The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed,with a mean of 3.2 pounds and a 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 know that the sample mean is also normally distributed with mean = population mean = 3.2 pounds and standard deviation = standard deviation of population / square root of sample size = 0.8 / sqrt(4) = 0.4 pounds. 
To find the percentage of samples with means between 3.0 and 4.0 pounds, we need to find the z-scores for these values and use the standard normal distribution table. 
z1 = (3.0 - 3.2) / 0.4 = -0.5 
z2 = (4.0 - 3.2) / 0.4 = 2 
The area between these z-scores represents the percentage of samples with means between 3.0 and 4.0 pounds. Using a standard normal distribution table, we find that the area between -0.5 and 2 is approximately 0.67 or 67%. Therefore, the answer is B.

The Owner of a Fish Market Has an Assistant Who