Answer:
The correct answer to the question is
C) The mean and range of the fingerling catfish are both greater than those of the large catfish.
Explanation:
From the attached chart and table, it is seen that the mean = (∑ f×m)/n
Where Mean x = Mean estimate of frequency distribution
f = Class frequency
m = midpoint of each class
∑ f×m = the sum of the product of each midpoint value and its respective frequency
n = The sum of the frequencies
Therefore from the tables we have
For the Fingerling catfish
(∑ f×m)/n = 11037.5/1.375 = 8027.3
and the range = (Maximum value) - (minimum value) in the data set
= 10500 - 3500 = 7000
For the Large catfish we have
(∑ f×m)/n = 2931.25/2.1 = 1395.83
and the range = (Maximum value) - (minimum value) in the data set
= 2125 - 375 = 1750
As seen above, the mean and range of the fingerling catfish are both greater than those of the large catfish