Question

QUESTION B

An ice-cream vendor on the beachfront knows from long experience that the average rate of ice-cream sales is 12 per hour. If, with two hours to go at work, she finds herself with only five ice-creams in stock, what are the probabilities that (a) She runs out before the end of the day; (b) She sells exactly what she has in stock by the end of the day without any excess demand after she sells the last one; and (c) She doesn't sell any?

Answer 1

Solution

Let X = number of ice-creams demanded per hour.

We assume X ~ Poisson (λ), where λ = average number of ice-creams demanded per hour.

Given λ = 12 (given) …………………………………………………………………(1)

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!) …………..(2)

where x = 0, 1, 2, ……. , ∞

Values of p(x) for various values of λ and x can be obtained by using Excel Function, POISSON(x,Mean,Cumulative) ……………………………………………………(2a)

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) …………….. (3)

Now, to work out the answer,

Preparatory Work

Let Y = number of ice-creams demanded per 2 hours. Then, vide (3) and (1),

Y ~ Poisson (24) …………………………………………………………………(4)

Part (a)

With two hours to go at work, if the ice-cream vendor finds herself with only five ice-creams in stock, the probabilities that she runs out before the end of the day

= P(Y > 5)

= 1 – 0.00000313 [vide (2a)]

= 0.99999687 ANSWER

Part (b)

With two hours to go at work, if the ice-cream vendor finds herself with only five ice-creams in stock, the probabilities that she sells exactly what she has in stock by the end of the day without any excess demand after she sells the last one

= P(Y = 5)

= 0.0000025 [vide (2a)] ANSWER

Part (c)

Probability she doesn't sell any

= P(Y = 0)

= 3.78E-11 [vide (2a)] ANSWER

DONE