Understanding Averages: From Student Grades to Speed Calculations

Quiz Results Analysis

In a class of five students, here are the results of a quiz on 10 marks:

• $s_1 = 8$
• $s_2 = 6$
• $s_3 = 9$
• $s_4 = 4$
• $s_5 = 7$

What’s the average performance in this class?

You do the total number of marks (calculation) divided by the total number of students:

$$\text{Average} = \frac{8 + 6 + 9 + 4 + 7}{5} = \frac{34}{5} = 6.8$$

Which gives 6.8 marks.

What Did You Actually Compute?

You know how to compute the average, but do you know what you computed?

What happens when you do average times frequency (here frequency = 1 for each student):

$$6.8 \times 5 = 34$$

You got the total marks of students! You had a range of performances - some aced the test and some failed. From the average student mark, you recreated the total perfectly.

Visualization: Individual vs Average Students

Student Type	Performance Distribution
Real Class	$s_1=8$, $s_2=6$, $s_3=9$, $s_4=4$, $s_5=7$
Average Class	$6.8$, $6.8$, $6.8$, $6.8$, $6.8$

The average homogenizes data that were different. It encompasses the performance of everyone in the class. To create that perfect average student, you have to encompass everyone’s performance.

So your distance from that average is very important to guide your learning. That distance tells you almost how you’re doing against your whole class.

Student Performance Analysis

Student	Score	Distance from Average	Performance Level
$s_1$	8	$8 - 6.8 = +1.2$	Above average
$s_2$	6	$6 - 6.8 = -0.8$	Below average
$s_3$	9	$9 - 6.8 = +2.2$	Well above average
$s_4$	4	$4 - 6.8 = -2.8$	Well below average
$s_5$	7	$7 - 6.8 = +0.2$	Slightly above average

Speed Example: The Weighted Average Problem

Example of a race where you move $x = 100$ meters in time $t_1 = 10$ seconds, and then you move to $x + 150 = 250$ meters total in $t_2 = 5$ more additional seconds.

Given Data:

• First interval: Distance = 100m, Time = 10s
• Second interval: Distance = 150m, Time = 5s
• Total distance = 250m, Total time = 15s

Speed Calculations:

Speed in first interval: $$v_1 = \frac{100}{10} = 10 \text{ m/s}$$

Speed in second interval: $$v_2 = \frac{150}{5} = 30 \text{ m/s}$$

Total average speed: $$v_{avg} = \frac{250}{15} = 16.67 \text{ m/s}$$

The False Average Problem

What happens if we take the simple average of the two speeds?

$$\text{False Average} = \frac{v_1 + v_2}{2} = \frac{10 + 30}{2} = 20 \text{ m/s}$$

But this is wrong! The correct average is 16.67 m/s, not 20 m/s.

Why This Happens: Unit Contribution Analysis

Scenario	Unit Contribution	Average Method
Student Grades	Each student = 1 unit	Simple average works
Speed Intervals	Each interval ≠ equal time	Weighted average needed

Student Frequency Table

Student	Score	Frequency	Contribution
$s_1$	8	1	Equal unit
$s_2$	6	1	Equal unit
$s_3$	9	1	Equal unit
$s_4$	4	1	Equal unit
$s_5$	7	1	Equal unit

Speed Interval Table

Interval	Speed (m/s)	Time (s)	Time Percentage	Weighted Contribution
1	10	10	$\frac{10}{15} = 66.67%$	$10 \times 0.667 = 6.67$
2	30	5	$\frac{5}{15} = 33.33%$	$30 \times 0.333 = 10.00$

The Weighted Average Solution

For speed, we need to weight each speed by its time contribution:

$$v_{weighted} = v_1 \times \frac{t_1}{t_1 + t_2} + v_2 \times \frac{t_2}{t_1 + t_2}$$

$$v_{weighted} = 10 \times \frac{10}{15} + 30 \times \frac{5}{15} = 6.67 + 10.00 = 16.67 \text{ m/s}$$

This matches our total average speed calculation perfectly!

Key Takeaway

In student grades, each student contributed the same unit (one grade per student). Average of class assumed that each student mark came from one student - a unit contribution.

But for speed, the intervals don’t contribute equally. One was 10 seconds, the other was 5 seconds. They don’t have equal weight in the total journey.

The solution: Take the weighted average by multiplying each speed by its percentage contribution to get the true average speed.