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How do I calculate the weighted average between a percentage and a number?

For $120\%$ and $88$, assuming that $120\%$ weighs $60\%$ and $88$ weighs $40\%$.

For context for 120% it is based off this formula , 6/5 =1.2 which is 120%

So basically in a contact center environment 120% is the average handling time which is 6 minutes divide by 5 tasks worked on and 88 is another metric which I would like to weigh in which is the satisfaction score out of a 100

Kcv
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  • Welcome to stackexchange. The short answer is that you cannot do this. In order to average two quantities they must have the same units. If you [edit] the question to tell us the context - where does this question come from? - we may be able to make some suggestions. (Do [edit] the question. Don't try to clarify in comments.) – Ethan Bolker Apr 06 '21 at 16:15
  • Thanks I have modified the question – Kcv Apr 06 '21 at 16:20
  • That does not help. What do the numbers mean? If they are just numbers and you just want a weighted average between $1.2$ and $88$, forget about percentages and just calculate $0.6 \times 1.2 + 0.4 \times 88$. If you want this average to say something useful (maybe about a product or movie rating) then you have to describe the whole situation. – Ethan Bolker Apr 06 '21 at 16:27
  • You need to know % for 88. – herb steinberg Apr 06 '21 at 16:31
  • Modified the question again – Kcv Apr 06 '21 at 16:46
  • Why don't you use $88%$ instead of $88$? Unlike the $120%$, it really is a percentage. – saulspatz Apr 06 '21 at 16:50

1 Answers1

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I think that what you call "weighted average" really should be called "multivariable criteria".

Suppose I have three different types: houses (H), cars (C) and trees (T). There are several groups, each having a different number of H,C,T.

I have two groups:
(a) $2H, 1C, 5T$
(b) $3H, 4C, 2T$

Now I select "importances" (a.k.a. "weights") to my criteria:
Number of houses is $30$% of my decision, number of cars is $10$% and trees is the rest ($60$%). You see, having more H is better than having more C, but worst than more T.

Now I calculate points:
For (a): $2·30 + 1·10 + 5·60 =\;\;370$
For (b): $3·30 + 4·10 + 2·60 =\;\;250$
So te winner is (a)

This kind of "average" allows to mix different types into a unique criterium.

Ripi2
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