Let X be the weight of fish in the lake.
Given that the mean and standard deviation as 20 and 6 respectively,
[tex]\begin{gathered} \mu=20 \\ \sigma=6 \end{gathered}[/tex]Consider the formula,
[tex]z=\frac{x-\mu}{\sigma}[/tex]The proportion of fish weighing less than 15 pounds is calculated as,
[tex]\begin{gathered} P(X<15)=P(z<\frac{15-20}{6}) \\ P(X<15)=P(z<-0.83) \\ P(X<15)=P(z<0)-P(-0.83From the Standard Normal Distribution Table,[tex]\emptyset(-0.83)=0.2967[/tex]Substitute the value,
[tex]\begin{gathered} P(X<15)=0.5-0.2967 \\ P(X<15)=0.2033 \end{gathered}[/tex]Mulyiply by 100 to get the equivalent percent,
[tex]P(X<15)=20.33\text{ percent}[/tex]Thus, 20.33% of fish in the lake will weigh less than 15 pounds.