Any source for this formula?

Question

I have never learned math through English as a medium of instruction, so my English math vocabulary is very poor.

My problem: I had some data for different neighborhoods. For example:

Neighborhood A: 10 square miles
Neighborhood B: 8 square miles
Neighborhood C: 7 square miles

What I needed to do is express these values in a 0 to 100 range. So, I thought about it and I came up with this formula:

v – value in the 0 to 100 range
n – given value
max – maximum value of the input range
min – minimum value of input range

I think this is the correct formula to represent the data under an even range. The problem is the professor (a geography professor) is asking why I chose this formula. Moreover, he is asking for a source. Anyone knows how this formula is called and what source to use for it?

Answer 1

I don't think this formula has a name. One can however prove it does what you wanted.

You just need to prove that $\frac{n-\min}{\max-\min}$ sends all the values correctly on $[0,1]$ since after you're multiplying by $100$.

The formula is equivalent to $n=x(\max-\min)+\min$ where $x$ is the new value in $[0,1]$.

$x(\max-\min)$ gives the distance between $x$ and $\min$ (for instance if $x=0.5$,$x(\max-\min)$ is half the length of the interval)

Adding $\min$ offsets $x$ so that it starts at the right point.

Answer 2

Here is how you rationalize it. We want a curve that represents your data from $0..100$ so lets use a straight line first. $$ y = md + C $$ $y$ is the representation of the distance on the $0..100$ scale, $m$ is the gradient, $d$ is the distance you want to scale, and $C$ is the intercept.

so in particular, we want the points to be defined as the following $$ (d_{\text{min}},0),\\ (d_{\text{max}},100) $$ thus we have $$ 0 = md_{\text{min}} + C\\ 100 = md_{\text{max}} + C\\ $$ you have a simultaneous equation for $m,C$. thus we get $$ 100-0 = md_{\text{max}} - md_{\text{min}} = m\left(d_{\text{max}} -d_{\text{min}}\right) $$ this leads to $$ m = \frac{100}{\left(d_{\text{max}} -d_{\text{min}}\right)} $$ now lets find the intercept $$ md_{\text{min}} +C = \frac{100}{\left(d_{\text{max}} -d_{\text{min}}\right)}d_{\text{min}} + C = 0 $$ thus $$ C = -\frac{100}{\left(d_{\text{max}} -d_{\text{min}}\right)}d_{\text{min}} $$ thus we get $$ y = \frac{100}{\left(d_{\text{max}} -d_{\text{min}}\right)}x -\frac{100}{\left(d_{\text{max}} -d_{\text{min}}\right)}d_{\text{min}} = \frac{x-d_{\text{min}}}{\left(d_{\text{max}} -d_{\text{min}}\right)}100 $$ note, $x$=$n$

Answer 3

I hope I have understand what you are looking for; an ''accepted'' formula which transforms values of some set of data into others data ranking between 0 and 100 and which preserves the ranking of the initial data. If it is so, then I suggest to you another formula rather then that you did find which gives the same output for the same input. However this formula is largely documented in statistics books. It is related to the percentile (in descriptive statistics).

First, you should know that your input is the value of the data, say $n$, in your notation and the output is in your notation, the $v$-th percentile (but in statistics books, it is called the $P$-th percentile). The well known formula is $$ the\ ordinal\ rank\ of\ a\ data=\frac{its\ P^{th}-percentile}{100}% \times the\ number\ of\ data $$ which may be found in http://en.wikipedia.org/wiki/Percentile. From this formula you can deduce that $$% the\ P^{th}-percentile\ of\ a\ data=\frac{the\ ordinal\ rank\ of\ a\ data}{the\ number\ of\ data}\times 100. %$$ which in your notation reads as follows

$$ v=\frac{the\ ordinal\ rank\ of\ n}{the\ number\ of\ data}\times 100. $$ You can use this last formula since it provides the same output as yours for the same input. In fact $$ \frac{the\ ordinal\ rank\ of\ n}{the\ number\ of\ data}=\frac{n-\min }{\max -\min }. $$ EDIT: Actually, I have used the formula $$ n=\left( \frac{P}{100}\times N\right) $$ instead of $$ n=\left\lceil \frac{P}{100}\times N\right\rceil . $$ since when rounding off your computations, it leads to almost the same numbers.