# How do I 'reverse engineer' the standard deviation? My problem is fairly concrete and direct. My company loves to do major business decisions based on many reports available on the media. These reports relates how our products are fairing in comparison to the competitor's offerings. The latest report had these scores (as percentages): Input values (%) 73.5, 16.34, 1.2, 1.15, 0.97, 0.94, 0.9, 0.89, 0.81, 0.31 Our product in the 'long' tail of this list. I argued with them that besides spots #2 and #1 all the other following the tail are on a 'stand', since, probably, the standard deviation will be much bigger that the points that separates everyone in the tail. So the question is: How may I calculate the standard deviation having only these percentual values available?

How do I 'reverse engineer' the standard deviation?
My problem is fairly concrete and direct.
My company loves to do major business decisions based on many reports available on the media. These reports relates how our products are fairing in comparison to the competitor's offerings.
The latest report had these scores (as percentages):
Input values (%) 73.5, 16.34, 1.2, 1.15, 0.97, 0.94, 0.9, 0.89, 0.81, 0.31
Our product in the 'long' tail of this list. I argued with them that besides spots #2 and #1 all the other following the tail are on a 'stand', since, probably, the standard deviation will be much bigger that the points that separates everyone in the tail.
So the question is: How may I calculate the standard deviation having only these percentual values available?
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Without more information, you can't say much. Suppose you knew there was a survey question "Which product do you like best?" and the above numbers are the shares of products named.
If, in addition, you knew the sample size, e.g., N=1000, then you could compute the standard deviation of the mean (and consequently standard errors and confidence intervals):
$sd\left(\text{x is best}\right)=\frac{1}{N-\mathrm{#}+1}\sum _{i=1}^{N}\left({x}_{i}-\overline{x}{\right)}^{2},$
where you have $\overline{x}\ast N$ times ${x}_{i}=1$ and $\left(1-\overline{x}\right)\ast N$ times $100\ast \overline{x}$ is the percentage from the survey, # the number of percentages computed from the data).
Without N, there's nothing you can say. Indeed, if they asked the whole population, then even a 0.01 difference is meaningful. But you are right, for small sample sizes the difference is likely not significantly different from zero.