Question

A researcher is interested in the lengths of brook trout, which are known to be approximately Normally distributed with mean 80 centimeters and standard deviation 5 centimeters. To help preserve brook trout populations, some regulatory standards need to be set for limiting the size of fish that can be caught. What is the probability of catching a brook trout less than 72 centimeters in length

Answer 1

`P(X<72)=P((X-\mu)/(\sigma)<(72-\mu)/(\sigma))=P(Z<(72-80)/(5))=P(z<-1.6)`

And we can find this probability with the normal standard distirbution or excel and we got:

`P(z<-1.6)=0.055`

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 lenghts of a population, and for this case we know the distribution for X is given by:

`X \sim N(80,5)`

Where
`\mu=80` and
`\sigma=5`

We are interested on this probability

`P(X<72)`

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<72)=P((X-\mu)/(\sigma)<(72-\mu)/(\sigma))=P(Z<(72-80)/(5))=P(z<-1.6)`

And we can find this probability with the normal standard distirbution or excel and we got:

`P(z<-1.6)=0.055`