Question 15:

If n = 10, x̄ = 12 and Σ x² = 1530,then calculate the coefficient of variation.

Solution:

The formula to find coefficient of variation is (C.V) = (σ/x̄) x 100

σ = √(Σ x²/n) - (Σ x/n)²

= √(1530/10) - (12)²

= √153 - 144

= √9

= 3

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

= (3/12) x 100

= (1/4) x 100

= 25

Question 16:

Calculate the coefficient of variation of the following data 20,18,32,24,26.

Solution:

First let us write the given data in ascending order

18,20,24,26,32

 x d = x - 24 d² 18 18 - 24 = -6 36 20 20 - 24 = -4 16 24 24 - 24 = 0 0 26 26 - 24 = 2 4 32 32 - 24 = 8 64 Σ x = 120 Σd²=120
 Mean = Σ x/n        = 120/5     x̄ = 24 σ = √(Σ d²/n)     = √(120/5)    = √24    = 4.9

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

=  (4.9/24) x 100

=   490/24

=   20.416

=   20.42

Question 17:

If the coefficient of variation of a collection of data is 57 and its standard deviation is 6.84, then find the mean.

Solution:

coefficient of variation (C.V) = 57

Standard deviation (σ) = 6.84

(σ/x̄) x 100 = 57

(6.84/x̄) x 100 = 57

x̄ = 684/57

x̄ = 12

Question 18:

A group of 100 candidates have their average height 163.8 cm with coefficient of variation 3.2. What is the standard deviation of their heights?

Solution:

mean of height of 100 candidates(x̄) = 163.8

coefficient of variation (C.V) = 3.2

(σ/x̄) x 100 = 3.2

(σ/163.8) x 100 = 3.2

σ = (3.2 x 163.8)/100

σ = 5.2416

Question 19:

Given Σ x = 99, n = 9 and Σ (x - 10)² = 79. Find Σ x² and Σ (x - x̄)².

Solution:

x̄ = (Σ x/n)

= (99/9)

= 11

Σ (x - 10)² = 79

Σ (x²  + 10² - 2 x (10)) = 79

Σ (x²  + 100 - 20 x) = 79

Σ x²  + 100 Σ - 20 Σ x = 79

Σ x²  + 100 (9) - 20 (99) = 79

Σ x²  + 900 - 1980 = 79

Σ x² - 1080 = 79

Σ x² = 79 + 1080

Σ x² = 1159

Σ (x - x̄)² = Σ (x - 11)²

= Σ (x²  + 11² - 2 x (11))

= Σ (x²  + 121 - 22 x)

= Σx² + 121Σ - 22 Σx

= 1159 + 121(9) - 22 (99)

= 1159 + 1089 - 2178

=  2248 - 2178

= 70