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Hello, here's my basic problem - I would appreciate any help. We use a weighted system based on 1 (developing) ,2 (performing) ,3 (leading) to evaluate an employee's rating using various metrics. Each metric has a weighted value.

Example:

Apples Goal 50% w: result 2 (performing) Oranges Goal 20% w: result 3 (leading) Grapes Goal: 30% w: result 1 (developing)

2(.50)+ 3(.2)+ 1(.30) = 1.9 Final Score

We try to set each indvidual metric to meet a simple distribution: 15% of employees will likely hit 1 (dev), 70% 2 (perf), and 15% 3(leading)

So here's where my knowledge of stats ends - how do I figure out what final score range would also lead to an overall 15%,70%,15% distribution?

example Final Score Thresholds (incorrect)

...>= 2.5 leading ...>= 1.5 perf ...< 1.5 dev

If it was just a single stat like "Apples" I could use 3,2,1 because we already know this dist. But I can't use 3.0 for leading when several stats are involved, because it seems to me the odds have now decreased that the top 15% of employees could exist in all 3 metrics of apples, oranges, and grapes?

Thank you for your time,

  • Tim / Dublin,Ohio USA
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closed as off-topic by Ricardo Andrade, Stefan Kohl, Lucia, Neil Strickland, Ian Morris Nov 24 '14 at 9:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Stefan Kohl, Lucia, Neil Strickland, Ian Morris
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Casting a vote against closing $\endgroup$ – Yemon Choi Dec 16 '10 at 20:34
  • $\begingroup$ If you wish to cast a vote to close, could you please instead leave a comment along the lines of: "I vote to close and hence cancel Yemon's previous comment"? Thanks. $\endgroup$ – Yemon Choi Dec 16 '10 at 20:36
  • $\begingroup$ @Yemon-Choi, is there something deeper in the question which I am missing? Normally, it is hard to compare apples and oranges (or apples, bananas, and cherries as I use in my example answer below), however this question already posits a weighting and a summing system which allows these disparate categories to be combined into a single weighted score, getting around the big problem of assigning a realistic weighting system for "comparing apples and oranges." $\endgroup$ – sleepless in beantown Dec 17 '10 at 0:03
  • $\begingroup$ I should have pointed Tim to mathoverflow.net/faq earlier $\endgroup$ – Yemon Choi Dec 17 '10 at 0:17
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This is a basic probability and distribution question, I believe, and I hope it is not homework. Based on Yemon Choi's comments, I'm posting this answer:

For the sake of example, I will relabel your fruit

  • $a$ = apple
  • $b$ = banana
  • $c$ = cherry

and I will relabel your scoring system

  • old $1$ = developing = $-1$
  • old $2$ = performing = $0$
  • old $3$ = leading = $+1$

Presume that you've set the scoring correctly so that the distribution of $a$ is 15% = | (-1) | , 70% = | (0) | , and 15% = | (+1) |, of $b$ is 15% = | (-1) | , 70% = | (0) | , and 15% = | (+1) }, and of $c$ is 15% = | (-1) | , 70% = | (0) | , and 15% = | (+1) |.

There are now 27 categories for the ordered triplet $(a,b,c)$; each of these 27 categories is $c_i \in$ {$-1, 0, 1$}$^3$. Calculate the distributions of each category. Draw as a histogram. Select the score ($85$th percentile) such that $15$% of your group exceeds it as the leading category. Select the score ($15$th percentile) such that 70%+15% of your group exceeds it as the lower-most value of your performing category, and use the score of the leading category as the upper-most boundary of your performing category. Use the $0$th to $15$th percentile as your bounds for your developing category.

Rescale the scores to {1, 2, 3} from {-1, 0, 1} which I used in this example to get the answer which you want. If this really is not homework, you won't have a problem doing that. If this is homework, you really should have asked somewhere else.

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  • $\begingroup$ Of course, this all assumes independence in the distributions of $a$, $b$, and $c$'s scorings. If there is any sort of dependence, this all falls apart. You've also got the problem that since scores only come as integer values, that the percentiles of the 27 subcategories may not fall in a way to allow you to split total percentiles at exactly 15%<bottom,15%<middle <85%,, top <85% $\endgroup$ – sleepless in beantown Dec 17 '10 at 0:07
  • $\begingroup$ @SIB: Regarding the homework comment: I think this is an example from someone outside math academia seeking help (from the OP's profile and the tone of his post). While I'm not a fan of too many questions along these lines, this kind of question annoys me less than "please do my grad school commutative algebra exercise for me", even if it is at a lower level. $\endgroup$ – Yemon Choi Dec 17 '10 at 0:16
  • $\begingroup$ @Yemon-Choi, actually I agree with you. I've got no problem with helping someone who's having difficulty formulating an approach to a question, because it's hard to search for answers when you don't know the key basic terms to enter as a query on search engines. A well formulated query from someone unfamiliar with a topic is also more acceptable to me than "this homework is hard" from a student. $\endgroup$ – sleepless in beantown Dec 17 '10 at 1:50
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    $\begingroup$ I will try to apply what is outline here.. I do appreciate the help! this is most certainly not homework I just simplified the real world variables to protect the actual metric names (company information). I'm part of an operations team for my company and wanted to learn the proper way to solve this, rather than wait for an answer "use these numbers" from our HQ. Though I do admit this is basic probability, the steps above were outside my ability for the time being. Many thanks $\endgroup$ – Tim C Dec 17 '10 at 3:38
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The overall distribution clearly depends on the dependence between the stats. For instance if everyone leading in Apples would be developing in Grapes, and vice versa, the limit for the top 15% would have to be lower than if everyone leading in Apples would also lead in Grapes and vice versa. As in the first case everyone gets at most 1.8 from Apples and Grapes so at most 2.4 at all, so the limit for the top 15% is at most 2.4, while in the second case the 15% having leading in Apples (and hence Grapes), would have a total score of at least 2.7 (3*0.5 + 1*0.2 + 3*0.3) and the limit hence has to be at least 2.7.

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