Question

A researcher is interested in the lengths of Salvelinus fontinalis (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 limiting the size of fish that can be caught. What proportion of fish are larger than 86 centimeters in length

Answer 1

`P(X>86)`

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>86)=P((X-\mu)/(\sigma)>(86-\mu)/(\sigma))=P(Z>(86-80)/(5))=P(z>1.2)`

And we can find this probability using the complement rule and with the normal standard distribution table:

`P(z>1.2)=1-P(z<1.2)= 1-0.885=0.115`

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>86)`

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>86)=P((X-\mu)/(\sigma)>(86-\mu)/(\sigma))=P(Z>(86-80)/(5))=P(z>1.2)`

And we can find this probability using the complement rule and with the normal standard distribution table:

`P(z>1.2)=1-P(z<1.2)= 1-0.885=0.115`