Question

Store A uses the newsvendor model to manage its inventory. Demand for its product is normally distributed with a mean of 500 and a standard deviation of 300. What is its stockout probability if Store A’s order quantity is 800 units?
A. 97.7%
B. 84.1%
C. 15.9%
D. 2.3%

Answer 1

15.9% is its stock out probability if Store A’s order quantity is 800 units

Solution:

z-score = $\frac{(Order quantity-Mean)}{Standard deviation}$

             = (800-500)/300 = 1

So , in-stock probability = NORM.S.DIST(1,TRUE) = 0.841345

Hence stock-out probability = 1 - 0.841345

                                               = 0.158655

                                               = 15.9% (Approximately)