Respuesta :
Answer:
a) [tex]P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-19}{25})=P(Z>2)[/tex]
And we can find this probability using the complement rule and the z table or excel:
[tex]P(Z>2)=1-P(Z<2) =1-0.977=0.023[/tex]
b) [tex]P(X<-56)=P(\frac{X-\mu}{\sigma}<\frac{-56-\mu}{\sigma})=P(Z<\frac{-56-19}{25})=P(Z<-3.12)[/tex]
And we can find this probability using the z table or excel:
[tex]P(Z<-3.12)=0.0009[/tex]
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".
Part a
Let X the random variable that represent the historical returns of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(19,25)[/tex]
Where [tex]\mu=19[/tex] and [tex]\sigma=25[/tex]
We are interested on this probability
[tex]P(X>69)[/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(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-19}{25})=P(Z>2)[/tex]
And we can find this probability using the complement rule and the z table or excel:
[tex]P(Z>2)=1-P(Z<2) =1-0.977=0.023[/tex]
Part b
We are interested on this probability
[tex]P(X<-56)[/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(X<-56)=P(\frac{X-\mu}{\sigma}<\frac{-56-\mu}{\sigma})=P(Z<\frac{-56-19}{25})=P(Z<-3.12)[/tex]
And we can find this probability using the z table or excel:
[tex]P(Z<-3.12)=0.0009[/tex]
