Question

"Wildlife biologists believe that the weights of adult trout can be described by a Normal model. They collect data from fishermen, finding that 22% of the trout caught were thrown back because they were below the 2-pound minimum, and only 6% weighed over 5 pounds. What mean and standard deviation should define the model?"I find the z values for 22% below and 6% above are -0.772 and 1.555. but i dont know what to do after that

Answer 1

The mean and standard deviation that should define the model are:

Standard Deviation: σ = 1.289

Mean: μ = 2.996

How to use the Central Limit theorem of Normal Distribution?

In probability theory, the central limit theorem (CLT) states that 68% of the population is within 1 standard deviation of the mean. 95% of the population is within 2 standard deviation of the mean. 99.7% of the population is within 3 standard deviation of the mean.

Now, we are told that:

22% of the trout caught were thrown back because they were below the 2-pound minimum, and only 6% weighed over 5 pounds.

Thus, the minimum difference in mean score is:

5 - 2 = 3

Now, the z values for 22% below and 6% above are -0.772 and 1.555.

Thus:

3 = (1.555 - (-0.772))σ

Where σ is standard deviation

3 = 2.327σ

σ = 3/2.327

σ = 1.289

We know that formula for z-score is:

z = (x - μ)/σ

1.555 = (5 - μ)/1.289

μ = 2.996