Respuesta :
Answer:[tex]P(28.5<X<31.5)=P(\frac{28.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{31.5-\mu}{\sigma})=P(\frac{28.5-32}{5}<Z<\frac{31.5-32}{5})=P(-0.7<z<-0.1)[/tex]And we can find this probability with this difference:
[tex]P(-0.7<z<-0.1)=P(z<-0.1)-P(z<-0.7)[/tex]
And in order to find these probabilities we use tables for the normal standard distribution, excel or a calculator.
[tex] P(-0.7<z<-0.1)=P(z<-0.1)-P(z<-0.1) =0.4602-0.2420=0.2182[/tex]
So then the best answer would be 21.82 %
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 time it takes to trim a tree of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(32,5)[/tex]
Where [tex]\mu=32[/tex] and [tex]\sigma=5[/tex]
We are interested on this probability
[tex]P(28.5<X<31.5)[/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(28.5<X<31.5)=P(\frac{28.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{31.5-\mu}{\sigma})=P(\frac{28.5-32}{5}<Z<\frac{31.5-32}{5})=P(-0.7<z<-0.1)[/tex]And we can find this probability with this difference:
[tex]P(-0.7<z<-0.1)=P(z<-0.1)-P(z<-0.7)[/tex]
And in order to find these probabilities we use tables for the normal standard distribution, excel or a calculator.
[tex] P(-0.7<z<-0.1)=P(z<-0.1)-P(z<-0.1) =0.4602-0.2420=0.2182[/tex]
So then the best answer would be 21.82 %
