Question

The mean number of arrival at an airport during rush hour is 20 planes per hour while the mean number of departures is 30 planes per hour. Let us suppose that the arrivals and departures can each be described by a respective poisson process. The number of passengers in each arrival or departure has a mean of 100 and a coefficient of variation of 40%.

a.) What is the probability that there will be a total of two arrivals and/or departures within a 6-minute period?

b.) Suppose that in the last hour there have been 25 plane-arrivals.

i.) What is the mean and variance of the total number of arriving passengers in the last hour?

ii.) What is the probability that the total number of arriving passengers exceeded 3000 in the last hour? State and justify all assumptions made.

Answer 1

Solution

Back-up Theory

If a random variable X ~ Poisson (λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ……...............................……..(1)

where x = 0, 1, 2, ……. , ∞

This probability can also be obtained by using Excel Function, Statistical, POISSON …................ (1a)

If X = number of times an event occurs during period t, Y = number of times the same

event occurs during period kt, and X ~ Poisson(λ), then Y ~ Poisson (kλ) …………....................….. (2)

Coefficient of variation (CV) = 100 x standard deviation/mean or

standard deviation = mean x CV/100 ............................................................................................(3)

CENTRAL LIMIT THEOREM

Let {X₁, X₂, …, X_n} be a sequence of n independent and identically distributed (i.i.d) random variables drawn from a distribution [i.e., {x₁, x_2, …, x_n} is a random sample of

size n] of expected value given by µ and finite variance given by σ². Then, as n gets

larger, the distribution of Z = {√n(Xbar − µ)/σ}, approximates the normal distribution

with mean 0 and variance 1 (i.e., Standard Normal Distribution)

Or symbolically, Z = {√n(Xbar − µ)/σ} ~ N(0, 1) i.e., sample average from any

distribution with mean µ and variance σ², follows Normal Distribution with mean µ

and variance σ²/n, if the sample size, n is large enough, say 30 or more........................ (4a)

and hence sample total follows Normal Distribution with mean nµ and variance nσ² ….....…… (4b)

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ²,

then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .……….....................………...…(5a)

The above probability can be directly read off from Standard Normal Tables........................... (5b)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ...........…(5c)

Now to work out the solution,

Let

X_A = number of planes arriving per hour

X_D = number of planes departing per hour

Then, we are given:

X_A ~ Poisson (20) ........................................................................................................... (6a)

X_D ~ Poisson (30) ........................................................................................................... (6b)

Part (a)

Let

Y_A = number of planes arriving within a 6-minute period

Y_D = number of planes departing within a 6-minute period

Then, vide (2), (6a) and (6b),

Y_A ~ Poisson (2) [6 minutes = 1/10^th of an hour] ............................................................ (7a)

Y_D ~ Poisson (3) [6 minutes = 1/10^th of an hour] ............................................................ (7b)

For 2 events, A and B,

P(A or B or both) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B), in general and …………………………..............…(8a)

= P(A) + P(B) – {P(A) x P(B)}, when A and B are independent. ................................................... (8b)

Now,

Probability that there will be a total of two arrivals and/or departures within a 6-minute period

= P[(Y_A = 2) ∪ (Y_D= 2]

= P[Y_A = 2] + P[Y_D - {P[Y_A = 2] x P[Y_D = 2]} [vide (8b)]

= 0.2707 + 0.2240 – (0.2707 x 0.2240) [vide (7a), (7b) and (1a)]

= 0.4333 Answer 1

Part (b)

Given that in the last hour there have been 25 plane-arrivals and the number of passengers in each arrival has a mean of 100 and a coefficient of variation of 40% i.e., standard deviation = 40 [vide (3)], if T = total number of arriving passengers in the

last hour, then vide (4b),

T ~ N(2500, 40000) [note 40000 is the variance or standard deviation = 200]..... (9).

Sub-part (i)

Vide (9)

mean of the total number of arriving passengers in the last hour = 2500 Answer 2

variance of the total number of arriving passengers in the last hour = 40000 Answer 3

Sub-part (ii)

Probability that the total number of arriving passengers exceeded 3000 in the last hour

= P(T >3000)

= P[Z > {(3000 - 2500)/200}] [vide (5a) and (9)]

= P(Z > 2.5)

= 0.0062 [vide (5c)] Answer 4

Justification and assumptions made

Central Limit theorem [vide (4a) and (4b)] Answer 5

DONE