Averages and Spread
The Skill
The Three Averages
Mean — Add all values, divide by how many there are. $$\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}$$
Median — The middle value when data is in order. For an even number of values, find the mean of the two middle values.
Mode — The most common value. There can be no mode, one mode, or multiple modes.
Range
The range measures spread: $$\text{Range} = \text{Highest} - \text{Lowest}$$
From Frequency Tables
For grouped or ungrouped frequency tables:
- Mean: Use midpoints for grouped data. $\text{Mean} = \frac{\sum fx}{\sum f}$
- Median: Find the position using $\frac{n+1}{2}$, then locate it in the cumulative frequency
- Modal class: The group with the highest frequency (grouped data)
The Traps
Common misconceptions and how to avoid them.
Adding highest and lowest instead of subtracting "The Range Reversal"
The Mistake in Action
Find the range of: 3, 7, 12, 4, 9
Student writes: "Range = 12 + 3 = 15"
Why It Happens
Students remember the range involves the highest and lowest values but use addition instead of subtraction. Sometimes they've confused it with another formula.
The Fix
Range is about the spread — how far apart the data stretches.
$$\text{Range} = \text{Highest} - \text{Lowest}$$
Range = 12 − 3 = 9
Think of it this way: If you stretch a rubber band from the lowest to highest value, how long is it? That's a "difference" — subtraction!
Spot the Mistake
Highest value = 12
Lowest value = 3
Range = 12 + 3 = 15
Click on the line that contains the error.
Averaging the values column instead of using frequencies "The Frequency Forgetter"
The Mistake in Action
Score: 1, 2, 3, 4, 5 Frequency: 3, 7, 5, 4, 1
Student writes: "Mean = (1+2+3+4+5) ÷ 5 = 3"
Why It Happens
Students see a column of values and instinctively average them, completely ignoring the frequency column. They treat it like a simple list.
The Fix
The frequency tells you how many times each value appears. You must:
- Multiply each value by its frequency (the fx column)
- Add up all the fx values
- Divide by the TOTAL frequency (not how many rows)
Correct working:
| Score | f | fx |
|---|---|---|
| 1 | 3 | 3 |
| 2 | 7 | 14 |
| 3 | 5 | 15 |
| 4 | 4 | 16 |
| 5 | 1 | 5 |
| Total | 20 | 53 |
Mean = $\frac{53}{20} = 2.65$
Spot the Mistake
Score × Frequency:
Mean = (1 + 2 + 3 + 4 + 5) ÷ 5
Click on the line that contains the error.
Finding median without ordering data first "The Middle Muddle"
The Mistake in Action
Find the median of: 7, 3, 9, 2, 5
Student writes: "There are 5 values, so median is the 3rd one = 9"
Why It Happens
Students see "middle value" and just pick the one in the middle position of the list as given. They skip the crucial first step of putting values in order.
The Fix
Always rewrite the data in order first — smallest to largest.
7, 3, 9, 2, 5 becomes 2, 3, 5, 7, 9
Now the middle (3rd) value is 5, not 9.
Memory trick: "ORDER before MIDDLE" — you can't find the middle of a queue if everyone's standing randomly!
Spot the Mistake
Data: 7, 3, 9, 2, 5
5 values, so median is the 3rd value
Median = 9
Click on the line that contains the error.
Giving the position as the answer instead of the value "The Position Problem"
The Mistake in Action
Find the median of 8 values where $\frac{n+1}{2} = 4.5$
Student writes: "The median is 4.5"
Why It Happens
Students correctly calculate the median position using $\frac{n+1}{2}$ but then give this position number as their answer, forgetting to actually look up what value is in that position.
The Fix
The formula $\frac{n+1}{2}$ tells you the position of the median, not the median itself!
For 8 ordered values, position = $\frac{8+1}{2} = 4.5$
This means: find the 4th and 5th values, then find their mean.
If the ordered data is: 2, 3, 5, 7, 8, 10, 12, 15
Median = $\frac{7+8}{2} = 7.5$
Spot the Mistake
8 values, so median position = (8+1) ÷ 2 = 4.5
Median = 4.5
Click on the line that contains the error.
Using class boundaries instead of midpoints "The Boundary Blunder"
The Mistake in Action
Time (t minutes): 0 < t ≤ 10, 10 < t ≤ 20, 20 < t ≤ 30 Frequency: 4, 8, 3
Student uses 10, 20, 30 as the values for calculating mean.
Why It Happens
Students use the upper boundary of each class (the numbers they can see clearly in the inequality) rather than finding the midpoint. Or they use the lower boundary.
The Fix
For grouped data, use the midpoint of each class:
- 0 < t ≤ 10 → midpoint is $\frac{0+10}{2} = 5$
- 10 < t ≤ 20 → midpoint is $\frac{10+20}{2} = 15$
- 20 < t ≤ 30 → midpoint is $\frac{20+30}{2} = 25$
Correct mean: $\frac{(5×4) + (15×8) + (25×3)}{4+8+3} = \frac{20+120+75}{15} = \frac{215}{15} = 14.3$
Spot the Mistake
Finding mean for grouped data:
Values: 10, 20, 30 (upper boundaries)
Mean = (10×4 + 20×8 + 30×3) ÷ 15 = 18.7
Click on the line that contains the error.
The Deep Dive
Apply your knowledge with these exam-style problems.
Level 1: Fully Worked
Complete solutions with commentary on each step.
Question
The table shows the number of pets owned by students in a class.
| Number of pets | Frequency |
|---|---|
| 0 | 5 |
| 1 | 8 |
| 2 | 6 |
| 3 | 4 |
| 4 | 2 |
Calculate the mean number of pets.
Solution
Step 1: Add an fx column (value × frequency)
| Pets | f | fx |
|---|---|---|
| 0 | 5 | 0 |
| 1 | 8 | 8 |
| 2 | 6 | 12 |
| 3 | 4 | 12 |
| 4 | 2 | 8 |
Step 2: Find the totals $$\sum f = 5 + 8 + 6 + 4 + 2 = 25$$ $$\sum fx = 0 + 8 + 12 + 12 + 8 = 40$$
Step 3: Calculate the mean $$\text{Mean} = \frac{\sum fx}{\sum f} = \frac{40}{25} = 1.6$$
Answer: 1.6 pets
Question
The table shows the time taken by students to complete a puzzle.
| Time, t (seconds) | Frequency |
|---|---|
| 30 < t ≤ 40 | 4 |
| 40 < t ≤ 50 | 9 |
| 50 < t ≤ 60 | 7 |
| 60 < t ≤ 70 | 5 |
Estimate the mean time.
Solution
Step 1: Find the midpoint of each class
| Time | Midpoint | f | fx |
|---|---|---|---|
| 30 < t ≤ 40 | 35 | 4 | 140 |
| 40 < t ≤ 50 | 45 | 9 | 405 |
| 50 < t ≤ 60 | 55 | 7 | 385 |
| 60 < t ≤ 70 | 65 | 5 | 325 |
Midpoint of 30 < t ≤ 40 is $\frac{30 + 40}{2} = 35$, and so on.
Step 2: Calculate totals $$\sum f = 4 + 9 + 7 + 5 = 25$$ $$\sum fx = 140 + 405 + 385 + 325 = 1255$$
Step 3: Calculate the mean $$\text{Mean} = \frac{1255}{25} = 50.2 \text{ seconds}$$
Answer: 50.2 seconds
Note: This is an estimate because we don't know exact values within each group.
Question
| Score | Frequency |
|---|---|
| 1 | 3 |
| 2 | 5 |
| 3 | 8 |
| 4 | 6 |
| 5 | 3 |
Find the median score.
Solution
Step 1: Find the total frequency $$n = 3 + 5 + 8 + 6 + 3 = 25$$
Step 2: Find the median position $$\text{Position} = \frac{n + 1}{2} = \frac{25 + 1}{2} = 13$$
We need the 13th value.
Step 3: Use cumulative frequency to locate it
| Score | f | Cumulative f |
|---|---|---|
| 1 | 3 | 3 |
| 2 | 5 | 8 |
| 3 | 8 | 16 |
| 4 | 6 | 22 |
| 5 | 3 | 25 |
The 13th value falls in the "Score = 3" row (positions 9-16).
Answer: Median = 3
Question
Five numbers have a mean of 12. Four of the numbers are 8, 15, 11, and 14.
Find the fifth number.
Solution
Step 1: Use the mean formula $$\text{Mean} = \frac{\text{Sum of values}}{\text{Number of values}}$$
Let the missing number be $x$.
Step 2: Set up the equation $$12 = \frac{8 + 15 + 11 + 14 + x}{5}$$
Step 3: Simplify the numerator $$12 = \frac{48 + x}{5}$$
Step 4: Multiply both sides by 5 $$60 = 48 + x$$
Step 5: Solve for x $$x = 60 - 48 = 12$$
Answer: The fifth number is 12
Check: (8 + 15 + 11 + 14 + 12) ÷ 5 = 60 ÷ 5 = 12 ✓
Question
Class A test scores: Mean = 65, Range = 28 Class B test scores: Mean = 58, Range = 12
Compare the two distributions.
Solution
Compare the averages: Class A has a higher mean (65) than Class B (58).
This tells us that, on average, Class A performed better on the test.
Compare the spread: Class A has a larger range (28) than Class B (12).
This tells us that Class A's scores were more spread out — there was greater variation in how students performed. Class B's scores were more consistent.
Full answer: "On average, Class A scored higher than Class B (mean of 65 compared to 58). However, Class A had more variation in their scores (range of 28 compared to 12), suggesting that Class B performed more consistently."
Level 2: Scaffolded
Fill in the key steps.
Question
| Goals | Frequency |
|---|---|
| 0 | 6 |
| 1 | 10 |
| 2 | 7 |
| 3 | 2 |
Calculate the mean number of goals.
Level 3: Solo
Try it yourself!
Question
Here are the ages of 9 people at a party:
23, 19, 25, 31, 19, 28, 22, 19, 34
Find: (a) the mode (b) the median (c) the mean (d) the range
Show Solution
(a) Mode The most common value is 19 (appears 3 times).
(b) Median First, order the data: 19, 19, 19, 22, 23, 25, 28, 31, 34
Position = $\frac{9+1}{2} = 5$
The 5th value is 23.
(c) Mean $$\text{Mean} = \frac{23+19+25+31+19+28+22+19+34}{9} = \frac{220}{9} = 24.4̄$$
Mean = 24.4 (1 d.p.)
(d) Range Range = 34 − 19 = 15
Examiner's View
Mark allocation patterns:
- Finding mean from a list: 2 marks (1 for sum, 1 for division)
- Mean from frequency table: 3 marks (1 for fx column, 1 for totals, 1 for division)
- Median from frequency table: 2 marks (1 for position, 1 for correct value)
- Comparing distributions: 2 marks (1 for average comparison, 1 for spread comparison)
Common errors examiners see:
- Forgetting to order data before finding median
- Using class boundaries instead of midpoints for grouped mean
- Confusing median position $\frac{n+1}{2}$ with the actual median value
- Not using frequencies — just averaging the values column
What gains full marks:
- Show the fx column clearly in tables
- State your totals before dividing
- For comparisons, always mention BOTH average AND spread
AQA Notes
AQA often asks you to compare two distributions. Always give one comparison about average (e.g., "Team A has a higher mean score") AND one about spread (e.g., "Team A has a larger range, showing more variation").