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
and standard deviation
, the zscore of a measure X is given by
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 = 10](https://tex.z-dn.net/?f=X%20%3D%2010)
X = 30
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{30 - 22}{4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B30%20-%2022%7D%7B4%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772
X = 10
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{10 - 22}{4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B10%20-%2022%7D%7B4%7D)
![Z = -3](https://tex.z-dn.net/?f=Z%20%3D%20-3)
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.