Notes · StatisticsUK · A-Levels

Numerical measures, graphs and diagrams

This topic covers the summary of univariate data by measures of location (mean, median, mode) and spread (range, interquartile range, variance and standard deviation), including grouped data and the effect of linear coding. It then treats skewness, the detection of outliers, and the choice, construction and interpretation of the standard statistical diagrams: histograms with frequency density, box plots and cumulative frequency curves.

5 sections~14 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

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Exam profile

AO1 · Calculate measures of location and spread, including from grouped data and using codingAO2 · Interpret and compare distributions by shape, location, spread and skewnessAO3 · Select and construct appropriate diagrams and identify and treat outliers in context

Operators:calculateestimatedetermineinterpretcomparecommentconstruct

basic level

The AS foundation expects the measures of location and spread, standard deviation from the summation formulae, and reading the standard diagrams.

higher level

The full A-Level expects fluent use of coding, both coefficients of skewness, outlier rules, and the critical comparison of two distributions.

Depth

Text

Contents · 5 sections

Numerical measures, graphs and diagrams

Types of data and measures of location#

FoundationPearson Edexcel 9ST0, Topic 1 (Paper 1)

Data are classified before they are summarised. Categorical (qualitative) data record a category such as colour or brand; quantitative data are numerical and are either discrete, taking separate values such as counts, or continuous, taking any value in an interval such as height or time. The classification governs which averages and which diagrams are legitimate: a mean is meaningful for quantitative data but not for unordered categories, and a histogram is for continuous data whereas a bar chart is for categorical data.

A measure of location, or average, is a single value representing the centre of the data. The mean

\bar{x} = \frac{\sum x}{n}

uses every value and is the basis of later theory, but it is sensitive to extreme values. The median is the middle value when the data are ordered and is resistant to outliers, so it is preferred for skewed data. The mode is the most frequent value, the only average available for purely categorical data.

For data in a frequency table the mean is

\bar{x} = \frac{\sum f x}{\sum f}

, weighting each value by its frequency. For grouped data the individual values are unknown, so each class is represented by its midpoint and the mean is estimated as

\frac{\sum f x}{\sum f}

with

x

the midpoints; the result is an estimate because it assumes the values are evenly spread within each class.

Choosing the right average is an examined judgement. For roughly symmetric quantitative data the mean is the most informative; for markedly skewed data (incomes, house prices) the median better represents a typical value because the mean is dragged towards the long tail. Being able to justify the choice in context is as important as the calculation itself.

\bar{x} = \frac{\sum x}{n}, \qquad \bar{x} = \frac{\sum f x}{\sum f}

Mean of raw and frequency data

The second form weights each value (or class midpoint) by its frequency.

Worked example

Estimating a grouped mean

Journey times (minutes) are recorded: 0-10 (8 people), 10-20 (15), 20-40 (24), 40-70 (9). Estimate the mean journey time.

Midpoints
The class midpoints are 5, 15, 30 and 55 minutes.
Sum of fx
$8(5) + 15(15) + 24(30) + 9(55) = 40 + 225 + 720 + 495 = 1480$ .
Divide by total frequency
The total frequency is $8+15+24+9 = 56$ , so the estimated mean is $\frac{1480}{56} = 26.4$ minutes (3 s.f.).
$\bar{x} = \frac{1480}{56} = 26.4\text{ min}$
Interpret
It is an estimate because we assumed each person's time equals their class midpoint.

Result: The estimated mean journey time is 26.4 minutes (3 s.f.), an estimate because grouped data hide the exact values.

Typical mistakes

Using class widths or boundaries instead of midpoints when estimating a grouped mean.
Quoting the mean for strongly skewed data where the median is the more representative average.

Active revision

The times (minutes) taken by 40 pupils to complete a task are grouped. Estimate the mean using the class midpoints and explain why your answer is only an estimate.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Measures of spread, variance and coding#

StandardPearson Edexcel 9ST0, Topic 1 (Paper 1)

Spread measures how dispersed the data are about the centre. The range (largest minus smallest) is simple but uses only two values and is dominated by extremes. The interquartile range

\text{IQR} = Q_3 - Q_1

measures the spread of the middle half and is resistant to outliers. The variance and standard deviation use every value and underpin all later inference.

The variance is the mean squared deviation from the mean. In practice it is computed from the summation quantity

S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n}

, giving population variance

\sigma^2 = \frac{S_{xx}}{n}

and standard deviation

\sigma = \sqrt{\sigma^2}

; the standard deviation carries the units of the data. For grouped data the same formula is used with

\sum f x^2

and

\sum f x

in place of

\sum x^2

and

\sum x

Coding rescales the data to simplify arithmetic. If a new variable is defined by

y = \frac{x - a}{b}

, then the mean transforms as

\bar{x} = b\bar{y} + a

and, crucially, the standard deviation transforms as

\sigma_x = b\,\sigma_y

. Adding a constant

a

shifts the data without changing the spread, while multiplying by

b

scales both the centre and the spread; this is why the

a

disappears from the standard-deviation relationship.

The distinction between the population divisor

n

and the sample divisor

n-1

matters and returns in estimation. When the data ARE the whole population, or we are simply describing a data set, divide

S_{xx}

n

. When the data are a sample used to estimate a population variance, divide by

n-1

to obtain an unbiased estimate. Descriptive questions here generally use

n

; the

n-1

version is developed in the estimation topic.

S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n}, \qquad \sigma^2 = \frac{S_{xx}}{n}, \qquad \sigma = \sqrt{\sigma^2}

Variance and standard deviation

$S_{xx}$ is the corrected sum of squares; the standard deviation has the units of the data.

y = \frac{x - a}{b} \;\Rightarrow\; \bar{x} = b\bar{y} + a, \quad \sigma_x = b\,\sigma_y

Effect of linear coding

A shift $a$ leaves the spread unchanged; a scale $b$ multiplies both centre and spread.

Worked example

Standard deviation from summation quantities

Five values have $\sum x = 60$ and $\sum x^2 = 910$ . Find the mean and the (population) standard deviation.

Mean
$\bar{x} = \frac{60}{5} = 12$ .
Corrected sum of squares
$S_{xx} = 910 - \frac{60^2}{5} = 910 - 720 = 190$ .
$S_{xx} = 910 - \frac{3600}{5} = 190$
Variance
$\sigma^2 = \frac{190}{5} = 38$ .
Standard deviation
$\sigma = \sqrt{38} = 6.16$ (3 s.f.).

Result: The mean is 12 and the standard deviation is $\sqrt{38} = 6.16$ (3 s.f.).

Typical mistakes

Forgetting to take the square root, so quoting the variance as the standard deviation.
Multiplying the standard deviation by the shift constant $a$ ; only the scale factor $b$ affects the spread.

Active revision

For a data set with $\sum x = 60$ , $\sum x^2 = 910$ and $n = 5$ , find the mean and standard deviation.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Quartiles, percentiles and outliers#

StandardPearson Edexcel 9ST0, Topic 1 (Paper 1)

Box-and-whisker plot with an outlier

Fig. 1Five-number summary 15, 30, 38, 46, 58; the value 72 lies beyond the upper fence

Q_3 + 1.5\,\text{IQR} = 70

and is plotted as an outlier.

Quartiles divide ordered data into four equal parts:

Q_1

(lower quartile) has one quarter of the data below it,

Q_2

is the median, and

Q_3

(upper quartile) has three quarters below it. Percentiles generalise this to hundredths, so the 90th percentile has 90% of the data below it. For grouped data these are read from a cumulative frequency curve or found by linear interpolation within the relevant class.

The interquartile range

Q_3 - Q_1

and the interpercentile ranges (for example the 10th-to-90th) describe spread while ignoring the extreme tails, which makes them robust. They are the natural spread measures to quote alongside the median for skewed data, just as the standard deviation is quoted alongside the mean for symmetric data.

An outlier is a value far enough from the bulk of the data to warrant attention. Two standard rules are used. The quartile (fence) rule flags any value below

Q_1 - 1.5\,\text{IQR}

or above

Q_3 + 1.5\,\text{IQR}

. The mean rule flags any value more than

k

standard deviations from the mean, commonly

k = 2

3

. A value flagged as an outlier is not automatically deleted: it should be investigated as a possible error, but a genuine extreme observation is part of the data.

The box plot below encodes the five-number summary (minimum,

Q_1

, median,

Q_3

, maximum) with outliers plotted individually beyond the whiskers. It makes the median, the IQR and any skewness immediately visible and is the standard tool for comparing two distributions at a glance.

\text{Outlier if } x < Q_1 - 1.5\,\text{IQR} \;\text{ or }\; x > Q_3 + 1.5\,\text{IQR}

Quartile (fence) rule for outliers

With $\text{IQR} = Q_3 - Q_1$ ; an alternative rule flags values more than 2 or 3 standard deviations from the mean.

Worked example

Testing a value with the fence rule

For a data set $Q_1 = 30$ and $Q_3 = 46$ . Determine whether the value 72 is an outlier by the quartile rule.

Interquartile range
$\text{IQR} = 46 - 30 = 16$ .
Upper fence
$Q_3 + 1.5 \times 16 = 46 + 24 = 70$ .
$Q_3 + 1.5\,\text{IQR} = 46 + 24 = 70$
Compare
Since $72 > 70$ , the value lies beyond the upper fence.
Conclude
72 is an outlier by the quartile rule and should be plotted separately and investigated.

Result: The upper fence is 70, and because 72 exceeds it, 72 is an outlier.

Typical mistakes

Deleting an outlier automatically; it should be investigated, and genuine extremes are kept.
Drawing whiskers to the outliers instead of to the most extreme values inside the fences.

Active revision

A data set has $Q_1 = 30$ , $Q_3 = 46$ and includes the value 72. Determine, using the $1.5\,\text{IQR}$ rule, whether 72 is an outlier.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Skewness and comparing distributions#

AdvancedPearson Edexcel 9ST0, Topic 1 (Paper 1)

Skewness measures the asymmetry of a distribution. In a positively (right) skewed distribution the tail stretches to the right and typically

\text{mean} > \text{median} > \text{mode}

; in a negatively (left) skewed distribution the tail stretches to the left and

\text{mean} < \text{median} < \text{mode}

. A symmetric distribution has mean, median and mode coinciding. Recognising the direction of skew from the ordering of the averages, or from the length of the whiskers on a box plot, is a core interpretation skill.

Two coefficients quantify skewness. The Pearson coefficient of skewness is

\frac{3(\text{mean} - \text{median})}{\text{standard deviation}}

, and the quartile coefficient of skewness is

\frac{Q_3 - 2Q_2 + Q_1}{Q_3 - Q_1}

. Both are zero for a symmetric distribution, positive for right skew and negative for left skew; the sign is what carries the interpretation, and the magnitude indicates how pronounced the skew is.

Comparing two distributions demands a comparison of location, spread and shape, each supported by a figure. State which has the higher centre (comparing medians or means), which is more variable (comparing IQRs or standard deviations), and which is more skewed and in which direction. Crucially, comparisons must be made in the context of the data, not left as bare numbers.

The choice of summary follows the shape. For an approximately symmetric distribution, report the mean and standard deviation; for a skewed distribution, report the median and interquartile range, because the mean and standard deviation are distorted by the tail. Matching the summary to the shape, and justifying it, is a frequent AO2/AO3 demand.

\text{Sk} = \frac{3(\bar{x} - \text{median})}{\sigma}

Pearson coefficient of skewness

Positive for right skew, negative for left skew, zero for symmetry.

\text{Sk}_Q = \frac{Q_3 - 2Q_2 + Q_1}{Q_3 - Q_1}

Quartile coefficient of skewness

Uses only the quartiles, so it is robust to outliers.

Worked example

Quartile coefficient of skewness

A distribution has $Q_1 = 30$ , $Q_2 = 38$ and $Q_3 = 46$ . Find the quartile coefficient of skewness and interpret it.

Numerator
$Q_3 - 2Q_2 + Q_1 = 46 - 2(38) + 30 = 46 - 76 + 30 = 0$ .
Denominator
$Q_3 - Q_1 = 46 - 30 = 16$ .
Coefficient
$\text{Sk}_Q = \frac{0}{16} = 0$ .
$\text{Sk}_Q = \frac{0}{16} = 0$
Interpret
The quartiles are symmetric about the median, so the distribution shows no skew by this measure.

Result: The quartile coefficient of skewness is 0, indicating a symmetric middle half of the distribution.

Typical mistakes

Reading the direction of skew backwards; a right (positive) skew has the longer tail to the right.
Comparing distributions on only one feature (for example location) and ignoring spread and shape.

Active revision

A distribution has mean 52, median 55 and standard deviation 12. Calculate the Pearson coefficient of skewness and describe the skew.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content (Statistics) (Department for Education / Ofqual)

Histograms, box plots and cumulative frequency curves#

StandardPearson Edexcel 9ST0, Topic 1 (Paper 1)

Histogram with unequal class widths

Fig. 2Journey times: classes 0-10, 10-20, 20-40, 40-70 with frequencies 8, 15, 24, 9 give frequency densities 0.8, 1.5, 1.2, 0.3; the area of each bar is its frequency.

A histogram represents continuous grouped data with adjacent bars whose AREA is proportional to frequency. When classes have unequal widths, the bar height must be the frequency density

\frac{\text{frequency}}{\text{class width}}

, not the raw frequency, so that area encodes frequency correctly. The histogram below uses classes of widths 10, 10, 20 and 30 and therefore plots frequency density; the area of each bar equals the frequency in its class.

A cumulative frequency curve plots the cumulative frequency against the upper class boundary and rises to the total frequency. It is the tool for reading off the median (at half the total frequency), the quartiles (at a quarter and three quarters) and any percentile: go up to the required cumulative frequency, across to the curve, and down to the value axis. The curve for the same journey-time data is shown; the median is read at a cumulative frequency of 28.

A box plot displays the five-number summary and any outliers and is unrivalled for comparing distributions. Choosing the right diagram is itself examined: histograms for the shape of continuous data, box plots for comparison and outliers, cumulative frequency curves for medians, quartiles and percentiles, and scatter diagrams for the relationship between two variables.

Every diagram must be read honestly. A histogram bar's height is a density, not a frequency, so a tall narrow bar can hold fewer observations than a short wide one; a cumulative frequency read gives an estimate because grouping hides the exact values; and a box plot compresses the data to five numbers and can hide multimodality. Stating these limitations is part of a full interpretation.

\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

Frequency density (histogram)

The bar height for a histogram, so that area = frequency even with unequal class widths.

Cumulative frequency curve

Fig. 3Cumulative frequencies at the upper class boundaries; the median is read across from a cumulative frequency of 28 (half of 56).

Worked example

Estimating the median from a cumulative frequency curve

For the journey-time data with total frequency 56 and cumulative frequencies 8, 23, 47, 56 at 10, 20, 40, 70 minutes, estimate the median.

Locate the median position
The median is at cumulative frequency $\frac{56}{2} = 28$ .
Find the class
Cumulative frequency reaches 23 at 20 minutes and 47 at 40 minutes, so the median lies in the 20-40 class.
Interpolate
$20 + \frac{28 - 23}{47 - 23} \times (40 - 20) = 20 + \frac{5}{24} \times 20 = 20 + 4.17 = 24.2$ minutes.
$\text{median} \approx 20 + \frac{28 - 23}{24}\times 20 = 24.2\text{ min}$
Interpret
About half of the journeys take under 24 minutes; it is an estimate because grouping hides the exact times.

Result: The estimated median journey time is about 24.2 minutes (from linear interpolation in the 20-40 class).

Typical mistakes

Plotting raw frequency as the histogram bar height when class widths are unequal.
Plotting cumulative frequency against the midpoint rather than the upper class boundary.

Active revision

Using the journey-time data (classes 0-10, 10-20, 20-40, 40-70 with frequencies 8, 15, 24, 9), estimate the median from the cumulative frequency curve.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

References & sources

Sources

Pearson Edexcel

Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification

Department for Education / Ofqual

GCE AS and A level subject content (Statistics)

Numerical measures, graphs and diagrams

5 sections~14 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

Numerical measures, graphs and diagrams

Key points

Estimating a grouped mean

Key points

Standard deviation from summation quantities

Key points

Testing a value with the fence rule

Key points

Quartile coefficient of skewness

Key points

Estimating the median from a cumulative frequency curve

Sources

Numerical measures, graphs and diagrams

Key points

Estimating a grouped mean

Key points

Standard deviation from summation quantities

Key points

Testing a value with the fence rule

Key points

Quartile coefficient of skewness

Key points

Estimating the median from a cumulative frequency curve

Sources