Question

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?

Answer 1

The z-score for a sample mean of 3.6 pounds when the population mean is 3.2 pounds with a standard deviation of 0.8 pound, based on a sample size of 25, is 2.5. This means that the sample mean is 2.5 standard deviations above the population mean.

Step-by-step explanation:

The calculation of the z-score for a given sample mean when compared to the population mean is a typical question in statistics, which falls under the Mathematics category. To find the z-score, you use the formula:

z = (Xbar - μ) / (σ / √n)

Where Xbar is the sample mean, μ is the population mean, σ is the standard deviation of the population, and n is the sample size. In this case, Xbar = 3.6 pounds, μ = 3.2 pounds, σ = 0.8 pound, and n = 25.

So, the calculation would be:

z = (3.6 - 3.2) / (0.8 / √25) = (0.4) / (0.8 / 5) = (0.4) / (0.16) = 2.5

The z-score is 2.5, which means that the sample mean is 2.5 standard deviations above the population mean.

Answer 2

3.6 - 3.2
z = ----------------
0.8

= 0.4/0.8 = 0.50 (answer)