We find the average, or mean, of a set of values by adding up the values and dividing the sum by the number of values. Thus, the average, \(\overline{x}\text{,}\) of the numbers \(x_1, x_2, \ldots , x_n\) is given by
\begin{equation*}
\overline{x} = \frac{x_1 + x_2 + \cdot \cdot \cdot +x_n}{n}
\end{equation*}
In a weighted average, the numbers being averaged occur with different frequencies or are weighted differently in their contribution to the average value. For instance, suppose a biology class of 12 students takes a 10point quiz. Of the 12 students, 2 receive 10s, three receive 9s, 5 receive 8s, and 2 receive scores of 6. The average score earned on the quiz is then
\begin{equation*}
\overline{x} = \frac{\alert{2}(10) + \alert{3}(9) + \alert{5}(8) + \alert{2}(6)}{12}=8.25
\end{equation*}
The numbers in color are called the weights—in this example they represent the number of times each score was counted. Note that \(n\text{,}\) the total number of scores, is equal to the sum of the weights:
\begin{equation*}
12 = 2 + 3 + 5 + 2
\end{equation*}
Example A.27.
Kwan’s grade in his accounting class will be computed as follows: Tests count for \(50\%\) of the grade, homework counts for \(20\%\text{,}\) and the final exam counts for \(30\%\text{.}\) If Kwan has an average of \(84\) on tests and \(92\) on homework, what score does he need on the final exam to earn a grade of \(90\text{?}\)
Solution.
Let \(x\) represent the final exam score Kwan needs.
Kwan’s grade is the weighted average of his test, homework, and final exam scores.
\begin{equation*}
\frac{\alert{0.50}(84) + \alert{0.20}(92) + \alert{0.30}x}{1.00}=90
\end{equation*}
(The sum of the weights is 1.00, or 100% of Kwan’s grade.) Multiply both sides of the equation by \(1.00\) to get
\begin{equation*}
0.50(84) + 0.20(92) + 0.30x = 1.00(90)
\end{equation*}
Solve the equation. Simplify the left side first.
\begin{align*}
60.4 + 0.30x \amp = 90\amp\amp\blert{\text{Subtract 60.4 from both sides.}}\\
0.30x \amp = 29.6\amp\amp\blert{\text{Divide both sides by 0.30.}}\\
x \amp = 98.7
\end{align*}
Kwan needs a score of \(98.7\) on the final exam to earn a grade of \(90\text{.}\)
Weighted Average.
The sum of the weighted values equals the sum of the weights times the average value. In symbols,
\begin{equation*}
w_1x_1 + w_2x_2 + \cdots + w_nx_n = W\, \overline{x}
\end{equation*}
where \(W\) is the sum of the weights.
This form is particularly useful for solving problems involving mixtures.
Example A.28.
The vet advised Delbert to feed his dog Rollo with kibble that is no more than \(8\%\) fat. Rollo likes JuicyBits, which are \(15\%\) fat. LeanMeal is more expensive, but it is only \(5\%\) fat. How much LeanMeal should Delbert mix with \(50\) pounds of JuicyBits to make amixture that is \(8\%\) fat?
Solution.
Let \(p\) represent the number of pounds of LeanMeal needed.

In this problem, we want the weighted average of the fat contents in the two kibbles to be \(8\%\text{.}\) The weights are the number of pounds of each kibble we use. It is often useful to summarize the given information in a table.
\(\) 
\(\%\) fat 
Total pounds 
Pounds of fat 
Juicy Bits 
\(15\%\) 
\(50\) 
\(0.15(50)\) 
LeanMeal 
\(5\%\) 
\(p\) 
\(0.05p\) 
Mixture 
\(8\%\) 
\(50+p\) 
\(0.08(50+p)\) 
The amount of fat in the mixture must come from adding the amounts of fat in the two ingredients. This gives us an equation,
\begin{equation*}
0.15(50) + 0.05p = 0.08(50 + p)
\end{equation*}
This equation is an example of the formula for weighted averages.
Simplify each side of the equation, using the distributive law on the right side, then solve.
\begin{align*}
7.5 + 0.05p \amp = 4 + 0.08p \amp\amp\blert{\text{Subtract}~4+0.05p~\text{from both sides.}}\\
3.5 \amp = 0.03p \amp\amp \blert{\text{Divid both sides by}~ 0.03.}\\
p \amp = 116.\overline{6}
\end{align*}
Delbert should mix \(116\frac{2}{3}\) pounds of LeanMeal with \(50\) pounds of JuicyBits to make a mixture that is \(8\%\) fat.