RELATIVE STANDARD DEVIATION is to express standard deviation in terms of % instead of standard unit. RSD = (SD(100)) / mean.

PRECISION. there is no precision. standard deviation is not a tool for precision measurement. SD in any form, i.e. RSD, is a means to provide a range of the expected value or the mean. with your observed 30% or 60% RSD, it means it is that % off about the mean. When reading SD, we speak in terms of big or small SD, not in the language of "precision." If couched in that language, then 30% and 60% about the mean tells us that it is not "precise" for being too far off---if we use, say, 5% as acceptable level of error.

LINK:

https://www.statisticshowto.datasciencecentral.com/relative-standard-deviation/

Article Australian Bureau of Statistics

https://www.statisticshowto.datasciencecentral.com/relative-standard-deviation/

Ahmed- Thanks for your question. This YouTube video explain it in simple way by using Excel. Good luck

https://youtu.be/adaXHXtKVb8

RELATIVE STANDARD DEVIATION is to express standard deviation in terms of % instead of standard unit. RSD = (SD(100)) / mean.

PRECISION. there is no precision. standard deviation is not a tool for precision measurement. SD in any form, i.e. RSD, is a means to provide a range of the expected value or the mean. with your observed 30% or 60% RSD, it means it is that % off about the mean. When reading SD, we speak in terms of big or small SD, not in the language of "precision." If couched in that language, then 30% and 60% about the mean tells us that it is not "precise" for being too far off---if we use, say, 5% as acceptable level of error.

LINK:

https://www.statisticshowto.datasciencecentral.com/relative-standard-deviation/

RSD = (SD(100)) / mean . There are examples of interpretation of its meaning on the site

https://www.educba.com/relative-standard-deviation-formula/

Ahmed - 

Well, your question was unexpected to me.  If you want an uncertainty measure relative to an estimated mean or an estimated total, I'd say the coefficient of variation, cv, in Cochran, W.G.(1977), Sampling Techniques, 3rd ed, John Wiley & Sons, on page 54 (also page 153), is what you want, which some call a relative standard error, RSE.  So that is what I thought you wanted.  However, in the reference given by Cristian, in Statistics How To, "Relative Standard Deviation: Definition & Formula," it appears that the RSD is actually something some people use, and that reference explains it as follows:

"The RSD tells you whether the 'regular' std dev is a small or large quantity when compared to the mean for the data set." 

So from that reference, unlike the RSE which can indicate how well you have estimated a mean, perhaps expressed as a percent, the RSD tells you something about the compactness of the population relative to the mean, which may also be expressed as a percent.  That is quite a different concept.  From an example in that Statistics How To reference, it says "This result tells you that your standard deviation is 2.3% of the mean of 4.4, which is pretty small. In other words, the data is tightly clustered around the mean."   

On page 51 in volume II, Theory, in Hansen, M.H., Hurwitz W.N. and Madow, W.G.(1953), Sample Survey Methods and Theory, 2 vols, Republished 1993, Wiley, New York, they show a coefficient of variation which appears to match that in the Statistics How To reference, except that only the positive sign is allowed, which matches the RSD. 

Apparently then, the definitions for 'coefficient of variation' in Cochran, and in Hansen, Hurwitz, and Madow do not match.  (I have both books and did not know that.)  The former is the RSE, and the latter is the RSD, though neither use those terms. 

To review then, a standard error is a standard deviation for a mean or total, for example, rather than for the population.  One may use the RSE to tell how accurately one is estimating the mean, where one may use the RSD to tell how little (or how much) the population varies from the mean. 

Interestingly, a larger sample size will decrease an estimated RSE, but a larger sample will only improve how well you estimate an RSD, which is a fixed quantity. 

Having worked for a statistical agency that published official statistics, I had only worked with RSEs, which are more convenient to publish in large volume than confidence or prediction intervals, as we were trying to indicate accuracy of the published estimates/predictions (where a "prediction" just means you used a regression to estimate).  The idea of looking at RSDs to see a relative spread of sorts for a population did not occur to me.  We were only interested in the accuracy of published results (which would also mean attention to any source of bias - a word that many may misuse but has meaning in statistics). 

So perhaps you need to consider both the RSE and the RSD, and decide for your application, which is appropriate. 

Cheers - Jim 

Another meaning of the CV might, perhaps, relate more closely to whatever you have in mind by "precision." It is discussed on a UCLA website:

https://stats.idre.ucla.edu/other/mult-pkg/faq/general/faq-what-is-the-coefficient-of-variation/

Some key points from that website follow.

"coefficient of variation (CV) can be calculated and interpreted in two different settings: analyzing a single variable and interpreting a model.  The standard formulation of the CV, the ratio of the standard deviation to the mean, applies in the single variable setting. In the modeling setting, the CV is calculated as the ratio of the root mean squared error (RMSE) to the mean of the dependent variable. In both settings, the CV is often presented as the given ratio multiplied by 100. The CV for a single variable aims to describe the dispersion of the variable in a way that does not depend on the variable’s measurement unit. The higher the CV, the greater the dispersion in the variable. The CV for a model aims to describe the model fit in terms of the relative sizes of the squared residuals and outcome values.  The lower the CV, the smaller the residuals relative to the predicted value.  This is suggestive of a good model fit. 

... The most obvious problem arises when the mean of a variable is zero.  In this case, the CV cannot be calculated.  Even if the mean of a variable is not zero, but the variable contains both positive and negative values and the mean is close to zero, then the CV can be misleading.  The CV of a variable or the CV of a prediction model for a variable can be considered as a reasonable measure if the variable contains only positive values."

I would add that, because a CV (defined in either of the above ways) is formed by taking the ratio of 2 values, both of which are measured in the units of the variable of interest, a strong argument can be made that the CV should only be used when that variable is measured on a ratio scale, with an actual real-world zero point. (Example: Consider a sample (or population) of interval-level IQ scores, with no true zero point, that is arbitrarily set to have a range of 200-800. If its mean is 500, and SD=100, then CV= 100/500, or 20%. Now subtract 200 from each IQ score, re-scoring the same data to have an arbitrary range of 0-600; the SD will still be 100, but the mean will drop to 300 so the CV becomes 100/300, or 33.3%. Which CV value is "right"? Arguably, neither, because the CV should not be used for a non-ratio variable like this one.) Calculating a CV for a variable that is not only non-ratio but also non-interval, such as an ordinal Likert item, is especially suspect.