For the coming season, Specialty Toys, Inc., plans to introduce a new product ca
ID: 3170484 • Letter: F
Question
For the coming season, Specialty Toys, Inc., plans to introduce a new product called Weather Teddy. Specialty’s managers claimed Weather Teddy gave predictions that were as good as many local TV weather forecasters!
Now, Specialty faces the decision of how many Weather Teddy units to order from manufacturers for the coming holiday season. Members of the management team suggested four order quantities of 15,000, 18,000, 24,000, or 28,000 units. The team asks you for an analysis of the stock-out probabilities for various order quantities, an estimate of the profit potential, and to help make an order recommendation.
Specialty expects to sell Weather teddy for $24 based on the unit cost of $20. If inventory remains after holiday season, Specialty will sell all surplus inventory for $5 per unit. After reviewing the sales history of similar products, Specialty’s senior sales forecaster predicted that the demand would be normally distributed with the mean as 20,000 units and that demand would be between 10,000 and 30,000 units with a 90% probability.
Compute the probability of a stock-out (i.e. quantity demanded is greater than the order quantity) for each order quantity suggested by the management team
Explanation / Answer
Solution
Back-up Theory
If X ~ N(µ, 2), P(X or t) = P[Z or {(t - µ))}], where Z ~ N(0, 1) whose probability values can be readily read off from Standard Normal Tables. …………………….(1)
Now, to work out the solution,
Let X = demand.
‘demand would be normally distributed with the mean as 20,000 units and that demand would be between 10,000 and 30,000 units with a 90% probability’ =>
P(10000 X 30000) = 0.90 given mean 20000
=> P[{(10000 – 20000)/} Z {(30000 – 20000)/}) = 0.90 [vide (1)]
=> P[{(– 10000)/} Z {(10000)/}) = 0.90
=> from Standard Normal Tables, (10000)/} = 1.645 or = 6079.03 ……… (2)
P(stock-out) = P(X > Q), where Q = order quantity
Part (a): Q = 15000
P(stock-out) = P(X > 15000) = P(Z > - 5000/6079.03) [vide (1)]
= P(Z > - 0.8225) = 0.7946 ANSWER [Standard Normal Tables]
Part (b) Q = 18000
P(stock-out) = P(X > 18000) = P(Z > - 2000/6079.03) [vide (1)]
= P(Z > - 0.329) = 0.6290 ANSWER [Standard Normal Tables]
Part (c) Q = 24000
P(stock-out) = P(X > 24000) = P(Z > 4000/6079.03) [vide (1)]
= P(Z > 0.658) = 0.2552 ANSWER [Standard Normal Tables]
Part (d) Q = 28000
P(stock-out) = P(X > 28000) = P(Z > 8000/6079.03) [vide (1)]
= P(Z > 1.316) = 0.0941 ANSWER [Standard Normal Tables]