Continuous data — heights, times, masses — take too many different values to list individually, so they are collected into class intervals.
$20 \leq t \lt 25$ means "$20$ or more, but less than $25$".
A time of exactly $25$ goes in the next class, not this one.
| Class | Width | Midpoint |
|---|---|---|
| $0 \leq t \lt 10$ | $10$ | $5$ |
| $10 \leq t \lt 20$ | $10$ | $15$ |
| $20 \leq t \lt 40$ | $20$ | $30$ |
| $40 \leq t \lt 70$ | $30$ | $55$ |
Area of each bar $=$ frequency
- Add a "class width" column to the table.
- Add a "frequency density" column: frequency $\div$ class width.
- Draw the horizontal axis as a continuous scale.
- Label the vertical axis frequency density.
- Draw each bar spanning its class, at the height of its frequency density.
Complete the table and describe how to draw the histogram.
| Time $t$ (min) | Frequency | Class width | Frequency density |
|---|---|---|---|
| $0 \leq t \lt 10$ | $10$ | $10$ | $1$ |
| $10 \leq t \lt 20$ | $30$ | $10$ | $3$ |
| $20 \leq t \lt 40$ | $40$ | $20$ | $2$ |
| $40 \leq t \lt 70$ | $30$ | $30$ | $1$ |
Note that the class $20$–$40$ has the largest frequency ($40$) but not the tallest bar — because it is twice as wide.
A histogram bar spans $15 \leq x \lt 35$ with height $4.5$. How many values are in that class?
A histogram bar covers $20 \leq x \lt 40$ with frequency density $3$. Estimate how many values lie between $25$ and $40$.
- Add a cumulative frequency column and fill it with running totals.
- Plot each cumulative frequency against the upper class boundary.
- Start the curve at the lowest boundary with a cumulative frequency of $0$.
- Join the points with a smooth curve.
- The final point must reach the total frequency.
Complete the cumulative frequency column for these $60$ test scores.
| Score $s$ | Frequency | Cumulative frequency |
|---|---|---|
| $0 \leq s \lt 20$ | $4$ | $4$ |
| $20 \leq s \lt 40$ | $11$ | $15$ |
| $40 \leq s \lt 60$ | $22$ | $37$ |
| $60 \leq s \lt 80$ | $17$ | $54$ |
| $80 \leq s \lt 100$ | $6$ | $60$ |
Using the table above, estimate the median, the quartiles and the interquartile range.
Using the same data, estimate how many students scored more than $70$.
| Diagram | Best for | What you can read off |
|---|---|---|
| Histogram | Showing the shape of a distribution | Frequencies (as areas), modal class, skew |
| Cumulative frequency curve | Finding medians and quartiles | Median, $Q_1$, $Q_3$, IQR, "how many under $x$" |
| Frequency polygon | Comparing two distributions | General shape; plotted at midpoints |
A researcher has grouped data on journey times with unequal class widths and wants to (a) show the shape of the distribution, (b) find the median. Which diagram for each?
Frequency density
$\dfrac{\text{frequency}}{\text{class width}}$ — the height of a histogram bar.
Area
Area of a histogram bar $=$ frequency.
Frequency back
Frequency $=$ density $\times$ width.
Bars touch
Histograms have no gaps — the axis is continuous.
Part of a class
Density $\times$ the width of the part you want.
Cumulative frequency
A running total; plot at the upper boundary.
Median
Read across at $\dfrac{n}{2}$.
Quartiles
$\dfrac{n}{4}$ and $\dfrac{3n}{4}$; IQR $= Q_3 - Q_1$.
"More than"
Subtract the reading from the total.
Frequency polygon
Plot at midpoints, not boundaries.
A class $10 \leq x \lt 25$ has frequency $45$. Find the frequency density.
▶ Show solution
Class width $= 25 - 10 = 15$
Frequency density $= \dfrac{45}{15} = 3$
A histogram bar spans $30 \leq t \lt 50$ with height $6.5$. Find the frequency.
▶ Show solution
Width $= 20$
Frequency $= 6.5 \times 20 = 130$
Complete the frequency density column.
| Mass $m$ (g) | Frequency |
|---|---|
| $0 \leq m \lt 5$ | $8$ |
| $5 \leq m \lt 15$ | $30$ |
| $15 \leq m \lt 40$ | $25$ |
▶ Show solution
Widths: $5$, $10$, $25$.
Densities: $\dfrac{8}{5} = 1.6$; $\dfrac{30}{10} = 3$; $\dfrac{25}{25} = 1$
Explain why a histogram bar for a wide class can be shorter than one for a narrow class, even though it represents more data.
▶ Show solution
In a histogram the area represents the frequency, not the height.
A wide class spreads its frequency over a greater width, so the frequency density (and therefore the height) is lower — but the total area can still be larger.
Complete the cumulative frequency column for these $50$ values.
| Length $L$ (cm) | Frequency |
|---|---|
| $0 \leq L \lt 5$ | $7$ |
| $5 \leq L \lt 10$ | $13$ |
| $10 \leq L \lt 15$ | $19$ |
| $15 \leq L \lt 20$ | $11$ |
▶ Show solution
Running totals: $7$, $20$, $39$, $50$.
Plot at the upper boundaries: $(5, 7)$, $(10, 20)$, $(15, 39)$, $(20, 50)$.
The final value equals the total of $50$ ✓
Using the data from Question 5, state the values of cumulative frequency at which you would read the median, $Q_1$ and $Q_3$.
▶ Show solution
$n = 50$.
Median at $\dfrac{50}{2} = 25$
$Q_1$ at $\dfrac{50}{4} = 12.5$
$Q_3$ at $\dfrac{3 \times 50}{4} = 37.5$
A cumulative frequency curve for $200$ people gives a reading of $145$ at a height of $180$ cm. How many people are taller than $180$ cm?
▶ Show solution
$145$ people are shorter than $180$ cm.
Taller than $180$ cm: $200 - 145 = 55$ people.
A histogram bar covers $10 \leq x \lt 30$ with frequency density $4$. Estimate how many values lie between $10$ and $22$.
▶ Show solution
Width of the part wanted $= 22 - 10 = 12$
Frequency $= 4 \times 12 = 48$ values
This assumes the values are evenly spread across the class.
The histogram of $180$ journey times has these bars.
| Time $t$ (min) | Frequency density |
|---|---|
| $0 \leq t \lt 10$ | $2$ |
| $10 \leq t \lt 20$ | $5$ |
| $20 \leq t \lt 40$ | $3$ |
| $40 \leq t \lt 80$ | $?$ |
(a) Find the frequency of the first three classes. (b) Find the missing frequency density. (c) State the modal class.
▶ Show solution
(a) $2 \times 10 = 20$; $5 \times 10 = 50$; $3 \times 20 = 60$.
(b) So far $20 + 50 + 60 = 130$. The last class has $180 - 130 = 50$.
Width $= 40$, so density $= \dfrac{50}{40} = 1.25$.
(c) The modal class is the one with the highest frequency density, which is $10 \leq t \lt 20$ (density $5$).
Note: the class $20 \leq t \lt 40$ has the largest frequency ($60$), but the modal class in a histogram is judged by density.
The table shows the masses of $120$ parcels.
| Mass $m$ (kg) | Frequency |
|---|---|
| $0 \leq m \lt 2$ | $14$ |
| $2 \leq m \lt 4$ | $32$ |
| $4 \leq m \lt 6$ | $41$ |
| $6 \leq m \lt 8$ | $23$ |
| $8 \leq m \lt 10$ | $10$ |
(a) Build the cumulative frequency table. (b) Estimate the median. (c) Estimate the IQR. (d) Estimate how many parcels weigh more than $7$ kg.
▶ Show solution
(a) Running totals: $14$, $46$, $87$, $110$, $120$.
Plot at $(2, 14)$, $(4, 46)$, $(6, 87)$, $(8, 110)$, $(10, 120)$.
(b) Median at the $60$th value. This lies between cf $46$ (at $4$ kg) and cf $87$ (at $6$ kg).
Interpolating: $60 - 46 = 14$ into a class of $41$, so $4 + \dfrac{14}{41} \times 2 = 4 + 0.68 = \mathbf{4.7}$ kg (1 d.p.).
(c) $Q_1$ at the $30$th value: between cf $14$ (at $2$) and cf $46$ (at $4$).
$2 + \dfrac{16}{32} \times 2 = 3.0$ kg
$Q_3$ at the $90$th value: between cf $87$ (at $6$) and cf $110$ (at $8$).
$6 + \dfrac{3}{23} \times 2 = 6.3$ kg
IQR $\approx 6.3 - 3.0 = \mathbf{3.3}$ kg
(d) Reading the curve at $7$ kg: between cf $87$ (at $6$) and cf $110$ (at $8$), roughly $87 + \tfrac{1}{2}(23) = 98.5$, say $99$.
More than $7$ kg: $120 - 99 = \mathbf{21}$ parcels.