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Weighted average problem with negative numbers in data set

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
by saintforlife » Thu Nov 29, 2012 10:41 pm
Let's say I have 4 values A, B, C and D. Let's say they each contribute 30%, 30%, 20% and 20% towards some final value X.

So my weighted average formula is:

X = [0.3(A) + 0.3(B) + 0.2(C) + 0.2(D)] / [0.3+0.3+0.2+0.2]

But what if the situation is such that values C and D contribute negatively to the over all value X?

As a sports analogy using a hypothetical situation, let's say I am calculating a Football Quarterback's rating using my own method as follows:

A = Yards per Attempt (I want this to be high, so this is a positive value) - 30% weight
B = Number of Touchdowns (I want this to be high, so this is a positive value) - 30% weight
C = Number of Incomplete passes (I want this number to be low/miniumum and it contributes negatively to the overall rating) - 20% weight
D = Number of Interceptions (Needs to be minimized and contributes negatively to the overall rating) - 20% weight

So for my overall weighted average QB rating can I use this formula?

X = [0.3(A) + 0.3(B) - 0.2(C) - 0.2(D)] / [0.3+0.3+0.2+0.2]

If this is wrong, how to I account of positive and negative numbers in the same weighted average formula? Do the positives and negatives have to add up to 100% individually? But what if they contribute different amounts to the end result?
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Source: — Quantitative Reasoning |

by Jim@StratusPrep » Fri Nov 30, 2012 8:19 am
2 things.

1) The negative values will get included in the weighted average.

2) You don't need to divide by the sum of .3, .3, .2 and .2 - this is equal to 1, as will be the case with all percent oriented weighted average problems.
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by saintforlife » Fri Nov 30, 2012 9:07 am
Jim@StratusPrep wrote:2 things.
1) The negative values will get included in the weighted average.

2) You don't need to divide by the sum of .3, .3, .2 and .2 - this is equal to 1, as will be the case with all percent oriented weighted average problems.
Thanks for the reply Jim. Can you elaborate on this a bit? Are you saying I can just use the negative numbers as-is in the formula for X?

Is it correct to write X = [0.3(A) + 0.3(B) - 0.2(C) - 0.2(D)] in this fashion?

Also does it matter if the values of A, B, C and D have greatly different values? Take some hypothetical numbers in my example:

A = Yards per Attempt = 7
B = Number of TDs = 3
C = Number of incompletions = 15
D = Number of Interceptions = 1

As you can see the numbers vary by quite a bit. With my formula X = [0.3(A) + 0.3(B) - 0.2(C) - 0.2(D)] am I unintentionally giving extra weight to certain values just because they numbers themselves are much larger than the others?

What would be the best way to go about solving such a problem where some values contribute and other take away from the final value, each number having its own contribution in the whole scheme of things?
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