Question

The final scores from the last five games of two basketball teams are given. Jumping Jackrabbits: 70, 65, 72, 80, 73 Leaping Lizards: 61, 47, 70, 63, 54
(a) Determine the MAD of each data set. In your work, show the mean and the absolute deviation of each data value in each set.
(b) Compare and interpret the MADs in context of the situation.

Answer 1

Using the Mean Absolute Deviation(MAD) concept, it is found that:

a) The Jumping Jackrabbits have a MAD of 3.6 and the Leaping Lizards of 6.8.

b) The MAD is lower for the Jumping Jackrabbits then for the Leaping Lizards, hence their scores are more consistent.

What is the mean absolute deviation of a data-set?

• The mean of a data-set is given by the sum of all observations divided by the number of observations.

• The mean absolute deviation(MAD) of a data-set is the sum of the absolute value of the difference between each observation and the mean, divided by the number of observations.

• The mean absolute deviation represents the average by which the values differ from the mean.

Item a:

For the Jumping Jackrabbits, the mean and MAD are given by:

M = (70 + 65 + 72 + 80 + 73)/5 = 72

MAD = (|70 - 72| + |65 - 72| + |72 - 72| + |80 - 72| + |73 - 72|)/5 = 3.6.

For the Leaping Lizards, the mean and the MAD are given by:

M = (61 + 47 + 70 + 63 + 54)/5 = 59

MAD = (|61 - 59| + |47 - 59| + |70 - 59| + |63 - 59| + |54 - 59|)/5 = 6.8.

The Jumping Jackrabbits have a MAD of 3.6 and the Leaping Lizards of 6.8.

Item b:

The MAD is lower for the Jumping Jackrabbits then for the Leaping Lizards, hence their scores are more consistent.

More can be learned about the Mean Absolute Deviation(MAD) concept at  brainly.com/question/3250070

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