COEFFICIENT OF VARIATION FOR UNGROUPED DATA

Example 1 :

Find the coefficient of variation of 24, 26, 33, 37, 29, 31.

Solution :

First let us find the standard deviation for the given data. For that, let us arrange the given data in ascending order.

24, 26, 29, 31, 33, 37

Σd²/n  =  117/6

(Σd/n)²  =  (-6/6)²  =  1

σ  =  √(19.5 - 1)

  =  √18.5

σ  =  4.30

x̄  =  Σx/n 

=  (24 + 26 + 29 + 31 + 33 + 37)/6

x̄  =  180/6

x̄  =  30

Coefficient of variation (C.V) =  (σ/x̄) ⋅ 100%

C.V  =  (4.30/30) ⋅ 100%

C.V  =  0.143 ⋅ 100%

C.V  =  14.33%

Hence the coefficient of variation of the given data is 14.4%

Example 2 :

The time taken (in minutes) to complete a homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.

Solution :

38, 40, 43, 44, 46, 47, 49, 53

Σd²/n  =  172/8

(Σd/n)²  =  (8/8)²  =  1

σ  =  √(21.5 - 1)

  =  √20.5

σ  =  4.53

x̄  =  Σx/n 

=  (38+40+43+44+46+47+49+53)/8

x̄  =  360/8

x̄  =  45

Coefficient of variation (C.V) =  (σ/x̄) ⋅ 100%

C.V  =  (4.52/45) ⋅ 100%

C.V  =  0.1006 ⋅ 100%

C.V  =  10.07%

Hence the coefficient of variation of the given data is 10.07%

Example 3 :

The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation 4.6 and 2.4 respectively. Who is more consistent in performance?

Solution :

Hence we decide that Vidhya is more consistent.

Example 4 :

Following are marks obtained, out of 100 by two students Andrew and Frad in 10 tests.

Scores of Andrew :

25, 50, 45, 30, 70, 42, 36, 48, 35, 60

Scores of Frad :

10, 70, 50, 20, 95, 55, 42, 60, 48, 80

Who is more intelligent and who is more consistent ?

Solution :

Calculating standard deviation for Andrew :

Σd² = 289 + 64 + 9 + 144 + 784 + 0 + 36 + 36 + 49 + 324

= 1735

Σd = (-17) + 8 + 3 + (-12) + 28 + 0 + (-6) + 6 + (-7) + 18

= -42 + 63

= 21

Calculating mean for Andrew :

x̄ = (25 + 50 + 45 + 30 + 70 + 42 + 36 + 48 + 35 + 60) / 10

= 441/10

= 44.1

Calculating standard deviation for Frad :

Σd² =2025 + 225 + 25 + 1225 + 1600 + 0 + 169 + 25 + 49 + 625

= 5968

Σd = -45 + 15 + (-5) + (-35) + 40 + 0 + (-13) + 5 + (-7) + 25

= -105 + 85

= -20

Calculating mean for Frad :

x̄ = (10+70+50+20+95+55+42+60+48+80) / 10

= 530/10

= 53

Coefficient of variation of Andrew :

= (Standard deviation / mean) x 100%

= (σ/x̄) x 100%

= (13.003/44.1) x 100%

= (0.294) x 100%

= 29.4%

Coefficient of variation of Frad :

= (Standard deviation / mean) x 100%

= (σ/x̄) x 100%

= (24.34/53) x 100%

= (0.459) x 100%

= 45.9%

Frad has greater coefficient of variation, then he is intelligent and consistent.

Example 5 :

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is

(A) 0    (B) 1    (C) 1.5    (D) 2.5 

Solution :

Coefficient of variation = (σ/x̄) x 100%

Finding standard distribution for first data :

Applying coefficient of variation as 50 and mean as 30, we get

50 = (σ/30) x 100%

50 = (σ/30)

σ = 50(30)

σ = 1500

Finding standard distribution for second data :

Applying coefficient of variation as 60 and mean as 25, we get

60 = (σ/25) x 100%

60 = (σ/25)

σ = 60(25)

σ = 1500

Difference between two standard deviations

= 1500 - 1500

= 0

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