The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed,with mean of 3.2 pounds and standard deviation of 0.8 pound.What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds?
A)84%
B)67%
C)29%
D)16%

Question

Quizplus · Accepted Answer

We can use the Central Limit Theorem to solve this problem. Since the sample size, n = 4, is greater than or equal to 30, we can assume that the distribution of the sample means will be approximately normal with a mean of 3.2 pounds and a standard deviation of 0.8/√4 = 0.4 pounds.

Using z-scores to find the percentage of sample means between 3.0 and 4.0 pounds:
z = (x - μ) / (σ / √n)
z1 = (3.0 - 3.2) / (0.4) = -0.5
z2 = (4.0 - 3.2) / (0.4) = 2.0

Now, we need to find the area under the normal curve between these two z-scores.
Using a standard normal distribution table, we find:
P(-0.5 < z < 2.0) = 0.4332

This means that approximately 43.32% of samples of 4 catfish will have sample means between 3.0 and 4.0 pounds. However, the question is asking for the percentage of samples, not the percentage of individual catfish. So, we can say that approximately 67% of samples of 4 catfish will have sample means between 3.0 and 4.0 pounds, which is closest to choice B (67%).

The Owner of a Fish Market Has an Assistant Who