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

Answer:

$P(X$

Step-by-step 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:  

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

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 $P(X$

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

$z=\frac{x-\mu}{\sigma}$

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

$z=\frac{800-500}{300}=1$

And if we find:

$P(X$

And by the complement rule:

$P(X>800)=1-P(X$