Question

The UDairy ice cream truck plans to sell ice cream at an upcoming alumni event in Central Park (NYC). From past events, demand for ice cream is highly weather dependent. On rainy days, demand is normally distributed with a mean of 120 scoops and standard deviation of 35 scoops. On non-rainy days, demand is normally distributed with a mean of 200 scoops and standard deviation of 60 scoops. The ice cream truck will be loaded two days before the event and travel to NYC. If the UDairy truck plans to bring enough ice cream to maintain a 90% service level for a rainy day, but the day surprisingly turns out to be non-rainy, what is the probability of the ice cream truck stocking out? For simplicity, assume the number of scoops demanded is a continuous variable. Enter probability as a three-digit decimal (e.g. 0.500, 0.275, 0.942).

Answer 1

Let X be the demand for icecream on a rainy day. X is normally distributed with mean scoops and standard deviation scoops.

To maintain a 90% serice level on a rainy day is the same as to have a probability of not running out of iceceam on a rainy day as 0.90. That means we want to load the truck with quantity q such that P(X<q) = 0.90.

From the standard normal table we can get the z value corresponding to P(Z<z) = 0.90. For z=1.28 we get P(Z<1.28) = 0.5+0.3997=0.90

By equating the z-score of q to 1.28 we get

If the truck brings 164.8 scoops then it can maintain a 90% service level for a rainy day.

But it turns out to be a non rainy day.

Let Y be the demand on a non-rainy day. Y has a normal distribution with mean scoops and standard deviation of scoops.

The truck will runout of stock if the demand Y is greater than q=164.8 scoops.

The  probability of the ice cream truck stocking out is

The  probability of the ice cream truck stocking out is 0.722