Question

The owner of a fish market has an assistant who has determined that the weights of fish are normally distributed, with mean of 3.2 kgs and standard deviation of 0.8 kgs. If a sample of 64 fish yields a mean of 3.4 kgs, what is probability of obtaining a sample mean this large or larger

Answer 1

Answer:

0.0228

Explanation:

If a sample of 64 fish yields a mean of 3.2 kg, what is the Z-score for this observation?

Z = (3.4 – 3.2)/(0.8/square root of 64) = 2.0

If a sample of 64 fish yields a mean of 3.4 Kg

Z = (3.4 – 3.2)/(0.8/square root of 64) = 2

P(Z > 2) = 1 – P(Z < 2) = .0228