Mathematics · AQA 8300 §S3
Averages and Spread
- Calculate mean, median, mode and range from raw data
- Find averages from frequency tables using $\bar{x} = \dfrac{\Sigma fx}{\Sigma f}$
- Estimate the mean from grouped frequency tables using class midpoints
- Identify the modal class and the class interval containing the median
- Compare two distributions using appropriate averages and measures of spread
🔑 Core Concepts
1. The Mean
The mean is the arithmetic average. It works by distributing the total sum equally across all data points. Every value contributes to the mean, which makes it the most mathematically powerful average — but also the most sensitive to extreme values (outliers).
DEFINITION — Mean
The mean of $n$ values is the sum of all values divided by the number of values:
$$\bar{x} = \dfrac{\Sigma x}{n}$$
Mean from Raw Data
$$\bar{x} = \dfrac{\Sigma x}{n} = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$$
$\bar{x}$ = mean (x-bar)
$\Sigma x$ = sum of all data values
$n$ = number of data values
Mean from a Frequency Table
$$\bar{x} = \dfrac{\Sigma fx}{\Sigma f}$$
$f$ = frequency (how many times each value occurs)
$x$ = each data value
$\Sigma f$ = total number of data points (NOT number of rows)
Estimated Mean from Grouped Frequency Data
$$\bar{x} \approx \dfrac{\Sigma fm}{\Sigma f}$$
$m$ = midpoint of each class interval: $m = \dfrac{a+b}{2}$ for class $a \leq x \lt b$
$f$ = frequency of that class
Result is always an estimate — we assume values are evenly spread within each class
COMMON MISTAKE — Wrong Denominator
When calculating the mean from a frequency table, divide by $\Sigma f$ (total frequency), NOT by the number of rows. If a table has 5 rows and $\Sigma f = 30$, you divide by 30, not 5.
IMPORTANT — Missing Value Problems (Grade 9)
If you know the mean and all values except one, rearrange the mean formula:$$\text{Missing value} = (\bar{x} \times n) - \Sigma(\text{known values})$$ Step 1: Find the total sum $= \bar{x} \times n$. Step 2: Subtract the sum of known values.
EXAM TIP — Grouped Mean is an Estimate
Always write “estimated mean” when working from grouped data. Write the symbol $\approx$ rather than $=$. This signals to the examiner that you understand the limitation of the calculation.
2. The Median
The median is the middle value of an ordered data set. Because it only depends on the position of data (not the actual values at the extremes), it is robust to outliers — an extreme value at either end has little or no effect on the median.
DEFINITION — Median
The median is the middle value when data is arranged in ascending order. For $n$ values, the median is at position $\dfrac{n+1}{2}$.
Median Position
$$\text{Median position} = \dfrac{n+1}{2}$$
$n$ = total number of data values
If $n$ is odd: position is a whole number — that value is the median
If $n$ is even: position ends in .5 — average the two surrounding values
Order all data values smallest to largest
→
Calculate position: $\dfrac{n+1}{2}$
→
Whole number? Read off that value. Ends in .5? Average the two middle values
COMMON MISTAKE — Forgetting to Order Data
You must order the data before finding the median. Finding the middle item of an unordered list gives the wrong answer. This is worth an automatic method mark — skip it and you lose it.
EXAM TIP — Median from a Frequency Table
Do not write out all the data. Instead, add a cumulative frequency column. The median is in the row where the cumulative frequency first reaches (or passes) the median position.
3. The Mode
The mode is the most frequently occurring value. It is the only average that can be used with categorical (non-numerical) data. For grouped data, you cannot give a single mode value; instead you give the modal class — the class interval with the highest frequency.
DEFINITION — Mode and Modal Class
Mode: The value that appears most often. A data set may be bimodal (two modes) or have no mode.Modal class: For grouped frequency data, the class interval with the highest frequency. Always state the full interval (e.g. $20 \leq t \lt 30$).
EXAM TIP — State the Full Class Interval
For grouped data, write the entire class interval as your modal class, e.g. “$20 \leq t \lt 30$ minutes”. Writing only “30” or the midpoint “25” will not earn the mark.
4. The Range
The range is a measure of spread, not an average. It quantifies how variable the data is. A small range means data is consistent and clustered together; a large range indicates high variability.
DEFINITION — Range
$$\text{Range} = x_{\max} - x_{\min}$$
The range is the difference between the largest and smallest values in the data set.
Range
$$\text{Range} = x_{\max} - x_{\min}$$
$x_{\max}$ = largest value in the data set
$x_{\min}$ = smallest value in the data set
Larger range = more spread out / variable; smaller range = more consistent
COMMON MISTAKE — Range is Not an Average
Never describe the range as “the average spread”. Say “the range shows how spread out the data is” or “a larger range indicates greater variability”. The range is a single measure of spread, not a typical value.
5. Comparing Distributions
When comparing two data sets, you must always comment on both an average (mean or median) and a measure of spread (range). Comparing only the average earns half marks at best.
IMPORTANT — The Comparison Framework
Structure every comparison answer in three parts:1. Compare averages: “The mean of A ($x$) is higher/lower than the mean of B ($y$)…”
2. Compare spread: “The range of A ($p$) is larger/smaller than the range of B ($q$)…”
3. Interpret in context: “…which means the values in A are more variable/consistent.”
Always quote actual values from the question. Vague comparisons do not earn the interpretation mark.
MEMORY TRICK — ACS
Average comparison → Context (what does it mean?) → Spread comparison → Context again. Four sentences, four marks.
6. Choosing the Appropriate Average
Each average has strengths and weaknesses. Choosing the right one — and justifying why — is a key Grade 9 skill.
| Average | Best used when… | Advantage | Disadvantage |
|---|---|---|---|
| Mean | Data is numerical, symmetrical, no outliers | Uses every value; most powerful statistically | Strongly distorted by outliers |
| Median | Data is skewed or contains outliers (e.g. salaries, house prices) | Resistant to outliers; unaffected by extremes | Ignores most data values in its calculation |
| Mode | Data is categorical, or you need the most popular item (e.g. shoe sizes) | Only average usable with categories; easy to find | May not exist; can be non-representative |
EXAM TIP — Always Justify Your Choice
Never just say “I would use the median.” Say: “I would use the median because the data contains an outlier (£95,000) which would distort the mean upwards, making it unrepresentative of the typical salary.” The justification earns the second mark.
WORKED EXAMPLE — Outlier Effect
Five employees earn: £18,000 £22,000 £19,000 £24,000 £95,000Mean = £(18+22+19+24+95) ÷ 5 = £178,000 ÷ 5 = £35,600
Ordered: 18, 19, 22, 24, 95 → Median = £22,000 (3rd value)
The mean (£35,600) is much higher than four out of five salaries, pulled upwards by the £95,000 outlier. The median (£22,000) is far more representative of the “typical” salary.
🗺️ Visual Notes
Averages & Spread
Mean
- $\bar{x} = \Sigma x \div n$ (raw data)
- $\bar{x} = \Sigma fx \div \Sigma f$ (freq table)
- Estimated: use class midpoints $m$
- Sensitive to outliers
Median
- Middle value when data is ordered
- Position $= (n+1) \div 2$
- Resistant to outliers
- Use cumulative freq. for tables
Mode
- Most frequently occurring value
- Modal class for grouped data
- Only average for categorical data
- Can be bimodal or not exist
Range (Spread)
- Range $= x_{\max} - x_{\min}$
- Measures spread, not average
- Larger range = more variable
- Sensitive to outliers
Grouped Data
- Modal class: highest frequency row
- Median class: use cumulative $f$
- Midpoint: $m = (a + b) \div 2$
- Estimated mean $\approx \Sigma fm \div \Sigma f$
Comparing Two Datasets
- Always compare average AND spread
- Quote actual values from data
- Interpret each comparison in context
- Four comparisons = four marks
Averages at a Glance
| Property | Mean | Median | Mode |
|---|---|---|---|
| Definition | Sum ÷ count | Middle value when ordered | Most frequent value |
| Formula | $\Sigma x \div n$ | Position $\dfrac{n+1}{2}$ | By inspection (highest frequency) |
| Affected by outliers? | Yes — strongly | Slightly or not at all | Not at all |
| Works with categorical data? | No | No | Yes |
| Always uniquely defined? | Yes | Yes | No — may not exist |
| Best for… | Symmetrical data, no outliers | Skewed data or outliers present | Most common item or categories |
Process: Median from a Frequency Table
Find total frequency $n = \Sigma f$
→
Calculate median position: $\dfrac{n+1}{2}$
→
Add a cumulative frequency column
→
Find the row where cumulative freq first reaches the median position
→
That row's $x$ value (or class interval) is the median
Process: Estimated Mean from Grouped Frequency Data
List the class intervals from the table
→
Find midpoint $m = \dfrac{a+b}{2}$ for each class
→
Calculate $f \times m$ for each row
→
Total the column: find $\Sigma fm$ and $\Sigma f$
→
Estimated mean $\approx \dfrac{\Sigma fm}{\Sigma f}$
✎️ Worked Examples
Grade 4–5
Question: The ages (years) of 8 members of a youth club are:
14 17 13 17 11 18 17 15
Find the (a) mean, (b) median, (c) mode, and (d) range of their ages.
14 17 13 17 11 18 17 15
Find the (a) mean, (b) median, (c) mode, and (d) range of their ages.
1
(a) Calculate the Mean
Add all values: $14 + 17 + 13 + 17 + 11 + 18 + 17 + 15 = 122$Divide by $n = 8$: $$\bar{x} = \dfrac{122}{8} = 15.25 \text{ years}$$
2
(b) Find the Median
Order the data first: 11, 13, 14, 15, 17, 17, 17, 18Median position $= \dfrac{8+1}{2} = 4.5$
Average the 4th and 5th values: $\dfrac{15 + 17}{2} = 16$ years
3
(c) Find the Mode
17 appears 3 times — more than any other value.Mode $= 17$ years
4
(d) Calculate the Range
Range $= \text{max} - \text{min} = 18 - 11 = 7$ years
(a) Mean = 15.25 years | (b) Median = 16 years | (c) Mode = 17 years | (d) Range = 7 years
Grade 6–7
Question: The frequency table shows the number of goals scored in 25 football matches.
Find (a) the mean, (b) the median, and (c) the mode number of goals scored.
| Goals scored ($x$) | Frequency ($f$) |
|---|---|
| 0 | 3 |
| 1 | 5 |
| 2 | 8 |
| 3 | 6 |
| 4 | 3 |
Find (a) the mean, (b) the median, and (c) the mode number of goals scored.
1
Extend the table: add $fx$ and cumulative frequency columns
| $x$ | $f$ | $fx$ | Cumulative $f$ |
|---|---|---|---|
| 0 | 3 | 0 | 3 |
| 1 | 5 | 5 | 8 |
| 2 | 8 | 16 | 16 |
| 3 | 6 | 18 | 22 |
| 4 | 3 | 12 | 25 |
| Total | 25 | 51 | — |
2
(a) Calculate the Mean
$$\bar{x} = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{51}{25} = 2.04 \text{ goals}$$
3
(b) Find the Median
Median position $= \dfrac{25+1}{2} = 13\text{th value}$Using the cumulative frequency column:
Cumulative up to $x = 1$: 8 Cumulative up to $x = 2$: 16
Since $8 \lt 13 \leq 16$, the 13th value falls in the $x = 2$ row.
Median = 2 goals
4
(c) Identify the Mode
The highest frequency is 8, corresponding to $x = 2$.Mode = 2 goals
(a) Mean = 2.04 goals | (b) Median = 2 goals | (c) Mode = 2 goals
Grade 9
Question: The grouped frequency table shows the times taken by 40 students to complete a puzzle. One frequency value, $p$, is unknown.
(a) Show that $p = 12$. [1]
(b) Estimate the mean time taken. [3]
(c) Write down the modal class. [1]
(d) A second group of 40 students completed the same puzzle. Their estimated mean was 13.4 minutes and their range was 22 minutes. Compare the times for both groups. [3]
| Time $t$ (minutes) | Frequency $f$ |
|---|---|
| $0 \leq t \lt 5$ | 3 |
| $5 \leq t \lt 10$ | $p$ |
| $10 \leq t \lt 15$ | 14 |
| $15 \leq t \lt 20$ | 9 |
| $20 \leq t \lt 25$ | 2 |
(a) Show that $p = 12$. [1]
(b) Estimate the mean time taken. [3]
(c) Write down the modal class. [1]
(d) A second group of 40 students completed the same puzzle. Their estimated mean was 13.4 minutes and their range was 22 minutes. Compare the times for both groups. [3]
1
(a) Find the value of $p$
The total frequency must equal 40:
$$3 + p + 14 + 9 + 2 = 40$$
$$28 + p = 40$$
$$p = 12 \checkmark$$
2
(b) Extend the table with midpoints and $fm$ column
| Class | $f$ | Midpoint $m$ | $fm$ |
|---|---|---|---|
| $0 \leq t \lt 5$ | 3 | 2.5 | 7.5 |
| $5 \leq t \lt 10$ | 12 | 7.5 | 90 |
| $10 \leq t \lt 15$ | 14 | 12.5 | 175 |
| $15 \leq t \lt 20$ | 9 | 17.5 | 157.5 |
| $20 \leq t \lt 25$ | 2 | 22.5 | 45 |
| Total | 40 | — | 475 |
3
(c) Identify the Modal Class
The class with the highest frequency is $f = 14$, corresponding to $10 \leq t \lt 15$.Modal class: $10 \leq t \lt 15$ minutes
4
(d) Compare the Two Distributions
Average: Group 1’s estimated mean (11.9 min) is lower than Group 2’s mean (13.4 min), so Group 1 students were typically faster at completing the puzzle.Spread: Group 1’s data spans $0$ to $25$ minutes (class range = 25 min), whilst Group 2’s range is 22 minutes. Group 2’s times are slightly less spread out, indicating marginally more consistent performance.
Conclusion: Overall, Group 1 performed better (lower mean time), though Group 2 was slightly more consistent in their times.
(a) $p = 12$ | (b) Estimated mean $\approx 11.9$ min | (c) $10 \leq t \lt 15$ min | (d) Group 1 faster on average; Group 2 slightly more consistent
❓ Exam Questions
Question 1
1 mark
The number of siblings of 7 students are: 2 0 3 2 1 2 4
Write down the mode.
Mode = 2 [B1]
Mark scheme: B1 for 2 (appears 3 times, more than any other value). No method required — answer only.
Mark scheme: B1 for 2 (appears 3 times, more than any other value). No method required — answer only.
Question 2
3 marks
The heights (cm) of 7 plants are: 23 31 19 27 42 31 16
Find (a) the mean height and (b) the range of heights. [3 marks]
(a) Mean:
Sum $= 23 + 31 + 19 + 27 + 42 + 31 + 16 = 189$ [M1 for summing all 7 values]
Mean $= 189 \div 7 = 27$ cm [A1]
(b) Range:
Range $= 42 - 16 = 26$ cm [B1]
Mark scheme: M1A1 for mean (M1 for dividing their sum by 7). B1 for range = 26.
Sum $= 23 + 31 + 19 + 27 + 42 + 31 + 16 = 189$ [M1 for summing all 7 values]
Mean $= 189 \div 7 = 27$ cm [A1]
(b) Range:
Range $= 42 - 16 = 26$ cm [B1]
Mark scheme: M1A1 for mean (M1 for dividing their sum by 7). B1 for range = 26.
Question 3
4 marks
The table shows the number of books read by 20 students over the summer.
| Books read ($x$) | Frequency ($f$) |
|---|---|
| 0 | 2 |
| 1 | 5 |
| 2 | 8 |
| 3 | 4 |
| 4 | 1 |
(a) Mean:
$\Sigma fx = (0)(2) + (1)(5) + (2)(8) + (3)(4) + (4)(1) = 0 + 5 + 16 + 12 + 4 = 37$ [M1 for at least 4 correct $fx$ products]
Mean $= \dfrac{37}{20} = 1.85$ books [A1]
(b) Median:
Median position $= \dfrac{20+1}{2} = 10.5$th value [M1 for finding position using $(n+1)/2$]
Cumulative frequencies: 2, 7, 15, 19, 20
Both the 10th and 11th values lie in the $x = 2$ row (cumulative goes from 7 to 15 in that row).
Median $= 2$ books [A1]
Mark scheme: M1A1 for mean. M1 for using cumulative frequency with correct position. A1 for median = 2.
$\Sigma fx = (0)(2) + (1)(5) + (2)(8) + (3)(4) + (4)(1) = 0 + 5 + 16 + 12 + 4 = 37$ [M1 for at least 4 correct $fx$ products]
Mean $= \dfrac{37}{20} = 1.85$ books [A1]
(b) Median:
Median position $= \dfrac{20+1}{2} = 10.5$th value [M1 for finding position using $(n+1)/2$]
Cumulative frequencies: 2, 7, 15, 19, 20
Both the 10th and 11th values lie in the $x = 2$ row (cumulative goes from 7 to 15 in that row).
Median $= 2$ books [A1]
Mark scheme: M1A1 for mean. M1 for using cumulative frequency with correct position. A1 for median = 2.
Question 4
4 marks
The grouped frequency table shows the heights of 35 students.
| Height $h$ (cm) | Frequency |
|---|---|
| $140 \leq h \lt 150$ | 3 |
| $150 \leq h \lt 160$ | 7 |
| $160 \leq h \lt 170$ | 12 |
| $170 \leq h \lt 180$ | 8 |
| $180 \leq h \lt 190$ | 5 |
(b) Calculate an estimate for the mean height. [3 marks]
(a) Modal class: $160 \leq h \lt 170$ cm [B1 — must give full class interval]
(b) Estimated Mean:
Midpoints: 145, 155, 165, 175, 185 [M1 for correct use of midpoints]
$\Sigma fm = (3)(145) + (7)(155) + (12)(165) + (8)(175) + (5)(185)$
$= 435 + 1085 + 1980 + 1400 + 925 = 5825$ [M1 for correct $\Sigma fm$]
Estimated mean $= \dfrac{5825}{35} \approx 166.4$ cm [A1]
Mark scheme: B1 for modal class. M1 for using midpoints. M1 for $\Sigma fm = 5825$. A1 for 166.4 cm (accept 166 cm or $\frac{5825}{35}$).
(b) Estimated Mean:
Midpoints: 145, 155, 165, 175, 185 [M1 for correct use of midpoints]
$\Sigma fm = (3)(145) + (7)(155) + (12)(165) + (8)(175) + (5)(185)$
$= 435 + 1085 + 1980 + 1400 + 925 = 5825$ [M1 for correct $\Sigma fm$]
Estimated mean $= \dfrac{5825}{35} \approx 166.4$ cm [A1]
Mark scheme: B1 for modal class. M1 for using midpoints. M1 for $\Sigma fm = 5825$. A1 for 166.4 cm (accept 166 cm or $\frac{5825}{35}$).
Question 5
6 marks
Eight values have a mean of 7.5. Seven of the values are: 4 9 6 11 8 5 7
(a) Find the missing eighth value. [2 marks]
The salaries of employees at two companies are summarised below:
Company A: Mean salary = £34,200 Range = £38,000
Company B: Mean salary = £36,500 Range = £12,000
(b) Compare the salaries at Company A and Company B. Write two comparisons, each with a contextual interpretation. [4 marks]
(a) Missing value:
Total sum $= \bar{x} \times n = 7.5 \times 8 = 60$ [M1]
Sum of seven known values $= 4 + 9 + 6 + 11 + 8 + 5 + 7 = 50$
Missing value $= 60 - 50 = \mathbf{10}$ [A1]
(b) Comparison:
Average: Company B has a higher mean salary (£36,500 > £34,200), so employees at Company B earn more on average — approximately £2,300 per year more. [B1 for comparison + B1 for contextual interpretation]
Spread: Company A has a much larger range (£38,000 compared to £12,000 at Company B), meaning salaries at Company A are far more spread out and inconsistent. Company B offers more consistent pay across employees. [B1 for comparison + B1 for contextual interpretation]
Mark scheme: M1 for $7.5 \times 8$. A1 for 10. B1B1 for average comparison with contextual comment. B1B1 for spread comparison with contextual comment.
Total sum $= \bar{x} \times n = 7.5 \times 8 = 60$ [M1]
Sum of seven known values $= 4 + 9 + 6 + 11 + 8 + 5 + 7 = 50$
Missing value $= 60 - 50 = \mathbf{10}$ [A1]
(b) Comparison:
Average: Company B has a higher mean salary (£36,500 > £34,200), so employees at Company B earn more on average — approximately £2,300 per year more. [B1 for comparison + B1 for contextual interpretation]
Spread: Company A has a much larger range (£38,000 compared to £12,000 at Company B), meaning salaries at Company A are far more spread out and inconsistent. Company B offers more consistent pay across employees. [B1 for comparison + B1 for contextual interpretation]
Mark scheme: M1 for $7.5 \times 8$. A1 for 10. B1B1 for average comparison with contextual comment. B1B1 for spread comparison with contextual comment.
⭐ Grade 9 Model Answers
The following is a fully annotated Grade 9 answer for Question 5(b). Each sentence is justified with the specific mark it earns and why.
GRADE 9 MODEL ANSWER — Q5(b) Comparison [4 marks]
Context: Company A mean = £34,200, range = £38,000. Company B mean = £36,500, range = £12,000.
Sentence 1: “The mean salary at Company B (£36,500) is higher than at Company A (£34,200).”
→ B1: Explicit numerical comparison of the two means, with values stated, and direction given (higher/lower).
Sentence 2: “On average, employees at Company B earn approximately £2,300 more per year than those at Company A.”
→ B1: Contextual interpretation — explains what the difference in means actually means for the employees in the real situation.
Sentence 3: “However, Company A has a much larger range (£38,000) compared to Company B (£12,000).”
→ B1: Explicit numerical comparison of the two ranges, with values stated, and direction given (larger/smaller).
Sentence 4: “This means salaries at Company A are far more spread out and unequal — some employees earn significantly more or less than the average. Company B offers more consistent pay.”
→ B1: Contextual interpretation — explains what the larger range means in terms of pay inequality and consistency.
Total: 4/4 marks
GRADE 4 RESPONSE — What Loses Marks
“Company B pays more. Company A has a bigger range.”
Marks earned: 1/4
This response has no numerical values, no direction words, and no contextual interpretation. Only one mark is available (B1 for identifying Company B has a higher mean — but even this is borderline without stating actual values).
GRADE 9 STRATEGY — Comparison Questions
Every comparison earns two marks: one for the statistical comparison (with values) and one for the contextual interpretation. Ask yourself after every sentence: “Have I said which one is bigger/smaller AND explained what that means in this real situation?”
GRADE 9 STRATEGY — Missing Value Problems
For the missing value problem in Q5(a), the key insight is that $\bar{x} = \frac{\Sigma x}{n}$ can be rearranged to $\Sigma x = \bar{x} \times n$. Calculate the total first, then subtract the sum of known values. This approach works even when the missing value appears in an algebraic expression — just form an equation and solve.
📋 Revision Sheet
Key Definitions
| Term | Definition |
|---|---|
| Mean | Sum of all values divided by the count of values |
| Median | Middle value when data is arranged in order |
| Mode | Most frequently occurring value in a data set |
| Range | Maximum value minus minimum value; measures spread |
| Modal class | Class interval with the highest frequency in grouped data |
| Midpoint | $(a + b) \div 2$ for a class $a \leq x \lt b$ |
| Outlier | A value much larger or smaller than the rest of the data |
| Estimated mean | Approximation of the mean using midpoints of grouped classes |
Essential Formulae
$$\bar{x} = \dfrac{\Sigma x}{n} \quad \text{(raw data)}$$
$$\bar{x} = \dfrac{\Sigma fx}{\Sigma f} \quad \text{(frequency table)}$$
$$\bar{x} \approx \dfrac{\Sigma fm}{\Sigma f} \quad \text{(grouped data, } m = \text{midpoint)}$$
$$\text{Median position} = \dfrac{n+1}{2}$$
$$\text{Range} = x_{\max} - x_{\min}$$
$$\text{Missing value} = (\bar{x} \times n) - \Sigma(\text{known values})$$
$$\text{Class midpoint: } m = \dfrac{a + b}{2}$$
Memory Hooks
- Mean = Add then Divide — “add them all, divide by all”
- Median = Middle — M-e-d is the middle of “me-di-an”
- Mode = Most — both “Mo” words
- Range = Room — how much “room” from smallest to largest
- $\Sigma fm \div \Sigma f$ — “Sigma FM over Sigma F”
- Midpoint — “add the two boundaries and halve”
- ACS rule — Average, then Contextual meaning, then Spread
- Missing value — “mean times n, subtract the rest”
Exam Tips
- Always order the data before finding the median — this step earns a method mark
- Divide by $\Sigma f$ in frequency tables, not the number of rows
- Always write “estimated mean” for grouped data answers
- Modal class requires the full class interval with inequality signs
- Comparison questions need both average AND spread — two comparisons, four marks
- Always quote numerical values when comparing distributions
- For missing value: total $= \bar{x} \times n$, then subtract
- Outlier effect: mean is pulled towards the outlier; median is largely unaffected
- If asked which average to use, give a reason involving outliers, skew, or data type
🔄 Flashcards
Click any card to reveal the answer. Work through all 15 before your exam — try to answer aloud before flipping!
✗ Common Mistakes
MISTAKE 1 — Not Ordering Data Before Finding the Median
What students do: Find the item in the middle position of the original, unordered list.Why marks are lost: The median is only the middle value after ordering. An unordered “middle” is meaningless and will almost always be wrong.
How to avoid it: As soon as you see “find the median”, write the numbers in order first. This takes 10 seconds and prevents a guaranteed error.
MISTAKE 2 — Dividing by the Number of Rows Instead of $\Sigma f$
What students do: In a frequency table with, say, 5 rows, students divide $\Sigma fx$ by 5 instead of by the total frequency.Why marks are lost: The number of rows is the number of distinct values, not the sample size. The mean formula requires the total count of data points.
How to avoid it: Always find $\Sigma f$ first, write it clearly, then divide. Double-check: $\Sigma f$ should equal the total number of items surveyed.
MISTAKE 3 — Using Class Boundaries Instead of Midpoints
What students do: Multiply frequency by the lower boundary, upper boundary, or the class width rather than the midpoint when building the $fm$ column.Why marks are lost: The midpoint is the best single representative value for each class. Using any other value yields an incorrect estimate of the mean.
How to avoid it: For any class $a \leq x \lt b$, the midpoint is $m = (a + b) \div 2$. Write a midpoint column explicitly before calculating $fm$.
MISTAKE 4 — Comparing Only the Average and Ignoring Spread
What students do: Write one sentence comparing means and nothing about range, thinking this is a complete answer.Why marks are lost: Comparison questions are typically 4 marks — 2 for average (comparison + context) and 2 for spread (comparison + context). Missing spread halves your mark.
How to avoid it: Use the ACS rule: Average comparison, Contextual meaning, Spread comparison, contextual meaning. Four points, four marks.
MISTAKE 5 — Writing the Midpoint as the Modal Class
What students do: Write a single number (e.g. 25 or 30) as the modal class instead of the full class interval.Why marks are lost: The modal class is an interval, not a single value. The mark scheme expects something like “$20 \leq t \lt 30$” with inequalities.
How to avoid it: Copy the full interval directly from the table, including the inequality symbols and units.
MISTAKE 6 — Comparing Without Context or Numerical Values
What students do: Write “Group A’s range is larger so it is more spread out” without quoting actual values, or without explaining what “more spread out” means in the real situation.Why marks are lost: Each comparison has two marks: one for the numerical comparison (must quote values) and one for contextual interpretation (must relate to the scenario).
How to avoid it: After every comparison, ask: (1) Did I quote the actual numbers? (2) Did I explain what this means for the people or things described?
✅ Final Checklist
Click each item when you are confident with it. Your progress is saved automatically.
- I can calculate the mean from a list of raw data values using $\bar{x} = \Sigma x \div n$
- I can find the median by ordering data and applying the position formula $(n+1) \div 2$
- I can correctly handle an even number of data values when finding the median (average two middle values)
- I can identify the mode from a list of values (including recognising bimodal data or no mode)
- I can calculate the range as maximum minus minimum
- I can set up an $fx$ column and calculate the mean from a frequency table using $\Sigma fx \div \Sigma f$
- I can use cumulative frequencies to locate the median value from a frequency table
- I can identify the modal class from a grouped frequency table and state it as a full interval
- I can calculate class midpoints and build the $fm$ column to estimate the mean from grouped data
- I can use cumulative frequencies to identify which class interval contains the median
- I can find a missing value given the mean and all other values using $\text{missing} = (\bar{x} \times n) - \Sigma(\text{known})$
- I can compare two distributions by making at least one comment on average AND one on spread
- I can interpret statistical comparisons in context — not just stating numbers but explaining their meaning
- I can explain how an outlier affects the mean (distorts it) but not the median (largely unaffected)
- I can justify the most appropriate average to use for a given data set, with a full written reason