Question

g The salinity, or salt content, in the ocean is expressed in parts per thousand (ppt). The number varies with depth, rainfall, evaporation, river runoff, and ice formation. The mean salinity of the oceans is 35 ppt. Suppose the distribution of salinity is normal and the standard deviation is 0.59 ppt, and suppose a random sample of ocean water from a region in a specific ocean is obtained.

Answer 1

a)
`P(X>36)=P((X-\mu)/(\sigma)>(36-\mu)/(\sigma))=P(Z>(36-35)/(0.59))=P(z>1.695)`

And we can find this probability using the complement rule:

`P(z>1.695)=1-P(z<1.695)=1-0.955=0.045`

b)
`P(X<33.5)=P((X-\mu)/(\sigma)<(33.5-\mu)/(\sigma))=P(Z<(33.5-35)/(0.59))=P(z<-2.542)`

And we can find this probability using the normal standard table or excel:

`P(z<-2.542)=0.0055`

c)
`P(33<X<35)=P((33-\mu)/(\sigma)<(X-\mu)/(\sigma)<(35-\mu)/(\sigma))=P((33-35)/(0.59)<Z<(35-35)/(0.59))=P(-3.389<z<0)`

And we can find this probability with this difference:

`P(-3.389<z<0)=P(z<0)-P(z<-3.389)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-3.389<z<0)=P(z<0)-P(z<-3.389)=0.5-0.00035=0.4996`

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:

`X \sim N(35,0.59)`

Where
`\mu=35` and
`\sigma=0.59`

We are interested on this probability

`P(X>36)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>36)=P((X-\mu)/(\sigma)>(36-\mu)/(\sigma))=P(Z>(36-35)/(0.59))=P(z>1.695)`

And we can find this probability using the complement rule:

`P(z>1.695)=1-P(z<1.695)=1-0.955=0.045`

(b) What is the probability that the salinity is less than 33.5 ppt? (Round your answer to four decimal places.)

`P(X<33.5)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<33.5)=P((X-\mu)/(\sigma)<(33.5-\mu)/(\sigma))=P(Z<(33.5-35)/(0.59))=P(z<-2.542)`

And we can find this probability using the normal standard table or excel:

`P(z<-2.542)=0.0055`

(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.)

`P(33<X<35)=P((33-\mu)/(\sigma)<(X-\mu)/(\sigma)<(35-\mu)/(\sigma))=P((33-35)/(0.59)<Z<(35-35)/(0.59))=P(-3.389<z<0)`

And we can find this probability with this difference:

`P(-3.389<z<0)=P(z<0)-P(z<-3.389)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-3.389<z<0)=P(z<0)-P(z<-3.389)=0.5-0.00035=0.4996`