Question

irline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 8 passengers per minute. a. Compute the probability of no arrivals in a one-minute period (to 6 decimals). b. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals). c. Compute the probability of no arrivals in a -second period (to 4 decimals). d. Compute the probability of at least one arrival in a -second period (to 4 decimals).

Answer 1

Answer:

A. 0.000045

B. 0.010336

C. 0.082085

D. 0.917915

Step-by-step explanation:

A. Compute the probability of no arrivals in a one-minute period.

P(x=o) = 0.000045

B. Compute the probability that three or fewer passengers arrive in a one-minute period.

p(x = 3) = 0.007567

p(x = 2) = 0.002270

p(x = 1) = 0.000454

p(x ≤ 3) = 0.000045 + 0.007567 + 0.002270 + 0.000454 = .010336

C. Compute the probability of no arrivals in a 15-second period.

p(x = 0) = 0.082085

D. Compute the probability of at least one arrival in a 15-second period.

p(x ≥ 1) = 1.0 - 0.082085 = 0.91791561

Answer 2

Answer:

a) 0.000335

b) 0.0424

c) 0.8751

d) 0.1248

Step-by-step explanation:

This is the case of Poisson's Distribution. Here,

μ = mean = 8 passengers/min

The Poisson's formula is:

P(X) = (e^-μ)(μ^x)/(x!)

a)

For no arrival x= 0

P(X = 0) = (e^-8)(8^0)/(0!)

P(X = 0) = 0.000335

b)

For probability of three or fewer, X ≤ 3

P (X ≤ 3) = ∑[(e^-8)(8^x)/(x!)]   (from x = 0 to x = 3)

P (X ≤ 3) = 0.0424

c)

Now, for second time we convert the mean to seconds:

μ = mean = (8 passengers/min)(1 min/60 sec)

μ = mean = 0.13333 passengers/sec

For no arrival x= 0

P(X = 0) = (e^-0.13333)(0.13333^0)/(0!)

P(X = 0) = 0.8751

d)

For at least one arrival X ≥ 1

P(X ≥ 1) = 1 - P(X = 0)

P(X ≥ 1) = 1 - 0.8751

P(X ≥ 1) = 0.1248