Respuesta :
Answer:
[tex]P(1<X<5)=P(\frac{1-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{1-4}{1.5}<Z<\frac{5-4}{1.5})=P(-2<z<0.667)[/tex]
And we can find this probability with thi difference:
[tex]P(-2<z<0.667)=P(z<0.667)-P(z<-2)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<0.667)=P(z<0.667)-P(z<-2)=0.748-0.023=0.725[/tex]
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:
[tex]X \sim N(4,1.5)[/tex]
Where [tex]\mu=4[/tex] and [tex]\sigma=1.5[/tex]
We are interested on this probability
[tex]P(1<X<5)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(1<X<5)=P(\frac{1-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{1-4}{1.5}<Z<\frac{5-4}{1.5})=P(-2<z<0.667)[/tex]
And we can find this probability with thi difference:
[tex]P(-2<z<0.667)=P(z<0.667)-P(z<-2)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<0.667)=P(z<0.667)-P(z<-2)=0.748-0.023=0.725[/tex]
So then we expect about 72.5% of the current shipment times between 1 and 5 days
