Question

Emily and Hunter tracked the average temperature in their city for the past 8 days. They recorded their temperature findings in this list 66°F, 57°F, 61°F, 65°F, 54°F, 62°F, 59°F, 56°F

What is the mean absolute deviation of this data set?


Enter your answer in the box.​

Answer 1

The mean absolute deviation = 3.5°F

Explanation:

The sum of the given temperatures, ∑T, is found as follows;

∑T = 66°F + 57°F + 61°F + 65°F + 54°F + 62°F + 59°F + 56°F = 480°F

The number of days, n, over which the temperature was tracked = 8 days

Therefore, the mean,
`\bar x`, of the data set = ∑T/n = 480°F/8 = 60°F

The mean absolute deviation is given by the following formula;

`Mean \ absolute \ deviation = (\sum \limits_(i = 1)^(n) \left | x_i - \bar x \right | )/(n)`

Therefore, we have;

66 - 60 =
`\left | 6\right |` = 6

57 - 60 =
`\left | -3\right |` = 3

61 - 60 =
`\left | 1\right |` = 1

65 - 60 =
`\left | 5\right |` = 5

54 - 60 =
`\left | -6\right |` = 6

62 - 60 =
`\left | 2\right |` = 2

59 - 60 =
`\left | -1\right |` = 1

56 - 60 =
`\left | -4\right |` = 4

The mean absolute deviation is therefore;

The mean absolute deviation = (6 + 3 + 1 + 5 + 6 + 2 + 1 + 4)/8 = 3.5°F.