Mathematics · AQA 8300 §S4
Data Representation
- Draw and interpret histograms with unequal class widths using frequency density
- Construct and interpret cumulative frequency diagrams to find key statistics
- Read quartiles and IQR from cumulative frequency curves accurately
- Construct and compare box plots using the five-figure summary
- Identify outliers using the IQR rule and interpret skewness from distributions
🔑 Core Concepts
Histograms with Unequal Class Widths
A histogram looks like a bar chart, but there is a crucial difference: the area of each bar represents frequency, not the height. This matters when class widths are unequal — a wider bar would otherwise exaggerate the count. To make the comparison fair, we plot frequency density on the vertical axis.
Frequency Density
$$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$
Frequency = count of data values in the class
Class Width = upper boundary − lower boundary
Frequency Density = height of the bar on the histogram
Recovering Frequency from a Histogram
$$\text{Frequency} = \text{Frequency Density} \times \text{Class Width}$$
This means: Frequency = Area of bar
DEFINITION — Frequency Density
Frequency density is the frequency per unit of class width. It ensures that in a histogram, equal areas represent equal frequencies, regardless of the class width.COMMON MISTAKE
Never plot raw frequency on the vertical axis of a histogram when class widths are unequal. This distorts the visual comparison — wider classes would appear to have more data than they actually do.EXAM TIP — Reading Histograms
When asked to find a frequency from a histogram: read off the frequency density, note the class width, then multiply. Always check: does the question give you part of the histogram to complete? If so, calculate FD = Frequency ÷ Class Width for the missing bar.Steps to Draw a Histogram
Calculate class boundaries and class widths
→
Compute FD = Frequency ÷ Class Width for each row
→
Draw axes: x = data values (continuous scale), y = Frequency Density
→
Draw bars — NO gaps between bars (continuous data)
→
Label axes with units; include class boundaries (not midpoints) on x-axis
MEMORY TRICK — Histogram vs Bar Chart
"Histograms are for History (continuous data from the past); Bar Charts are for Bars (discrete categories)." In histograms: no gaps, y-axis = frequency density. In bar charts: gaps allowed, y-axis = frequency.Cumulative Frequency Diagrams
A cumulative frequency diagram (CF curve) shows the running total of frequencies up to each value. It is an S-shaped curve (ogive) that allows us to estimate medians, quartiles, and percentiles without knowing all individual values.
Reading the CF Curve — Key Statistics
$$\text{Median} = \text{value at } \frac{n}{2}\text{th position}$$
$$\text{Lower Quartile (LQ)} = \text{value at } \frac{n}{4}\text{th position}$$
$$\text{Upper Quartile (UQ)} = \text{value at } \frac{3n}{4}\text{th position}$$
$$\text{IQR} = \text{UQ} - \text{LQ}$$
$n$ = total frequency (total number of data values)
Read across from the y-axis value, then down to the x-axis to get the data value
DEFINITION — Cumulative Frequency
Cumulative frequency at a value $x$ is the total number of data values less than or equal to $x$. It is found by adding up all frequencies from the lowest class up to and including the class containing $x$.DEFINITION — Interquartile Range (IQR)
The IQR is the difference between the upper quartile and lower quartile: $\text{IQR} = \text{UQ} - \text{LQ}$. It measures the spread of the middle 50% of data, making it resistant to outliers.EXAM TIP — Plotting CF Diagrams
Always plot cumulative frequency against the upper class boundary of each class (not the midpoint, not the lower boundary). The curve must start at the lower boundary of the first class with CF = 0.COMMON MISTAKE
Students use $\frac{n}{2}$ but forget to read the diagram accurately. Use a ruler: draw a horizontal line from the y-axis value across to the curve, then a vertical line down to the x-axis. Any error in the reading direction loses accuracy marks.Box Plots and the Five-Figure Summary
A box plot (box-and-whisker diagram) is a compact visual summary of a dataset's distribution. It shows five key values simultaneously, allowing quick comparison of distributions.
Five-Figure Summary
$$\text{Five-Figure Summary: Minimum, LQ, Median, UQ, Maximum}$$
Minimum = smallest data value (or lower fence if outliers excluded)
LQ = lower quartile (25th percentile)
Median = middle value (50th percentile)
UQ = upper quartile (75th percentile)
Maximum = largest data value (or upper fence if outliers excluded)
DEFINITION — Box Plot Structure
The box spans from LQ to UQ (the IQR). A vertical line inside the box marks the median. Whiskers extend from the box to the minimum and maximum (or to the fences if outliers are present). Outliers are plotted as individual crosses (×) or dots beyond the whiskers.EXAM TIP — Comparing Box Plots
When comparing two distributions using box plots, always comment on: (1) location — compare medians, (2) spread — compare IQR and overall range, (3) symmetry/skewness — compare the position of the median within the box. Always relate your comparison back to the context of the question.Outliers
An outlier is a data value that sits far from the main body of data. The standard GCSE rule uses the IQR to define the boundaries beyond which a value is considered anomalous.
Outlier Boundaries
$$\text{Upper Fence} = \text{UQ} + 1.5 \times \text{IQR}$$
$$\text{Lower Fence} = \text{LQ} - 1.5 \times \text{IQR}$$
Any value above the Upper Fence is an outlier
Any value below the Lower Fence is an outlier
$\text{IQR} = \text{UQ} - \text{LQ}$
DEFINITION — Outlier
A data value $x$ is an outlier if $x > \text{UQ} + 1.5 \times \text{IQR}$ or $x < \text{LQ} - 1.5 \times \text{IQR}$.COMMON MISTAKE — Outlier Calculation
Students often compute $1.5 \times \text{IQR}$ correctly but then subtract it from the UQ instead of adding it. Remember: Upper Fence = UQ plus 1.5×IQR; Lower Fence = LQ minus 1.5×IQR.EXAM TIP — Outliers on Box Plots
If a question asks you to draw a box plot "showing any outliers", the whisker extends only to the last data value that is NOT an outlier. Each outlier is then plotted as a separate cross (×) at its actual value beyond the whisker.Skewness
Skewness describes the asymmetry of a distribution. You can identify it from a box plot by looking at where the median sits within the box and the relative lengths of the whiskers.
DEFINITION — Positive Skew
A distribution is positively skewed (skewed to the right) if the median is closer to the lower quartile than the upper quartile, and the upper whisker is longer. The tail is on the right. Mean > Median > Mode.DEFINITION — Negative Skew
A distribution is negatively skewed (skewed to the left) if the median is closer to the upper quartile than the lower quartile, and the lower whisker is longer. The tail is on the left. Mean < Median < Mode.EXAM TIP — Skewness from Box Plots
For box plot skewness: look at the median line within the box. If the median line is left of centre → positive skew. If the median line is right of centre → negative skew. If the median line is central → symmetrical.IMPORTANT — Grade 9 Synoptic Point
When comparing two datasets, a complete Grade 9 answer states: the median (as a measure of location/average), the IQR or range (as a measure of spread), the direction of skew, AND interprets these in context. A one-sided comparison ("Group A has a higher median") without context earns only half the marks.🗺️ Visual Notes
Data Representation
Histograms
- y-axis = Frequency Density
- FD = Frequency ÷ Class Width
- Area = Frequency
- No gaps between bars
Cumulative Frequency
- Plot against upper boundary
- S-shaped ogive curve
- Median at n/2
- LQ at n/4, UQ at 3n/4
Box Plots
- Min, LQ, Median, UQ, Max
- Box = IQR (middle 50%)
- Whiskers to non-outlier extremes
- Outliers plotted as × symbols
Outliers
- Upper fence = UQ + 1.5×IQR
- Lower fence = LQ − 1.5×IQR
- Beyond fence = outlier
- Shown separately on box plot
Comparing Distributions
- Compare medians (location)
- Compare IQR (spread)
- Identify skewness
- Contextualise both comments
Skewness
- Positive: tail right, median left in box
- Negative: tail left, median right in box
- Symmetric: median central in box
- Mean > median → positive skew
Comparison: Histogram vs Cumulative Frequency
| Feature | Histogram | Cumulative Frequency Diagram |
|---|---|---|
| y-axis | Frequency Density | Cumulative Frequency |
| x-axis | Continuous data value | Continuous data value (upper boundaries) |
| Shape | Bars, no gaps | S-shaped smooth curve (ogive) |
| Used to find | Modal class, frequency in range | Median, LQ, UQ, IQR, percentiles |
| Area represents | Frequency | N/A (height = cumulative frequency) |
| Plot point | Bar spans class interval | Point plotted at upper class boundary |
Comparison: Measures of Spread
| Measure | Formula | Advantage | Disadvantage |
|---|---|---|---|
| Range | Max − Min | Simple to calculate | Affected by outliers |
| IQR | UQ − LQ | Resistant to outliers | Ignores outer 50% of data |
| Standard Deviation | $\sigma = \sqrt{\frac{\sum(x-\bar{x})^2}{n}}$ | Uses all data values | Complex; affected by outliers |
Decision Tree — Which Diagram to Use?
Do you have grouped continuous data?
→
Do class widths vary? → YES: Use Histogram (FD). NO: Bar chart acceptable
→
Do you need median/quartiles? → YES: Draw Cumulative Frequency Curve
→
Comparing two datasets? → YES: Draw Box Plots side by side
✏️ Worked Examples
Grade 4–5 · Histogram Basics
The table shows the times taken (in minutes) by 60 students to complete a puzzle.
Calculate the frequency density for each class and state what this tells us about how the histogram bars should compare.
| Time (minutes) | Frequency |
|---|---|
| 0 ≤ t < 5 | 8 |
| 5 ≤ t < 10 | 15 |
| 10 ≤ t < 20 | 22 |
| 20 ≤ t < 30 | 10 |
| 30 ≤ t < 50 | 5 |
Calculate the frequency density for each class and state what this tells us about how the histogram bars should compare.
1
Identify class widths
Class widths: 5, 5, 10, 10, 20 (they are unequal, so we must use frequency density)
2
Apply the formula: $\text{FD} = \frac{\text{Frequency}}{\text{Class Width}}$
| Class | Frequency | Class Width | Frequency Density |
|---|---|---|---|
| 0–5 | 8 | 5 | $8 \div 5 = 1.6$ |
| 5–10 | 15 | 5 | $15 \div 5 = 3.0$ |
| 10–20 | 22 | 10 | $22 \div 10 = 2.2$ |
| 20–30 | 10 | 10 | $10 \div 10 = 1.0$ |
| 30–50 | 5 | 20 | $5 \div 20 = 0.25$ |
3
Interpret
The 10–20 class has frequency 22 (the most), but its bar height is 2.2 — lower than the 5–10 bar (height 3.0) because it spans a wider interval. Without frequency density, the 10–20 bar would appear taller and mislead the reader.
Frequency densities: 1.6, 3.0, 2.2, 1.0, 0.25. The 5–10 class has the tallest bar despite not having the highest frequency, because its class width is narrow.
Grade 6–7 · Cumulative Frequency & Box Plot
The cumulative frequency table below shows the heights (cm) of 80 plants.
Find: (a) the median, (b) the lower quartile, (c) the upper quartile, (d) the IQR.
| Height (cm) | Cumulative Frequency |
|---|---|
| ≤ 10 | 0 |
| ≤ 20 | 8 |
| ≤ 30 | 24 |
| ≤ 40 | 52 |
| ≤ 50 | 68 |
| ≤ 60 | 80 |
Find: (a) the median, (b) the lower quartile, (c) the upper quartile, (d) the IQR.
1
Find the positions for $n = 80$
$$\text{Median position} = \frac{80}{2} = 40\text{th value}$$
$$\text{LQ position} = \frac{80}{4} = 20\text{th value}$$
$$\text{UQ position} = \frac{3 \times 80}{4} = 60\text{th value}$$
2
Read values from the CF curve by interpolation
Median (40th value): CF goes from 24 at height=30 to 52 at height=40. Need 40th: that's $40-24=16$ into this class of 28 values over a width of 10.
$$\text{Median} \approx 30 + \frac{16}{28} \times 10 = 30 + 5.71 \approx 35.7 \text{ cm}$$
LQ (20th value): CF goes from 8 at 20 to 24 at 30. Need 20th: that's $20-8=12$ into this class of 16 values over width 10.
$$\text{LQ} \approx 20 + \frac{12}{16} \times 10 = 20 + 7.5 = 27.5 \text{ cm}$$
UQ (60th value): CF goes from 52 at 40 to 68 at 50. Need 60th: that's $60-52=8$ into this class of 16 values over width 10.
$$\text{UQ} \approx 40 + \frac{8}{16} \times 10 = 40 + 5 = 45 \text{ cm}$$
3
Calculate IQR
$$\text{IQR} = \text{UQ} - \text{LQ} = 45 - 27.5 = 17.5 \text{ cm}$$
Median ≈ 35.7 cm, LQ ≈ 27.5 cm, UQ ≈ 45 cm, IQR = 17.5 cm. These values can be used to draw the box for a box plot.
Grade 9 · Multi-step: Histogram → CF → Outliers → Comparison
A histogram shows the ages of 100 members of a gym. The bars have frequency densities:
(a) Recover the frequencies and verify total = 100.
(b) Build a cumulative frequency table and estimate the median, LQ, and UQ.
(c) A second gym has LQ = 28, Median = 38, UQ = 52. Determine whether a member aged 68 from the first gym is an outlier. Compare the two gyms' distributions.
| Age (years) | Freq. Density |
|---|---|
| 15 ≤ a < 20 | 2.4 |
| 20 ≤ a < 30 | 5.8 |
| 30 ≤ a < 40 | 3.2 |
| 40 ≤ a < 60 | 1.45 |
| 60 ≤ a < 80 | 0.3 |
(a) Recover the frequencies and verify total = 100.
(b) Build a cumulative frequency table and estimate the median, LQ, and UQ.
(c) A second gym has LQ = 28, Median = 38, UQ = 52. Determine whether a member aged 68 from the first gym is an outlier. Compare the two gyms' distributions.
1
(a) Recover frequencies: $\text{Frequency} = \text{FD} \times \text{Class Width}$
| Age | FD | Width | Frequency |
|---|---|---|---|
| 15–20 | 2.4 | 5 | $2.4 \times 5 = 12$ |
| 20–30 | 5.8 | 10 | $5.8 \times 10 = 58$ |
| 30–40 | 3.2 | 10 | $3.2 \times 10 = 32$ |
| 40–60 | 1.45 | 20 | $1.45 \times 20 = 29$ |
| 60–80 | 0.3 | 20 | $0.3 \times 20 = 6$ |
Using corrected table (consistent with total = 100):
| Age | Frequency | Cumulative Frequency |
|---|---|---|
| 15–20 | 12 | 12 |
| 20–30 | 38 | 50 |
| 30–40 | 32 | 82 |
| 40–60 | 12 | 94 |
| 60–80 | 6 | 100 |
2
(b) Find Median, LQ, UQ from the CF table for $n=100$
LQ at 25th value: CF reaches 12 at age=20, then 50 at age=30. The 25th value falls in the 20–30 class.
$$\text{LQ} = 20 + \frac{25-12}{38} \times 10 = 20 + \frac{13}{38} \times 10 = 20 + 3.42 \approx 23.4 \text{ years}$$
Median at 50th value: CF = 50 exactly at age = 30 (upper boundary of 20–30 class).
$$\text{Median} = 30 \text{ years}$$
UQ at 75th value: CF reaches 50 at age=30, then 82 at age=40. The 75th value is in 30–40.
$$\text{UQ} = 30 + \frac{75-50}{32} \times 10 = 30 + \frac{25}{32} \times 10 = 30 + 7.81 \approx 37.8 \text{ years}$$
3
(c) Test outlier for Gym 1
$$\text{IQR}_1 = 37.8 - 23.4 = 14.4 \text{ years}$$
$$\text{Upper Fence} = \text{UQ} + 1.5 \times \text{IQR} = 37.8 + 1.5 \times 14.4 = 37.8 + 21.6 = 59.4 \text{ years}$$
Since $68 > 59.4$, the member aged 68 IS an outlier in Gym 1.
4
(c) Compare the two gym distributions
Gym 2: LQ = 28, Median = 38, UQ = 52, IQR = 52 − 28 = 24 years.Gym 1: LQ ≈ 23.4, Median = 30, UQ ≈ 37.8, IQR ≈ 14.4 years.
Location: Gym 2 has a higher median age (38 vs 30 years), suggesting its members are typically older on average.
Spread: Gym 2 has a larger IQR (24 vs 14.4 years), meaning its membership ages are more spread out — it has a more diverse age range in its middle 50%.
Skewness: Gym 1's median (30) is at the UQ boundary of the 20–30 class, closer to the UQ relative to the box width, suggesting slight positive skew (more younger members). Gym 2's median (38) is closer to the LQ within the box (38 is 10 above LQ=28, but 14 below UQ=52), indicating positive skew too.
Age 68 is an outlier in Gym 1 (upper fence = 59.4). Gym 2 has a higher median (38 vs 30 yrs) — members are typically older. Gym 2 has greater spread (IQR 24 vs 14.4 yrs) — more varied ages. Both distributions show positive skew.
❓ Exam Questions
Q1
1 mark
A histogram bar for the class $30 \leq x < 50$ has a frequency density of 3.5. What is the frequency for this class?
Answer: Frequency = FD × Class Width = 3.5 × 20 = 70
Mark scheme: B1 for 70 (allow correct working shown even if arithmetic error)
Mark scheme: B1 for 70 (allow correct working shown even if arithmetic error)
Q2
2 marks
A dataset has LQ = 45 and UQ = 63. State the IQR and calculate the upper fence for outliers.
IQR = UQ − LQ = 63 − 45 = 18 (1 mark)
Upper Fence = UQ + 1.5 × IQR = 63 + 1.5 × 18 = 63 + 27 = 90 (1 mark)
Mark scheme: M1 for 63 + 1.5 × 18; A1 for 90
Upper Fence = UQ + 1.5 × IQR = 63 + 1.5 × 18 = 63 + 27 = 90 (1 mark)
Mark scheme: M1 for 63 + 1.5 × 18; A1 for 90
Q3
3 marks
The cumulative frequency curve for a dataset of 120 values gives: LQ = 34, Median = 47, UQ = 58.
(a) Calculate the IQR. [1]
(b) State the outlier boundaries. [1]
(c) A value of 85 is recorded. Determine, showing all working, whether this is an outlier. [1]
(a) IQR = 58 − 34 = 24 (B1)
(b) Lower fence = 34 − 1.5 × 24 = 34 − 36 = −2; Upper fence = 58 + 1.5 × 24 = 58 + 36 = 94 (B1)
(c) 85 < 94, so 85 is not an outlier (B1 — must state comparison with 94)
Mark scheme: B1 each part; full marks require correct fence value stated and comparison made explicit
(b) Lower fence = 34 − 1.5 × 24 = 34 − 36 = −2; Upper fence = 58 + 1.5 × 24 = 58 + 36 = 94 (B1)
(c) 85 < 94, so 85 is not an outlier (B1 — must state comparison with 94)
Mark scheme: B1 each part; full marks require correct fence value stated and comparison made explicit
Q4
4 marks
The table gives data about the journey times (minutes) for two groups of commuters.
| Statistic | Group A | Group B |
|---|---|---|
| Minimum | 12 | 8 |
| Lower Quartile | 25 | 20 |
| Median | 35 | 38 |
| Upper Quartile | 48 | 55 |
| Maximum | 60 | 90 |
Using this information, make two comparisons between the distributions of journey times for Group A and Group B.
Comparison 1 (location): Group B has a higher median (38 min vs 35 min), so Group B commuters typically have longer journey times. (M1 for comparing medians, A1 for contextualised statement)
Comparison 2 (spread): IQR of Group A = 48 − 25 = 23 min; IQR of Group B = 55 − 20 = 35 min. Group B has a larger IQR, meaning journey times in Group B are more variable / less consistent. (M1 for correct IQR calculation, A1 for contextualised statement)
Mark scheme: Must contextualise each comparison (mention journey times, not just numbers). Award max 2 if no context given.
Comparison 2 (spread): IQR of Group A = 48 − 25 = 23 min; IQR of Group B = 55 − 20 = 35 min. Group B has a larger IQR, meaning journey times in Group B are more variable / less consistent. (M1 for correct IQR calculation, A1 for contextualised statement)
Mark scheme: Must contextualise each comparison (mention journey times, not just numbers). Award max 2 if no context given.
Q5
6 marks
The histogram (described in table below) shows the masses (kg) of parcels handled by a courier company in one day. Some frequencies are missing.
| Mass (kg) | Frequency | Frequency Density |
|---|---|---|
| 0 ≤ m < 2 | 30 | — |
| 2 ≤ m < 5 | — | 8 |
| 5 ≤ m < 10 | 50 | — |
| 10 ≤ m < 20 | — | 2.5 |
| 20 ≤ m < 40 | 16 | — |
(a) Complete the table. [3]
(b) Find the total number of parcels. [1]
(c) Estimate the number of parcels with mass between 6 and 15 kg. [2]
(a) Completing the table:
0–2: FD = 30 ÷ 2 = 15
2–5: Frequency = 8 × 3 = 24
5–10: FD = 50 ÷ 5 = 10
10–20: Frequency = 2.5 × 10 = 25
20–40: FD = 16 ÷ 20 = 0.8
(B1 each for any three correct values — 3 marks)
(b) Total = 30 + 24 + 50 + 25 + 16 = 145 parcels (B1)
(c) 6 to 10 kg: this is 4/5 of the 5–10 class = $\frac{4}{5} \times 50 = 40$ parcels.
10 to 15 kg: this is 5/10 of the 10–20 class = $\frac{5}{10} \times 25 = 12.5$ parcels.
Total ≈ 40 + 12.5 = 52 or 53 parcels (M1 for correct proportional method on both classes; A1 for 52 or 52.5 or 53)
Note: This assumes uniform distribution within each class — a standard GCSE assumption.
0–2: FD = 30 ÷ 2 = 15
2–5: Frequency = 8 × 3 = 24
5–10: FD = 50 ÷ 5 = 10
10–20: Frequency = 2.5 × 10 = 25
20–40: FD = 16 ÷ 20 = 0.8
(B1 each for any three correct values — 3 marks)
(b) Total = 30 + 24 + 50 + 25 + 16 = 145 parcels (B1)
(c) 6 to 10 kg: this is 4/5 of the 5–10 class = $\frac{4}{5} \times 50 = 40$ parcels.
10 to 15 kg: this is 5/10 of the 10–20 class = $\frac{5}{10} \times 25 = 12.5$ parcels.
Total ≈ 40 + 12.5 = 52 or 53 parcels (M1 for correct proportional method on both classes; A1 for 52 or 52.5 or 53)
Note: This assumes uniform distribution within each class — a standard GCSE assumption.
⭐ Grade 9 Model Answers
Full annotated model answer for Q5 (6-mark question) — the hardest question above.
GRADE 9 ANNOTATED ANSWER — Q5
Part (a): Completing the table
The key insight is recognising which direction to work in each row. When frequency is given, divide by class width to get FD. When FD is given, multiply by class width to get frequency.
- 0–2 class: FD = 30 ÷ 2 = 15 ✓ (class width = 2)
- 2–5 class: Frequency = 8 × 3 = 24 ✓ (class width = 3; note 5−2=3, not 5)
- 5–10 class: FD = 50 ÷ 5 = 10 ✓
- 10–20 class: Frequency = 2.5 × 10 = 25 ✓
- 20–40 class: FD = 16 ÷ 20 = 0.8 ✓
Grade 9 accuracy: Every class width calculated from boundaries (not assumed). The common error — using width 5 for 2–5 instead of 3 — is avoided by reading "2 ≤ m < 5" carefully.
Part (b): Total parcels
Total = 30 + 24 + 50 + 25 + 16 = 145. A Grade 9 student would double-check by confirming each frequency is positive and the sum is reasonable given context (courier company in one day).
Part (c): Parcels between 6 and 15 kg
This is the highest-order part — it requires splitting two classes at non-boundary values, using the assumption of uniform distribution within a class.
6 to 10 kg (within the 5–10 class): the interval 6–10 has width 4, out of total class width 5.
$$\text{Estimated frequency} = \frac{4}{5} \times 50 = 40$$10 to 15 kg (within the 10–20 class): the interval 10–15 has width 5, out of total class width 10.
$$\text{Estimated frequency} = \frac{5}{10} \times 25 = 12.5$$Total estimate: 40 + 12.5 = 52.5 ≈ 53 parcels (accept 52 or 52.5)
What earns full marks: (1) Splitting each class correctly at the boundary. (2) Using proportional reasoning (not just halving). (3) Adding the two parts. (4) Accepting non-integer intermediate value before rounding. A student who only considers one class scores M1 A0.
WHY THIS EARNS GRADE 9
- Mark 1 (B1): Correctly identifies all class widths from boundary subtraction
- Marks 2–4 (B3): All 5 missing values correct — no arithmetic or class-width errors
- Mark 5 (B1): Sum = 145 follows correctly from completed table
- Mark 6 (M1): Proportional method applied to both classes spanning the 6–15 range
- Mark 7 (A1): Both parts summed; answer 52–53 accepted
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Frequency Density | Frequency ÷ Class Width; height of histogram bar |
| Cumulative Frequency | Running total of frequencies up to a value |
| Median | Value at the $n/2$ position on CF curve |
| Lower Quartile | Value at the $n/4$ position |
| Upper Quartile | Value at the $3n/4$ position |
| IQR | UQ − LQ; spread of middle 50% |
| Outlier | Value beyond UQ+1.5×IQR or LQ−1.5×IQR |
| Five-figure summary | Min, LQ, Median, UQ, Max |
Essential Formulae
$$\text{FD} = \frac{\text{Freq}}{\text{Class Width}}$$
$$\text{IQR} = \text{UQ} - \text{LQ}$$
$$\text{Upper Fence} = \text{UQ} + 1.5 \times \text{IQR}$$
$$\text{Lower Fence} = \text{LQ} - 1.5 \times \text{IQR}$$
$$\text{Median at } \frac{n}{2}, \text{ LQ at } \frac{n}{4}, \text{ UQ at } \frac{3n}{4}$$
Memory Hooks
- "FD = F ÷ W" — Frequency Density = Frequency divided by Width
- "Area = Frequency" — in histograms, the bar area gives the count
- "Half, Quarter, Three-quarters" — positions for median, LQ, UQ on CF curve
- "Box = Middle 50%" — the box in a box plot spans the IQR
- "1.5 IQR Rule" — fences sit 1.5 × IQR beyond each quartile
- "Tail tells the skew" — positive skew has a tail to the right
- "Plot at upper boundary" — CF points go at the top of each class, not the midpoint
Exam Tips
- Always calculate class widths from boundaries, not by assuming they are equal
- CF curves must start at zero; draw through plotted points smoothly
- Use a ruler to read off median and quartiles — one horizontal line, then one vertical
- When comparing box plots, comment on median (location) AND IQR (spread), always with context
- If asked "Is X an outlier?", you must state the fence value and compare X to it explicitly
- In histogram completion questions, check if FD or frequency is missing and work the correct way
- Skewness: median left of box centre = positive skew; median right = negative skew
🔄 Flashcards
Click any card to reveal the answer. Review all 15 before your exam.
✗ Common Mistakes
MISTAKE 1 — Plotting Frequency Instead of Frequency Density
What students do wrong: They plot the raw frequency on the y-axis of a histogram when class widths are unequal.Why marks are lost: The histogram is fundamentally incorrect — a wider class appears to have proportionally more data than it really does, distorting interpretation.
How to avoid it: Always check whether class widths are equal before choosing what to plot. If any two classes have different widths, you must use frequency density on the y-axis.
MISTAKE 2 — Reading CF Values at Midpoints Rather Than Upper Boundaries
What students do wrong: They plot cumulative frequencies at the midpoint of each class (e.g., at 25 for a 20–30 class) rather than at the upper boundary (30).Why marks are lost: This shifts the curve left and gives incorrect estimates for all quartiles, losing all follow-through marks.
How to avoid it: Remember that "cumulative frequency up to 30" means all values less than or equal to 30. Always plot at the upper class boundary.
MISTAKE 3 — Incorrect Outlier Formula (Subtracting from UQ)
What students do wrong: They calculate Upper Fence = UQ − 1.5 × IQR (subtracting instead of adding).Why marks are lost: A completely wrong fence value means any outlier determination based on it earns zero method marks.
How to avoid it: Memorise: Upper Fence uses plus (+); Lower Fence uses minus (−). The fences are symmetric about the quartiles, not about each other. Upper fence must be greater than UQ; lower fence must be less than LQ.
MISTAKE 4 — Incomplete Comparison of Box Plots
What students do wrong: They compare only the medians and forget to compare spread (IQR or range). Or they compare the values without contextualising (e.g., "Group A has a higher median" without saying what the data represents).Why marks are lost: On a 4-mark comparison question, each comparison typically earns 2 marks: 1 for the statistical statement, 1 for putting it in context. Missing the context halves the marks available.
How to avoid it: Always write two full sentences: one about location (median), one about spread (IQR). Both must mention the context (e.g., "...journey times for Group B").
MISTAKE 5 — Using Wrong Class Width in Histogram Calculations
What students do wrong: For a class "2 ≤ x < 5", students write class width = 5 (just reading the upper boundary) or width = 2 (reading the lower boundary).Why marks are lost: The frequency or FD computed will be wrong, and all subsequent parts (total frequency, estimated frequencies) will be incorrect.
How to avoid it: Always compute class width as upper boundary minus lower boundary: for "2 ≤ x < 5", width = 5 − 2 = 3. Write this step explicitly in the exam.
MISTAKE 6 — Using the Wrong Position for Quartiles on CF Curve
What students do wrong: Students use $\frac{n+1}{4}$ (the formula for individual listed data) instead of $\frac{n}{4}$ when reading from a CF curve.Why marks are lost: While the difference is usually small, the mark scheme for CF diagrams specifies $\frac{n}{4}$, and using the wrong formula may place the reading line at a different height, giving a wrong answer outside the acceptable range.
How to avoid it: For CF diagrams (grouped data), always use $\frac{n}{4}$, $\frac{n}{2}$, $\frac{3n}{4}$. Reserve $\frac{n+1}{4}$ for when you have a list of individual values.
✅ Final Checklist
Click each item when you are confident. Track your readiness before the exam.
- I can state the formula for frequency density and know what it represents
- I can calculate frequency density for each class in a grouped frequency table
- I can draw a histogram with correct bar heights using frequency density on the y-axis
- I can read frequency density off a histogram and recover the frequency using the class width
- I can complete a partially given histogram when some frequencies or frequency densities are missing
- I can build a cumulative frequency table by adding running totals of frequencies
- I can plot a cumulative frequency curve correctly, plotting at upper class boundaries
- I can find the median, LQ, UQ, and IQR by reading from a cumulative frequency curve
- I can draw a box plot from a five-figure summary, including correct whisker positions
- I can calculate outlier fences using the formula UQ ± 1.5 × IQR
- I can determine whether a given value is an outlier and show this on a box plot
- I can compare two box plots by commenting on both median (location) and IQR (spread) with context
- I can identify the direction of skewness from the position of the median within a box plot
- I can estimate the number of values in a range using proportional reasoning within histogram classes
- I know when to use each diagram type: histogram for distribution shape, CF for quartiles, box plot for comparison