# Can a standard deviation of raw scores be reported as a standard deviation of percentages?

Suppose we have a test consisting of 30 questions, and 10 people take this test. The mean test score of these 10 people is 17, and the standard deviation of all the scores in the sample is 4. When reporting the descriptive statistics at school, we use these raw scores and write (M=17, SD=4); but in some cases I have the feeling that reporting percentages would be better. Because I think that we have a more intuitive grasp of what it means to score 56.7 over 100 than to score 17 over 30 (probably because we are accustomed to the decimal system).

So, for the example given above, would it be possible to report the mean and standard deviation as (M=56.7%, SD=13.3%)?

Does it make sense to say that the exam scores in a sample have the standard deviation of 13.3%?

These percentages are the arithmetic equivalent of the raw scores I made up and given above, but I am not sure whether it is good practice to directly convert them into percentages like that.

• AFAIK you can transform continuous variables to another scale and show the distribution on that scale as long as you are clear how you got there. In your case however, you can wonder whether the raw score attained from 30 questions offers enough information to transform to a continuous scale ranging from 0-100 (%) (because the data supports only increments of 3.33%). – IWS Oct 4 '17 at 10:18
• Yes, that's true. A score over 30 is not as informative as the converted score (over 100), because the increments of 100 are smaller (1) and therefore a test over 100 points would be more "sensitive" provided that all grades are integers. Still, given your answer, I think it will not be considered "malpractice" in my case to report them in this way. (I am actually preparing an assignment for school, and I will report the raw mean & SD in the text, but I just believe it will make more sense to show these scores as percentages in the tables&graphs). From what I understand, this would be doable. – Freya Oct 4 '17 at 10:29
• And thank you very much for your answer. – Freya Oct 4 '17 at 10:31
• Just to be complete: do note that some transformations might actually change the distribution's shape, so you should not just apply the transformation you want to apply to your data directly to your measure location and spread. Instead apply the transformation to your data and then assess the measures of location and spread as if this was the original data (i.e. redefine the mean and sd in your case). – IWS Oct 4 '17 at 10:35
• Technically, as a measure of distance rather than location, the SD would be percentage points (pp) rather than percent. I'd also be wary of interpreting it in this context since the error model should take into account that the scale is discrete as @freya mentioned. – James Oct 4 '17 at 13:04

The standard deviation is just a statistical property that you can measure for a set of data points. The standard deviation does not itself make any assumptions that your data is normally distributed or has/has not passed through any transformations, linear or otherwise.

Therefore, it's perfectly acceptable to use the standard deviation on any data, including the percentage scores.

Note that, in your particular case, the transformation you are applying is a linear transform, of the form:

$$y = Ax + b$$

i.e. an affine transform. So you can calculate the standard deviation on the original, untransformed data and then multiply by A to get the standard deviation after the transform. There seems to be no particular advantage to doing this rather than simply calculating the standard deviation on the already transformed data, but it might be reassuring.

We can see that an affine transformation will transform the standard deviation linearly by $A$, as follows:

Given we have input data $\{X_1, X_2, ..., X_n\}$, the original standard deviation, $\sigma$, will be given by:

$$\sigma_X^2 = \frac{1}{n}\sum_{i=1}^n \left(X_i - \frac{1}{n}\sum_{j=1}^n X_j\right)^2$$

Let's apply the transform $Y = AX + b$. Then we have

$$\sigma_Y^2 = \frac{1}{n}\sum_{i=1}^n \left( AX_i + b - \frac{1}{n} \sum_{j=1}^n \left( AX_j + b \right) \right)^2$$

$$= \frac{1}{n}\sum_{i=1}^n \left( AX_i + b - n\frac{1}{n}b - \frac{1}{n} \sum_{j=1}^n \left( AX_j \right) \right)^2$$

$$= \frac{1}{n}\sum_{i=1}^n \left( AX_i - \frac{1}{n} \sum_{j=1}^n \left( AX_j \right) \right)^2$$

$$= A^2 \left( \frac{1}{n}\sum_{i=1}^n \left( X_i - \frac{1}{n} \sum_{j=1}^n \left( X_j \right) \right)^2 \right)$$

$$= A^2 \sigma_X^2$$

Therefore

$$\sigma_Y = A \sigma_X.$$

• This is actually very helpful and illuminating. Thank you. – Freya Oct 4 '17 at 12:42
• @freya Thanks! Dont suppose... if you feel this answer is ok-ish, do you mind clicking on the green 'tick', on the left of the answer, to mark it as 'accepted'? – Hugh Perkins Oct 5 '17 at 9:22