Answer:
0.3991
Step-by-step 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
[tex]n=15[/tex]
Standard error of sample = [tex]\frac{\sigma}{\sqrt{n} } \\=2.065[/tex]
the probability that the mean weight will be between 16.6 and 21.6 lb
[tex]=P(16.6<x<21.6)\\=P(-0.30<z<0.77)\\=0.3991[/tex]
So we get the required probability as
0.3991