The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established?
A) 7.36 pounds
B) 1.56 pounds
C) 5.20 pounds
D) 4.84 pounds

Question

Quizplus · Accepted Answer

First, we need to find the z-score for the top 2% weight, which corresponds to the area of 0.02 in the right tail of the standard normal distribution. From the standard normal distribution table or calculator, we find that the z-score for this area is approximately 2.05 (rounded to two decimal places).

Next, we use the formula z = (x - μ) / σ to convert the z-score into the corresponding weight value x, where μ is the mean weight and σ is the standard deviation. In this case, we have z = 2.05, μ = 3.2, and σ = 0.8, so we can solve for x as follows:

2.05 = (x - 3.2) / 0.8

2.05 * 0.8 = x - 3.2

1.64 + 3.2 = x

x = 4.84

Therefore, the citation catfish should be established at 4.84 pounds to qualify as one of the top 2% in weight. The answer is D.

The Owner of a Fish Market Determined That the Average