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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 Determine the MAD of each data set. In your work, show the mean and the absolute deviation of each data value in each set. Compare and interpret the MADs in context of the situation.
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The mean absolute deviation for the Jumping Jackrabbits was of 3.8, which was lower than the 6.8 for the Leaping Lizards, hence they had more consistent scores.

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.

For the Jumping Jackrabbits and the Leaping Lizards, the mean and the MAD are given as follows:

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

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

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

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

The mean absolute deviation for the Jumping Jackrabbits was of 3.8, which was lower than the 6.8 for the Leaping Lizards, hence they had more consistent scores.

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

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