This is partly related to the question at Visualising the variance which I haven't accepted an answer yet, partly because I have become unsure if I'm asking the correct question.

The case is this, we have a population P that consists of 50 people. These have answered questions on a scale of 1-5, which our software normalizes to a 0-100 scale. We also have a reference population R which consists of several hundred, if not thousands of people. The reference data is "pre-rendered" to have an average score, a standard deviation, and an average standard deviation (ASD). The last value is found by doing the standard deviation per survey, then the average of those standard deviations. (The reference population contains a representative collection of previous surveys)

We then generate a value for what we directly translated call "variation", but I have started to believe is more a comparison of spread... This is where my issues lie. The formula related to this is

(((((stdev(P) - ASD) / ASD) * 100) + 100) / 2)

(The last part is to adapt to a 0-100 scale, where 50 means that the spread in your result is the same as the spread in the reference population)

What are we really calculating with this formula?

(I have taken over this system as a developer, and statistics are not my strong side)

  • $\begingroup$ I'm a little confused. How does that formula guarantee that the result is between 0 and 100? e.g. if P was three times ASD, then the result would be 150 wouldn't it? Is there something about the setup that makes that impossible? $\endgroup$
    – Glen_b
    Commented Jan 27, 2014 at 23:45
  • $\begingroup$ A choice was made that all values over 100 and under 0 just gets cut off. These numbers are not for scientific research, but just to identify problem areas in the customers organization. And we're not even saying that a high or low number necessarily is a problem, just that your spread is higher/lower than what is usual in our reference population. $\endgroup$ Commented Jan 28, 2014 at 7:39
  • $\begingroup$ I note that you begin by defining P to be a population, but in your formula presumable you intend it to be a number. Please make your notation consistent. $\endgroup$
    – Glen_b
    Commented Jan 28, 2014 at 8:29
  • $\begingroup$ I'm sorry, I've updated the formula. The formula should be correct now. (Ignore my noobish notation) $\endgroup$ Commented Jan 28, 2014 at 8:58

1 Answer 1


Note that (P - ASD) / ASD = P/ASD - 1

Hence ((P - ASD) / ASD * 100) = 100 P/ASD - 100

Hence (((P - ASD) / ASD * 100)+ 100) = 100 P/ASD

Hence (((((P - ASD) / ASD * 100)+ 100) / 2) = 50 P/ASD

So the formula you give is nothing more than 50 times P / ASD.

That's what you're really calculating.

You'll note that this is only restricted to the range 0-100 if 0 < P < 2 ASD.

Note that while this relates to scaling percentages to a range (and so is arguably on topic), the actual problem you have here is simply a trivial algebra issue.

(A lack of statistics shouldn't stop you from simplifying a simple algebraic formula -- it really doesn't require any statistics background to do that.)

  • $\begingroup$ Well yes. The remainder over 100 or under 0 is cut off. But my question was actually if it was a valid formula to calculate some sort of relative spread in the first place. The last part is more of a cosmetic solution for explaining the number. $\endgroup$ Commented Jan 28, 2014 at 7:38
  • $\begingroup$ If the remainder is cut off for both P and ASD, then they're both restricted to 0-100, in which case that calculation is apparently not restricted to the range you say it is. If P and ASD are comparable measures of spread then their ratio is relative spread. But multiplying by 50 seems bizarre. $\endgroup$
    – Glen_b
    Commented Jan 28, 2014 at 8:27
  • $\begingroup$ Again, I don't know why this was done. But someone sometime thought that a 0-100 scale was more layman than a 0+/- scale. I also now see an additional problem as using this formula a high standard deviation in the population against a low reference can yield higher than 1, but a low standard deviation in the population against a high reference can never be lower than -1, so in reality the normalized scale goes towards infinity as the reference goes towards 0... This is now officially above my paygrade ;) $\endgroup$ Commented Jan 28, 2014 at 9:14
  • $\begingroup$ It doesn't matter how low the population SD is, the calculation you give is still restricted to be positive, as my post already showed. It can never, in any circumstances, be negative. The formula simplifies to a constant times a ratio - did you not follow the algebra? $\endgroup$
    – Glen_b
    Commented Jan 28, 2014 at 11:35

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