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?

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:

[tex]X \sim N(\mu=500,\sigma=300)[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

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:

[tex]P(X>800)[/tex] and [tex]P(X<800)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]z=\frac{800-500}{300}=1[/tex]

And if we find:

[tex]P(X<800)=P(Z<1)=0.841[/tex]

And by the complement rule:

[tex]P(X>800)=1-P(X<800)= 1-0.841=0.159[/tex]