0
$\begingroup$

Suppose now I have a group of students, and for each student two measurements are given: one is the height of the student and the other is the weight of the student. Then my question is how I can select a student that have both decent height (the higher, the better) and decent weight (the heavier, the better)? One solution I can give is as follows:

Step 1: Normalize the weight and the height for each student, and the normalization is performed in the following way:

1) Calculate the average m and standard derivation s of the height (or the weight) of all the students.

2) The up-threshold and down-threshold is defined as:

upThreshold = m+3*s

downThreshold = m-3*s

So if the student's height is higher than upThreshold, then its normalization value will become 1. If the student's height is lower than downThrehold, then its normalization value will become 0.

3) For the height in between, I use the following formula to normalize the height:

 (height-m)/(3*s)*0.5+0.5 

Step 2: calculate the sum of the normalized height and normalized weight for each student, and select the one with the maximum sum.

I do not know whether there are some problems in my solution, and any comments will be appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$
  1. Simply speaking, you normalize weights and heights to mean=1/2 std.deviation=1/6, then force them into [0,1] interval. The same thing could be achieved simpler if you'd normalize them to mean=0, std.deviation=1, and force them into [-3,3] interval.

  2. The problem with you approach is: suppose that one student has very high weight, but height below the average. He may be selected over all others. Another problem is, you have an arbitrary constant 3.

  3. Therefor, if you want "both decent height and decent weight", I suggest to normalize weights and heights to mean=0, std.deviation=1, then just select the student with highest max(weight, height). If the value is same for two students, select one with highest weight+height.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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