Representation of Data

— Probability & Statistics 1 · AS & A Level Maths (9709) · Chapter 1 —

★ Exam Tip: Chapter 1 is where the exam loves histograms from a table (frequency density!) and cumulative frequency graphs (plot at upper boundaries, read off estimates). Two real 9709 Paper 62 questions are worked at the end of these notes.

Why collect, display & analyse data?

Collect by counting, surveys, questionnaires, measuring.
Display + analyse to describe what we see, find answers that are not obvious, and spot questions for further investigation.
Also trains us to judge inaccurate / misleading ways data can be shown, and to check the validity of other people's research (media, adverts, weather, stock reports…).

Prerequisite check (from /)

Check your skillsbounds · histograms · cumulative frequency

1. Plot 20 m × 12 m, each to the nearest metre ⇒ length ∈ [19.5, 20.5), width ∈ [11.5, 12.5).

least perimeter = 2(19.5 + 11.5) = 62 mupper boundary of area = 20.5 × 12.5 = 256.25 m²

2. Histogram columns: widths 3, 4 and heights 8, 6 ⇒ areas 24 and 24 ⇒ the two frequencies are equal (area ∝ frequency).

3. 50 trees: 17 below 3 m, 44 below 4 m, all below 5 m.

between 3 and 4 m: 44 − 17 = 274 m or more: 50 − 44 = 6

1. Types of Data

two families of data; quantitative splits again into discrete / continuous

Qualitative (categorical): described by words, non-numerical — e.g. blood types, colours.

Quantitative: takes numerical values; either discrete or continuous.

Trick: Counted ⇒ discrete. Measured ⇒ continuous.

Discrete data: can take only certain values; usually counted and can't be made more precise.

No. of letters in the words of a book: 1, 2, 3, 4… — never 3⅓ or 4.75.
Discrete does not have to be whole numbers! US coins $0.01, 0.05, 0.10, 0.25, 0.50, 1.00; shoe sizes 6½, 7, 7½ (UK, Canada…).

Continuous data: can take any value (possibly within a limited range); measurements given to a chosen degree of accuracy.

Times of athletes in a 100 m race: to nearest second, 0.1 s, or more accurately if equipment allows. Range limited to positive real numbers.

DISCRETE: separate points

CONTINUOUS: unbroken stretch

★ Key Point 1.1: Discrete data can take only certain values. Continuous data can take any value, possibly within a limited range.

2. Stem-and-Leaf Diagrams (discrete data)

Stem-and-leaf diagram: a table best suited to small amounts of discrete data.

Leaf = last digit of each value. Stem = all the other digits.
Stem digits ordered vertically; leaves ordered horizontally, smallest nearest the stem.
Each row = a class; rows must have equal-width intervals.
A key with units is compulsory (e.g. Key: 5 | 3 means 53%).
Big advantage: raw data can still be seen; two related sets go back-to-back for comparison.

Steps — Physics scores of 15 students

Raw data (%): 58, 55, 58, 61, 72, 79, 97, 67, 61, 77, 92, 64, 69, 62, 53

Group into equal-width classes (width 10 here ⇒ stems 5, 6, 7, 8, 9).
Write leaves in the order they appear (draft).
Re-write with each row in ascending order.
Add the key.

unordered (draft)
5	8 5 8 3
6	1 7 1 4 9 2
7	2 9 7
8
9	7 2

ordered ✓
5	3 5 8 8
6	1 1 2 4 7 9
7	2 7 9
8
9	2 7

Key: 5 | 3 represents a score of 53%

Tip: the diagram should have a bar-chart-like shape — keep leaves aligned in columns. Redraw (or use pencil) if you spot an error. Row 8 is empty but must still appear (equal intervals!).

Back-to-back & long rows

Back-to-back: right leaves ascend left → right; left leaves ascend right → left (both increase as you move away from the stem).
Key for back-to-back names both sides, e.g. Key: 3 | 5 | 0 = 53 passengers on A and 50 on Z.
Rows too long (say 30 leaves in one row)? Repeat the stem with narrower classes 0–4, 5–9, 10–14, 15–19, 20–24 ⇒ stems 0, 0, 1, 1, 2.

Reading one: most common value = repeated leaf · number below a value = count leaves · % = count ÷ total × 100 · which row has fewest = shortest row · compare: totals, typical value, spread.

★ Key Point 1.2: Data in a stem-and-leaf diagram are ordered in rows of equal widths.

Worked Example 1.1back-to-back stem-and-leaf · rainfall

Days of rain per month in a town. 2016: 17, 20, 13, 12, 10, 8, 0, 1, 5, 11, 16, 9. 2017: 9, 13, 11, 8, 6, 3, 1, 2, 2, 4, 8, 7. Display back-to-back and compare.

Group into classes 0–9, 10–19, 20–29 ⇒ stems 0, 1, 2; order each side; add key.

2016		2017
9 8 5 1 0	0	1 2 2 3 4 6 7 8 8 9
7 6 3 2 1 0	1	1 3
0	2

Key: 5 | 0 | 6 represents 5 days in a month of 2016 and 6 days in a month of 2017

2016 total = 122 days2017 total = 74 days

⇒ It rained on more days in 2016 than 2017.

Common mistake: no information is given about the amount of rain that fell, so it is wrong to say more rain fell in 2016 — only that it rained on more days.

3. Histograms (continuous data)

① Rounding, gaps and class boundaries

Continuous data are given to some accuracy (3 s.f., 2 d.p., nearest 10…) = rounding. Written classes show gaps that do not really exist.

e.g. heights to the nearest cm: 146–150, 151–155, 156–160 (gap of 1 cm between classes). Using h for height, the true classes are:

145.5 ≤ h < 150.5 | 150.5 ≤ h < 155.5 | 155.5 ≤ h < 160.5

gaps vanish once you use boundaries: upper boundary of one class = lower boundary of the next

Boundaries: lower: 145.5, 150.5, 155.5
upper: 150.5, 155.5, 160.5

Width = upper − lower = 5, 5, 5
Mid-value = lower + upper2 = 148, 153, 158

Half-the-gap rule: go half the gap down/up. Gap 1 ⇒ ±0.5 · gap 0.1 ⇒ ±0.05 (2.6–2.8 becomes 2.55–2.85, width 0.3 min) · nearest 10 ⇒ ±5.

② Frequency diagram vs histogram — area ∝ frequency

A histogram suits continuous data best (also OK for discrete). Data may be given in a grouped frequency table. Ages of 100 students — three classes of equal width 2:

Age (A years)	16 ≤ A < 18	18 ≤ A < 20	20 ≤ A < 22
No. students (f)	34	46	20
Column area	2 × 34 = 68	2 × 46 = 92	2 × 20 = 40

equal widths ⇒ column heights can just be the frequencies. Area ratio 68 : 92 : 40 = 34 : 46 : 20

Area ratio = frequency ratio even when widths are unequal. So area ∝ frequency

No gaps between columns (upper boundary = next lower boundary). A gap only when a class has zero frequency.
Horizontal axis = continuous number line; column width = class width.
Unequal widths ⇒ height is not the frequency; calculate it so that area ∝ frequency.

Remember: ∝ means "is proportional to". A zigzag (concertina) on an axis shows part of the scale has been omitted.

③ Frequency density

The vertical axis is labelled frequency density = frequency per standard interval (simplest: 1 unit of measurement).

★ Key Point 1.3 (standard interval = 1 unit):
Frequency density = class frequencyclass width ⇒ Class frequency = class width × frequency density

Example:

85 objects with masses 50 to 60 kg: fd = 85 objects(60 − 50) kg = 8.5 objects per kg = 0.0085 objects per gram.

How to draw a histogram

Find class boundaries (half-gap rule); widths = upper − lower.
fd = f ÷ width for each class (label what it measures, e.g. children per kg).
Pick sensible scales (1 cm for 1, 5, 10, 20, 50 units — never 3, 7, 23); use most of the page; check it fits first.
Draw axes and columns with a ruler — never freehand; columns touch; label axes ("Frequency density", variable + units).

Worked Example 1.2histogram + estimate

Masses m kg of 100 children: 40 ≤ m < 50: 40 children, 50 ≤ m < 70: 60 children. (a) Draw the histogram. (b) Estimate the number with masses between 45 and 63 kg.

Mass (m kg)	40 ≤ m < 50	50 ≤ m < 70
No. children (f)	40	60
Class width (kg)	10	20
Frequency density	40 ÷ 10 = 4	60 ÷ 20 = 3

shaded = the part of each column between 45 and 63 kg

45–50 kg (half of class 1)½ × 40 = 20

50–63 kg (inside class 2)freq = width × fd= (63 − 50) × 3= 13 × 3 = 39

estimate = 20 + 39 = 59 children

Why it works: we assume the values in each class are spread evenly over the whole class interval, so a part of a class has frequency = (part width) × fd. Column area = frequency, e.g. (50 − 40) × 4 = 40. (If heights were 8 and 6, fd would mean children per 2 kg — area still 40 children.)

Another example — 36 athletes, race times (nearest minute)

Time (min)	13	14–15	16–18
No. athletes (f)	4	14	18
Boundaries	12.5 ≤ t < 13.5	13.5 ≤ t < 15.5	15.5 ≤ t < 18.5
Class width (min)	1	2	3
Frequency density	4 ÷ 1 = 4	14 ÷ 2 = 7	18 ÷ 3 = 6

Tip: the class with the highest frequency (18) does not have the highest fd (7)!

Common mistake: use boundaries (not the rounded 14–15) to get widths, otherwise fd is wrong (14–15 would give width 1 not 2).

1 small square = ½ min × 1 fd = ½ athlete ⇒ 2 squares = 1 athlete

Worked Example 1.3reading estimates off a histogram

Use the race-time histogram to estimate the number of athletes who took: (a) less than 13.0 min (b) between 14.5 and 17.5 min (c) the slowest three — find their time range.

Trick: 2 squares = 1 athlete, so count squares ÷ 2 instead of using fd each time.

(a)4 squares left of 13.04 ÷ 2 = 2 athletes

(b)14 + 24 = 38 squares38 ÷ 2 = 19 athletes

Same via areas: (b) 1 × 7 (14.5–15.5) + 2 × 6 (15.5–17.5) = 7 + 12 = 19 ✓

Explore 1.1 — what if the table has gaps? (scores of 100 students)

Score (%)	10–29	30–59	60–79	80–99
No. students (f)	6	21	60	13
Width	20	30	20	20
Freq. density	0.3	0.7	3	0.65

Close the gaps by choosing boundaries: whole-number scores ⇒ 9.5, 29.5, 59.5, 79.5, 99.5 (half-gap rule); if scores were rounded down ⇒ 10, 30, 60, 80, 100. More than one way — but the widths and fd are the same either way; only the position of the columns moves.

4. Cumulative Frequency Graphs

Cumulative frequency (cf): the total frequency of all values less than a given value. Used for continuous data.

Plot cf against the UPPER class boundary — we only know the total up to those points exactly.

Join points with straight lines ⇒ cumulative frequency polygon; join with a smooth curve ⇒ cf curve.
e.g. 100 values below 7.5 and 200 below 9.5 ⇒ plot (7.5, 100) and (9.5, 200).
Use it to estimate the number/proportion of values above, below, or between given values.
Polygon: no uncertainty (we know exactly where the segments go). Curve: two people never draw the same one. No rule for which is better.

Common mistake: plotting at class mid-values. The total frequency up to a mid-value is not precise — only an estimate.

A polygon makes exactly the same assumption as a histogram (even spread within each class) ⇒ polygon estimates = histogram estimates, exactly.

How to draw one

List upper boundaries and the running total of frequencies (cf).
Include the lowest lower boundary with cf = 0 and plot it.
Sensible scales; plot points with crosses; join in order with a ruler (polygon) / smooth curve.
Read estimates with dotted lines: value → up to graph → across to cf (or the reverse).

Worked Example 1.4cf polygon & curve · 80 leaves

Lengths (cm, nearest cm): 1–2 (8), 3–4 (20), 5–7 (38), 8–9 (10), 10–11 (4). Draw cf curve + polygon; estimate (a) the number of leaves shorter than 3.7 cm (b) the lower boundary of the longest 22 leaves.

Length (l cm)	l < 0.5	l < 2.5	l < 4.5	l < 7.5	l < 9.5	l < 11.5
Running total	0	0+8	0+8+20	28+38	66+10	76+4
cf	0	8	28	66	76	80

plot at (0.5,0) (2.5,8) (4.5,28) (7.5,66) (9.5,76) (11.5,80); dotted lines = workings for (a) and (b)

(a) leaves < 3.7 cmpolygon ⇒ ≈ 20 leaveshand-drawn curve ≈ 18

(b) longest 22 ⇒ cf = 80 − 22 = 58polygon ⇒ ≈ 6.9 cmhand-drawn curve ≈ 6.7

Note: polygon and curve estimates differ — they only coincide at the plotted points. All answers are estimates (the true shape between plotted points is unknown). Dotted lines show the workings.

Reading a cf graph without drawing it — interpolation (Explore 1.3)

150 objects: m < 0: 0 · m < 12: 24 · m < 30: 60 · m < 53: 106 · m < 70: 138 · m < 80: 150. Polygon estimates: fewer than 20 g = 40 objects; the 100th object = 50 g. Same answers by proportion along a straight segment (similar triangles):

Value for a given cf (target) lying between (x₁, cf₁) and (x₂, cf₂)x = x₁ + target − cf₁cf₂ − cf₁ × (x₂ − x₁)

cf at m = 20 (between 12 and 30)24 + 20 − 1230 − 12 × (60 − 24)24 + 818 × 36 = 40

100th object (cf 100, between 30 and 53)30 + 100 − 60106 − 60 × (53 − 30)30 + 4046 × 23 = 50 g

Histogram shape ⇄ cf graph shape (Explore 1.2)

Gradient of the cf graph = frequency density (column height). Taller columns ⇒ cf rises steeper; the cf graph never goes down.

decreasing columns ⇒ cf steep at first then flattens (concave)

equal columns (uniform) ⇒ straight line

tall middle ⇒ S-shaped (steep in the middle)

increasing columns ⇒ cf gets steeper and steeper (convex)

Why impossible? cf is a running total of frequencies (never negative), so the graph can never decrease.

a dip like this can never be a cf graph

5. Choosing & Comparing Representations

Pictograms, vertical line graphs, bar charts, pie charts = easy ways to show qualitative and ungrouped quantitative data.
Small grouped sets ⇒ stem-and-leaf (raw values still visible).
Large data sets ⇒ individual values lose significance; group them in a frequency table, then use a histogram (absolute/relative frequencies) or cf graph (cumulative). Both allow classes of any and varied widths (some detail lost by grouping).
The choice depends on: type + quantity of data · audience · objectives. Above all it must show the data clearly and not be misleading.

guide to the usual choice of representation

Misleading or poor representations — watch for

Broken / truncated axis: Tom's cf graph has a vertical scale starting at 49 (not 0), so the rise from 49 to 52 weeks looks dramatic — and it hides that most of the year is already counted by the first point.
Wrong conclusion from the data: rain days ≠ amount of rain (WE 1.1). Say only what the data support.
No key / no labels or units: the reader cannot interpret the diagram (a key with units is compulsory in stem-and-leaf).
Absolute vs relative: compare using counts and proportions (below).

Ex 1D Q9 ideaabsolute vs relative poverty

People in poverty (× 10⁵) and population (× 10⁶), World Bank 2011/12:

Country	Chile	Sri Lanka	Malaysia	Georgia	Mongolia
In poverty (× 10⁵)	24.8	18.4	10.8	7.91	7.50
Population (× 10⁶)	17.25	20.65	28.33	4.47	2.74
Poverty rate	14.4%	8.9%	3.8%	17.7%	27.4%

Counts: Chile highest … Mongolia lowest. Rates: Mongolia highest, then Georgia, Chile, Sri Lanka, Malaysia — the order almost reverses! Rates are fairer for countries of very different size; show both (e.g. bar chart of counts with the % marked on each bar).

Did you know? Bar charts first appeared in William Playfair's Commercial and Political Atlas (1786); Florence Nightingale used them in 1859 to compare army and civilian mortality. Image histograms (colour tones) help enhance medical scans.

6. Exam-Style Questions Worked

9709 Paper 62 Q4 (Nov 2011, adapted)cf table → histogram

220 sausages: cf — <20: 0 · <30: 20 · <40: 50 · <45: 100 · <50: 160 · <60: 210 · <70: 220. (i) How many weighed between 50 g and 60 g? (ii) Draw a histogram.

(i) cf(60) − cf(50) = 210 − 160 = 50

(ii) Subtract consecutive cf to get f, then divide by width:

Weight (g)	20–30	30–40	40–45	45–50	50–60	60–70
Frequency	20	30	50	60	50	10
Width	10	10	5	5	10	10
Freq. density	2	3	10	12	5	1

sausage histogram: tallest column is 45–50 g (fd 12)

Ratio problem (end-of-chapter Q7)widths : fd ⇒ frequencies

Histogram with 3 columns, widths in ratio 1 : 2 : 4, frequency densities in ratio 16 : 12 : 3, total frequency 390. (a) Find each frequency. (b) The two classes with the highest frequencies are merged; the merged column is 30 cm high. Find the correct height of the remaining column.

frequency = width × fd ⇒ frequency ratio = (1×16) : (2×12) : (4×3).

ratio = 16 : 24 : 12 (52 parts)1 part = 390 ÷ 52 = 7.5frequencies: 120, 180, 90

(b) Highest two = 120 and 180 (classes 1 and 2, widths 1 + 2 = 3 units):

merged fd = 300 ÷ 3 = 100 ⇒ 100 fd = 30 cm ⇒ 0.3 cm per unit fdremaining class: fd = 90 ÷ 4 = 22.5height = 22.5 × 0.3 = 6.75 cm

9709 Paper 62 Q5 (Jun 2011)cf graph · hotel rooms

Hotel has 90 rooms; rooms occupied per day over 200 days: 1–20 (10), 21–40 (32), 41–50 (62), 51–60 (50), 61–70 (28), 71–90 (18). (i) Draw a cf graph. (ii) Estimate days with more than 30 rooms occupied. (iii) On 75% of days at most n rooms were occupied — estimate n.

Boundaries (half-gap rule): 0.5, 20.5, 40.5, 50.5, 60.5, 70.5, 90.5.

Rooms <	0.5	20.5	40.5	50.5	60.5	70.5	90.5
cf	0	10	42	104	154	182	200

hotel cf polygon with the readings for (ii) and (iii)

(ii) more than 30 roomscf(30.5) = 10 + 1020 × 32 = 26200 − 26 = ≈ 174 daysreading at 30 gives ≈ 175 — a small graph-reading difference

(iii) 75% of 200 = 150 days150 lies between cf 104 (50.5) and 154 (60.5)n = 50.5 + 4650 × 10n = 59.7 ⇒ n ≈ 60

Quick Recap — Full Chapter Cheat-Sheet

✓ Types of data

Qualitative (categorical) = words. Quantitative = numbers: discrete (counted, certain values only) or continuous (measured, any value in a range).

✓ Stem-and-leaf

Small amounts of discrete data · rows of equal width · leaves ascending away from the stem · always a key with units.
Back-to-back for comparing two sets. Long rows ⇒ repeat stems (0–4, 5–9…). Keep empty rows.

✓ Class boundaries

Rounded data ⇒ gaps are fake. Boundary = ± half the gap. Width = upper − lower. Mid-value = (lower + upper) ÷ 2.

✓ Histograms

Area ∝ frequency; vertical axis = frequency density.
fd = class frequencyclass width

frequency = width × fd
No gaps (except zero-frequency); ruler; sensible scales; axis zigzag if scale omitted; label units.
Estimates for part of a class: part-width × fd (values assumed evenly spread). Highest frequency ≠ highest fd.
Ratios: frequency ratio = width ratio × fd ratio.

✓ Cumulative frequency

cf = total frequency below a value. Plot at UPPER boundaries + (lowest lower boundary, 0). Polygon = straight lines; curve = smooth.
Number between a and b = cf(b) − cf(a). Value for a given cf: read across then down.
Interpolation: x = x₁ + target − cf₁cf₂ − cf₁ × (x₂ − x₁). Polygon estimates = histogram estimates.
Gradient of cf graph = fd; cf never decreases. Mid-values are the common mistake.

✓ Choosing a representation

Qualitative / ungrouped: pictogram, line graph, bar chart, pie chart · small grouped: stem-and-leaf · large grouped / continuous: histogram, cf graph.
Must be clear, fit the audience/objective, and not be misleading (truncated axes, wrong comparisons, counts vs proportions).

Last-minute checklist: key included? · boundaries not rounded values? · fd calculated (not f)? · cf plotted at upper boundaries incl. start point 0? · ruler + sensible scale? · answers marked as estimates?