Question

When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwise, it must be returned to the waters. In a region of the Gulf of Mexico, the lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches. What is the probability that a fisherman catches a spotted seatrout within the legal limits?

Answer 1

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by

`Z = (X - \mu)/(\sigma)`

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so
`\mu = 22, \sigma = 4`.

What is the probability that a fisherman catches a spotted seatrout within the legal limits?

They must be between 10 and 30 inches.

So, this is the pvalue of the Z score of
`X = 30` subtracted by the pvalue of the Z score of
`X = 10`

X = 30

`Z = (X - \mu)/(\sigma)`

`Z = (30 - 22)/(4)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

X = 10

`Z = (X - \mu)/(\sigma)`

`Z = (10 - 22)/(4)`

`Z = -3`

`Z = -3` has a pvalue of 0.00135.

This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.