Question 5.3: Web-based Marketing Study A web-based travel agency uses its......

(a) The random variable x = number of web-based enquiries received per day ‘fits’ the Poisson process for the following reasons:

• The random variable is discrete. The agency can receive {0, 1, 2, 3, 4, …} web-based enquiries per day. Any number of enquiries can be received per day.
• The random variable is observed in a predetermined time interval (i.e. one day).
• The average number of web-based enquiries per day, a, is known. Here λ = 5.

Find P(x = 3) when λ (average number of web-based enquiries per day) = 5.

P(x = 3) = \frac{e^{-5}\cdot 5^3}{3!} = 0.006738 × 20.833 = 0.1404

Thus there is only a 14.04% chance that the travel agency will receive only three web-based enquiries on a given day when the average number of web-based enquiries per day is five.

(b) In terms of the Poisson question, ‘at most two web-based enquiries’ implies either 0 or 1 or 2 enquiries on a given day.

These possible outcomes (0, 1, 2) are mutually exclusive, and the combined probability can be found using the addition rule of probability for mutually exclusive events. Thus:

P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)

Each probability is calculated separately using Formula 5.3.

P(x = 0 web-based enquiries):

P(x = 0) = \frac{e^{-5}\cdot 5^0}{0!} = 0.006738(1) = 0.00674                (Recall 0! = 1.)

P(x = 1 web-based enquiry):

P(x = 1) = \frac{e^{-5}\cdot 5^1}{1!} = 0.006738(5) = 0.0337

P(x = 2 web-based enquiries):

P(x = 2) = \frac{e^{-5}\cdot 5^2}{2!} = 0.006738(12.5) = 0.0842

Then P(x ≤ 2) = 0.00674 + 0.0337 + 0.0842 = 0.12464.

Thus there is only a 12.5% chance that at most two web-based enquiries for travel packages will be received by this travel agency on a given day, when the average number of web-based enquiries received per day is five.

(c) The Poisson question requires us to find P(x > 4) when λ = 5.

Since x is a discrete random variable, the first integer value of x above four is x = 5.

Hence the problem becomes one of finding:

P(x ≥ 5) = P(x = 5) + P(x = 6) + P(x = 7) + …

To solve this problem, the complementary rule of probability must be used. The complement of x ≥ 5 is all x ≤ 4. Thus:

P(x ≥ 5) = 1 − P(x ≤ 4)

= 1 − [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)]

= 1 − [0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755]            (Formula 5.3)

= 1 – 0.4405 = 0.5595

Thus there is a 55.95% chance that the travel agency will receive more than four web-based enquiries for their travel packages on a given day, when the average number of web-based enquiries received per day is five.

(d) Notice that the time interval over which the web-based enquiries are received has changed from one day to two days. Thus λ = 5 per day, on average, must be adjusted to λ = 10 per two days, before the Poisson formula can be applied.

Then find P(x > 4) where λ = 10.

P(x > 4) = P(x ≥ 5) = [P(x = 5) + P(x = 6) + P(x = 7) + … + P(x = ∞)]

This can be solved using the complementary rule as follows:

P(x > 4) = 1 − P(x ≤ 4)

= 1 − [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)]

Each Poisson probability is calculated separately using Formula 5.3 with λ = 10.

P(x > 4) = 1 − [0.0000454 + 0.000454 + 0.00227 + 0.00757 + 0.01892]

= 1 – 0.02925 = 0.97075

Thus there is a 97.1% chance that the travel agency will receive more than four webbased enquiries for their travel packages in any two-day period when the average number of web-based enquiries received is 10 per two-day period.