# How to Find a Coefficient of Variation

Probability and Statistics > Basic Statistics > How to Find a Coefficient of Variation

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## What is the Coefficient of Variation?

The coefficient of variation (CV) is a measure of relative variability. It is the ratio of the standard deviation to the mean (average). For example, the expression “The standard deviation is 15% of the mean” is a CV.

The CV is particularly useful when you want to compare results from two different surveys or tests that have different measures or values. For example, if you are comparing the results from two tests that have different scoring mechanisms. If sample A has a CV of 12% and sample B has a CV of 25%, you would say that sample B has more variation, relative to its mean.

### Formula

The formula for the coefficient of variation is:

Coefficient of Variation = (Standard Deviation / Mean) * 100.
In symbols: CV = (SD/) * 100.

Multiplying the coefficient by 100 is an optional step to get a percentage, as opposed to a decimal.

### Coefficient of Variation Example

A researcher is comparing two multiple-choice tests with different conditions. In the first test, a typical multiple-choice test is administered. In the second test, alternative choices (i.e. incorrect answers) are randomly assigned to test takers. The results from the two tests are:

 Regular Test Randomized Answers Mean 59.9 44.8 SD 10.2 12.7

Trying to compare the two test results is challenging. Comparing standard deviations doesn’t really work, because the means are also different. Calculation using the formula CV=(SD/Mean)*100 helps to make sense of the data:

 Regular Test Randomized Answers Mean 59.9 44.8 SD 10.2 12.7 CV 17.03 28.35

Looking at the standard deviations of 10.2 and 12.7, you might think that the tests have similar results. However, when you adjust for the difference in the means, the results have more significance:
Regular test: CV = 17.03

The coefficient of variation can also be used to compare variability between different measures. For example, you can compare IQ scores to scores on the Woodcock-Johnson III Tests of Cognitive Abilities.

Note: The Coefficient of Variation should only be used to compare positive data on a ratio scale. The CV has little or no meaning for measurements on an interval scale. Examples of interval scales include temperatures in Celsius or Fahrenheit, while the Kelvin scale is a ratio scale that starts at zero and cannot, by definition, take on a negative value (0 degrees Kelvin is the absence of heat).

## How to Find a Coefficient of Variation: Overview.

Watch the video, or read the article below:

Use the following formula to calculate the CV by hand for a population or a sample.

σ is the standard deviation for a population, which is the same as “s” for the sample.
μ is the mean for the population, which is the same as XBar in the sample.

In other words, to find the coefficient of variation, divide the standard deviation by the mean and multiply by 100.

## How to find a coefficient of variation in Excel.

You can calculate the coefficient of variation in Excel using the formulas for standard deviation and mean. For a given column of data (i.e. A1:A10), you could enter: “=stdev(A1:A10)/average(A1:A10)) then multiply by 100.

## How to Find a Coefficient of Variation by hand: Steps.

Sample question: Two versions of a test are given to students. One test has pre-set answers and a second test has randomized answers. Find the coefficient of variation.

 Regular Test Randomized Answers Mean 50.1 45.8 SD 11.2 12.9

Step 1: Divide the standard deviation by the mean for the first sample:
11.2 / 50.1 = 0.22355

Step 2: Multiply Step 1 by 100:
0.22355 * 100 = 22.355%

Step 3: Divide the standard deviation by the mean for the second sample:
12.9 / 45.8 = 0.28166

Step 4: Multiply Step 3 by 100:
0.28166 * 100 = 28.266%

That’s it! Now you can compare the two results directly.

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How to Find a Coefficient of Variation was last modified: October 15th, 2017 by

# 43 thoughts on “How to Find a Coefficient of Variation”

1. Liam

I have data sets from repeat examinations on 22 patients. I’m looking for some way to quantify the reliability of my measurements. Can i calculate an individual CV for each patient (based on just two values each time) and then compute “the average CV”? Or would the resulting number be meaningless? Thanks in advance for any help.

2. Andale

The result would probably be meaningless. First, you’re looking to quantify “reliability”, but the CV is a way to compare data sets that have different measurement criteria, like a different scale. Second, you need the SD, which would be impossible to calculate meaningfully from two data points. Why not calculate the CV for the entire group of 22 people? That gives you the “average” for the set.

3. Ryan

What a great article. It helped me so much in my job. Clear, concise, informative. And it applies the knowledge to MS Excel which is what I really needed.

4. Andale

Could you expand on what you mean by mixed sets? Is there no way to separate them? How about you find the CV for the mixed sample. Would that work for your purposes?

5. Mohamed fikry

How I solve this plz?

Find out the coefficient of variation of a series for which the following results are given: N=50, €X=25, €X(square) = 500
Where X = deviation from the assumed average 5

6. george

Am so happy with this please I need a private contact or someone should contact me on here +2348162055748 I really wanna learn more

7. Mohannas

Hi thank u
Whats your comment if there is two disease
Disease A: and disease B

But disease A have more larger value of mean , vairance , sd , and cv than disease B
What dose that mean ??

8. Heather

How do you find the coefficient of variation if you are given a set of data that includes the standard deviation and amount of sales but NO mean? So I have 30 lines that have their own standard deviation and other info….I’m looking for the item with the highest Cv.

9. Andale

If you don’t have the mean you can’t find the cov. Is there any way of figuring out the mean from the “other info”?

10. Odoom Bryan Samuel

Pls given 200 as sample packages,26kg as the average weight(mean), 3.9kg as standard deviation and 8.8 cubic feet as another mean, 2.2 cubic feet as the other standard deviation. How can you compare the variation of weight and the volume?

11. C Joyce

What exactly does “multiply by 100%” mean?
How does it differ from “multiply by 100”?

The second “=” in the Excel formula should be deleted. Also, why do you not multiply the Excel formula by 100?

13. Mark

Please I want to know how to solve for the coefficient of variation when given just the mean. How will you solve it ?

14. Andale

You can’t. It’s like asking what’s the range of allowances for kids if you only know the average is \$10. Are you given any other information?

15. arnold komba

if you are given a table with years and respective profit what do we take as classmark and frequence when computing mean standard deviation and coefficient of variation

16. Andale

It depends on what you mean by “better.” CV is a measure of variance, much like a standard deviation. A smaller standard deviation (or CV) doesn’t mean it’s better than a bigger one, just that it varies more. If you are looking for the smallest possible variation in your results then yes, the smaller CV would be better.

17. Clancy

This is very helpful. I want to clarify your response in #30 above. I believe you are saying that a smaller CV means there is less variation than a larger CV. Is that correct? For example, if Sample A has a CV of 12% and Sample B has a CV of 19% I would say that there is less variation in Sample A. Do I understand correctly?

Thanks

18. Andale

I just added a little to the article to help clarify this point (I hope :) ).
If sample A has a CoV of 12% and sample B has a CoV of 19%, you would say that sample B has more variation, relative to its mean (rather than just “more variation”).

19. Andale

I think you’ll find your answer here. “n some cases, the coefficient of variation and the RSD are the same thing. However, the RSD cannot be negative while the Coefficient of Variation can be positive or negative. This is because the two formulas differ in a minor way: the Coefficient of Variation divides by the mean while the RSD divides by the absolute value of the mean.”

20. Simon O

Hi Andale,

You seem to take your time to answer people, which I very much appreciate, so I’d like to ask two questions.

1. I just wanted a clarification of your response (#30 and #32) regarding whether a CV value is “better” or not compared to another. You said “If sample A has a CoV of 12% and sample B has a CoV of 19%, you would say that sample B has more variation, relative to its mean (rather than just “more variation”)”.
So if I understand this correctly, a smaller CV doesn’t necessarily mean a smaller spread, since it is correlated to the mean value. For example, two sets: A) Mean is 100, SD is 6 –> CV is 6%. B) Mean is 50, SD is 4 –> CV is 8%. So, by only judging by the CV value, set A would have “smaller variation”. But in reality, since it is relative to the mean, the spread is actually wider for set A and smaller for set B, even though the CV is higher. In other words, if I would want to look for the data set with better precision (small spread), set B would be preferred, since the SD is smaller?
Does this imply that comparing CV between two sets can be misleading regarding variation, if the mean is not the same?

2. Secondly, also concerning CV (precision). I have a machine that needs to perform within specifications. I have a reference value (R), along with trueness (T) and precision (CV) specifications that would be the criteria for maximum deviations allowed. Let’s say that R = 1000 units, T = 10%, CV = 5%. How do I know what the lowest acceptable measured value is?
I apologize in advance if I may be totally wrong with this. I was thinking that it could be 850 or 855, depending on the corresponding unit value of CV. First, T would allow the average measured value to be between 900 and 1100 (is this correct?) Then, CV, as you said, is based on the mean value, but is it calculated from the reference value (1000) or the actually measured mean value, e.g. 900? Because 5% would then correspond to either 50 or 45, respectively. And then subtract this value from the lowest value of trueness deviation to get the absolute lowest acceptable measured value. Is this anywhere near the correct solution?
My argument for 50 would be that the value should be fixed, as I cannot raise any logical or rather physical reason why the variation (spread) would differ depending on the measured mean value. In other words, in my mind I cannot reason that a systematic error (trueness; deviation of the measured mean value from the reference value) would affect the random error (precision; variation around the measured mean value).
But my argument for 45 is, as discussed in my first question, the CV of the measurements would not be comparable to the specifications, since they most probably will be based on different mean values. Since the second criteria for the machine is to have a acceptable precision, I have to compare CVs. But to be a truly fair comparison, the specified CV and the CV of the measurements must both be based on the same mean value, to represent the same variation. For example, if I had a measured mean value of 1100 (within T specification) and a SD of 55, this would give 5% CV, which would be within CV specification. However, the variation (SD) would be larger than for 5% CV calculated for 1000, which would be 50. Thus, the variation would technically be too much, while on paper it is accepted, and as I wrote earlier, I can’t understand why they would accept a larger spread for larger volumes (but maybe that’s because I’m not a chemist or a physicist). Perhaps this is a natural phenonema I didn’t know about?
In any case, this is why I’m thinking that using CV in this case could be misleading or confusing. Had they specified a hard value (like 50 units) for the measurement deviation (random error) it would have been so clear to apply. I just can’t make sense of it.

I’m really baffled by this situation, so any help would be greatly appreciated!

Thank you!

Best Regards,
Simon

21. Andale

1. “For example, two sets: A) Mean is 100, SD is 6 –> CV is 6%. B) Mean is 50, SD is 4 –> CV is 8%. So, by only judging by the CV value, set A would have “smaller variation”.
Yes.
“But in reality, since it is relative to the mean, the spread is actually wider for set A and smaller for set B”
No. You could only say that the standard deviation is larger.
2. “ow do I know what the lowest acceptable measured value is?” Sounds like that would be test-specific, so I wouldn’t know. For example, the lowest acceptable measured value of an artificial heart valve would be a lot smaller than the lowest acceptable measured value of a pylon.

22. Simon O

Hi Andale,

Just some follow-up inquiries:

1. So, I cannot use the expression that a larger SD is the same as a wider spread? In this case, I’m referring to “spread” as in the numerical interval wherein the majority of measurements are found (for instance, 1 SD). I’m sorry if I’m using the wrong terms here, but I hope you understand my question.

2. Yes, I realize there’s a difference between situations and objects. That is why I wanted to give a specific example with clear numbers. But, to make it a shorter question, my main issue here is regarding whether the acceptable CV (in this case, 5%) is based only on the reference value, or can be applied to any mean value derived from the test measurements, regardless how off this mean value is from the reference value. I’m confused, since the same CV for different mean values implies different SD, which seems unnatural to me in regards of test criteria since the accepted spread (in the sense I described above) would differ depending on the measured mean value.

Thank you!

Best Regards,
Simon

23. Andale

1. You can say that, yes. But trying to compare SD and CV is really like trying to compare apples and oranges.
2. “my main issue here is regarding whether the acceptable CV (in this case, 5%) is based only on the reference value, or can be applied to any mean value derived from the test measurements, regardless how off this mean value is from the reference value”. The CV is usually used to compare two different sets of data, not to compare to a reference value. But assuming there was a reference value, surely it would be a CV? for example, if that 5% was a reference value, then any subsequent results would have to be under 5%, correct?
What you’re suggesting though (I think), is that there’s a reference mean, and that the results have to be within a certain amount of that mean. In that case, what you’re really talking about here sounds like a margin of error, not a coefficient of variation. And again, you can’t compare the CV and M of E in the same sentence — they are different measurements and tell you different things.

24. Andale

Not sure what you mean by “how to give interpretation of a CV”. In general, it is the ratio of the standard deviation to the mean (average).