Question

Data from the past shows that on average, a ready-mixed concrete plant receives 100 orders for concrete every year. The maximum number of orders that the plant can fulfil each week is 2. (a) What is the probability that in a given week the plant cannot fulfil all the placed orders? (b) Assume the answer to part (a) is 20% (It is not; I just want to make sure that everybody uses the same number for part (b)). Suppose there are 5 of such plants. What is the probability that in a given week 2 of the plants cannot fulfill their orders?

Answer 1

ANSWER::

number of orders received annually = 100

Assuming there are 52 weeks in a year.

Number of Orders recieved per week = 100/52 = 1.923

Maximum number of order a plant can fulfill in a week = 2.

So, in any random week the plant will fail to fulfill the orders received if the number of orders received per week exceeds 2.

Now, assuming the number of orders received per week (X) follows poisson distribution with mean(lambda) equal to 1.923 orders per week.

Part b)

Now,

probability that a plant fails to fulfill orders in a week = p = 0.2

q = 1- p = 0.8

Number of trials = number of plants = n = 5

Since there are only two possible outcomes either the plant fails to fulfill orders or fulfill the orders and also the the trails are independent from each other.

Hence this is a typical binomial probability case.

So, proabability that 2 of 5 plants in a given week fails to fulfill the orders received

= P(X = 2)

= 5C2 × ( 0.2)2 × (0.8)3

= 0.2048 or 20.48 %

So, there is 20.48% probability that in a given week 2 out of 5 plants will fail to fulfill their Orders that they received.

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