for the flight.

So, X ~Binomial (n = 176, p = 0.93)

Since the sample size, n is large(176>30), we can use the Central Limit Theorem(CLT) and approximate a binomial distribution with a normal distribution.

So, X ~Normal (µ = np = 176*0.93 =163.68, σ ^{2} =np(1 –
p) =163.68(0.07) =11.4576).

The probability that everyone who appears for the flight will get a seat means we need to find the probability of "170 or less passengers" because the available seats are 170.

So, P(X ≤ 170) =P[X < 170+0.5] =P[X < 170.5]

(because of 'continuity correction factor').

P[X < 170.5] =P[Z < (X - )/] =P[Z <
(170.5 – 163.68)/]
=P[Z ≤ 2.0148] =**0.978**

Since option C. 0.9778 can be rounded to 0.978, it is correct.

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