The mean absolute deviation for the number of hours students practiced the violin is 6.4.
The average absolute deviation of the collected data set is the average of absolute deviations from a center point of the data set.
Given
Students reported practicing violin during the last semester for 45, 38, 52, 58, and 42 hours.
The given data set is;
45, 38, 52, 58, 42
Mean Deviation = Σ|x − μ|/N.
μ = mean, and N = total number of values
|x − μ| = |45 − 47| = 2
|38− 47| = 9
|52− 47| = 5
|58− 47| = 11
|42− 47| = 5
The mean absolute deviation for the number of hours students practiced the violin is;
[tex]\rm Mean =\dfrac{2+9+5+11+5}{5}\\\\Mean=\dfrac{32}{5}\\\\Mean = 6.4[/tex]
Hence, the mean absolute deviation for the number of hours students practiced the violin is 6.4.
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