Answer: 0.0222
Step-by-step explanation:
Given : The wildlife department has been feeding a special food to rainbow trout fingerlings.
Based on a large number of observations, the weight of the trout are normally distributed with
Mean : [tex]\mu=402.7\text{ grams}[/tex]
Standard deviation : [tex]\sigma=8.8\text{ grams}[/tex]
Sample size : = 40
Let x be the random variable that represents the weight of the trout .
Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x=405.5 grams
[tex]z=\dfrac{405.5-402.7}{\dfrac{8.8}{\sqrt{40}}}\approx2.01[/tex]
By using standard normal distribution table , the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams :-
[tex]P(405.5<X)=P(z>2.01)=1-P(z<2.01)\\\\=1-0.9777844=0.0222156\appprox0.0222[/tex]
