Answer: 5.94%
Step-by-step explanation:
Let X be the random variable that represents the weights of the fish.
Given: Weights of the fish in a certain lake are normally distributed with a mean of 10.1 lb ([tex]\mu[/tex]) and a standard deviation of 2.9 lb ([tex]\sigma[/tex]).
sample size : n= 32
The probability that the mean weight will be more than 10.9 lb:
[tex]P(\overline{X}>10.9)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{10.9-10.1}{\dfrac{2.9}{\sqrt{32}}})\\\\=P(Z>1.56)\\\\=1-P(Z<1.56)\\\\=1- 0.9406\ \ [\text{By p-value table}]\\\\=0.0594=5.94\%[/tex]
Hence, the required probability = 5.94%.
