Respuesta :
Complete question is;
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute.
a. Compute the probability of no arrivals in a one-minute period. Round your answer to six decimal places.
b. Compute the probability that three or fewer passengers arrive in a one-minute period. Round your answer to four decimal places.
c. Compute the probability of no arrivals in a 15-second period. Round your answer to four decimal places.
d. Compute the probability of at least one arrival in a 15-second period. Round your answer to four decimal places.
Answer:
A) 0.000045
B) 0.0099
C) 0.0821
D) 0.9179
Step-by-step explanation:
This is a Poisson probability question. We are given; μ = 10
Formula from Poisson probability is;
f(k) = [μ^(k) × e^(-μ)]/k!
A) probability of no arrivals in a one-minute period will be;
f(0) = [10^(0) × e^(-10)]/0!
f(0) = 0.00004539993
To six decimal places gives;
f(0) = 0.000045
B) the probability that three or fewer passengers arrive in a one-minute period is given by;
P(x ≤ 3) = f(0) + f(1) + f(2) + f(3)
We already have f(0) = 0.000045
f(1) = [10^(1) × e^(-10)]/1!
f(1) = 0.000454
f(2) = [10^(2) × e^(-10)]/2!
f(2) = 0.00227
f(3) = [10^(3) × e^(-10)]/3!
f(3) = 0.007567
Thus;
P(x ≤ 3) = 0.000045 + 0.00227 + 0.007567 = 0.009882
To four decimal places gives;
P(x ≤ 3) = 0.0099
C) no arrivals in a 15-second period means that;
μ = 10 × 15/60
μ = 2.5
Thus;
the probability of no arrivals in a 15-second period ia;
f(0) = [2.5^(0) × e^(-2.5)]/0!
f(0) = 0.082085
To four decimal places is;
f(0) = 0.0821
D) the probability of at least one arrival in a 15-second period will be gotten from the complement rule. Thus;
P(x ≥ 1) = 1 - f(0)
P(x ≥ 1) = 1 - 0.0821
P(x ≥ 1) = 0.9179
