Question

A shopkeeper hires vacuum cleaners to the general public at 5$ per day. The mean daily demand is 2.6. Suppose the demand follows Poisson distribution. (a) Calculate the expected daily income from this activity assuming an unlimited number of vacuum cleaners is available (b) Find the probability that the demand on a particular day is: 0, exactly 1, exactly 2, exactly 3 or more than 3; (c) If only 3 vacuum cleaners are available for hire calculate the mean of the daily income. A nearby large store is willing to lend vacuum cleaners at short notice to the shopkeeper, so that she is able to meet the demand. However, this shop will charge £2 per day for this service regardless of how many, if any, cleaners are borrowed. Would you advice the shopkeeper to take this offer?

Answer 1

Solution

NOTE: The question gives two monetary units:

‘A shopkeeper hires vacuum cleaners to the general public at 5$ per day.’

‘this shop will charge £2 per day for this service’

$ is taken as the common monetary unit.

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, ……. , ∞

Values of p(x) for various values of λ and x can be obtained by using Excel Function,

Statistical, POISSON…………………………………………………………….…...……. (1a)

Mean = λ ……………………………………………………………………………....…….. (2)

Expected value, E(X) = Σx.p(x), where p(x) is probability of x. ………………………… (3)

E(aX) = aE(X) ……………………………………………………………………...……….. (3a)   

Now to work out the solution,

Let X = daily demand for vacuum cleaners. Then, we are given X ~ Poisson (λ), where λ = 2.6 given [The mean daily demand is 2.6. ] ………………………………………………………….. (4)

The daily income, I = $5X ……………………………………………………………...…… (5)

[given, ‘hires vacuum cleaners to the general public at 5$ per day’]

Part (a)

Expected daily income = E(I)

= 5E(X) [vide (3a) and (5)]

= 5 x 2.6 [vide (2) and (4)]

= $13 Answer 1

Part (b)

Probability that the demand on a particular day is 0

= P(X = 0)

= 0.0743 [vide (1a)] Answer 2

Probability that the demand on a particular day is exactly 1

= P(X = 1)

= 0.1931 [vide (1a)] Answer 3

Probability that the demand on a particular day is exactly 2

= P(X = 2)

= 0.2510 [vide (1a)] Answer 4

Probability that the demand on a particular day is exactly 3

= P(X = 3)

= 0.2176 [vide (1a)] Answer 5

Probability that the demand on a particular day is more than 3

= P(X > 3)

= 1 – {P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)}

= 0.2640 [vide Answers 2 to 5] Answer 6

Part (c)

If only 3 vacuum cleaners are available for hire calculate the mean of the daily income

= 5 x Σx.p(x), x = 0, 1, 2, 3 [Demand of more than 3 cannot be met.]

= (0 x 0.0734) + (1 x 0.1931) + (2 x 0.2510) + (3 x 0.2176)

= 5 x 1.3479

= $6.74 Answer 7

Part (d)

Vide Answer 6, P(X > 3) = 0.2640

This implies that with just 3 cleaners, expected loss = 5 x 0.2640 = $1.320

But, lending changes per day is $2 which is more than the loss. So, it is not worthwhile taking the lending offer. Answer 8

DONE