Passengers of the area lines arrive at random and independently to the documenta
ID: 3220492 • Letter: P
Question
Passengers of the area lines arrive at random and independently to the documentation section at the airport, the average arrival frequency is 0.4 passengers per minute.a. What is the probability that three or fewer passengers arrive within five minutes? b. What is the probability of three or more passengers arriving within fifteen minutes. c. What is the expected value of passengers arriving within five hours? Passengers of the area lines arrive at random and independently to the documentation section at the airport, the average arrival frequency is 0.4 passengers per minute.
a. What is the probability that three or fewer passengers arrive within five minutes? b. What is the probability of three or more passengers arriving within fifteen minutes. c. What is the expected value of passengers arriving within five hours?
a. What is the probability that three or fewer passengers arrive within five minutes? b. What is the probability of three or more passengers arriving within fifteen minutes. c. What is the expected value of passengers arriving within five hours? a. What is the probability that three or fewer passengers arrive within five minutes? b. What is the probability of three or more passengers arriving within fifteen minutes. c. What is the expected value of passengers arriving within five hours? b. What is the probability of three or more passengers arriving within fifteen minutes. c. What is the expected value of passengers arriving within five hours? c. What is the expected value of passengers arriving within five hours?
Explanation / Answer
Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where
= parameter of the distribution.
x = is the number of independent trials
the average arrival frequency is 0.4 passengers per minute
a.
expected number of passengers for 5 minuete the average arrival rate is 5 * 0.4 = 2
P( X < = 3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= e^-2 * 2 ^ 3 / 3! + e^-2 * 4 ^ 2 / 2! + e^-2 * ^ 1 / 1! + e^-2 * ^ 0 / 0!
= 0.8571
b.
expected number of passengers for 5 minuete the average arrival rate is 15 * 0.4 = 6
P( X < 3) = P(X=2) + P(X=1) + P(X=0)
= e^-6 * 4 ^ 2 / 2! + e^-6 * ^ 1 / 1! + e^-6 * ^ 0 / 0!
= 0.062
P( X > = 3 ) = 1 - P (X < 3) = 0.938
c.
expeccetd number of passenger for 5 hours is = 300/0.4 = 750