Weighted AveragesDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/