Answer:
The mean absolute deviation = 3.5°F
Step-by-step 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, [tex]\bar x[/tex], of the data set = ∑T/n = 480°F/8 = 60°F
The mean absolute deviation is given by the following formula;
[tex]Mean \ absolute \ deviation = \frac{\sum \limits_{i = 1}^{n} \left | x_i - \bar x \right | }{n}[/tex]
Therefore, we have;
66 - 60 = [tex]\left | 6\right |[/tex] = 6
57 - 60 = [tex]\left | -3\right |[/tex] = 3
61 - 60 = [tex]\left | 1\right |[/tex] = 1
65 - 60 = [tex]\left | 5\right |[/tex] = 5
54 - 60 = [tex]\left | -6\right |[/tex] = 6
62 - 60 = [tex]\left | 2\right |[/tex] = 2
59 - 60 = [tex]\left | -1\right |[/tex] = 1
56 - 60 = [tex]\left | -4\right |[/tex] = 4
The mean absolute deviation is therefore;
The mean absolute deviation = (6 + 3 + 1 + 5 + 6 + 2 + 1 + 4)/8 = 3.5°F.
