Question

POISSON PROBABILITY DISTRIBUTION

Airline passengers arrive randomly and independently at the passenger check-in counter. The average arrival rate is 10 passengers per minute.

to. Calculate the probability that no passenger will arrive within one minute.

    μ = 10 in 1 minute
x = 0 in 1 minute

b. Find the probability that three or fewer passengers will arrive within one minute.

    μ = 10 in 1 minute
x ≤ 3 in 1 minute >>> P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)

c. That no passenger arrives within 15 seconds.

    μ = 2.5 in 15 seconds
x = 0 in 15 seconds

Answer 1

Answer:-

Given That:-

Airline passengers arrive randomly and independently at the passenger check-in counter. The average arrival rate is 10 passengers per minute.

Let X denote the arrival time passengers such that X ~ poission (10)

i.e, Arival rate is 10 per minute

So,

(a). Calculate the probability that no passenger will arrive within one minute.

μ = 10 in 1 minute
x = 0 in 1 minute

We need to find P[No passenger within 1 minute] = P[X = 0]

P(X = 0) = 0.000045

(b). Find the probability that three or fewer passengers will arrive within one minute.

μ = 10 in 1 minute
x ≤ 3 in 1 minute >>> P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)

We need to find P[3 or less passengers within 1 minute] =

Now,

P(X = 0) = 0.000045 From part (a)

= 0.00045

= 0.00227

= 0.007567

= 0.000045 + 0.00045 + 0.00227 + 0.007567

= 0.010332

(c). That no passenger arrives within 15 seconds.

μ = 2.5 in 15 seconds
x = 0 in 15 seconds

We need to find P(no passenger with 15 seconds)

Arrival rate for 60 seconds = 10

So,

Arrival rate for 15 seconds = 10/60 * 15 = 2.5

= exp(-2.5)

P(X = 0) = 0.082085