Question

An airline makes 200 reservations for a flight who holds 185 passengers. The probability that a passenger arrives for the flight is 0.9 and the passengers are assumed to be independent. Use the normal approximation of Binomial distribution to answer the following questions.
a. Approximate the probability that all the passengers who arrives can be seated.
b. Approximate the probability that there are empty seats.
c. Approximate the number of reservations that the airline should make so that the probability that everyone who arrives can be seated is 0.95.

Answer 1

Answer:

The answer is below

Step-by-step explanation:

given that:

n = sample size = 200, p = probability of a cancelled reservation = 1 - 0.9 = 0.1

The z score is given by:

$z=\frac{x-mean}{standard\ deviation} =\frac{x-np}{\sqrt{np(1-p)} }$

a) The probability that all the passengers who arrives can be seated can only occur if atleast 15 or more passengers cancelled their reservations.

P(X ≥ 15) = 1 - P(X ≤ 14.5)

$z=\frac{x-np}{\sqrt{np(1-p)} }=\frac{14.5-(200*0.1)}{\sqrt{(200*0.1)(1-0.1)} }=-1.3$

From the normal distribution table:

P(X ≥ 15) = 1 - P(X ≤ 14.5) = 1 - P(z < -1.3) = 1 - 0.0968 = 0.9

b) The probability that there are empty seat only occur if atleast 16 or more passengers cancelled their reservations.

P(X ≥ 16) = 1 - P(X ≤ 15.5)

$z=\frac{x-np}{\sqrt{np(1-p)} }=\frac{15.5-(200*0.1)}{\sqrt{(200*0.1)(1-0.1)} }=-1.06$

From the normal distribution table:

P(X ≥ 16) = 1 - P(X ≤ 15.5) = 1 - P(z < -1.06) = 1 - 0.1446 = 0.8554

c)

$0.95=P(X\geq n-185)=1-P(X\leq n-185)\\\\P(X\leq n-185)=1-0.95\\\\P(X\leq n-185)=0.05\\\\0.05=P(Z$