Respuesta :
Answer:
[tex]P(-2.06<X<-1.43)=P(\frac{-2.06-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.43-\mu}{\sigma})=P(\frac{-2.06-0}{1}<Z<\frac{-1.43-0}{1})=P(-2.06<z<-1.43)[/tex]
And we can find this probability with this difference:
[tex]P(-2.06<z<-1.43)=P(z<-1.43)-P(z<-2.06)= 0.0765 -0.0197=0.0567 [/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".
Solution to the problem
Let X the random variable that represent the temperatures of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(0,1)[/tex]
Where [tex]\mu=0[/tex] and [tex]\sigma=1[/tex]
We are interested on this probability
[tex]P(-2.06<X<-1.43)[/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(-2.06<X<-1.43)=P(\frac{-2.06-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.43-\mu}{\sigma})=P(\frac{-2.06-0}{1}<Z<\frac{-1.43-0}{1})=P(-2.06<z<-1.43)[/tex]
And we can find this probability with this difference:
[tex]P(-2.06<z<-1.43)=P(z<-1.43)-P(z<-2.06)= 0.0765 -0.0197=0.0567 [/tex]
