Question: Find the mean deviation about the mean for the following data:
Mark | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
Number Of Students | 3 | 7 | 12 | 15 | 8 | 3 | 2 |
Solution:
Arranging data in a table,
Marks | Frequency (f) | Mid Value(x) | fx | |x-x̄| | f|x-x̄| |
30-40 | 3 | 35 | 105 | 27 | 81 |
40-50 | 7 | 45 | 315 | 17 | 113 |
50-60 | 12 | 55 | 660 | 7 | 84 |
60-70 | 15 | 65 | 975 | 3 | 45 |
70-80 | 8 | 75 | 600 | 13 | 104 |
80-90 | 3 | 85 | 255 | 23 | 69 |
90-100 | 2 | 95 | 190 | 33 | 66 |
N= 50 | $\sum$fx = 3100 | $\sum$f|x-x̄| = 568 |
Mean (x̄) =$\dfrac{\sum fx}{N}$
$= \dfrac{3100}{50}$
$= 62$
And,
Mean Deviation from Mean (M.D.) = $\dfrac{\sum f|x-x̄|}{N}$
$= \dfrac{568}{50}$
$= 11.36$
Hence, the required mean deviation about the mean of the given data is 11..36
Related Notes and Solutions:
Here is the website link to the notes of Statistics of Class 10.
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