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?

Respuesta :

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

   = 0.4/0.8 = 0.50 (answer)