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 in-stock probability if Store A’s order quantity is 800 units?

Answer 1

`P(X<800)=P(Z<1)=0.841`

Explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the Demand for its product on this case, and for this case we know the distribution for X is given by:

`X \sim N(\mu=500,\sigma=300)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

What is its in-stock probability if Store A’s order quantity is 800 units?

We are looking for this probability:

What is its in-stock probability if Store A’s order quantity is 800 units?

So we can find the following values:

`P(X>800)` and
`P(X<800)`

Sor this problem we can use the z score formula given by:

`z=(x-\mu)/(\sigma)`

If we find the z score for the value 800 we got:

`z=(800-500)/(300)=1`

And if we find:

`P(X<800)=P(Z<1)=0.841`

And by the complement rule:

`P(X>800)=1-P(X<800)= 1-0.841=0.159`