Question

You have been hired as a consultant by the Roi du Poisson Seafood Market to work on problems related to inventory. Your first assignment is to determine the size of their weekly order of pickled crappie balls. Suppose that the store has thousands of customers, but only sells an average of eight jars of pickled crappie balls per week. Assume the store is open the same number of hours every day.

(a) Assuming six jars are in stock at the beginning of the week, what is the probability of not having enough jars to meet demand that week?

(b) What is the probability the first sale will occur in the second half of the week?

(c) Assuming no jars are sold in a particular week, what is the average amount of time until the first sale the following week?

Answer 1

We have been hired as consultants by the Roi du Poisson Seafood Market to work on their problems related to inventory.

We have been told that the average number of jars sold a week is eight (8). Thus,

Assumption: The store is open the same number of hours every day.

a.

We are told that we have six jars stocked up. Find the probability that we will not meet the demand that week, that is, the probability that the sale is greater than six jars.

Or, we can find the probability of 6 or less than 6 and subtract it from 1 to get the probability of sale greater than 6. The formula for a poisson distribution is:

PMF ,P (X=r)= (e^{- *r} ) / r!

CDF, P(X<=r)=  e^- * _iⁱ ) / i !

Therefore, P( X<=6)=0.0003354626* ( 8^0)/1 + (8^1)/1 + (8^2)/2 + (8^3)/6 + (8^4)/24 + (8^5)/120 + (8^6)/720)

= 0.0003354626* (1+ 8+ 32+ 85.33333+ 170.6667+ 273.0667+ 364.0889)

= 0.0003354626* 934.1556= 0.3133743

Thus, there is a 31.33743 % of chance that the demand will be 6 or less than 6.

Thus, the probability that we will run out of jars is 100% - 31.33743%= 68.66257%

b.

We know that the average calls in the first half of the week is 8/2= 4. This we know because the store is open for the same number of hours everyday.

Thus, the probability of failure is the probability of getting 0 calls in the first half of the week, and the probability of success is (1- the probability of failure). Thus,

q= probability of failure= getting 0 calls when the average is 4

p= probability of success= 1-q

Thus, q= (e^{- *r} ) / r!= (e^{-4 *}4⁰ ) / 0!= 0.01831564*1= 0.01831564

Thus, p= 1-0.01831564= 0.9816844

Thus, the probability that the first call will be in the second half of the week is 0.01831564* 0.9816844

= 0.01798018= 1.79%= 1.8%