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Weighted Averages

Author: Sophia

what's covered
In this lesson, you will use the statistical concept of weighted average to determine the concentration of a mixed solution and how to calculate a class grade based on a variety of assignments, projects, and tests. You will also see how weighted averages can enhance your results driven skills, along with your problem solving skills. Specifically, this lesson will cover:

Table of Contents

1. Average/Mean

Before we begin discussing weighted average, it may be helpful to review how to calculate the simple average or mean of a data set. In order to find the average, we find the sum of all data values, and then divide by the number of data values in our set.

EXAMPLE

Find the average of the following numbers: left curly bracket 4 comma space 4 comma space 8 comma space 12 comma space 14 right curly bracket.

mean equals fraction numerator 4 plus 4 plus 8 plus 12 plus 14 over denominator 5 end fraction equals 42 over 5 equals 8.4blank

The average, or mean, of this set of numbers is 8.4


2. Weighted Average

Weighted average is different from simple average in that certain data values carry more weight, or influence the average more heavily. With simple average, every data point is equally represented in the calculation for the mean, but in weighted averages, some data points are multiplied by a number in its calculation.

We use this formula for weighted average:

formula to know
Weighted Average
weighted space average equals fraction numerator sum space of space left parenthesis value times weight right parenthesis space of space each space item over denominator sum space of space weights end fraction

Results Driven: Skill in Action
Weighted averages can help you drive toward results with your academics. Each semester, your instructors track grades of all types related to homework, quizzes, projects, and exams. Since some grades are worth more than others, you can use weighted average to make sense of the numbers. Knowing about weighted average will help you predict your final grade and make adjustments throughout the semester.

Let's see exactly how weighted average pertains to class grades.

EXAMPLE

Suppose that tests and projects account for 50% of your grade, daily assignments represent 30%, and participation is worth 20% of your grade. To calculate your final grade for the class, we multiply your test/project score by its weight of 50%, your assignment score by its weight of 30%, and your participation grade by its weight of 20%, all before adding these figures together. This is illustrated below:

Test/Project Grade: 83% Assignment Grade: 94% Participation Grade: 80%
Weight: 50% Weight: 30% Weight: 20%

weighted space average equals fraction numerator sum space of space left parenthesis value times weight right parenthesis space of space each space item over denominator sum space of space weights end fraction Use the formula for weighted average.
weighted space average equals fraction numerator open parentheses 0.83 times 0.50 close parentheses plus open parentheses 0.94 times 0.30 close parentheses plus open parentheses 0.80 times 0.20 close parentheses over denominator 0.50 plus 0.30 plus 0.20 end fraction Calculate the weighted grade for test/projects, assignments, and participation.
weighted space average equals fraction numerator 0.415 plus 0.282 plus 0.16 over denominator 0.50 plus 0.30 plus 0.20 end fraction Multiply the weight by each value in the numerator.
weighted space average equals fraction numerator 0.857 over denominator 1 end fraction Evaluate numerator and denominator.
weighted space average equals 0.857 Divide 0.857 by 1.
weighted space average equals 85.7 percent sign Change to a percent.

The final grade for this class is 85.7%.

big idea
In general, data values in the denominator of the fraction are multiplied by their corresponding weight. The corresponding weights are then added together to form the denominator of the fraction.

In the example above, the weights summed to 100% because the weights of tests/projects, assignments, and participation need to reflect 100% of your grade in total.


3. Mixture Problems

Next, we will use the concept of weighted average to set up and solve a mixture problem from the field of chemistry. We'll use this adapted formula as we work through these types of problems:

formula to know
Weighted Average for Mixture Problems
fraction numerator open parentheses C subscript 1 close parentheses open parentheses Q subscript 1 close parentheses plus open parentheses C subscript 2 close parentheses open parentheses Q subscript 2 close parentheses over denominator Q subscript 1 plus Q subscript 2 end fraction equals C subscript 3

In this formula,

  • C subscript 1 is the concentration of Item 1
  • Q subscript 1 is the quantity of Item 1
  • C subscript 2 is the concentration of Item 2
  • Q subscript 2 is the quantity of Item 2
  • C subscript 3 is the concentration of the combined items
hint
If you notice, there is no Q subscript 3 defined. Q subscript 3 would be the quantity of al the combined items. We technically do see this in our formula in the denominator, Q subscript 1 plus Q subscript 2.

Problem Solving: Skill Reflect
Problem solving is about exploring new ideas. It's one of the cornerstones of chemistry and chemical engineering. Chemists use their analytical skills and math know-how to create mixtures and solutions for industry and consumers. They balance ingenuity with safety to design products that can improve our quality of life.

EXAMPLE

To prepare for a lab experiment, you mix two concentrations of HCl (hydrochloric acid): 30 mL of 15% HCl solution, and 20 mL of 40% HCl solution. What is the concentration of the mixed solution?

To solve this problem, we need to multiply the quantity of each solution by its concentration (or its weight to be included in our calculation). This will represent the numerator of our fraction for weighted average. As for the denominator, we divide by the total quantity of the solution (so we add the individual amounts to get a total amount of liquid). The solution is worked out below:

fraction numerator open parentheses C subscript 1 close parentheses open parentheses Q subscript 1 close parentheses plus open parentheses C subscript 2 close parentheses open parentheses Q subscript 2 close parentheses over denominator Q subscript 1 plus Q subscript 2 end fraction equals C subscript 3 Use the formula for weighted average for mixture problems.
fraction numerator open parentheses 0.15 times 30 space mL close parentheses plus open parentheses 0.40 times 20 space mL close parentheses over denominator 30 space mL plus 20 space mL end fraction equals C subscript 3 Plug in the concentrations and quantities for the two known solutions:
C subscript 1 equals 0.15
Q subscript 1 equals 30 space mL
C subscript 2 equals 0.40
Q subscript 2 equals 20 space mL
fraction numerator 4.5 space mL plus 8 space mL over denominator 50 space mL end fraction equals C subscript 3 Multiply each quantity by its weight in numerator. Add quantities in denominator.
fraction numerator 12.5 space mL over denominator 50 space mL end fraction equals C subscript 3 Add 4.5 to 8.
0.25 equals C subscript 3 Divide by 50 mL.
25 percent sign equals C subscript 3 Change to a percent

The concentration of the mixed solution will be 25%.

Some mixture problems will ask us to find specific amounts of each solution that must be mixed together in order to yield a certain amount of a specific concentration. Let's again use the HCl example.

EXAMPLE

We have two kinds of solutions already made: a 15% solution, and a 40% solution. To prepare for a lab experiment, we are going to need 120 mL of a 30% solution. How many mL of each solution must be mixed in order to create this mixture?

Let's define variables for this situation:
  • x: mL of 15% solution
  • y: mL of 40% solution
We have an immediate relationship between x and y, that x plus y equals 120. How does this affect our equation for weighted average?

fraction numerator open parentheses C subscript 1 close parentheses open parentheses Q subscript 1 close parentheses plus open parentheses C subscript 2 close parentheses open parentheses Q subscript 2 close parentheses over denominator Q subscript 1 plus Q subscript 2 end fraction equals C subscript 3 Use the formula for weighted average for mixture problems.
fraction numerator 0.15 x plus 0.40 y over denominator x plus y end fraction equals 0.30 Plug in the concentrations for the known solutions: C subscript 1 equals 0.15 comma space C subscript 2 equals 0.40 comma space C subscript 3 equals 0.30
fraction numerator 0.15 x plus 0.40 y over denominator 120 end fraction equals 0.30 Substitute x plus y equals 120
0.15 x plus 0.40 y equals 36 Multiply both sides by 120

We have simplified our equation, but it still has two variables. If this equation could be expressed using only one variable, we could solve for that variable, and then use substitution again to solve for the other variable. Let's return to the relationship x plus y equals 120. We can write this equivalently as y equals 120 minus x.

Now that we have an expression for y, we can write 120 minus x into the original equation instead of y. The result will be a single-variable equation, and we'll be able to solve for x.

0.15 x plus 0.40 y equals 36 The simplified weighted average equation.
0.15 x plus 0.40 left parenthesis 120 minus x right parenthesis equals 36 Substitute 120 minus x in for y.
0.15 x plus 48 minus 0.40 x equals 36 Distribute 0.40 into left parenthesis 120 minus x right parenthesis.
short dash 0.25 x plus 48 equals 36 Combine like terms.
short dash 0.25 x equals short dash 12 Subtract 48.
x equals 48 Divide by -0.25.

This tells us that we must use 48 mL of the 15% solution in our mixture. Since we know we need a total of 120 mL, we can deduce that we'll need 72 mL of the 40% solution. This combination will yield 120 mL of 30% solution.

summary
Recall that to find the average/mean of a data set, we find the sum of all data values, and then divide by the number of data values in our set. A weighted average gives you the average of a set of values that may carry different weights. The more weight that a value has, the more it is accounted for in the calculation. We can use the concept of weighted average to calculate mixture problems.

Best of luck in your learning!

Source: THIS TUTORIAL WAS AUTHORED BY SOPHIA LEARNING. PLEASE SEE OUR TERMS OF USE.

Formulas to Know
Weighted Average

weighted space average equals fraction numerator sum space of space left parenthesis value times weight right parenthesis space of space each space item over denominator sum space of space weights end fraction

Weighted Average for Mixture Problems

fraction numerator open parentheses C subscript 1 close parentheses open parentheses Q subscript 1 close parentheses plus open parentheses C subscript 2 close parentheses open parentheses Q subscript 2 close parentheses over denominator Q subscript 1 plus Q subscript 2 end fraction equals C subscript 3