Data Representation

Q: What is the difference between a bar chart and a histogram?

Bar charts are for categorical/discrete data with gaps between bars, and height equals frequency. Histograms are for continuous grouped data, no gaps, and area equals frequency.

Q: How do I calculate class width?

Class width = upper class boundary lower class boundary. Do not use the difference between stated class limits, which will give the wrong value for whole-number grouped data.

CIE A-Level Mathematics· Unit 4: Probability & Statistics 1· 34 min read

1. Frequency Distributions★☆☆☆☆⏱ 5 min

A frequency distribution organises raw data into non-overlapping classes, showing how many observations (frequency) fall into each class. For ungrouped discrete or categorical data, each value/category is its own class. For grouped continuous data, we use class boundaries to eliminate gaps between adjacent classes.

📘 Definition

Class Boundaries

The actual upper and lower limits of a grouped data class, adjusted to avoid gaps between classes

Example:

A class written as $10 - 19$ for whole-number data has boundaries $9.5$ and $19.5$

📐 Worked Example

The following are heights (cm) of 20 students: 152, 156, 161, 162, 164, 165, 167, 168, 169, 170, 171, 172, 173, 175, 176, 178, 180, 181, 183, 185. Construct a grouped frequency distribution with 4 equal-width classes.

1
Calculate the range of the data and find class width:
2
$185 - 152 = 33, \quad \text{round up to width } 9$
3
Define classes and their boundaries, then count frequencies:
4
Stated Class Class Boundaries Frequency
150-158 149.5-158.5 2
159-167 158.5-167.5 5
168-176 167.5-176.5 8
177-185 176.5-185.5 5

Stated Class	Class Boundaries	Frequency
150-158	149.5-158.5	2
159-167	158.5-167.5	5
168-176	167.5-176.5	8
177-185	176.5-185.5	5

Exam tip:

Always write down class boundaries even if not asked, to avoid mistakes in later histogram calculations.

2. Histograms★★☆☆☆⏱ 6 min

Unlike bar charts, the area of each bar in a histogram equals frequency, not the height. This is critical for unequal class widths. The height of each bar (frequency density) is calculated with the standard formula:

\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}

📘 Definition

Frequency Density

The height of a histogram bar, scaled so that area of the bar equals the class frequency

📐 Worked Example

A grouped frequency distribution has a class $20 - 30$ with frequency 12, and a class $31 - 50$ with frequency 28. Calculate the height of each bar for a histogram.

1
Find class boundaries and widths for each class:
2
$20-30: \text{ width } = 30.5 - 19.5 = 11 \\ 31-50: \text{ width } = 50.5 - 30.5 = 20$
3
Calculate frequency density (height) for each bar:
4
$\text{Height}_1 = \frac{12}{11} \approx 1.09, \quad \text{Height}_2 = \frac{28}{20} = 1.4$
5
The areas check out: $1.09 \times 11 = 12$ and $1.4 \times 20 = 28$ , matching the class frequencies.

Exam tip:

CIE almost always uses unequal class widths for histogram questions, so always check class width before calculating frequency density.

3. Stem-and-Leaf Plots★☆☆☆☆⏱ 4 min

A stem-and-leaf plot is a simple visual display that retains all original data values, making it easy to find the median and quartiles directly from the plot. Back-to-back stem-and-leaf plots are used to compare two distributions.

📐 Worked Example

Draw an ordered stem-and-leaf plot for the following test scores out of 50: 12, 45, 23, 17, 31, 25, 29, 35, 41, 19, 22, 32.

1
Use the tens digit as the stem, so stems are 1, 2, 3, 4. Sort the units digits (leaves) for each stem in ascending order:
2
Stem Leaves
1 2 7 9
2 2 3 5 9
3 1 2 5
4 1 5
3
Add a key to explain the plot: Key: $1∣2 = 12$ marks. The median can be read directly as the average of the 6th and 7th values: $\frac{25 + 29}{2} = 27$ .

Stem	Leaves
1	2 7 9
2	2 3 5 9
3	1 2 5
4	1 5

Exam tip:

In CIE marking a missing key usually costs an accuracy mark, so add a clear key after every stem-and-leaf plot.

4. Box-and-Whisker Plots★★☆☆☆⏱ 5 min

A box plot displays the five-number summary of a data set: minimum, lower quartile ( $Q_{1}$ ), median ( $Q_{2}$ ), upper quartile ( $Q_{3}$ ), maximum. In the CIE convention the whiskers are drawn all the way out to the smallest and largest values, so the total span of the whiskers shows the range. Some questions give the rule that a value is an outlier if it is less than $Q_{1} - 1.5 \times I QR$ or greater than $Q_{3} + 1.5 \times I QR$ (where $I QR = Q_{3} - Q_{1}$ ); this is a separate calculation used only when a question asks you to identify outliers, and it does not change how the standard box plot is drawn.

📐 Worked Example

A data set has $Q_{1} = 20$ , median $= 28$ , $Q_{3} = 36$ , minimum $= 12$ , maximum $= 52$ . Draw the box plot and identify any outliers.

1
Calculate the interquartile range:
2
$IQR = Q_3 - Q_1 = 36 - 20 = 16$
3
Calculate outlier boundaries:
4
$\text{Lower} = 20 - (1.5 \times 16) = -4 \\ \text{Upper} = 36 + (1.5 \times 16) = 60$
5
All values fall between $- 4$ and $60$ , so there are no outliers. Draw a box from 20 to 36 with a line at 28, and whiskers to 12 and 52.

Exam tip:

When comparing two box plots, always make two comments: one about central tendency (median) and one about spread (IQR/range).

5. Cumulative Frequency Graphs★★★☆☆⏱ 6 min

A cumulative frequency graph shows the running total of frequencies. Each cumulative frequency is plotted against the UPPER class boundary of its class, and the points are joined with a smooth curve. The graph is used to estimate the median, quartiles and percentiles, and to find the number or proportion of values above or below a given value (or between two values).

📘 Definition

Cumulative Frequency

The running total of all frequencies up to and including the upper boundary of a class.

Example:

If the classes $0 - 9$ , $10 - 19$ , $20 - 29$ have frequencies $4, 11, 7$ , the cumulative frequencies at boundaries $9.5, 19.5, 29.5$ are $4, 15, 22$ .

📐 Worked Example

The table shows the time (t minutes) taken by 80 people to complete a task.

Time (min)	Frequency
$0 \leq t < 10$	6
$10 \leq t < 20$	14
$20 \leq t < 30$	24
$30 \leq t < 40$	20
$40 \leq t < 50$	16

Estimate the median and interquartile range, and the proportion of people taking less than 25 minutes.

1
Form the cumulative frequencies at the upper class boundaries $10, 20, 30, 40, 50$ :
2
Time $<$ 10 20 30 40 50
Cumulative frequency 6 20 44 64 80
3
Plot the points $(10, 6), (20, 20), (30, 44), (40, 64), (50, 80)$ and join with a smooth curve.
4
Read the median at cumulative frequency $\frac{n}{2} = 40$ :
5
$\text{median} \approx 20 + \frac{40 - 20}{44 - 20}\times 10 \approx 28.3 \text{ min}$
6
Read the quartiles at $\frac{n}{4} = 20$ and $\frac{3 n}{4} = 60$ , giving $Q_{1} \approx 20$ min and $Q_{3} \approx 38$ min, so $IQR \approx 18$ min.
7
For the proportion under 25 min, read up from $t = 25$ : the cumulative frequency is about $32$ , so roughly $\frac{32}{80} = 40%$ of people take less than 25 minutes.

Time $<$	10	20	30	40	50
Cumulative frequency	6	20	44	64	80

Exam tip:

Always plot cumulative frequency against the UPPER class boundary, never the midpoint, and read the median at $\frac{n}{2}$ (not $\frac{n + 1}{2}$ ) when estimating from the curve.

6. Measures of Central Tendency★★☆☆☆⏱ 5 min

The three averages, or measures of central tendency, are the mean, the median and the mode. For a frequency distribution the mean is $\overset{x}{ˉ} = \frac{Σ f x}{Σ f}$ ; for grouped data you use the class midpoint as $x$ . The median is the middle value when the data are in order, and the mode (or modal class) is the value (or class) with the highest frequency.

\bar{x} = \frac{\Sigma x}{n} \quad \text{or} \quad \bar{x} = \frac{\Sigma f x}{\Sigma f}

📐 Worked Example

The number of goals scored by a team in 24 matches is summarised below. Find the mean, median and mode.

Goals $x$	0	1	2	3	4
Frequency $f$	5	8	6	3	2

1
Mean: form $Σ f x$ and divide by $Σ f = 24$ :
2
$\bar{x} = \frac{(0)(5)+(1)(8)+(2)(6)+(3)(3)+(4)(2)}{24} = \frac{37}{24} \approx 1.54$
3
Median: with $n = 24$ , the median is the mean of the 12th and 13th values. The cumulative frequencies are $5, 13, \dots$ , so both the 12th and 13th values are $1$ , giving a median of $1$ .
4
Mode: the highest frequency is $8$ , at $x = 1$ , so the mode is $1$ . (For grouped continuous data you would instead state the modal class.)

Exam tip:

For grouped data the mean is only an estimate, because you replace each class by its midpoint; give answers to 3 significant figures unless told otherwise.

7. Measures of Variation: Range, IQR and Standard Deviation★★★☆☆⏱ 8 min

Spread (variation) is measured by the range ( $max - min$ ), the interquartile range ( $IQR = Q_{3} - Q_{1}$ ) and — the most heavily examined measure in this unit — the standard deviation. The standard deviation is the root-mean-square distance of the values from the mean; its square is the variance.

\text{sd} = \sqrt{\frac{\Sigma x^2}{n} - \bar{x}^2} \quad \text{or} \quad \text{sd} = \sqrt{\frac{\Sigma f x^2}{\Sigma f} - \bar{x}^2}

📘 Definition

Standard Deviation (S1 convention)

A measure of spread found by dividing by $n$ (or $Σ f$ ), never by $n - 1$ . The $n - 1$ (unbiased estimator) version is not used in this unit.

Example:

Both formulae above are given in the MF19 list; you only need to substitute the totals $Σ x$ and $Σ x^{2}$ .

📐 Worked Example

A set of 10 values has $Σ x = 250$ and $Σ x^{2} = 6850$ . Find the mean and standard deviation.

1
Mean:
2
$\bar{x} = \frac{\Sigma x}{n} = \frac{250}{10} = 25$
3
Standard deviation (divide by $n = 10$ , not $9$ ):
4
$\text{sd} = \sqrt{\frac{6850}{10} - 25^2} = \sqrt{685 - 625} = \sqrt{60} \approx 7.75$

Coding makes large numbers easier: subtracting a constant $a$ and working with the totals $Σ (x - a)$ and $Σ (x - a)^{2}$ shifts the mean by $a$ but leaves the standard deviation unchanged, since $\overset{x}{ˉ} = a + \frac{Σ ( x - a )}{n}$ and $sd = \frac{Σ ( x - a ) ^{2}}{n} - (\frac{Σ ( x - a )}{n})^{2}$ .

📐 Worked Example

For the five values $204, 207, 201, 208, 210$ , use the coding $y = x - 200$ to find the mean and standard deviation.

1
Coded values $y = x - 200$ are $4, 7, 1, 8, 10$ , so $Σ y = 30$ and $Σ y^{2} = 230$ with $n = 5$ .
2
$\bar{x} = 200 + \frac{30}{5} = 206$
3
$\text{sd} = \sqrt{\frac{230}{5} - \left(\frac{30}{5}\right)^2} = \sqrt{46 - 36} = \sqrt{10} \approx 3.16$
4
To combine two data sets, add their totals: $n = n_{1} + n_{2}$ , $Σ x = Σ x_{1} + Σ x_{2}$ and $Σ x^{2} = Σ x_{1}^{2} + Σ x_{2}^{2}$ , then use the same formulae.

Exam tip:

The mean and standard deviation formulae are given in MF19, so marks come from clean substitution: show $Σ x$ and $Σ x^{2}$ , keep full accuracy, and always divide by $n$ (or $Σ f$ ), never $n - 1$ .

8. Common Pitfalls

Wrong move:

Treating the height of a histogram bar as equal to frequency for unequal class widths

Why:

This leads to incorrect bar areas, so any estimates from the histogram will be wrong

Correct move:

Always calculate height as frequency density = frequency / class width, regardless of class width equality

Wrong move:

Forgetting to add a key to a stem-and-leaf plot

Why:

CIE marking expects a key to confirm what the values represent, so a missing key usually costs an accuracy mark

Correct move:

Always add a simple key like $2∣3 = 23$ after drawing your plot

Wrong move:

Calculating class width from stated class limits instead of boundaries

Why:

For the class 10-19, the difference between 19 and 10 is 9, but the actual width is 10, leading to wrong frequency density

Correct move:

Always calculate width as upper boundary minus lower boundary

Wrong move:

Stopping the whiskers before the smallest or largest value and marking outliers as separate crosses

Why:

That is the modified box plot used by some other boards; in the CIE convention the whiskers must reach the smallest and largest data values

Correct move:

Draw the whiskers to the minimum and maximum, so they show the range; only identify outliers separately when a question explicitly asks

Wrong move:

Only commenting on one feature when comparing two box plots

Why:

Most comparison questions are worth two marks, one for each feature

Correct move:

Always comment on both the difference in median (centre) and difference in spread (IQR/range)

9. Quick Reference Cheatsheet

Display Type	Key Rule	Required Check
Grouped frequency distribution	Always calculate class boundaries first	No overlapping classes
Histogram	Area = frequency	Height = frequency ÷ class width
Stem-and-leaf plot	Retains all raw data values	Must include a clear key
Box plot	Five-number summary; whiskers to min & max	Whiskers show the range; mark outliers only if asked
Cumulative frequency graph	Plot cumulative frequency at the upper class boundary	Median at n/2, quartiles at n/4 and 3n/4
Mean / median / mode	Mean = Σfx ÷ Σf	Use class midpoints for grouped data
Standard deviation	sd = √(Σx² ÷ n − mean²)	Divide by n (or Σf), never n − 1
Comparing data sets	Two comments: one location, one spread	e.g. median/mean + IQR/sd, in context

10. Frequently Asked

What is the difference between a bar chart and a histogram?

Bar charts are for categorical/discrete data with gaps between bars, and height equals frequency. Histograms are for continuous grouped data, no gaps, and area equals frequency.

How do I calculate class width?

Class width = upper class boundary $-$ lower class boundary. Do not use the difference between stated class limits, which will give the wrong value for whole-number grouped data.

Going deeper

syllabusOfficial CIE 9709 SyllabusCheck latest Unit 4 requirements

What's Next

Data representation is the foundation for all further statistics topics in CIE A-Level, as you need to be able to organise and interpret raw data before you can calculate probabilities, analyse relationships or draw valid statistical conclusions. Mastery of the construction rules for histograms, stem-and-leaf plots, cumulative frequency graphs and box plots, together with the summary measures (mean, median, mode and standard deviation), is required for both standalone questions and combined questions that ask you to interpret results alongside measures of central tendency and spread, or that lead into probability and the normal distribution. This sub-topic is almost always examined for 5-10 marks per paper, so consistent accuracy here translates directly to higher overall grades.