Answer:
D) Westfield's data shows greater variability, since Westfield's MAD is approximately 2.9 times greater than Eastfield's MAD.
Explanation:
Westfield's data shows greater variability, since Westfield's MAD is approximately 2.9 times greater than Eastfield's MAD.
Westfield's MAD = 6.56
Eastfield's MAD = 2.28
Therefore,
6.56
2.28
= 2.877
MAD =
∑|x − X|
n
, where x = Data value, X = Mean, and n = Number of values