Question

(a) Suppose that in the long run, 5% of ticket holders with a certain airline do not show up for their flight. If the airline sells 181 tickets for a flight that only has seats for 175 passengers, estimate the probability that everyone who shows up for this flight will get a seat. (Use the continuity correction). (b) The amount of time that a bank teller spends dealing with a random customer follows an exponential distribution with a mean of 5 minutes. Estimate the probability that the total amount of time that the teller spends dealing with 40 customers is under 3 hours.

Answer 1

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then probability mass function (pmf) of X is given by p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n ..(1)

[The above probability can also be directly obtained using Excel Function: Statistical, BINOMDIST………………………...(1a)

If X ~ B(n, p), np ≥ 5 and np(1 - p) ≥ 5, then Binomial probability can be approximated by Standard Normal probabilities by Z = (X – np)/√{np(1 - p)} ~ N(0, 1) ……….............................................................................................………………………....(3)

Continuity Correction

If X ~ B(n, p), P(X ≤ k) can be approximated by the Normal probability by

P[Z ≤ (k + 0.5 – np)/√{np(1 - p)}], where Z ~ N(0, 1) ……....................................................………………………………….. (4)

Adding 0.5 to the k value is referred as Continuity Correction

If X ~ Exponential with parameter β (average inter-event time), the pdf (probability density function) of X is given by f(x) = (1/β)e^-x/β, 0 ≤ x < ∞ …………………………….............................................................................................……….…(6)

CDF (cumulative distribution function), F(t) = P(X ≤ t) = 1- e^-t/β ………………………......................................................…(7)

From (2), P(X > t) = e^-t/β ……………………….....................................................….………………………………………….(8)

If X ~ Exp(β) and Y = kX, then Y Exp(kβ) …….………........................................................………………………………….(9)

Now to work out the solution,

Part (a)

Let X = Number of ticket holders out of 181 tickets sold, who show up. Then, X ~ B(181, p), where

p = probability of a ticket holder showing up.

Given, ‘in the long run, 5% of ticket holders do not show up’, p = 1 – 0.05 = 0.95.

Given the aircraft capacity is 175, probability everyone showing up gets a seat

= P(X ≤ 175)

= P[Z ≤ (175.5 – 171.95)/2.9321] [vide (4), also noting, n = 181, p = 0.95 and hence (1 - p) = 0.05]

= P(Z ≤ 1.211)

= 0.8871 [Using Excel Function: Statistical NORMSDIST] Answer

Part (b)

Let Y = Time the bank teller spends dealing with a customer and T = Total time the bank teller spends dealing with 40 customers.

Given X ~ Exp(5 min), vide (9), Y ~ Exp(3.3333 hours) [40 x 5 = 200 minutes = 3.3333 hours]

So, probability the total time the bank teller spends dealing with 40 customers is under 3 hours

= P(T < 3)

= 1- e^-3/3.3333 [vide (7)]

= 1- e^-0.9

= 0.5934 Answer

DONE