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?

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

Step-by-step explanation:

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

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

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 .

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 subtracted by the pvalue of the Z score of

X = 30

has a pvalue of 0.9772

X = 10

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.

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent