Page 237 - DSOC404_METHODOLOGY_OF_SOCIAL_RESEARCH_ENGLISH

Unit–28: Measures of Dispersion: Standard Deviation

17 1 17 6 6 36 notes
16 2 32 5 10 50

15 3 45 4 12 48
14 2 28 3 6 18
13 3 39 2 6 12

12 4 48 1 4 4
11 6 66 0 0 0
10 8 80 –1 –8 8
9 7 63 –2 –14 28
8 5 40 –3 –15 45

7 2 14 –4 –8 32
6 1 6 –5 –5 25
5 3 15 –6 –18 108
N (Sf ) = 50 Sfx = 550 Sfd = 612
2
S fx
Arithmetic mean, M =
N
Sfx = 550, N = 50
550
M =
50
M = 11

S fd 2
Standard deviation σ =
N
2
Sfd = 612, N = 50
612
(σ) =
50
(σ) = 12.4
σ = 3.5 (Approx.)
solving by using short-cut method
formula

(σ) = S fd 2 −  '  S fd '   2
 N  N

Let us consider that the given scores have an assumed mean (A) = 10.

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