Meow help me pls :(((
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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
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.
<span>Which expression is equivalent to (5z^2+3z+2)^2</span>
