In our series on averages, we’ve looked at mean/median/mode, then at details of the (arithmetic) mean, and then at different kinds of mean (arithmetic, geometric, harmonic, quadratic). Next, I want to look at the **weighted mean** concept. In checking what we’ve said about this, I found a useful series of explanations of one application of it: grade averages. As many students are more familiar with this than any other application, it’s a good place to start, before we expand to look at other cases of weighted averages.

## What is a weighted average?

We can start with this 1998 question:

Weighted Averages 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?

### Final counts as three

I answered, starting with a basic example:

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 saysthe 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, 10070 + 80 + 90 + 100 + 100 + 100 540 average = ------------------------------ = --- = 90 6 6

Here we are taking the description literally, by including three copies of the final grade, as if it were in fact three final tests, making a total of 6. This idea underlies everything we will be seeing here, though it becomes harder to take literally as we proceed.

If we want to calculate this directly, we can justmultiply the final score by 3when we add them up, but we have to remember also tocount 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, youdivide the sum of the weighted values by the sum of the weights themselves.

This is the essential idea of any weighted average, though we’ll be seeing it in several forms:

$$\text{Weighted average}=\frac{\text{sum of weights times values}}{\text{sum of weights}}=\frac{\sum\limits_{i=1}^n w_ix_i}{\sum\limits_{i=1}^n w_i}$$

### Combining a new test into the old average

Next week we’ll be looking at how weights can help when we need to combine averages of different groups. I introduced this idea with a common thing we may need to do with grade averages:

A similar problem arises if you've calculated the average of some set of things, and want to figure out thenew 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 toweight the old average proportionallyto 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

This is really very much like the idea of counting the final as three tests; we are counting the current average (of two tests) as exactly what it represents, two tests! And we’d do the same thing if we knew, say, that the average of the first two tests was 75 and the average of the remaining *three* tests was 90: weight the first average by 2 and the second by 3, so we get $$\text{Overall average}=\frac{2(75)+3(90)}{2+3}=\frac{420}{5}=84$$

### Weights that sum to 1

Since we have to divide by the sum of the weights, it can be convenient to choose weights that sum to 1:

Now, in general, you canassign 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.

In this case, dividing by 1 does nothing, so we can take the weighted average this way: $$\text{Weighted average}=\text{sum of weights times values}=\sum\limits_{i=1}^n w_ix_i$$ It’s just the sum of my last column. (If you have studied probability, you may notice that this is identical to the expected value of a probability distribution, where the weights are the probabilities.)

### Why “weighted”?

Incidentally, do you wonderwhy 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 ^

We saw this idea of balancing when we examined arithmetic means; giving scores different weights just expands the idea a little.

## Weights as percentages

At the end of my answer above, I referred to this 1995 answer, which gives a different perspective in a couple senses:

Grade Averages If Johnny has a 82% average for the 1st 9wks and a 75% average for the 2nd 9wks,what grade would he have to get on the finalto receive 80% semester grade in this class. The grades are weighted as follows: each 9wks 40 % and the semester test 20%.

Here the weights are neither integers nor fractions, as in my answer above, but percentages. And we have to reverse the process, finding a missing grade.

Doctor Ethan answered:

Okay, good question. Here is how it works. .4 + .4 + .2 = 1 So what we are going to do is to set up an equation involving the percentages of the present grades and how they add up. It looks like this: .4(82) + .4(75) + .2(X) = 80 So to solve for X we simplify to 32.8 + 30 + .2(X) = 80 SO..... .2(X) = 80 - 32.8 - 30 So .2(X) = 17.2 Now divide both sides by .2, and X = 86. Hope that helps.

The percentages, 40%, 40%, and 20% are the weights, and because they add to 1 (that is, 100%), we don’t have to divide by their sum. The result is a linear equation that can be solved for the missing grade.

## Calculating your grade average

The next question, from 1999, is explicitly about grades:

Calculating a Grade Average This year my dad wants me to keep track of all my scores of worksheets, tests, etc. to find out what my average is. Well here's my problem: If some things are worth more than others are, how can I do that?

Doctor Jeff answered, making up an example because no details of the grading system were given:

### Using a point system

Hello, Rachel. Thanks for writing to Dr. Math. Let's assume that your average is calculated on apoint system. That is, everything is worth a certain number of points, and certain things are worth more points than others are. A test might be worth 100 points; a paper, 50 points; and a quiz, 10 points. Here are a student's scores, using the above system, for the first quarter of the school year: Quizzes Papers Tests 9/10 or 90% 43/50 or 86% 79/100 or 79% 7/10 or 70% 41/50 or 82% 92/100 or 92% 10/10 or 100% 90/100 or 90% 8/10 or 80% 8/10 or 80%

Observe that the unsimplified fractions show the actual points over the possible points for each item, while the percentages, which are equal to the fractions, show no information about the point values. The first quiz, for example, counts as 9 points (out of 10) in the point system. We don’t just average the percentages!

If you want to calculate the student's average for the entire first quarter, the easiest way is toadd up the total number of pointsthe student earned and divide that by the total number of points the student could have earned. In this case, it's best toavoid dealing with the individual percentages, since, for example, a 90% score on a quiz does not count nearly as much as a 90% on a test. This student's average would then be: 9+ 7+10+ 8+ 8+43+41+ 79+ 92+ 90 387 -------------------------------- = --- = .86 = 86% 10+10+10+10+10+50+50+100+100+100 450

Incidentally, you might look at what he did here and think, this is the wrong way to add fractions! You don’t add the numerators and the denominators! But here we are *not* adding fractions, but averaging the point values in a special way that takes into account their specific denominators. In fact, if you simplified any of the fractions, you would not get the same result, so we are not actually working with the *values* of the fractions at all.

If we were using a weighted average method for the class above, we might say that the quizzes, together, count as much as one paper, and a test counts as much as two papers. The quizzes total 1/9 of the grade, each paper 1/9, and each test 2/9.

### Using a percent system

Of course,not all teachers use this simplified point system. Perhaps a more common method for determining a student's average goes something like this: "Quizzes are worth 20% of your grade; homework is worth another 10%; papers are worth 30%; the final is worth 30%; and class participation is worth the remaining 10%." In this scenario, students might only be given percentages for their work. Let's say a student received the following scores: Quizzes Homework Papers Final Participation 95% 84% 85% 88% 100% 80% 100% 89% 75% 100% 90% 90% 91% The first thing the student needs to do is come up with an average for each of the five categories above. Let's assume that each quiz is worth the same as every other quiz, that each homework assignment is worth the same as every other homework assignment, etc.

This time, each item gets only a percentage; its contribution to the grade is indicated separately by the percentages. If a point system were used, we might have said that each quiz is worth 5 points, each homework 2 points, each paper 15 points, the final 30 points, and participation 10 points, and the point value of an item would be the indicated percentage of its points. (Some of these items would have very odd numbers of points! That first quiz got 95% of 5 points, which is 4.75 points.)

To find anaverage for a category, simply add up the percentages and divide by the number of scores there are. For example, the quiz average would be: 95%+80%+75%+90% 340% --------------- = ---- = 85% 4 4 The category averages are: Quizzes = 85% Homework = 93% Papers = 87% Final = 88% Partic. = 100%

There is something important hidden here. We are finding the average of each category, and then taking a weighted average of those category averages! That’s appropriate, because each average can stand in for all the individual scores in that category in a sum.

Now, remembering that each category is worth a certain amount in calculating the average for the quarter,multiply the average for each categoryby the fraction of the quarter average that the category represents: Quizzes: 85% * .2 = 17% Homework: 93% * .1 = 9.3% Papers: 87% * .3 = 26.1% Final: 88% * .3 = 26.4% Partic.: 100% * .1 = 10% Now, just add up these new percents to get the student's weighted average for the entire quarter: 17% + 9.3% + 26.1% + 26.4% + 10% = 88.8%

I suspect that Doctor Jeff used .2, .1, etc. instead of the 20%, 10%, etc. from the grading system description because it sounds odd to take a percent (20%) of a percentage (the 85% average). It may help to think of the 85% as 85 points, so we are taking 20% of the 85 points for the quizzes, etc.

### Working backward: what grade do I need?

Related to this question of calculating averages is how to determine what a student needs to score on, say, a test, to get a certain grade. Let's pretend that our student has not yet taken her final. The question is, assuming the student has the rest of the grades listed above, what does she need to get on the final to raise her average to a 90%? We already saw that summing the products of each category's percent and its "category worth" will give us a student's average. We can use this same equation, but this time call the first quarter's average 90%, since that's what the student is aiming for. Then all we need do is solve for the grade on the final, which we'll call f for simplicity: average = 90% = 17% + 9.3% + 26.1% + (.3*f) + 10% 90% = 62.4% + (.3*f) 27.6% = .3*f 92% = f The student must, therefore, score at least 92% on the final to raise her average to the 90% level.

Another way to think of this would be that without the final, we have a total of \(17 + 9.3 + 26.1 + 10 = 62.4\) points, so we need to get \(90 – 62.4 = 27.6\) more points from the final. Since the points we get are 0.30 times the grade, the grade has to be \(27.6\div 0.30 = 92\).

There are about as many ways of calculating averages as there are teachers. At the beginning of the school year, make sure to find out how your teachers go about their calculations so you don't wind up with completely different grades when report cards are given out!

I’ll add one more thought: In the middle of the term, you can find your current average by adding everything up but dividing by the sum of the weights used so far. In the example above, before the final, this student’s average would beĀ $$\frac{85(0.20)+93(0.10)+87(.30)+100(0.10)}{0.20+0.10+0.30+0.10} = \frac{17 + 9.3 + 26.1 + 10}{0.70} = \frac{62.4}{0.70} = 89.1$$ It makes sense that he just needs a little over 90 to bring the average up to 90.

## A complete example

Next, we have a more detailed example from 2000, similar to the percent system we just saw:

Calculating a Weighted Average Grade The following grades make up 100%. 45% = 3 tests (85, 78, 92) 25% = final (94) 15% = project (91.9) 15% = 3 case questions each worth 25 points (25, 25, 23.5) What is the final grade to this point?

Doctor TWE answered, walking through the details:

Hi Sam - thanks for writing to Dr. Math. Here's how I'd figure out the final grade. First, I'd convertany non-percentile grades to percentages. In your case, that would be the case questions. To convert them to percentages, divide the points earned by the maximum number of points, then multiply by 100%. The third case question would be: 23.5 / 25 * 100% = 94% I'll let you do the other two (they're pretty easy).

Now we have these grades:

- Tests: \(85, 78, 92\)
- Final: \(94\)
- Project: \(91.9\)
- Case questions: \(100\), \(100\), \(94\)

Next, I'dfind the averagesfor all the items that have more than one grade. In your case, this would be the tests and the case questions. To find the average, add up the grades and divide by the number of grades. For the tests, the average would be: (85 + 78 + 92) / 3 = 85

Now we have these averages:

- Test average: \(\displaystyle\frac{85+78+92}{3} = \frac{255}{3} = 85\)
- Final: \(94\)
- Project: \(91.9\)
- Case question average: \(\displaystyle\frac{100+100+94}{3} = \frac{294}{3} = 98\)

Now you need tofind the "weighted average"of each of these items. Convert their percentage weight (the 45% for tests, for example) to a decimal by dividing by 100 (or simply moving the decimal point two places to the left) and removing the percent sign. Then multiply the grade for that item by this decimal. For example, for the tests you get: 45% = 0.45 so 85 * 0.45 = 38.25 Do this for each item and add the results. That's the final grade. (If a 90 is the cutoff for an A, I think congratulations are in order.)

Finishing up, we weight each average and add them:

- Tests: \(0.45\times 85 = 38.25\)
- Final: \(0.25\times 94 = 23.5\)
- Project: \(0.15\times 91.9 = 13.875\)
- Case questions: \(0.15\times 98 = 14.7\)
- Total: \(38.25+23.5+13.875+14.7 = 90.325\)

## Weighted averages in general

We’ll close with a question (from 2001) about weighted averages more generally, whose answer still emphasizes grades:

Examples of Weighted Averages Can you tell me what is meant by "weighted average," and give me some examples of when and how weighted averages are used? When should mean average be used, versus weighted average?

Doctor TWE answered again, providing three examples.

### Baseball: slugging average

Hi Robert - thanks for writing to Dr. Math. A weighted average is one in which different data in the data set are given different "weights." Here are a few examples:Slugging average in baseball:A batter's slugging average (also called slugging percentage) is computed by: SLG = (1*SI + 2*DO + 3*TR + 4*HR) / AB where: SLG = slugging percentage SI = number of singles DO = number of doubles TR = number of triples HR = number of home runs AB = total number of at-bats Here, each single has a "weight" of 1, each double has a "weight" of 2, etc. The average counts home runs four times as important as singles, and so on. An at-bat without a hit has a "weight" of zero! Slugging average is sometimes referred to as slugging percentage. The term is a misnomer, for it is actually a weighted average, not a percentage. As such, it's possible for a batter's slugging average to exceed 1.000 or 100%.

This is the **average number of bases per at-bat**. The weights in this case are entirely natural!

### Course grades

Course grades:Many teachers will use a "weighted average" when calculating a student's grade in a course. For example, a teacher might say the test average is 60% of the grade, quiz average is 30% of the grade, and a project is 10% of the grade. Suppose Mary got 90 and 78 on the tests; 100, 100 and 85 on the quizzes; and an 81 on the project. Her course grade would be: Test average = (90 + 78)/ 2 = 84 Quiz average = (100 + 100 + 85)/3 = 95 Course grade = .60*84 + .30*95 + .10*81 = 87 Here, the tests carry a "weight" of .60 (or .30 each), the quizzes carry a "weight" of .30 (or .10 each), and the project carries a weight of .10. Note that the test average and the quiz average are not weighted averages, but the course grade is.

This is just like what we’ve seen. Each item is weighted arbitrarily, according to whatever the teacher considers its relative importance.

### GPA (Grade-point average)

Grade point average (GPA):Most colleges assign "weights" to the individual course grades in the form of credits. A grade in a 4-credit course affects your GPA more by 33% than a grade in a 3-credit course. For example, suppose Joe took the following courses: COURSE CR GR Calculus 4 C Discr. Math 3 A English Lit. 3 A Chemistry 4 D Comp. Sci. 3 B Most colleges use the scale: A = 4, B = 3, C = 2, D = 1, F = 0. To compute Joe's GPA, we multiply each course grade (converted to the number equivalent) by the course credits, then divide the sum by the total number of credits: COURSE CR GR Calculus 4 C 4*2 = 8 Discr. Math 3 A 3*4 = 12 English Lit. 3 A 3*4 = 12 Chemistry 4 D 4*1 = 4 Comp. Sci. 3 B 3*3 = 9 ---- ---- 17 45 GPA = 45 / 17 = 2.65 If the grades had been unweighted, the GPA would have been: (2 + 4 + 4 + 1 + 3) / 5 = 2.80 Why is Joe's GPA lower? Because he did less well in the "more important" courses, i.e. those worth more credits.

The weights this time are not arbitrary, but are based on the number of hours you (nominally) put in each week. In effect, this averages the grade for each class hour, as if you got a C for each of your 4 hours in calculus, and an A for each of your 3 hours in Discrete Math, and so on: $$\frac{(2+2+2+2)+(4+4+4)+(4+4+4)+(1+1+1+1)+(3+3+3)}{17} = 2.65$$

Next week, we’ll look at other ways in which weighted means can arise.

Pingback: Weighted Averages: Averaging Averages or Rates – The Math Doctors