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Sagot :
The mean absolute deviation is given by:
[tex]\frac{\sum ^{}_{}\lvert x_i-\bar{x}\rvert}{n}[/tex]where xi represent each data, x bar the mean and n the number of data we have. Then:
[tex]\begin{gathered} \frac{\lvert1-6\rvert+\lvert7-6\rvert+\lvert7-6\rvert+\lvert7-6\rvert+\lvert8-6\rvert}{5} \\ =\frac{\lvert-5\rvert+\lvert1\rvert+\lvert1\rvert+\lvert1\rvert+\lvert2\rvert}{5} \\ =\frac{5+1+1+1+2}{5} \\ =\frac{10}{5} \\ =2 \end{gathered}[/tex]Therefore the mean absolute value is 2 and the answer is C.
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