(Poisson) The demand for a particular type of pump at an isolated mine is random

Question

(Poisson) The demand for a particular type of pump at an isolated mine is random and independent of previous occurrences, but the average demand in a week (7 days) is for 2.8 pumps. Further supplies are ordered each Tuesday morning and arrive on the weekly plane on Friday morning. Last Tuesday morning only one pump was in stock, so the storesman ordered six more to come in Friday morning. (a) What is the average demand for 1 day? (b) What is the average demand for 3 days? (c) Find the probability that one pump will still be in stock on Friday morning when new stock arrives. Hint: From Tuesday morning to Friday morning is 3 days.

Explanation / Answer

Let W = weekly demand (for 7 days); D = daily demand and T = demand for 3 days….. (A)     

Back-up Theory

If a random variable X ~ Poisson(), i.e., X has Poisson Distribution with mean then

probability mass function (pmf) of X is given by P(X = x) = e – .x/(x!) …………............(1)

where x = 0, 1, 2, ……. ,

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

If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(), then Y ~ Poisson (k) …………….. .....................................(2)

Part (a)

Given W ~ Poisson(2.8), vide (A) and (1) above, D ~ Poisson(0.4).

Hence, average demand per day = 0.4 ANSWER

Part (b)

Given W ~ Poisson(2.8), vide (A) and (1) above, T ~ Poisson(1.2).

Hence, average demand for 3 days = 1.2 ANSWER

Part (c)

One pump will be in stock on Friday morning, if during the 3-day period from Tuesday to Friday, there is no demand for the pump. Hence, the required probability = P(T = 0)

= e- 1.2

= 0.3012 ANSWER

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.