Page 72 - DMGT209_QUANTITATIVE_TECHNIQUES_II

Unit 4: Research Problem

Notes
84
= = 0.2896
290
Illustration 2: Calculate coefficient of mean deviation about (i) Median (ii) mean from the
following data

X 14 16 18 20 22 24 26
f 2 4 5 3 2 1 4

Solution:
X F Cf fx |x – x | |x – M| f| x – x | f| x – M|
14 2 2 28 5.71 4 11.42 8
16 4 6 64 3.71 2 14.84 8
18 5 11 90 1.71 0 8.55 0
20 3 14 60 0.29 2 0.87 6
22 2 16 44 2.29 4 4.58 8
24 1 17 24 4.29 6 4.29 6
26 4 21 104 6.29 8 25.16 32
21 414 69.71 68

 f x i 414
i
x =  = 19.71
N 21
N +1 22
= = 11  Median M = 18
2 2

 f |x - x| 69.71
i
i
Now (i) M.D. ( x ) =  = 3.32
N 21
MD(x) 3.32
Coefficient of MD( x ) =  = 0.16
x 19.71
 i f x  M 68
i
(ii) M.D. (M) =  = 3.24
N 21
MD(M) 3.24
Coefficient of MD (M) =  = 0.18
M 18
Illustration 3: A purchasing agent obtained a sample of incandescent lamps from two suppliers.
He had the sample tested in his laboratory for length of life with following results.
Length of Light in hours Sample A Sample B
700 – 900 10 3
900 – 1100 16 42
1100 – 1300 26 12
1300 – 1500 8 3
Which company's lamps are more uniform.

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