Hoping for some help with this. I need to calculate the the ranking for a number (eighty or so) groups, based on a number of component scores. Each component score is weighed. For example:

Component 1: 10% of score

Component 2: 20% of score

Component 3: 5% of score

Component 4: 7.5% of score

Think like- Component 1 is "Average Wait Time" and Component 2 is "Customer Satisfaction Rating" etc. That sort of thing.

Each component score is found using z-scores- because the components aren't a standard measure, z-scores standardize the measurement so you can compare performance weighing the different components.

Total score is found by adding up all your Component z-scores, and then scaling them so the top score gets 100, and the rest get their percent of top score.

Here's my problem. Obviously z-scores can be negative. This makes it very difficult to explain "ok so this component your score is -4.35, for this one your component score is -2.11" etc etc. I need a way to convert the z-scores into all positive values while maintaining their function. Any ideas? Would really appreciate any help, I'm not the lunatic who came up with this, just the poor bastard who has to figure it out...

  • $\begingroup$ Why do you want to force your data to have a normal distribution. That is essentially what you do when you use a z score. The problem you have artificially created for yourself is to model your data as coming from a normal distribution which always has a positive probability to be less than 0 even when the mean is a large positive value. $\endgroup$ – Michael R. Chernick Oct 2 '19 at 0:21
  • $\begingroup$ Unclear what you mean by % weights. What accounts for the remaining 67.5%? // Assuming you have %s that add to 100, how about putting each component onto a scale of 0-100, then weighting by %s to get a score that's btw 0 and 100. (Because each score is the sum of several components, the combined scores might be roughly normal.) $\endgroup$ – BruceET Oct 2 '19 at 0:32
  • $\begingroup$ A Z score of -4 means one is 4 standard deviation below the mean. So make the calculation above, take the absolute value and state where in reference to the mean. $\endgroup$ – Dave2e Oct 2 '19 at 2:44
  • $\begingroup$ re: why am I doing it this way, I didn't design it and I can't change the base formula. This is how it's done unfortunately. $\endgroup$ – Varianz Oct 2 '19 at 12:12
  • $\begingroup$ @DaveT, I don't think I can do that. The goal here is to explain how much of a percent each component score is- so for example Component A gets a Z score of -1.5. Multiply that by it's weight of, say, 33.333% and you get a Component Score of -0.5. But I can't describe a negative number as a percent, or represent it visually in say a pie chart etc. So I need some way to convert all component Z scores into positive integers. $\endgroup$ – Varianz Oct 2 '19 at 12:16

IQ Scores are formulated using 100 + 15 * Z-Score

This results in a nice clean number thats easy to explain because an 80 is 'barely qualifies for elementary school' and a 120 is 'genius'

You'll still have to explain to the managers, however, why there's so little variation between good and bad scores, or why someone doesn't get a 0 for 0 contribution.

  • $\begingroup$ My initial thought was to do a linear translation to, say, a 1 to 10 scale. But the problem with that is the lowest scores will be 1, which may be psychologically problematic. I like the IQ model. That way managers with a score of 80 can be told that they're --- in the words of George Carlin --- "marginally exceptional". And even the average department gets a score of 100... I'm actually being serious here. I thinks that's a good approach that doesn't come off as too critical. Let upper management decide what the goal score is. Don't stick your neck our giving people negative scores. $\endgroup$ – Sal Mangiafico Oct 17 '19 at 0:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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