Coefficient of Variation Formula

The process by which the data is collected and analyzed is known as statistics. The coefficient of deviation in statistics explain as the ratio of the standard deviation to the arithmetic mean, for instance, the expression standard deviation is 15 % of the arithmetic mean  is the coefficient variation

Coefficient of variation

The coefficient of deviation is the dimension of comparable variability. The coefficient of deviation is the percentage of the expected deviation to the standard.

It’s very useful if one wants to compare the results from the two different research or tests that consists of the two different results. For example, if comparing results of two different matches that have two completely different scoring methods, like if model X has a CV of 15% and model Y has a CV of 30%, it would be conveyed that model Y has more deviation, comparable to its mean. It enables us to supply with the relatively simple and quick tools that help us to compare the data of different series

Formula for the coefficient of variation

Coefficient of deviation = (Standard Deviation / Mean) × 100.

In symbols: CV = (SD/x̄) × 100

Coefficient of variation in the context of finance

It helps in the investment selection process that’s why it is important in terms of finance. In the financial matrix, it shows the risk-to-reward ratio means here the standard deviation/ volatility show as the risk of the investment, and the mean is shown as the expected reward of the investment. The investors in the company identify the risk-to-reward ratio of each one of the security to develop an investment decision. In this, the low coefficient is not favourable when the average expected return is below the value of zero

The Formula for the coefficient of variation in the context of finance,

Coefficient of variation = σ/μ × 100%

Where,

σ – the standard deviation

μ – the mean

Example for the coefficient of variation finance

An investor Sudhir wants to find new investments for his portfolio purpose. Where he was looking for secure investments that provides him a stable return. So, consider some of the following options for the investment.

• Stocks: Sudhir has gotten an offer for the stocks of XYZ.Pvt corporates is a very mature company with strong financial and operational performance. The volatility of this stock is 9% and the expected return in of 13%.
• Bonds: Sudhir is getting bonds with excellent credit ratings, which offered him an expected return of 5% with 3% of volatility.
• ETFs: Another option sudhir gets is an EXCHANGE-TRADED FUND which helps in the tracking of the S&P 500 index. An ETF offers him an expected return of 15% with a volatility of 8%.

Standard deviation 

The standard deviation formula helps us to find the values of a particular data that is dispersed. Simply, it is defined as the deviation of the data from an average mean. In this, higher values mean that the values are far from the average mean as well as the lower values mean that values are very close to their average mean. It is said that the value of standard deviation can never be negative.

Standard deviation is of two types

Population standard deviation 

σ = √∑(X − μ)²/n

Sample standard deviation 

s = √∑(X − x̄)²/n − 1

Notation for standard deviation,

σ = Standard Deviation

xi = Terms Given in the Data

x̄ = Mean

n = Total number of Terms

Steps taken for the calculation of the coefficient of variation are as follow

• Step 1: Firstly, get the standard deviation of the data provided.
• Step 2: After getting the standard deviation of the data we have to get the mean of the given data as we know that the coefficient of variation is the ratio between the standard deviation and the mean of the given data.
• Step 3: Now put both the standard deviation and the mean in the formula and multiply the coefficient by 100 is an optional step to get a percentage, as opposed to a decimal.

Sample Problems

Question 1: The standard deviation and mean of the data are 8.5 and 14.5 respectively. Find the coefficient of variation.

Solution:

SD/σ = 8.5                  

mean/μ = 14.5          

Coefficient of variation = σ/μ × 100%

= 5.5/14.5 × 100

Coefficient of variation =  58.6%

Question 2: The standard deviation and coefficient of variation of data are 1.4 and 26.5 respectively. Find the value of the mean.

Solution:

C.V = 26.5                    

SD/σ = 1.4

 Mean/x̄ = ?

 C.V =  σ/x̄ × 100

 26.5 = 1.4 / x̄ × 100

  x̄ = 1.4/26.5 × 100

  x̄ = 5.28

Question 3: If the mean and coefficient of deviation of data are 13 and 38 respectively, then locate the value of expected variation?

Solution: 

C.V = 38                    

SD/σ = ?

Mean/x̄ = 13

C.V =  σ/x̄ × 100

38 =  σ/13 × 100

σ = 13 × 38/100

σ = 4.9

Question 4: The mean and standard variation of marks received by 40 students of a class in three subjects Mathematics, English and economics are given below.

Subject	Mean	Standard deviation
Maths	65	10
English	60	12
Economics	57	14

Which of the three subjects indicates the most elevated deviation and which indicates the most subordinate variation in marks?

Solution:

Coefficient of variation for maths = σ/x̄ × 100

σ = 10.                

x̄ = 65

C.V = 10/65 × 100

Coefficient of variation for maths = 15.38%

Coefficient of variation for english = σ/x̄ × 100

σ = 12               

 x̄ = 60

C.V = 12/60 × 100

Coefficient of variation for english = 20%

Coefficient of variation for economics = σ/x̄ × 100

σ = 14                

x̄ = 57

C.V = 14/57 × 100

Coefficient of variation for economics = 24.56%

The highest variation is economics.

And the lowest variation in maths.

Question 5: The following table gives the values of mean and friction of heights and weights of the 10th common students of a school.

	Height	Weight
Mean	157 cm	56.50 kg
Variance	72.25 cm	28.09 kg

Which is more varying than the other?

Solution:

Coefficient of variation for heights

Mean x̄₁ = 157cm, variance σ1² = 72. 25 cm²

Therefore standard deviation σ₁ = 8. 5

Coefficient of variation

C.V₁ =  σ/x̄ × 100

= 8.5/157 × 100

C.V₁ = 5.41% (For heights)

Coefficient of variation for weights

Mean x̄₂ = 56.50kg, variance σ₂² = 28.09 kg²

Therefore standard deviation σ₂ = 5.3kg

Coefficient of variation

C.V₁ =  σ/x̄ × 100

= 5.3/56.50 × 100

C.V₂ = 9.38% (For weight)

C.V₁ = 5.41% and C.V₂  = 9.38%

Since C .V₂ > C.V₁, the weight of the students is more varying than the height.

Question 6: During a survey, 6 students were asked how many hours per day they study on average? Their answers were as follows: 3, 7, 5, 4, 3, 5. Evaluate the standard deviation.

Solution:

Find the mean of the data:

(3 + 7 + 5 + 4 + 3 + 5)/6

= 4.5

X1	X1 – x̄	(X1 – x̄)²
3	-1.5	2.25
7	2.5	6.25
5	0.5	0.25
4	-0.5	0.25
3	-1.5	2.25
5	0.5	0.25

= 11.5

Sample standard deviation formula:

s = √∑(X − x̄)²/n − 1

= √(11.5/[6 – 1])

= √[2.3]

= 1.516

Question 7: Four friends were comparing their scores on a recent essay. Calculate the standard deviation of their scores: 7, 3, 4, 2.

Solution:

Find the mean of the data

(7 + 3 + 4 + 2)/4

= 4

X1	X1 – x̄	(X1 – x̄)²
7	3	9
3	-1	1
4	0	0
2	-2	4

= 14

Population standard deviation

σ = √∑(X − x̄)²/n

= √(14/[4])

= √[10]

= 3.162