Question

Sample grade point averages for ten male students and ten female students are listed. Find the coefficient of variation for each of the two data sets. Then compare the results

Males 2.5 3.8 3.6 3.9 2.6 2.6 3.6 3.2 3.9 1.8

Females 2.8 3.5 2.1 3.7 3.5 4.1 2.1 3.9 3.9 2.3

The coefficient of variation for males is ​__%.

The Coefficient of variation for females Is __ %

Answer 1

Solution:- Given that data

  Males 2.5 3.8 3.6 3.9 2.6 2.6 3.6 3.2 3.9 1.8

Females 2.8 3.5 2.1 3.7 3.5 4.1 2.1 3.9 3.9 2.3

=> The coefficient of variation for males is ​23.2%.

Explanation:-

Mean = (2.5 + 3.8 + 3.6 + 3.9 + 2.6 + 2.6 + 3.6 + 3.2 + 3.9 + 1.8)/10
= 31.5/10
Mean = 3.15

Standard Deviation σ = √(1/10 - 1) x ((2.5 - 3.15)² + (3.8 - 3.15)² + (3.6 - 3.15)² + (3.9 - 3.15)² + (2.6 - 3.15)² + (2.6 - 3.15)² + (3.6 - 3.15)² + (3.2 - 3.15)² + (3.9 - 3.15)² + (1.8 - 3.15)²)
= √(1/9) x ((-0.65)² + (0.65)² + (0.45)² + (0.75)² + (-0.55)² + (-0.55)² + (0.45)² + (0.05)² + (0.75)² + (-1.35)²)
= √(0.1111) x ((0.4225) + (0.4225) + (0.2025) + (0.5625) + (0.3025) + (0.3025) + (0.2025) + (0.0025) + (0.5625) + (1.8225))
= √(0.1111) x (4.805)
= √(0.5338355)
= 0.7307

Cofficient of Varaiance = σ/μ
= 0.7307 / 3.15

Coefficient of Variance = 0.232

=> The Coefficient of variation for females Is 24.75 %

Explanation:-

Mean = (2.8 + 3.5 + 2.1 + 3.7 + 3.5 + 4.1 + 2.1 + 3.9 + 3.9 + 2.3)/10
= 31.9/10
Mean = 3.19

Standard Deviation σ = √(1/10 - 1) x ((2.8 - 3.19)² + (3.5 - 3.19)² + (2.1 - 3.19)² + (3.7 - 3.19)² + (3.5 - 3.19)² + (4.1 - 3.19)² + (2.1 - 3.19)² + (3.9 - 3.19)² + (3.9 - 3.19)² + (2.3 - 3.19)²)
= √(1/9) x ((-0.39)² + (0.31)² + (-1.09)² + (0.51)² + (0.31)² + (0.91)² + (-1.09)² + (0.71)² + (0.71)² + (-0.89)²)
= √(0.1111) x ((0.1521) + (0.0961) + (1.1881) + (0.2601) + (0.0961) + (0.8281) + (1.1881) + (0.5041) + (0.5041) + (0.7921))
= √(0.1111) x (5.609)
= √(0.6231599)
= 0.7894

Cofficient of Varaiance = σ/μ
= 0.7894/3.19
Coefficient of Variance = 0.2475