Answer: 0.0222
Step-by-step explanation:
Given : The distribution of trout weights is normally distributed with
Mean : [tex]\mu=402.7\text{ grams}[/tex]
Standard deviation : [tex]\sigma=8.8\text{ grams}[/tex]
Sample size : [tex]n=40[/tex]
The formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Let x be the weight of randomly selected trout.
Then for x = 405.5 , we have
[tex]z=\dfrac{405.5 -402.7}{\dfrac{8.8}{\sqrt{40}}}\approx2.01[/tex]
The p-value : [tex]P(405.5<x)=P(2.01<z)[/tex]
[tex]1-P(2.01)=1-0.9777844=0.0222156\approx0.0222[/tex]
Thus,the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams= 0.0222.
The probability that the mean weight for a sample of 40 trout exceeds 405.5 grams is 0.0222.
The distribution of trout weights is normally distributed with
We have given that
Mean=402.7 grams
Standard deviation =8.8 grams
Sample size (n)=40
We have to calculate
The probability that the mean weight for a sample of 40 trout exceeds 405.5 grams
Te formula of Z score is given by,
[tex]z=\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}[/tex]
n= the ample size
x=mean
sigma=standard deviation
So by using the formula we have,
Let x is the weight of randomly selected trout.
Then for x = 405.5
[tex]z=\frac{405.5-\402.7}{\frac{\8.8 }{\sqrt{40}}}\\\\\z=2.01[/tex]
we have
The p-value :(405.5<x)
(1-2.01)=1-0.97778
=0.02221
=0.0222
Therefore,the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams= 0.0222.
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