Answer:
0.3991
Explanation:
Given that the weights of the fish in a certain lake are normally distributed with a mean of 18 lb and a standard deviation of 18 .
we know that when a sample is taken mean would be the same but std error would vary
when a sample of size 15 is taken we have
![n=15](https://img.qammunity.org/2020/formulas/mathematics/college/zg05qrt3v3bnhoe1v93rbztzcdelf4v6ht.png)
Standard error of sample =
![(\sigma)/(√(n) ) \\=2.065](https://img.qammunity.org/2020/formulas/mathematics/high-school/a11khar7gyjep21pbpfdrtt9dwyddtdkzo.png)
the probability that the mean weight will be between 16.6 and 21.6 lb
![=P(16.6<x<21.6)\\=P(-0.30<z<0.77)\\=0.3991](https://img.qammunity.org/2020/formulas/mathematics/high-school/tnke8j7yucdwcnsuyje7n4y6k26emkpx9j.png)
So we get the required probability as
0.3991