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Let's say I am measuring a metric where the goal is to be $3$ or lower. The actual value is $4$. I want to be at least $95$% within the goal.

If the actual value is for e.g. $2$, then I can say I am $100$% within the goal, as it is below $3$. But what would be the best way to display an actual value of $4$ as a percentage in relation to $3$?

Currently I am doing $\frac{3}{4}$, which is backwards compared to what you would normally do, but the value it results in makes more sense ($75$%).

$\frac{4}{3}$ results in $133$%, which you could interpret as $33$% over the goal i.e. you are actually at $77$%, but this method isn't scalable; if we went with a larger value like $15$, the result would be $500$%, meaning $400$% over the goal, which seems exaggerated.

yui
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  • There is no pure mathematical answer to your question. Math can be used to model some reality, but you need to tell us what that reality is. Your problem statement has words such as "best way", "I want to be", "you would normally do", "seems exaggerated" etc. If all you want is something that will be 100% on $[0,3]$ and then decrease down to $0$ on $[3,\infty)$, and let's say (I am putting here my own bias) it is continuous ... there are very many continuous functions satisfying those properties. –  Dec 27 '19 at 17:10
  • Yes, your example matches my reality. Could you elaborate on some of the continuous functions? – yui Dec 27 '19 at 17:16
  • I suggest you just take a graph paper, and try to draw a graph of the function I mention here - then you will see what sorts of choices you have. You may decide to join the point $(3, 100%)$ with $(A, 0)$ with a straight line, for example, for some number $A>3$, and then continue with $0$ for any number bigger than $A$. The corresponding graph is $$\begin{cases}100&0\le x\lt 3\100-100\frac{x-3}{A-3}&3\le x\lt A\0&x\ge A\end{cases}$$ but it looks a bit zigzag, and again you can choose $A$ whatever number you want. –  Dec 27 '19 at 17:22
  • Or, you can make it more "smooth" by putting some other function, e.g. a hyperbola going through $(3,100)$: $$\begin{cases}100&0\le x\lt 3\100\frac{3-C}{x-C}&3\le x\end{cases}$$ where $C\lt 3$ is again an arbitrary number ($C$ closer to $3$ makes it "drop" quicker from $100$% to $0$% as $x$ goes over $3$). And ... there are many, many, other choices ... –  Dec 27 '19 at 17:28

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