Weighted Mean Formula - Definition, Steps, Examples

#Formula
TL;DR
The weighted mean formula divides the total of weight times value by the total of the weights, xˉw=∑wixi∑wi\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}. This article gives five worked examples.
BT
Bhanzu TeamLast updated on November 4, 20266 min read

Key Points

  • The weighted mean formula is xˉw=∑wixi∑wi\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}, dividing the weighted total by the total weight.

  • NIST's Dataplot documentation defines the weighted sum as ∑wixi\sum w_i x_i, pairing each weight with its observation, and notes the weights are not normalised.

  • When every weight is equal, the weighted mean reduces exactly to the ordinary arithmetic mean.

  • The US Census Bureau builds survey statistics this way, summing a variable multiplied by its weight and dividing by the summed weights.

  • No weight may be negative, and the weights do not need to add to 1 because the denominator divides by their total.

What Is the Weighted Mean Formula?

The weighted mean formula averages values that carry unequal importance:

$$\bar{x}w = \frac{\sum{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i}$$

Each value xix_i is multiplied by its weight wiw_i, the products are added, and the total is divided by the sum of the weights. NIST's Dataplot documentation defines the numerator as the weighted sum, with the weights variable paired to the response variable observation by observation.

Why the Denominator Is the Sum of the Weights

Dividing by nn would treat the weights as though they were part of the data rather than a scaling of it. The denominator has to be ∑wi\sum w_i so that the units and the scale cancel correctly.

A quick check makes this concrete. If all the weights equal 1, then ∑wi=n\sum w_i = n and the formula becomes ∑xin\frac{\sum x_i}{n}, the ordinary mean, exactly as it should.

Because the division handles the total, the weights need no normalising. Weights of 2, 3 and 5 give the same answer as 0.2, 0.3 and 0.5, which is why NIST can state that its weighted sum leaves the weights un-normalised.

Variable Key

Symbol

Meaning

Unit or type

Notes

xˉw\bar{x}_w

weighted mean

same units as the data

the quantity being computed

xix_i

the iith value

same units as the data

one observation

wiw_i

weight on the iith value

non-negative

need not sum to 1

∑wixi\sum w_i x_i

weighted total

weighted units

the numerator

∑wi\sum w_i

total weight

same as ww

the denominator

Examples of the Weighted Mean Formula

Example 1

A course weights homework 10 per cent, quizzes 20 per cent, the midterm 30 per cent and the final 40 per cent. A student scores 90, 75, 80 and 65. Find the weighted mean.

xˉw=0.10(90)+0.20(75)+0.30(80)+0.40(65)0.10+0.20+0.30+0.40=741=74\bar{x}_w = \frac{0.10(90) + 0.20(75) + 0.30(80) + 0.40(65)}{0.10+0.20+0.30+0.40} = \frac{74}{1} = 74

Final answer: 74

Example 2

Find the same student's average by taking the plain mean of the four marks.

The four scores average to 90+75+80+654=77.5\frac{90+75+80+65}{4} = 77.5, which looks like the answer.

It is wrong because it treats the homework as mattering as much as the final, when the final carries four times the weight. The student did best on the lightest component and worst on the heaviest, so ignoring the weights flatters the result.

Weighting it properly gives 74, as in Example 1.

Final answer: 74, not 77.5. The unweighted mean overstates the grade by 3.5 marks.

Example 3

Three shipments cost 4, 6 and 9 per unit, for 100, 300 and 600 units. Find the weighted mean cost per unit.

Here the quantities are the weights:

xˉw=100(4)+300(6)+600(9)100+300+600=400+1800+54001000=7.6\bar{x}_w = \frac{100(4) + 300(6) + 600(9)}{100+300+600} = \frac{400 + 1800 + 5400}{1000} = 7.6

Final answer: 7.60 per unit

Example 4

Find the weighted mean of 10, 20 and 30 with weights 2, 3 and 5. Then repeat with weights 0.2, 0.3 and 0.5.

2(10)+3(20)+5(30)2+3+5=23010=23\frac{2(10)+3(20)+5(30)}{2+3+5} = \frac{230}{10} = 23 0.2(10)+0.3(20)+0.5(30)0.2+0.3+0.5=231=23\frac{0.2(10)+0.3(20)+0.5(30)}{0.2+0.3+0.5} = \frac{23}{1} = 23

Final answer: 23 both times, because only the relative sizes of the weights matter.

Example 5

A survey records hours worked, with person weights. Two respondents report 40 hours with weight 1,500 and 20 hours with weight 500. Find the weighted mean hours.

xˉw=1500(40)+500(20)1500+500=60000+100002000=35\bar{x}_w = \frac{1500(40) + 500(20)}{1500+500} = \frac{60000+10000}{2000} = 35

Final answer: 35 hours

This mirrors how the Census Bureau produces survey statistics, summing a variable multiplied by its weight and dividing by the summed weights.

Don't Confuse the Weighted Mean With Its Relatives

Mean

Formula

Use it for

Weighted

∑wixi∑wi\frac{\sum w_i x_i}{\sum w_i}

values of unequal importance

Arithmetic

∑xin\frac{\sum x_i}{n}

equally important values

Geometric

x1⋯xnn\sqrt[n]{x_1 \cdots x_n}

quantities that compound

Grouped-data mean

∑fixi∑fi\frac{\sum f_i x_i}{\sum f_i}

the weighted mean with frequencies as weights

The grouped-data mean is the same formula under another name. NIST notes that weighted means are often used for frequency data, which is exactly that case.

Why Students Get the Weighted Mean Wrong (and the Fix)

Mistake 1: Dividing by the number of values instead of the total weight

Where it slips in: The habit of dividing by nn is strong, and with four components it is tempting to divide by 4.

Don't do this: ∑wixin\frac{\sum w_i x_i}{n}

The correct way: ∑wixi∑wi\frac{\sum w_i x_i}{\sum w_i}. When the weights happen to sum to 1 the two look similar, which hides the error until the weights are given as whole numbers.

Students who write the two totals side by side, the weighted total and the total weight, before dividing anything rarely slip here, because the denominator is then a number they have actually computed.

Mistake 2: Treating a weight as a value

Where it slips in: Both columns are numbers, and in a quantity-weighted problem the weights look like data.

Don't do this: Average the weights in with the values.

The correct way: Decide which column answers "how much of this" and which answers "how important". In Example 3 the unit counts are weights and the costs are values.

Mistake 3: Using a negative weight

Where it slips in: A modelling problem produces a negative coefficient and it is passed in as a weight.

Don't do this: Allow wi<0w_i < 0.

The correct way: Keep all weights non-negative, as NIST requires. A negative weight can make the denominator small or zero and send the result outside the range of the data, which is the clearest sign something is wrong.

The Short Version

  • The weighted mean formula is xˉw=∑wixi∑wi\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}.

  • The denominator is the total weight, never the count of values, which is the single most common error.

  • Weights need no normalising, because only their relative sizes affect the result.

  • Equal weights collapse the formula to the ordinary mean, which is a useful check on any implementation.

  • To work through averages with a teacher, explore a 7th grade math tutor or math classes online.

A Practical Next Step

Compute the weighted mean of 60, 70 and 95 with weights 1, 1 and 3. Then recompute with weights 0.2, 0.2 and 0.6 and confirm the answers match.

Want a live Bhanzu trainer to walk through more averages problems? Book a free demo class.

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Frequently Asked Questions

What is the weighted mean formula?
xˉw=∑wixi∑wi\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}. Multiply each value by its weight, add the products, and divide by the sum of the weights.
Do the weights in the weighted mean formula have to add up to 1?
No. The denominator divides by whatever the weights total, so only their relative sizes matter. Weights of 2, 3 and 5 give the same answer as 0.2, 0.3 and 0.5.
What happens if all the weights are equal?
The formula reduces to the ordinary arithmetic mean, since ∑wi=nw\sum w_i = nw and the common weight cancels.
Can the weighted mean fall outside the range of the data?
Not with non-negative weights. It always lies between the smallest and largest values, so an answer outside that range means a weight was negative or a total was miscomputed.
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Bhanzu Team
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Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.
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