Question

1. 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:

Answer 1

Using it's concept, it is found that:

a) For the Jumping Jackrabbits, the MAD is of 3.6 and for the Leaping Lizards, the MAD is of 6.8.

b) Due to the higher value of the MAD, the Leaping Lizards have less consistent scores than the Jumping Jackrabbits.

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 the MAD are given by:For the Jumping Jackrabbits, the MAD is of 3.6 and for the Leaping Lizards, the MAD is of 6.8.

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.

For the Jumping Jackrabbits, the MAD is of 3.6 and for the Leaping Lizards, the MAD is of 6.8.

Item b:

Due to the higher value of the MAD, the Leaping Lizards have less consistent scores than the Jumping Jackrabbits.

More can be learned about the Mean Absolute Deviation at https://brainly.com/question/3250070

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