The owner of a fish market has an assistant who has determined that the weights of rainbow trout are normally distributed, with mean of 3.2 kilograms and standard deviation of 0.8 kilogram. If a sample of 25 fish yields a mean of 3.6 kilograms, what is the Z-score for this observation?Question 13Select one:a.18.750b.2.500c.1.875d.0.750

The Z-score is a measure of how many standard deviations an element is from the mean. To calculate the Z-score, we use the formula:

Z = (X - μ) / σ

where:

• X is the value of the element
• μ is the mean
• σ is the standard deviation

In this case:

• X is the sample mean, which is 3.6 kg
• μ is the population mean, which is 3.2 kg
• σ is the standard deviation of the population, which is 0.8 kg

Substituting these values into the formula, we get:

Z = (3.6 - 3.2) / 0.8 = 0.5

However, because we are dealing with a sample of 25 fish, we need to adjust the standard deviation by dividing it by the square root of the sample size (n). This is because the standard deviation of the sampling distribution (also known as the standard error) is equal to the standard deviation of the population divided by the square root of the sample size.

So, the adjusted standard deviation (σ') is:

σ' = σ / sqrt(n) = 0.8 / sqrt(25) = 0.8 / 5 = 0.16

Substituting these values into the formula, we get:

Z = (3.6 - 3.2) / 0.16 = 2.5

So, the Z-score for this observation is 2.5. Therefore, the correct answer is b. 2.500.