1) I am trying to design a product for weekly ratings (1-5) of an entity. The entity's rating will be reset every year. I know most of the weeks the rating will be the same (3) and I want a way to give more weight to 4 (which can occur only 5 to 6 times a year) and a rating of 5, which can happen only 2-3 times a year. Is there a way I can ignore the arithmetic mean and use any other metric / strategy that highlights the outliers so that the entities which get the occasional 5 star stand out far than the regular 3s or 4s?

For I have an Object 1 which has got 3 rating for all 40 weeks and 4 rating for 12 weeks (Avg - 3.23) should be of less value than an Object 2 which has got 3 rating for 48 weeks and 5 rating for 4 weeks (Avg - 3.15). I want Object 2 to be ranked better with higher rating and the difference should be also significant.

2) Will it help my cause if I use a separate scale as in software user story sizing (1,2,3,5,8) instead of (1,2,3,4,5)?

  • 1
    $\begingroup$ The good news for you is that there are many ways to do this; correspondingly the results will be arbitrary. For example, squaring scores will push 3 to 9, 4 to 16, 5 to 25. Or you take the mean of the 5 or 10 highest scores, or whatever. Will anyone insist on means if you can make a case that another summary suits your purposes better? The question is who judges your work here and what will they expect. (I have no idea what you mean by "significant" here.) $\endgroup$
    – Nick Cox
    Oct 23, 2017 at 16:37

1 Answer 1


If arithmetic mean is your reference point, try to use its extension - the Ordered Weighted Averaging operator (OWA) defined as:

$$OWA_w(x) = \sum_{i=1}^n w_i*x_i^*$$ $\text{where} \sum_{i=1}^n w_i=1, w_i \in [0,1]$ and $x^*$ is $x$ sorted in the non-increasing order.

OWA is the arithmetic mean when $w_i=\frac{1}{n}$, for $i=1,\dots,n$.

Try to experiment with weight vector $w$ to suit your needs. Assume that the bigger the $x_i$ the bigger weight is used.

For example:

Given $n=5$, $x=(2,3,3,4,4,5)$ define $w=(\frac{6}{21},\frac{5}{21},\frac{4}{21},\frac{3}{21},\frac{2}{21},\frac{1}{21})$.

Then $x^*=(5,4,4,3,3,2)$ and your average is $$OWA_w(x)=\frac{6}{21}*5+\frac{5+4}{21}*4+\frac{3+2}{21}*3+\frac{1}{21}*2.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.