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Weighted Averages


Date: 11/02/98 at 21:09:00
From: Jacob Smith
Subject: Help with "weighted Averaging"

Dr. Math,

I am in the 9th grade, and our math teacher is explaining "weighted 
averaging." Could you help me by giving a simple but detailed 
description of this?

Thank you for your time.
Sincerely,
Jacob


Date: 11/03/98 at 12:18:47
From: Doctor Peterson
Subject: Re: Help with "weighted Averaging"

Hi, Jacob. I think I'll start at the beginning with an explanation of 
what weighted averaging means in a simple case, and then look at a more 
general case.

Suppose that your teacher says the final exam counts as much as three 
tests. Then if your scores are:

   tests: 70, 80, 90   final: 100

your average will be just as if you got:

   tests: 70, 80, 90, 100, 100, 100

             70 + 80 + 90 + 100 + 100 + 100   540
   average = ------------------------------ = --- = 90
                           6                   6

If we want to calculate this directly, we can just multiply the final 
score by 3 when we add them up, but we have to remember also to count 
it three times in the denominator, not just divide by 4. You can do 
this by writing it out this way:

   score    weight    value
   -----    ------    -----
     70        1        70
     80        1        80
     90        1        90
    100        3       300
              ---      ---
               6       540  --> average = 540/6 = 90

That is, you divide the sum of the weighted values by the sum of the 
weights themselves.

A similar problem arises if you've calculated the average of some set 
of things, and want to figure out the new average. It's natural to 
want to just average the old average with the new value:

   tests: 70, 80    average = (70 + 80)/2 = 75
   new test: 90     new average = (75 + 90)/2 = 82.5 ?

But this is wrong, since the new average really is:

   new average = (70 + 80 + 90)/3 = 240/3 = 80

What you have to do in this case is to weight the old average 
proportionally to the number of scores it represents, since this 
situation is just as if you had scores of 75, 75, and 90:

                 2 * 75 + 90   240
   new average = ----------- = --- = 80
                    2 + 1       3

Now, in general, you can assign any weights you want, not necessarily 
integers. Often we choose weights that add up to 1. In our first 
example, we can say that each test counts for 1/6 of the grade, and 
the final counts for 1/2 of the grade. Then we calculate this way:

   score    weight    value
   -----    ------    -----
     70       1/6     11.66
     80       1/6     13.33
     90       1/6     15
    100       1/2     50
              ---     -----
               1      89.99  --> average = 90/1 = 90

I hope that helps out.

Here's an explanation of weighted averages I  found in our archives, if 
you want another perspective:

  http://mathforum.org/dr.math/problems/riggins11.16.95.html   

Incidentally, do you wonder why it's called "weighted"? An average can 
be thought of as the place where a set of identical weights placed at 
different locations on the number line would balance:

                                  X   X   X   X
   ---+---+---+---+---+---+---+---+---+---+---+---
      0  10  20  30  40  50  60  70  80  90  100
                                        ^

If the weights are different, you get a weighted average:

                                  1   1   1   3
   ---+---+---+---+---+---+---+---+---+---+---+---
      0  10  20  30  40  50  60  70  80  90  100
                                            ^

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Statistics

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