Table of Contents |
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 set .Add terms and divide by the number of terms, 5 | |
Evaluate addition in numerator | |
Divide | |
Our Solution |
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:
Often times, we see weighted average in class grades.
EXAMPLE
Suppose you are taking a science course this semester. Tests and projects are worth 50% of your grade, daily assignments are worth 30% of your grade, and participation is worth 20% of your grade. You scored the following on these assessments:Assessment | Score | Weight |
---|---|---|
Tests and Projects | 83% | 50% |
Daily Assignments | 94% | 30% |
Participation | 80% | 20% |
Multiply 83% by 50%, 94% by 30%, 80% by 20% and divide by the sum of the weights | |
Evaluate multiplication in numerator | |
Simplify numerator and denominator | |
Divide 0.857 by 1 | |
Our Solution |
Next, we will use the concept of weighted average to set up and solve a mixture problem. We'll use this adapted formula as we work through these types of problems:
In this formula,
Consider the following example:
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?Plug in the concentrations and quantities for the two known solutions: | |
Multiply the quantity by its weight in numerator. | |
Simplify the numerator and denominator | |
Divide | |
Our Solution |
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 20% solution, and a 50% 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?Plug in the concentrations for the known solutions: | |
Substitute | |
Multiply both sides by 120 | |
Equivalent weighted average equation |
Substitute | |
Distribute 0.50 into | |
Combine like terms | |
Subtract 60 from both sides | |
Divide both sides by -0.30 | |
Our solution x, the quantity of 20% solution |