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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.

1 Answer

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Answer:


P(1<X<5)=P((1-\mu)/(\sigma)<(X-\mu)/(\sigma)<(5-\mu)/(\sigma))=P((1-4)/(1.5)<Z<(5-4)/(1.5))=P(-2<z<0.667)

And we can find this probability with thi difference:


P(-2<z<0.667)=P(z<0.667)-P(z<-2)

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


P(-2<z<0.667)=P(z<0.667)-P(z<-2)=0.748-0.023=0.725

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

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<X<5)

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


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

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


P(1<X<5)=P((1-\mu)/(\sigma)<(X-\mu)/(\sigma)<(5-\mu)/(\sigma))=P((1-4)/(1.5)<Z<(5-4)/(1.5))=P(-2<z<0.667)

And we can find this probability with thi difference:


P(-2<z<0.667)=P(z<0.667)-P(z<-2)

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


P(-2<z<0.667)=P(z<0.667)-P(z<-2)=0.748-0.023=0.725

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

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