Dan McClure owns a thriving independent bookstore in artsy New Hope, Pennsylvani
ID: 379298 • Letter: D
Question
Dan McClure owns a thriving independent bookstore in artsy New Hope, Pennsylvania. He must decide how many copies to order of a new book, Power and Self-Destruction, an exposé on a famous politician’s lurid affairs. Interest in the book will be intense at first and then fizzle quickly as attention turns to other celebrities. The book’s retail price is $20, and the wholesale price is $12. The publisher will buy back the retailer’s leftover copies at a full refund, but McClure Books incurs $4 in shipping and handling costs for each book returned to the publisher. Dan believes his demand forecast can be represented by a normal distribution with a mean of 200 and a standard deviation of 80.
a.
Dan will consider this book to be a blockbuster for him if it sells more than 400 units. What is the probability that Power and Self-Destruction will be a blockbuster?
b.
Dan considers a book a “dog” if it sells less than 50 percent of his mean forecast. What is the probability this exposé is a “dog”?
c.
What is the probability that demand for this book will be within 20 percent of the mean forecast?
d.
What order quantity maximizes Dan’s expected profit?
a.
Dan will consider this book to be a blockbuster for him if it sells more than 400 units. What is the probability that Power and Self-Destruction will be a blockbuster?
b.
Dan considers a book a “dog” if it sells less than 50 percent of his mean forecast. What is the probability this exposé is a “dog”?
c.
What is the probability that demand for this book will be within 20 percent of the mean forecast?
d.
What order quantity maximizes Dan’s expected profit?
Explanation / Answer
a)
to be a blockbuster: books sold (X) = 400
Mean M = 200
Standard deviation s = 80
zscore = (X-M)/s
z = (400-200)/80
z = 2.5
corresponding % value from z-table = 0.9938 = 99.38%
hence probability of selling more than 400 units = 100-99.38 = 0.62%
b)
less than 50% of mean forecast X = 200/2 = 100
Mean M = 200
Standard deviation s = 80
zscore = (X-M)/s
z = (100-200)/80
z = -1.25
corresponding % value from z-table = 0.1056 = 10.56%
Hence, probability of the book being a dog = 10.56%
c)
20% of mean forecast = 20%*200 = 40
Upper value = 200+40 = 240
lower value = 200-40 = 160
Lower value:
X= 160
zscore = (X-M)/s
z = (160-200)/80
z = -0.5
corresponding % value from z-table = 0.3085 = 30.85%
hence probability of sales below 160 = 30.85%
similarly the z score of upper value will be 0.5, the probability of sales above 240 = 30.85%
Now the probability of sales being within 20 percent of the mean forecast i.e. between 160 and 240
= 100-30.85-30.85 = 38.3%
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