L1StatisticsCore

Representing Data

A well-chosen diagram makes the shape of a data set visible in a way a table never does. Three appear in the specification — histograms, box plots and cumulative frequency curves — and each answers a different question about the data.

35 min Data Presentation

The big picture

The one idea worth carrying away is that in a histogram it is $\textbf{area}$ , not height, that represents frequency. That is not a technicality invented to catch people out: it is what allows classes of different widths to be compared honestly on the same axes. Plotting frequency as height when the classes are unequal exaggerates the wide ones and is genuinely misleading — the same trick that appears in badly drawn newspaper graphics. Once frequency density is second nature, histograms stop being fiddly and the rest of the chapter, including estimating from a diagram, follows straightforwardly.

What you'll be able to do

Draw and interpret histograms using frequency density
Estimate frequencies from a histogram, including for part of a class
Draw and compare box plots, identifying outliers by the $1.5\times\text{IQR}$ rule
Use cumulative frequency curves to estimate the median, quartiles and percentiles

Histograms and frequency density

A histogram represents frequency by $\textbf{area}$ . When class widths differ, plotting raw frequency as height would make a class twice as wide look twice as important for the same count — so the vertical axis carries $\textbf{frequency density}$ instead.

Bars touch, because the underlying variable is continuous. That is the visible difference from a bar chart, which has gaps and is for categorical data.

To read a frequency back off a histogram, multiply the height by the width. To estimate the frequency in part of a class, take the corresponding fraction of that bar’s area — which assumes an even spread within the class, the same assumption used for grouped means.

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

The vertical axis of a histogram.

\text{frequency} = \text{area of bar}

Height times width.

Example 1A class

20 \leq x < 40

has frequency 30. Find its frequency density.

1Class width

= 40 - 20 = 20

2Density

= \dfrac{30}{20}

Answer

1.5

Example 2On a histogram, the class

10 \leq x < 30

has frequency density

2.4

. Estimate the number of values between 15 and 30.

1Total frequency of the class:

2.4 \times 20 = 48

2The part from 15 to 30 is 15 wide out of 20, so three-quarters of the class.

\tfrac{15}{20}\times48 = 36

AnswerAbout 36 values.

Tip — Label the vertical axis "frequency density", not "frequency". Mislabelling loses a mark even when the bars are drawn correctly.

Box plots and outliers

A box plot shows a five-figure summary: minimum, $Q_{1}$ , median, $Q_{3}$ , maximum. The box spans the interquartile range with the median marked inside it, and whiskers reach out to the extremes.

An $\textbf{outlier}$ is conventionally any value more than $1.5\times\text{IQR}$ beyond a quartile. Outliers are plotted as separate points and the whisker is drawn only as far as the most extreme value that is not an outlier.

Because a box plot compresses a distribution into five numbers, two of them side by side make comparison very quick — which is exactly what exam questions ask for.

x < Q_{1} - 1.5\,\text{IQR} \;\text{or}\; x > Q_{3} + 1.5\,\text{IQR}

The standard outlier rule.

Example 1A data set has

Q_{1} = 24

and

Q_{3} = 40

. Determine whether 68 is an outlier.

1IQR

= 40 - 24 = 16

2Upper boundary:

40 + 1.5(16) = 40 + 24 = 64

368 exceeds 64.

AnswerYes — 68 lies beyond the upper fence of 64, so it is an outlier.

Why this works

The $1.5$ is a convention, not a law of nature. Some questions specify a different multiplier, or define an outlier as more than two standard deviations from the mean — always use the rule the question gives rather than the one you remember.

Cumulative frequency

A cumulative frequency curve plots the running total of frequencies against the $\textbf{upper class boundary}$ — not the midpoint, which is the standard error here. Points are joined with a smooth curve.

Reading across from a cumulative frequency and down to the axis estimates a value: use $\tfrac{n}{2}$ for the median, $\tfrac{n}{4}$ for $Q_{1}$ , $\tfrac{3n}{4}$ for $Q_{3}$ , and $\tfrac{pn}{100}$ for the $p$ th percentile.

Reading the other way — up from a value and across — estimates how many observations fall below it, which is how "how many scored under 50?" questions are answered.

Example 1A cumulative frequency curve is drawn for 80 values. Which cumulative frequencies locate the quartiles and the median?

Q_{1}

\tfrac{80}{4} = 20

2Median:

\tfrac{80}{2} = 40

Q_{3}

\tfrac{3\times80}{4} = 60

AnswerRead across at cumulative frequencies 20, 40 and 60.

Tip — Plot cumulative frequency against the $\textbf{upper}$ boundary of each class. Using midpoints shifts the whole curve and every estimate taken from it.

Reading shape and skew

Diagrams reveal $\textbf{skew}$ , which the averages alone can suggest but not confirm. In a positively skewed distribution the tail stretches to the right, and the mean is pulled above the median. Negative skew is the mirror image, with the tail and the mean to the left.

On a box plot, skew shows as an off-centre median: closer to $Q_{1}$ means positive skew, closer to $Q_{3}$ means negative. A roughly symmetric distribution has the median near the middle of the box with whiskers of similar length.

The quartile test states this numerically, and it is the version to quote when a question asks you to justify the skew rather than describe it.

Q_{3}-Q_{2} > Q_{2}-Q_{1} \Rightarrow \text{positive skew}

The longer half sits above the median.

\text{mean} > \text{median} \Rightarrow \text{positive skew}

Consistent with the tail direction.

Example 1A distribution has

Q_{1} = 12

Q_{2} = 15

Q_{3} = 25

. Describe the skew.

1Lower half:

Q_{2}-Q_{1} = 3

2Upper half:

Q_{3}-Q_{2} = 10

3The upper half is much longer.

AnswerPositive skew — the tail extends towards higher values.

Why this works

Positive skew is common in real data bounded below by zero — incomes, waiting times, rainfall. There is a floor but no ceiling, so the tail can only run one way.

Think like an examiner

Histogram axes must be frequency density, and bars must touch. Both are credited.
Plot cumulative frequency at upper class boundaries, never midpoints.
State the outlier rule you are applying and show the fence calculation.
When comparing two box plots, compare location and spread and interpret in context — not just "A is bigger".

Common misconceptions

Plotting frequency as the height of a histogram with unequal classes.

Height must be frequency density, so that area represents frequency.

Leaving gaps between histogram bars.

The variable is continuous, so bars touch. Gaps belong to bar charts.

Plotting cumulative frequency against class midpoints.

Use upper class boundaries — the running total is complete only at the top of each class.

Drawing the whisker out to an outlier.

The whisker stops at the most extreme non-outlier; outliers are marked separately.

Diagrams

\text{f.d.} = \tfrac{\text{frequency}}{\text{width}}

Histogram height.

\text{frequency} = \text{area}

Reading a histogram back.

Q_{1}-1.5\,\text{IQR},\; Q_{3}+1.5\,\text{IQR}

Outlier fences.

\tfrac{n}{4},\ \tfrac{n}{2},\ \tfrac{3n}{4}

Cumulative frequency positions.

Stretch yourself

A histogram has classes $0\!-\!10$ (frequency density 1.2), $10\!-\!20$ (3.5) and $20\!-\!50$ (0.8). Find the total frequency, and estimate how many values lie between 15 and 35.

Hint — Frequency is area. For the partial classes, take the matching fraction of each bar.

Reveal solution

SolutionFrequencies:

1.2 \times 10 = 12

;

3.5 \times 10 = 35

;

0.8 \times 30 = 24

. Total

= 12 + 35 + 24 = 71

. For 15 to 35: from the second class, the portion 15–20 is half its width, giving

\tfrac{5}{10}\times35 = 17.5

. From the third class, 20–35 is 15 of its 30 width, giving

\tfrac{15}{30}\times24 = 12

. Estimated total

= 17.5 + 12 = 29.5

, so about 30 values. The answer is an estimate because it assumes values are spread evenly within each class — the third class is 30 wide, so that assumption is doing a lot of work there and the true figure could differ noticeably.

\blacksquare

Questions students ask

Why must histograms use frequency density?

So that area represents frequency. With unequal class widths, plotting raw frequency as height makes wide classes look disproportionately large — a genuinely misleading picture.

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

Histograms show continuous data with touching bars and area representing frequency. Bar charts show categorical data with gaps, and height represents frequency directly.

Why plot cumulative frequency at upper boundaries?

Because the running total is only complete once the whole class has been counted, which happens at its upper boundary. Plotting at the midpoint claims the total is reached halfway through.

Is the

1.5\times\text{IQR}

rule the only definition of an outlier?

No. Some questions use two standard deviations from the mean, or a different multiplier. Always apply the definition the question gives.

Key takeaways

In a histogram, area represents frequency, so the height is frequency density.
Bars touch because the data is continuous.
Box plots show a five-figure summary; outliers lie beyond $1.5\times\text{IQR}$ from a quartile.
Cumulative frequency is plotted at upper class boundaries and read across for quartiles.
Skew is visible from an off-centre median and confirmed by comparing quartile gaps.