A data set's mean absolute deviation is the sum of its absolute departures from a central point. The second bowler is less consistent with his strikes.
What is the mean absolute deviation?
A data set's mean absolute deviation is the sum of its absolute departures from a central point.
The mean of 1st bowler is,
[tex]\rm Mean = \dfrac{8+5+5+6+8+7+4+7+6}{9} = \dfrac{56}{9} = 6.\bar 2[/tex]
Now, the Absolute deviation can be written as,
Absolute deviation
= |8-6.2|+|5-6.2|+|5-6.2|+|6-6.2|+|8-6.2|+|7-6.2|+|4-6.2|+|7-6.2|+|6-6.2|
= 1.8 + 1.2 + 1.2 + 0.2 + 1.8 + 0.8 + 2.2 + 0.8 + 0.2
[tex]\rm Mean\ Absolute\ Deviation = \dfrac{Absolute\ Deviation}{n} = \dfrac{10.2}{9}=1.13[/tex]
The mean of the 2nd bowler is,
[tex]\rm Mean = \dfrac{10+6+8+8+5+5+6+8+9}{9} = \dfrac{65}{9} = 7.\bar 2[/tex]
Now, the Absolute deviation can be written as,
Absolute deviation
= |10-7.2| + |6-7.2| + |8-7.2| + |8-7.2| + |5-7.2| + |5-7.2| + |6-7.2| + |8-7.2| + |9-7.2|
= 2.8+ 2.8 + 0.8+ 0.8+ 2.2+ 2.2+ 1.2 + 0.8 + 1.8
[tex]\rm Mean\ Absolute\ Deviation = \dfrac{Absolute\ Deviation}{n} = \dfrac{15.4}{9}=1.711[/tex]
As we can see that the mean absolute deviation of the second bowler is more therefore, the second bowler is less consistent with his strikes.
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