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. If a sample of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation?
A) 18.750
B) 1.875
C) 0.750
D) 2.500

Question

Quizplus · Accepted Answer

The Z-score is calculated as (x - mu) / sigma, where x is the sample mean, mu is the population mean, and sigma is the population standard deviation. In this case, x = 3.6, mu = 3.2, and sigma = 0.8. Plugging these values into the formula, we get (3.6 - 3.2) / 0.8 = 0.5 / 0.8 = 0.625. Therefore, the Z-score for this observation is 0.625, which is closest to choice D, 0.750.

The Owner of a Fish Market Has an Assistant Who