Respuesta :
Answer:
a) [tex]P(X>36)=P(\frac{X-\mu}{\sigma}>\frac{36-\mu}{\sigma})=P(Z>\frac{36-35}{0.59})=P(z>1.695)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>1.695)=1-P(z<1.695)=1-0.955=0.045[/tex]
b) [tex]P(X<33.5)=P(\frac{X-\mu}{\sigma}<\frac{33.5-\mu}{\sigma})=P(Z<\frac{33.5-35}{0.59})=P(z<-2.542)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-2.542)=0.0055[/tex]
c) [tex]P(33<X<35)=P(\frac{33-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{35-\mu}{\sigma})=P(\frac{33-35}{0.59}<Z<\frac{35-35}{0.59})=P(-3.389<z<0)[/tex]
And we can find this probability with this difference:
[tex]P(-3.389<z<0)=P(z<0)-P(z<-3.389)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-3.389<z<0)=P(z<0)-P(z<-3.389)=0.5-0.00035=0.4996[/tex]
Step-by-step explanation:
a) What is the probability that the salinity is more than 36 ppt? (Round your answer to four decimal places.)
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".
Let X the random variable that represent the salt content of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(35,0.59)[/tex]
Where [tex]\mu=35[/tex] and [tex]\sigma=0.59[/tex]
We are interested on this probability
[tex]P(X>36)[/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>36)=P(\frac{X-\mu}{\sigma}>\frac{36-\mu}{\sigma})=P(Z>\frac{36-35}{0.59})=P(z>1.695)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>1.695)=1-P(z<1.695)=1-0.955=0.045[/tex]
(b) What is the probability that the salinity is less than 33.5 ppt? (Round your answer to four decimal places.)
[tex]P(X<33.5)[/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(X<33.5)=P(\frac{X-\mu}{\sigma}<\frac{33.5-\mu}{\sigma})=P(Z<\frac{33.5-35}{0.59})=P(z<-2.542)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-2.542)=0.0055[/tex]
(c) A certain species of fish can only survive if the salinity is between 33 and 35 ppt. What is the probability that this species can survive in a randomly selected area? (Round your answer to four decimal places.)
[tex]P(33<X<35)=P(\frac{33-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{35-\mu}{\sigma})=P(\frac{33-35}{0.59}<Z<\frac{35-35}{0.59})=P(-3.389<z<0)[/tex]
And we can find this probability with this difference:
[tex]P(-3.389<z<0)=P(z<0)-P(z<-3.389)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-3.389<z<0)=P(z<0)-P(z<-3.389)=0.5-0.00035=0.4996[/tex]
