EXAMPLE 7: A gasoline distribu

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

Answer:

Let X be the number of trucks arrive in a day. Assuming Poissondistribution, 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. AssumingPoisson distribution, Y ~ Poisson( = 6)

Probability of three to five trucks arriving in the total of twodays = 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 toanother 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, youhave 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


 
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