Question

EXAMPLE 7: A gasoline distributor is capable of receiving, under current conditions, a maximum of three trucks per day. If more than three trucks arrive, the excess must be sent to another distributor, and in this case there is an average loss of $ 800.00 per day in which all trucks can not be accepted.
(a) What is the probability of three to five trucks arriving in the
total of two days?
b) What is the probability that, on a certain day, you have to send
trucks to another distributor?
c) What is the average monthly loss (thirty days) due to trucks that
could not be accepted?

Answer 1

Let X be the number of trucks arrive in a day. Assuming Poisson distribution, X ~ Poisson( = 3)

(a)

For 2 days, average number of trucks arrival, = 2 * 3 = 6

Let Y be the number of trucks arrive in two days. Assuming Poisson distribution, Y ~ Poisson( = 6)

Probability of three to five trucks arriving in the total of two days = P(X = 3) + P(X = 4) + P(X = 5)

= 6³ * exp(-6) / 3! + 6⁴ * exp(-6) / 4! + 6⁵ * exp(-6) / 5!

= 0.0892 + 0.1339 + 0.1606

= 0.3837

(b)

Probability that, on a certain day, you have to send trucks to another distributor = P(X > 3)

= 1 - P(X 3)

= 1 - [3⁰ * exp(-3) / 0! + 3¹ * exp(-3) / 1! + 3² * exp(-3) / 2!  + 3³ * exp(-3) / 3! ]

= 1 - (0.04978707 + 0.14936121 + 0.22404181 + 0.22404181)

= 0.3528

(c)

Expected Daily loss = Probability that, on a certain day, you have to send trucks to another distributor * loss per day

= 0.3528 * $ 800.00

= $282.24

Average monthly loss = 30 * Expected Daily loss = 30 * $282.24

= $8467.2