Question

I can find percentages and values using the 68-95-99.7 rule, z-scores, and the standard normal distribution.only answer please if you know the answer

Answer 1

Let X be the weight of fish in the lake.

Given that the mean and standard deviation as 20 and 6 respectively,

`\begin{gathered} \mu=20 \\ \sigma=6 \end{gathered}`

Consider the formula,

`z=(x-\mu)/(\sigma)`

The proportion of fish weighing less than 15 pounds is calculated as,

Substitute the value,

`\begin{gathered} P(X<15)=0.5-0.2967 \\ P(X<15)=0.2033 \end{gathered}`

Mulyiply by 100 to get the equivalent percent,

`P(X<15)=20.33\text{ percent}`

Thus, 20.33% of fish in the lake will weigh less than 15 pounds.