Question

In Big Bear Lake, the blue gill (fish) lengths average 8 inches and have a standard deviation of 2 inches. In Moose Jaw Lake, the blue gill lengths average 9.5 inches and have a standard deviation of 1 inch. If people are basing a successful fishing trip on the proportion of fish over 11 inches, they should head for

Answer 1

Explanation:

To determine whether one lake is a better choice for a successful fishing trip based on the proportion of fish over 11 inches, you can use z-scores to compare the fish sizes in both lakes.

First, find the z-scores for a fish length of 11 inches in both lakes using the z-score formula:

**Z = (X - μ) / σ**

Where:

- X is the fish length you want to find the z-score for (in this case, 11 inches).

- μ is the mean (average) fish length for the lake.

- σ is the standard deviation of fish lengths for the lake.

For Big Bear Lake:

- X = 11 inches

- μ = 8 inches (average length)

- σ = 2 inches (standard deviation)

**Z (Big Bear Lake) = (11 - 8) / 2 = 3 / 2 = 1.5**

For Moose Jaw Lake:

- X = 11 inches

- μ = 9.5 inches (average length)

- σ = 1 inch (standard deviation)

**Z (Moose Jaw Lake) = (11 - 9.5) / 1 = 1.5 / 1 = 1.5**

Now, both lakes have the same z-score of 1.5 for a fish length of 11 inches. This means that the proportion of fish over 11 inches in both lakes is equally favorable. Therefore, people can head to either lake, and they are equally likely to catch fish over 11 inches.