3. A flight hold 200 passengers. An airline makes 215 reservations. They estimat

Question

3. A flight hold 200 passengers. An airline makes 215 reservations. They estimate the probability that a passenger with a reservation shows up for the flight to be equal to 0.95, independently of all other passengers. a) Approximate (using normal approximation to binomial) the probability that all pas- sengers arriving for the flight can be seated. Compare the result to the exact one. b) Approximate the probability that the flight has empty seats. Compare the result to the exact one. c) Approximate the number of reservation that can be made so that the probability that every arriving passenger can be seated is at least 0.8. Compare the answer with that obtained without using the approximation

Explanation / Answer

x = number of passenger who show up
X follow binomial distribution with n = 215
p = 0.95
mean = np = 204.25
sd =sqrt(npq) = sqrt(10.2125) = 3.1957

a)
P(X<= 200)
= P(X < 200.5)
=P(Z< ( 200.5 - 204.25)/3.1957)
=P(Z< -1.173451)
= 0.1203

b)
P(X < 200)
= P(X<199.5)
=P(Z< ( 199.5 - 204.25)/3.1957)
=P(Z< -1.48637)
=0.0686

c)
let the required number be n
P(X <= 200) > 0.8
if n = 208
P(X<= 200) = 0.820569
if n = 209
then
P(X<= 200) = 0.7219
hence
n = 208

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