How to find the weighted mean (with GPA example)

The weighted mean formula

A weighted mean lets some values count more than others.

Read $\bar{x}_w$ as "weighted x-bar". Each value $x_i$ carries a weight $w_i$ that says how much say it gets. Multiply each value by its weight, add up those products, then divide by the total weight.

The ordinary mean is the special case where every weight is the same. Set every $w_i = 1$ and the numerator becomes $\sum x_i$ while the denominator becomes $n$, which is exactly $\bar{x} = \frac{\sum x_i}{n}$. So a weighted mean is not a different kind of average, it is the general version, and the plain mean is what you get when the values genuinely deserve equal say.

Weights can be counts (30 students), credits (4 credit hours), percentages (0.50 of the grade), or probabilities (0.35 chance). The formula does not care. If the weights already add to 1, the denominator is 1 and you can skip the division.

When the plain mean is the wrong answer

Three situations show up constantly.

• Groups of different sizes. You know each group's mean and each group's size, and you want the mean of everyone combined. Averaging the group means treats a class of 10 as equal in weight to a class of 30, and it is not.
• Items of different importance. A syllabus that says tests are 50 percent of the grade and homework is 15 percent is handing you weights directly.
• A value repeated many times. A frequency table where the value 2 appears 7 times is the same problem: the counts are the weights.

The common thread is the phrase "a mean of means". Whenever you are tempted to average numbers that are themselves already averages, stop and ask how many observations sit behind each one. If the answer is not "the same number for every group," the plain average is wrong. The situation you can rely on is all the groups being exactly the same size: with two sections of 20 each, the weighted mean and the plain average of the two means agree.

Why averaging two group means gives the wrong answer

Here is the failure with real numbers. Section A has 30 students with a mean test score of 82. Section B has 10 students with a mean of 92. What is the mean score for all 40 students?

The tempting move is $\frac{82 + 92}{2} = 87$. That is wrong, and you can see why by rebuilding the totals. Section A's 30 students scored $30 \times 82 = 2460$ points in total. Section B's 10 students scored $10 \times 92 = 920$ points. All 40 students together scored $2460 + 920 = 3380$ points, so the combined mean is

The correct answer is 84.5, not 87. The plain average is off by 2.5 points, and it is off in a predictable direction: it inflates the result by giving the small, high-scoring section the same say as a section three times its size.

Notice that the weighted mean here is just the total of everything divided by the count of everything. That is the sanity check to fall back on. If you can reconstruct the grand total and the grand count, you never need the formula at all. And notice where 84.5 lands: closer to 82 than to 92, because most of the students are in Section A. A weighted mean with non-negative weights always falls between the smallest and largest value being averaged, pulled toward whichever carries more weight.

GPA and course grades: weights you are handed

A grade point average is a weighted mean with credit hours as the weights. Suppose a student earns 4.0 in a 4-credit calculus course, 3.0 in a 3-credit statistics course, 3.7 in a 3-credit history course, and 2.0 in a 1-credit PE course.

Multiply each grade point by its credits: $4 \times 4.0 = 16$, $3 \times 3.0 = 9$, $3 \times 3.7 = 11.1$, and $1 \times 2.0 = 2$. The products add to 38.1 and the credits add to 11, so the GPA is $38.1 / 11 = 3.4636$, which reports as 3.46.

Averaging the four grades without weights gives $\frac{4.0 + 3.0 + 3.7 + 2.0}{4} = \frac{12.7}{4} = 3.175$. That is nearly three tenths of a point lower, because it lets the 1-credit PE grade of 2.0 count as much as the 4-credit calculus grade of 4.0.

Course grade weights work the same way but usually add to 1 already. If tests are 50 percent, quizzes 20 percent, homework 15 percent, and the final 15 percent, and a student earns 88, 92, 100, and 79, then the grade is $0.50(88) + 0.20(92) + 0.15(100) + 0.15(79) = 44 + 18.4 + 15 + 11.85 = 89.25$. The weights sum to 1, so there is nothing to divide by. The unweighted average of the same four scores is $359/4 = 89.75$, half a point higher, and it is worth seeing where that half point comes from, because the obvious answer is wrong. The perfect homework score is not the culprit. Dropping to equal weights promotes the 100 and the 79 by the same 10 percentage points each, and since they sit on opposite sides of 89.25 their effects very nearly cancel: together they account for only 0.05 of the 0.50 gap. Most of the gap is the test score of 88, which sits below the weighted average and has its say cut from 50 percent of the grade to 25 percent.

Frequency tables and expected value are weighted means

Two things you already do are weighted means wearing different clothes.

A frequency table. Suppose 20 households report the number of cars they own: four households with 1, seven with 2, five with 3, three with 4, and one with 5. The mean is not $\frac{1+2+3+4+5}{5} = 3$. The counts are weights, so

The answer is 2.5 cars, not 3. Reading the same idea from proportions instead of counts is relative frequency, and the running-total version of it is cumulative relative frequency. The distinction between a count and a proportion is covered in frequency vs relative frequency.

Expected value. The mean of a discrete random variable is $\mu_X = \sum x_i P(x_i)$, which is a weighted mean whose weights are probabilities. Those weights already add to 1, so again there is no division. That is why the AP course can call the expected value the long-run mean without any new machinery. See how to find expected value and is expected value the same as the mean.

Mistakes to avoid

• Averaging averages. The single most common error. Averaging two group means directly can only be relied on when the groups are the same size.
• Dividing by the number of values instead of the total weight. In the GPA example the denominator is 11 credits, not 4 courses.
• Forgetting to check that percentage weights add to 100. If they do not, you either missed a category or the denominator is not 1.
• Weighting by the wrong quantity. For a combined mean, the weights are the sample sizes, not the standard deviations and not the number of groups.
• Assuming a weighted mean also fixes the spread. It does not. You cannot combine two group standard deviations by averaging them either, and rebuilding a combined standard deviation needs the original data or the sum of squares. A weighted mean tells you nothing about shape, so pair it with a look at the distribution: see how to describe the shape of a distribution.
• Reporting a weighted mean without units or context. "The combined mean was 84.5 points on the 100-point test" earns credit; "84.5" does not.