Nine students of a class obtained following marks. Calculate mean deviation from median.

S. No. 1 2 3 4 5 6 7 8 9

Marks 68 49 32 21 54 38 59 66 41

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Solution

The given data series is discrete in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.
$21, 32, 38, 41, 49, 54, 59, 66, 68 Here, N = 9 (o d d) Median = size of (\frac{N + 1}{2}) = (\frac{9 + 1}{2}) = size of 5$ ^th item
Thus, we calculate the deviation of values from 49.

Marks Ascending Order |X- M| = |d_M|

68
49
32
21
54
38
59
66
41 21
32
38
41
49
54
59
66
68 28
17
11
8
0
5
10
17
19

∑|d_M| = 115

$Mean Deviation from Median (M D_{M}) = \frac{\sum_{} |d_{M}|}{N} = \frac{115}{9} = 12.77$

Suggest Corrections

Marks Obtained	1	2	3	4	5	6	7	8	9
No. of students	32	41	57	98	123	83	46	17	3

Marks Obtained	140−150	150−160	160−170	170−180	180−190	190−200
Frequency	4	6	10	18	9	3

Marks	0−4	4−8	8−12	12−16	16−20	20−24
No. of Students	7	9	16	8	6	4

S. No.	1	2	3	4	5	6	7	8	9	10
Marks	43	60	37	48	65	48	57	78	31	59