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Measures of Dispersion — Economics STD 11 Commerce — Question

CBSE BoardEnglish MediumSTD 11 CommerceEconomicsMeasures of Dispersion6 Marks
Question
  1. Define Mean deviation.
  2. Calculate Mean deviation in the following data using Median:
X 5 6 7 8 9 10 11 12 13
f 4 5 6 7 8 9 10 11 12
✓

Answer

  1. Mean deviation is defined as the arithmetic average of the absolute deviations of the various items from a measure of central tendency, either mean or median, ignoring the $\pm$ signs of the deviations. Symbolically,
M.D. $=\frac{\Sigma\text{f|D|}}{\Sigma\text{f}}$
In case of discrete and continuous series, the deviations are multiplied with their respective frequencies.
M.D. $=\frac{\Sigma\text{f|D|}}{\Sigma\text{f}}$
  1.  
X
f
c.f.
|D| = |X - M|
f|D|
5
4
4
5
20
6
5
9
4
20
7
6
15
3
18
8
7
22
2
14
9
8
30
1
8
10
9
39
0
0
11
10
49
1
10
12
11
60
2
22
13
12
72
3
36
 
$\Sigma\text{f}=72$
 
 
$\Sigma\text{f|D|}=148$
Median $=\Big(\frac{\text{N}+1}{2}\Big)^{\text{th}}\text{item}=\Big(\frac{72+1}{2}\Big)^{\text{th}}\text{item}$
= 36.5 = Size of $36$.
$5^{th}$ item = 10
M.D. $M=\frac{\Sigma\text{f|D|}}{\Sigma\text{f}}$
$=\frac{148}{72}=2.05$

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