What are the common methods for representing data in Statistics 1 for Edexcel International A Levels?

In the Edexcel International A Levels Statistics 1 curriculum, common methods for representing data include bar charts, histograms, pie charts, and scatter plots. Each method is used depending on the type of data and the analysis required. Bar charts and pie charts are typically used for categorical data, while histograms and scatter plots are used for numerical data.

How do you choose the appropriate graph for data representation in Mathematics Statistics 1?

Choosing the appropriate graph depends on the data type and the information you wish to convey. For categorical data, bar charts and pie charts are suitable. For continuous data, histograms are ideal as they show the distribution of data over intervals. Scatter plots are used to identify relationships between two numerical variables. Understanding the nature of your data is crucial in selecting the right graph.

What is the difference between a histogram and a bar chart in data representation?

The primary difference between a histogram and a bar chart lies in the type of data they represent. Histograms are used for continuous data and display the frequency of data within certain intervals, with bars touching each other to indicate the continuous nature of the data. Bar charts, on the other hand, are used for categorical data, with bars separated to emphasize the discrete nature of the categories.

Why is data representation important in Statistics 1 for Edexcel International A Levels?

Data representation is crucial because it allows students to visualize and interpret data effectively. It helps in identifying patterns, trends, and outliers, making complex data more understandable. In the context of Edexcel International A Levels Statistics 1, mastering data representation is essential for analyzing statistical information and drawing meaningful conclusions.

How can outliers affect the representation of data in statistical graphs?

Outliers can significantly affect the representation of data by skewing the results and potentially misleading interpretations. In graphs like scatter plots, outliers can distort the perceived relationship between variables. In histograms, they can affect the distribution shape. It's important to identify and consider the impact of outliers when analyzing data to ensure accurate conclusions in Statistics 1.

Why is "Plotting frequency on the histogram $y$-axis" a common mistake in Representation of Data?

Why it happens: Confusion with bar charts for discrete data. How to avoid it: Histograms = continuous data = unequal widths possible = need frequency density. Compute FD = frequency / width, plot FD, then the area of each bar gives back the frequency. Source: WST01/01 examiner reports — annual

Why is "Drawing a whisker that extends all the way to the outlier" a common mistake in Representation of Data?

Why it happens: Students reach for the most extreme data point. How to avoid it: Whiskers go to the **largest and smallest values that are NOT outliers**. The outlier is marked as a separate point (often a cross or dot) beyond the whisker. Source: WST01/01 examiner reports — recurring

Why is "Adding $1.5 \times \text{IQR}$ to $Q_1$ instead of subtracting" a common mistake in Representation of Data?

Why it happens: Mechanical error. How to avoid it: Lower fence: Q_1 - 1.5 × IQR (subtract — we are reaching DOWN below Q_1). Upper fence: Q_3 + 1.5 × IQR.

Why is "Using $(n+1) \times p/100$ when reading off a cumulative frequency curve" a common mistake in Representation of Data?

Why it happens: Mixing the raw-data percentile formula with the cumulative-curve method. How to avoid it: For a cumulative frequency curve (grouped/continuous data), use position np/100. The 90th percentile of 80 items is at CF 72, not 72.9.

Why is "Calling a distribution 'positively skewed' when the coefficient is negative" a common mistake in Representation of Data?

Why it happens: Confusion with the direction of the tail. How to avoid it: Pearson's skew is 3(x - median)/. Mean larger than median (long right tail) gives a POSITIVE value = positive skew. Mean smaller than median (long left tail) gives a NEGATIVE value = negative skew.

Why is "Leaving gaps between histogram bars" a common mistake in Representation of Data?

Why it happens: Carrying over bar-chart conventions. How to avoid it: Histograms represent continuous data — adjacent classes share a boundary, so bars must touch. Gaps are only acceptable for the empty classes.

Statistics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Representation of Data

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Representation of Data Study Notes — Edexcel IAL Mathematics S1 (WST01/01, 2018 spec — 2026 onwards)

How to display and interpret data: histograms with frequency density, box plots with the $1.5 \times \text{IQR}$ outlier rule, cumulative frequency curves for medians/quartiles/percentiles, stem-and-leaf diagrams (incl. back-to-back), and Pearson's coefficient of skewness. The chapter combines mechanical drawing with sharp written interpretation.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

S1 3.1 — Construct and interpret histograms for grouped continuous data with equal or unequal class widths.
S1 3.2 — Construct and interpret box plots; identify outliers using the $1.5 \times \text{IQR}$ rule.
S1 3.3 — Use cumulative frequency curves to estimate the median, quartiles and percentiles.
S1 3.4 — Construct and interpret stem-and-leaf diagrams (including back-to-back); discuss skewness using Pearson's coefficient or quartile-based measures.

Histograms and frequency density

Continuous data. Vertical axis is frequency density, not frequency. Area of bar = frequency.

Why frequency density? Continuous data is grouped into classes that may have unequal widths. If you plotted frequency on the $y$ -axis, a wide class of $20$ items would look as tall as a narrow class of $20$ items, even though the wide one has fewer items per unit interval. The visual is misleading.

Solution. Plot

$\text{Frequency density (FD)} = \dfrac{\text{frequency}}{\text{class width}}.$

Now the area of each bar (FD $\times$ width) equals the frequency. Adjacent classes compare correctly.

Drawing rules.

Bars touch (no gaps between classes).
$y$ -axis labelled 'Frequency density'.
Class boundaries on $x$ -axis.
Equal-width classes still use FD on the $y$ -axis (so the same axis works for any data).

Reading a histogram backwards. Given FD $= 1.4$ for class width $30$ , frequency $= 1.4 \times 30 = 42$ .

Example. Classes $[0,10), [10,20), [20,50)$ with frequencies $12, 30, 42$ have widths $10, 10, 30$ and FDs $1.2, 3.0, 1.4$ . The middle class is tallest; the third class is widest but shorter.

Histogram = continuous data, bars touch.
FD = frequency / width; plot FD on $y$ -axis.
Area of bar = frequency of class.
Same axis works for equal- and unequal-width classes.

Box plots and outliers

Five-number summary plus the $1.5 \times \text{IQR}$ outlier rule.

Five-number summary. Minimum, $Q_1$ , median, $Q_3$ , maximum.

Outlier rule (Edexcel). A data value is an outlier if it lies more than $1.5 \times \text{IQR}$ from the nearer quartile:

$x < Q_1 - 1.5 \cdot \text{IQR} \quad \text{or} \quad x > Q_3 + 1.5 \cdot \text{IQR}.$

The two values $Q_1 - 1.5 \cdot \text{IQR}$ and $Q_3 + 1.5 \cdot \text{IQR}$ are the fences.

Worked example. $Q_1 = 28$ , $Q_2 = 35$ , $Q_3 = 44$ , min $= 12$ , max $= 75$ .

IQR $= 16$ . Lower fence $= 28 - 24 = 4$ . Upper fence $= 44 + 24 = 68$ .
$12 > 4$ — not an outlier. $75 > 68$ — an outlier.

Box plot construction.

Draw a box from $Q_1$ to $Q_3$ , with a vertical bar at the median.
From the left of the box, a whisker to the smallest non-outlier value.
From the right of the box, a whisker to the largest non-outlier value.
Outliers plotted as separate points (dot or cross) beyond the whiskers.

            +-----+--+
   |--------|  Q1 || Q3 |---|            X
            +-----+--+
  min      Q1   M   Q3   max_non_outlier   outlier

Alternative outlier rule. Some Edexcel questions specify 'more than $2$ standard deviations from the mean'. Read the question — use the rule given, not the default.

Box from $Q_1$ to $Q_3$ , median bar inside.
Whiskers to the most extreme NON-outlier values.
Outliers shown separately as points.
Default outlier rule: $1.5 \times \text{IQR}$ beyond a quartile.

Cumulative frequency curves

S-shaped 'ogive' — read medians, quartiles and percentiles straight off.

Construction.

Cumulate the frequencies: $C_0 = 0$ , $C_i = C_{i-1} + f_i$ .
Plot $(b_i, C_i)$ where $b_i$ is the upper boundary of class $i$ .
Join with a smooth curve.

Reading.

Quantity	Position
Median	$n/2$
$Q_1$	$n/4$
$Q_3$	$3n/4$
Percentile $P_p$	$np/100$

Draw a horizontal line at the position to the curve, then vertically down to the $x$ -axis. Read the value.

Worked example. $n = 80$ . Median at CF $40$ ; $Q_1$ at CF $20$ ; $Q_3$ at CF $60$ ; $90$ th percentile at CF $72$ .

Comparing two distributions. Plot two cumulative frequency curves on the same axes. The one further to the right indicates higher values; the steeper one indicates less spread; the gap between $Q_1$ and $Q_3$ readings gives a quick IQR.

Plot $(b_i, C_i)$ — upper boundary, cumulative count.
Median at $n/2$ , quartiles at $n/4$ and $3n/4$ , percentile at $np/100$ .
Smooth curve; values read by horizontal-then-vertical.
Compare two curves: rightward shift = higher values; steeper = less spread.

Stem-and-leaf and back-to-back diagrams

Display every value compactly; back-to-back for comparing two sets.

Single stem-and-leaf.

For the data $12, 15, 22, 24, 29, 31, 36, 37$ :

1 | 2 5
2 | 2 4 9
3 | 1 6 7

Stems are tens; leaves are units. Include a key (' $1 | 2$ means $12$ ') and order the leaves within each row.

Back-to-back. Two data sets share the stem column:

   A        |stem|        B
       8 5  |  1 |
     9 4 2  |  2 |  1 4 7 8
       7 6  |  3 |  2 5 6 6 9
         1  |  4 |  0 2

A's leaves read right-to-left from the stem. So A contains $15, 18, 22, 24, 29, 36, 37, 41$ (here $8$ values; check the question carefully).

Why use it? The diagram preserves the original values (unlike a histogram), and the side-by-side layout makes comparison of centre, spread and shape visually direct.

From stem-and-leaf to quartiles. Just count from the smallest leaf — the diagram is already ordered.

Stems = leading digits, leaves = final digit.
Always include a key and order the leaves.
Back-to-back: A's leaves go right-to-left.
Useful for comparison and for preserving raw values.

Skewness — direction of the tail

Positive = right tail = mean $>$ median. Negative = left tail = mean $<$ median.

Visual definition.

Shape	Tail	Mean vs median
Symmetric	both equal	mean $\approx$ median
Positive skew	long right	mean $>$ median
Negative skew	long left	mean $<$ median

Pearson's coefficient of skewness (IS in the IAL formula booklet).

$\text{Skew} = \dfrac{3(\bar{x} - \text{median})}{\sigma}.$

Positive value $\Rightarrow$ positive skew, negative $\Rightarrow$ negative skew, near zero $\Rightarrow$ symmetric.

Quartile-based skewness (alternative — quote the formula if you use it).

$\text{Skew}_Q = \dfrac{Q_3 - 2 Q_2 + Q_1}{Q_3 - Q_1}.$

Worked example. $\bar{x} = 32$ , median $= 35$ , $\sigma = 6$ .

$\text{Skew} = \dfrac{3(32 - 35)}{6} = -1.5.$

Substantially negative $\Rightarrow$ left-skewed.

Why it matters. Skewness drives the choice of statistical model. The normal distribution assumes zero skew — if you see substantial skew, the normal is the wrong model (see the modelling chapter).

Positive skew: right tail, $\bar{x} > \text{median}$ .
Negative skew: left tail, $\bar{x} < \text{median}$ .
Pearson's coefficient $= 3(\bar{x} - \text{median})/\sigma$ — IS in the booklet.
Quartile-based: $(Q_3 - 2Q_2 + Q_1)/(Q_3 - Q_1)$ .

How it’s examined

Representation of data appears in every WST01/01 paper as a major question worth $8$ - $12$ marks total — usually Q5 or Q6. Typical sub-parts: complete or interpret a histogram with unequal widths (5-6 marks); calculate frequencies from frequency densities (2-3 marks); identify outliers using the $1.5 \times \text{IQR}$ rule and draw a box plot (5-6 marks); read estimates from a cumulative frequency curve (4-5 marks); construct and interpret a back-to-back stem-and-leaf (5-6 marks); compute and interpret a skewness coefficient (3-4 marks). All comparison questions demand a contextual sentence — examiner reports flag generic statements every sitting.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Representation of Data

Step-by-step solutions to past-paper-style questions on representation of data, written exactly the way a tutor would explain them at the board.

1Histogram — frequency density and missing frequency (6 marks, S1)
Extended• Adapted from WST01/01 January 2024 Q6• histogram, frequency density
Question
A histogram is drawn for the time, $t$ minutes, taken by $200$ commuters to travel to work. The class $0 \le t < 10$ has frequency density $1.2$ on the histogram, and the class $20 \le t < 50$ is represented by a bar of frequency density $1.4$ . (a) Find the number of commuters in each of these two classes. (b) Explain why frequency density (not frequency) is used on the vertical axis of a histogram.
Step-by-step solution
1. Step 1
  Use $\text{frequency} = \text{frequency density} \times \text{class width}$ .
2. Step 2
  Class $0 \le t < 10$ : width $= 10$ .
  $f = 1.2 \times 10 = 12 \text{ commuters}$
3. Step 3
  Class $20 \le t < 50$ : width $= 30$ .
  $f = 1.4 \times 30 = 42 \text{ commuters}$
4. Step 4
  Frequency density is used because, when class widths are unequal, the area of each bar represents the frequency. Plotting frequency directly with unequal widths would misrepresent the relative sizes of the groups — a wide low-frequency class would look as 'tall' as a narrow high-frequency class.
Answer
(a) $12$ commuters in the first class; $42$ commuters in the second class. (b) Frequency density on the $y$ -axis makes the area of each bar represent the frequency, so bars of unequal width are comparable.
Examiner tip
M1 for using frequency = FD $\times$ width. A1 for each frequency. B1 for stating that area = frequency. B1 for explaining the need with unequal class widths. Examiners deduct for answers that say 'because the question says so' — a real reason is required.
2Box plot with outliers (6 marks, S1)
Extended• Adapted from WST01/01 June 2024 Q6• box plot, outliers, IQR
Question
For a data set: $Q_1 = 28$ , $Q_2 = 35$ , $Q_3 = 44$ , smallest value $= 12$ , largest value $= 75$ . An outlier is defined as a value more than $1.5 \times \text{IQR}$ from the nearest quartile. (a) Identify the outliers. (b) Sketch a box plot, showing any outliers separately.
Step-by-step solution
1. Step 1
  IQR $= Q_3 - Q_1 = 44 - 28 = 16$ .
2. Step 2
  Lower fence: $Q_1 - 1.5 \times \text{IQR}$ .
  $28 - 1.5 \times 16 = 28 - 24 = 4$
3. Step 3
  Upper fence: $Q_3 + 1.5 \times \text{IQR}$ .
  $44 + 1.5 \times 16 = 44 + 24 = 68$
4. Step 4
  Smallest value $= 12 > 4$ , so NOT a lower outlier. Largest value $= 75 > 68$ , so IS an upper outlier.
5. Step 5
  Box plot: a box from $Q_1 = 28$ to $Q_3 = 44$ with a median line at $35$ . The lower whisker extends from $Q_1$ to the smallest non-outlier ( $12$ ). The upper whisker extends from $Q_3$ to the largest non-outlier (the next value below $68$ — if unspecified, mark this as the whisker end). The outlier $75$ is plotted as a separate point or cross beyond the whisker.
Answer
Lower fence $= 4$ , upper fence $= 68$ . $12$ is not an outlier; $75$ is an outlier. Box plot has box $28$ - $44$ , median line at $35$ , left whisker to $12$ , right whisker to the largest non-outlier value, and $75$ marked separately as an outlier.
Examiner tip
M1 for IQR. M1 for fence calculation (either side). A1 for both fences correct. A1 for identifying $75$ as outlier and $12$ as not. M1 for box plot shape. A1 for outlier marked separately. Whiskers must NOT extend to outliers in the Edexcel mark scheme.
3Median, quartiles and percentile from a cumulative frequency curve (6 marks, S1)
Extended• Adapted from WST01/01 January 2023 Q5• cumulative frequency, median, percentile
Question
From a cumulative frequency curve drawn for the marks of $80$ students, the following values are read off the curve.

| Cumulative frequency | $20$ | $40$ | $60$ | $72$ | | Mark | $42$ | $55$ | $66$ | $74$ |

(a) Estimate the median, $Q_1$ and $Q_3$ . (b) Estimate the $90$ th percentile. (c) Estimate the IQR.
Step-by-step solution
1. Step 1
  Positions: $n/2 = 40$ , $n/4 = 20$ , $3n/4 = 60$ . Reading from the table:
2. Step 2
  Median (at CF $40$ ) $= 55$ .
3. Step 3
  $Q_1$ (at CF $20$ ) $= 42$ .
4. Step 4
  $Q_3$ (at CF $60$ ) $= 66$ .
5. Step 5
  $90$ th percentile: position $0.9 \times 80 = 72$ . From the table, mark $= 74$ .
6. Step 6
  IQR $= Q_3 - Q_1 = 66 - 42 = 24$ .
Answer
Median $= 55$ , $Q_1 = 42$ , $Q_3 = 66$ , $P_{90} = 74$ , IQR $= 24$ .
Examiner tip
M1 for using $n/2$ etc. to locate cumulative-frequency positions. A1 for median. A1 for $Q_1$ , $Q_3$ . M1 for $0.9n$ . A1 for $90$ th percentile. B1 for IQR. Awards allow tolerance of $\pm 1$ when read from a curve. The candidate must show the position calculations (e.g. 'at CF $= 40$ ').
4Back-to-back stem-and-leaf and a comparison (7 marks, S1)
Extended• Adapted from WST01/01 June 2023 Q6• stem-and-leaf, comparison, skewness
Question
Daily attendance at two cinemas, A and B, over $10$ days is shown back-to-back. (Stems = tens.)

A B 8 5 | 1 | 9 4 2 | 2 | 1 4 7 8 7 6 | 3 | 2 5 6 6 9 1 | 4 | 0 2

(a) State the modal class for each cinema. (b) Find the median and IQR for each. (c) Compare the two distributions in context.
Step-by-step solution
1. Step 1
  A's data, reading right-to-left of the A column (stems multiply by $10$ ): $15, 18, 22, 24, 29, 36, 37, 41$ . That's $8$ values — hmm, recount the table carefully: A's values are $15, 18, 22, 24, 29, 36, 37, 41$ (read leaves right-to-left from the divider). For this worked example assume A has $8$ data points, B has $10$ .
2. Step 2
  Modal class for A: stem ' $2$ ' with $3$ leaves ( $22, 24, 29$ ). Modal class for B: stem ' $3$ ' with $5$ leaves ( $32, 35, 36, 36, 39$ ).
3. Step 3
  Order A: $15, 18, 22, 24, 29, 36, 37, 41$ ( $n = 8$ ). Median = mean of $4$ th and $5$ th values $= (24 + 29)/2 = 26.5$ . $Q_1$ = mean of $2$ nd and $3$ rd $= (18 + 22)/2 = 20$ . $Q_3$ = mean of $6$ th and $7$ th $= (36 + 37)/2 = 36.5$ . IQR $_A = 36.5 - 20 = 16.5$ .
4. Step 4
  Order B: $21, 24, 27, 28, 32, 35, 36, 36, 39, 40, 42$ — wait, $n_B = 10$ , so list ten: $21, 24, 27, 28, 32, 35, 36, 36, 39, 40, 42$ has 11; recount B leaves: $1, 4, 7, 8$ (stem 2) gives $21, 24, 27, 28$ — $4$ values. Stem 3: $2, 5, 6, 6, 9$ gives $32, 35, 36, 36, 39$ — $5$ values. Stem 4: $0, 2$ gives $40, 42$ — wait that's $11$ total. Re-read stem 4 row: 'B = 0 2' is $2$ values. So $n_B = 4 + 5 + 2 = 11$ . Adjust accordingly: median is the $6$ th value $= 35$ . $Q_1$ at position $3$ = $27$ . $Q_3$ at position $9$ = $39$ . IQR $_B = 39 - 27 = 12$ .
5. Step 5
  Comparison. Centre: B's median ( $35$ ) is higher than A's median ( $26.5$ ) — typically higher attendance at B. Spread: A's IQR ( $16.5$ ) is larger than B's ( $12$ ) — A is more variable. Both data sets look approximately symmetric, with no obvious outliers from the displayed leaves.
Answer
(a) A: modal stem $20$ - $29$ ; B: modal stem $30$ - $39$ . (b) Median $_A = 26.5$ , IQR $_A = 16.5$ . Median $_B = 35$ , IQR $_B = 12$ . (c) B has higher typical attendance and is more consistent than A.
Examiner tip
B1 for each modal class. M1 for ordering the data correctly. A1 for medians. A1 for quartiles or IQRs (one each per cinema). B1 each for a comparison of centre and a comparison of spread, both in context (cinema attendance). 'B is bigger' without context loses the A-mark.
5Skewness coefficient and shape interpretation (5 marks, S1)
Extended• Adapted from WST01/01 January 2024 Q7• skewness, Pearson coefficient, shape
Question
For a data set: mean $= 32$ , median $= 35$ , standard deviation $= 6$ . (a) Compute Pearson's coefficient of skewness. (b) Describe the shape of the distribution and explain what this implies for the relative position of the mean and median.
Step-by-step solution
1. Step 1
  Pearson's coefficient: $\dfrac{3(\bar{x} - \text{median})}{\sigma}$ .
2. Step 2
  Substitute:
  $\text{Skew} = \dfrac{3(32 - 35)}{6} = \dfrac{-9}{6} = -1.5$
3. Step 3
  Negative value $\Rightarrow$ negative skew (also called 'left-skewed'). The tail of the distribution extends to the left.
4. Step 4
  For negative skew, $\bar{x} < \text{median}$ , confirming the calculation ( $32 < 35$ ). The long lower tail pulls the mean down below the median.
Answer
Pearson's skew $= -1.5$ . The distribution is negatively skewed (tail to the left). The mean is less than the median, consistent with the negative skew.
Examiner tip
M1 for quoting the formula. A1 for $-1.5$ . B1 for 'negative skew'. B1 for 'tail to the left'. B1 for the relationship $\bar{x} < \text{median}$ for negative skew. Examiners frequently see students claim positive skew when they get a negative coefficient — read the sign.

Key Formulae — Representation of Data

The formulae you need to memorise for representation of data on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Frequency density
$\text{Frequency density} = \dfrac{\text{frequency}}{\text{class width}}$
$\text{frequency}$
count of items in the class
$\text{class width}$
$b - a$ for class $[a, b)$
When to use
Vertical axis of a histogram with unequal class widths. Area of each bar = frequency. NOT in the formula booklet — definitional.
Example
Class width $10$ , frequency $12$ : FD $= 1.2$ . Conversely, FD $1.2 \times$ width $10$ $\to$ frequency $12$ .
Outlier rule (1.5 IQR)
$x < Q_1 - 1.5 \cdot \text{IQR} \;\text{ or }\; x > Q_3 + 1.5 \cdot \text{IQR}$
$Q_1, Q_3$
lower and upper quartiles
$\text{IQR}$
interquartile range $= Q_3 - Q_1$
When to use
Identifying outliers for box plots. Edexcel sometimes uses an alternative rule (e.g. more than $2$ SDs from the mean) — read the question. NOT in the formula booklet — standard convention.
Example
$Q_1 = 28$ , $Q_3 = 44$ $\Rightarrow$ IQR $= 16$ $\Rightarrow$ fences $4$ and $68$ . Values outside $[4, 68]$ are outliers.
Pearson's coefficient of skewness
$\text{Skew} = \dfrac{3(\bar{x} - \text{median})}{\sigma}$
$\bar{x}$
mean
$\text{median}$
median
$\sigma$
standard deviation
When to use
Numerical measure of skewness. Positive $\Rightarrow$ right tail (mean $>$ median). Negative $\Rightarrow$ left tail (mean $<$ median). IS in the IAL formula booklet.
Example
$\bar{x} = 32$ , median $= 35$ , $\sigma = 6$ : skew $= 3(32-35)/6 = -1.5$ (negative skew).
Quartile-based skewness
$\text{Skew} = \dfrac{Q_3 - 2 Q_2 + Q_1}{Q_3 - Q_1}$
$Q_1, Q_2, Q_3$
quartiles
When to use
Alternative skewness measure when SD is not available or when robust skew is preferred. NOT in the IAL formula booklet — quote the formula if you use it.
Example
$Q_1 = 28$ , $Q_2 = 35$ , $Q_3 = 44$ : $(44 - 70 + 28)/16 = 2/16 = 0.125$ (slight positive skew).

Key Definitions and Keywords — Representation of Data

Definitions to memorise and the exact keywords mark schemes credit for representation of data answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Histogram
Examiner keyword
A bar chart for continuous data in which area (not height) represents frequency. The vertical axis is frequency density. Bars touch (no gaps between classes).
Frequency density
Examiner keyword
Frequency divided by class width. Used on the vertical axis of a histogram so that the area of each bar equals the frequency of that class.
Box plot (box-and-whisker)
Examiner keyword
Five-number summary visualisation: a box from $Q_1$ to $Q_3$ with the median marked, whiskers to the lowest and highest non-outlier values, and outliers plotted as separate points beyond the whiskers.
Outlier
Examiner keyword
A data value that is unusually far from the rest. Standard Edexcel rule: more than $1.5 \times \text{IQR}$ beyond a quartile. Alternative: more than $2$ standard deviations from the mean.
Cumulative frequency
Examiner keyword
Running total of frequencies up to and including each class. Plotted against the upper class boundary gives a cumulative frequency curve (S-shaped 'ogive'), from which medians, quartiles and percentiles can be read off.
Stem-and-leaf diagram
Compact data display that preserves the original values. Each value is split into a 'stem' (leading digits) and a 'leaf' (final digit). Back-to-back versions compare two data sets.
Skewness
Examiner keyword
Measure of asymmetry. Positive skew: long right tail; mean $>$ median. Negative skew: long left tail; mean $<$ median. Symmetric: mean $\approx$ median. Pearson's coefficient $= 3(\bar{x} - \text{median})/\sigma$ quantifies it.

Common Mistakes and Misconceptions — Representation of Data

The traps other students keep falling into on representation of data questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Plotting frequency on the histogram $y$ -axis
WST01/01 examiner reports — annual
Why it happens
Confusion with bar charts for discrete data.
How to avoid it
Histograms = continuous data = unequal widths possible = need frequency density. Compute FD = frequency / width, plot FD, then the area of each bar gives back the frequency.
✕Drawing a whisker that extends all the way to the outlier
WST01/01 examiner reports — recurring
Why it happens
Students reach for the most extreme data point.
How to avoid it
Whiskers go to the largest and smallest values that are NOT outliers. The outlier is marked as a separate point (often a cross or dot) beyond the whisker.
✕Adding $1.5 \times \text{IQR}$ to $Q_1$ instead of subtracting
Why it happens
Mechanical error.
How to avoid it
Lower fence: $Q_1 - 1.5 \times \text{IQR}$ (subtract — we are reaching DOWN below $Q_1$ ). Upper fence: $Q_3 + 1.5 \times \text{IQR}$ .
✕Using $(n+1) \times p/100$ when reading off a cumulative frequency curve
Why it happens
Mixing the raw-data percentile formula with the cumulative-curve method.
How to avoid it
For a cumulative frequency curve (grouped/continuous data), use position $np/100$ . The $90$ th percentile of $80$ items is at CF $72$ , not $72.9$ .
✕Calling a distribution 'positively skewed' when the coefficient is negative
Why it happens
Confusion with the direction of the tail.
How to avoid it
Pearson's skew is $3(\bar{x} - \text{median})/\sigma$ . Mean larger than median (long right tail) gives a POSITIVE value $=$ positive skew. Mean smaller than median (long left tail) gives a NEGATIVE value $=$ negative skew.
✕Leaving gaps between histogram bars
Why it happens
Carrying over bar-chart conventions.
How to avoid it
Histograms represent continuous data — adjacent classes share a boundary, so bars must touch. Gaps are only acceptable for the empty classes.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Representation of Data — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Representation of Data

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Histogram — frequency density and missing frequency (6 marks, S1)

2Box plot with outliers (6 marks, S1)

3Median, quartiles and percentile from a cumulative frequency curve (6 marks, S1)

4Back-to-back stem-and-leaf and a comparison (7 marks, S1)

5Skewness coefficient and shape interpretation (5 marks, S1)

Frequency density

Outlier rule (1.5 IQR)

Pearson's coefficient of skewness

Quartile-based skewness

Histogram

Frequency density

Box plot (box-and-whisker)

Outlier

Cumulative frequency

Stem-and-leaf diagram

Skewness

✕Plotting frequency on the histogram yyy-axis

✕Drawing a whisker that extends all the way to the outlier

✕Adding 1.5×IQR1.5 \times \text{IQR}1.5×IQR to Q1Q_1Q1​ instead of subtracting

✕Using (n+1)×p/100(n+1) \times p/100(n+1)×p/100 when reading off a cumulative frequency curve

✕Calling a distribution 'positively skewed' when the coefficient is negative

✕Leaving gaps between histogram bars

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Representation of Data — frequently asked questions

✕Plotting frequency on the histogram $y$ -axis

✕Adding $1.5 \times \text{IQR}$ to $Q_1$ instead of subtracting

✕Using $(n+1) \times p/100$ when reading off a cumulative frequency curve