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

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30 - 22}{4}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 10

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 22}{4}$

$Z = -3$

$Z = -3$ 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.