Question

A sales completion team, aiming to reduce the shipment time of urgent orders, studies the current process and finds that the current shipment time has a mean of 4 days with a standard deviation of 1.5 days. Industry expectations are for urgent shipments to be delivered between 1 and 5 days.

Answer 1

Answer:

$P(1$

And we can find this probability with thi difference:

$P(-2$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-2$

So then we expect about 72.5% of the current shipment times between 1 and 5 days

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

Let X the random variable that represent the current shipment time of a population, and for this case we know the distribution for X is given by:

$X \sim N(4,1.5)$  

Where $\mu=4$ and $\sigma=1.5$

We are interested on this probability

$P(1$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(1$

And we can find this probability with thi difference:

$P(-2$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-2$

So then we expect about 72.5% of the current shipment times between 1 and 5 days