Mathematics 9709 · Probability & Statistics 1

Histograms

A histogram is not just a bar chart for grouped data.
In a histogram, area is frequency.
The height is frequency density: how crowded the class is, not just how many values it holds.

Area = frequency

When classes have different widths, frequency alone is not enough.
A tall narrow bar and a short wide bar can hold the same frequency because frequency is area.
So the height must be frequency density, and histogram bars must touch.

\text{frequency density}=\dfrac{\text{frequency}}{\text{class width}}\qquad\Longrightarrow\qquad \text{frequency}=\text{width}\times\text{density}=\text{area}

If labels are rounded, convert each class to true boundaries (± half the rounding unit).
width = upper boundary − lower boundary (not the label span).
frequency density = frequency ÷ width.
Plot density (label the axis “frequency density”); bars touch.
Reverse: a part-class count is its proportional area.

100 children. The second bar is SHORTER (density 3 vs 4) yet holds MORE children (area 60 vs 40) — proof that height is not the count.

Worked example

Start simple: equal labels, unequal widths

100 children: $40\le m<50$ has 40; $50\le m<70$ has 60. Widths $10,\ 20$ ; densities $40\div10=4$ and $60\div20=3$ ; areas $40,\ 60$ , total 100. The wider class is shorter yet holds more.

Answer

\text{densities } 4 \text{ and } 3

Plot frequency density (= frequency ÷ width), never raw frequency. Each bar's area IS its frequency.

Worked example

Find a missing frequency, then the densities

690 medical samples: $4\le m<12$ (224), $12\le m<24$ (396), $24\le m<28$ ( $p$ ). Find $p$ and the three frequency densities.

Frequencies add to the total: $p=690-224-396=70$ .
These labels are already true boundaries (the $\le\,/\,<$ form), so widths are $8,\ 12,\ 4$ .
Densities: $224\div8=28$ , $396\div12=33$ , $70\div4=17.5$ .

Answer

p=70;\quad \text{densities } 28,\ 33,\ 17.5

True boundaries: the classic trap

Rounded labels hide the gaps. Times to the nearest minute labelled $14\text{–}15$ mean $13.5\le t<15.5$ — width $2$ , not 1. Always rebuild the boundaries before any width.

Worked example

Rounded to the nearest minute

36 athletes (race time to the nearest minute): $13$ has $f=4$ ; $14\text{–}15$ has 14; $16\text{–}18$ has 18. True boundaries $12.5,\ 13.5,\ 15.5,\ 18.5$ → widths $1,\ 2,\ 3$ , densities $4,\ 7,\ 6$ . The last class has the most athletes but not the tallest bar ( $7>6$ ). Area check $1{\times}4+2{\times}7+3{\times}6=36$ .

Answer

\text{densities } 4,\ 7,\ 6

Rounded labels hide half a unit each side: use TRUE boundaries for the width, then density = frequency ÷ width.

Worked example

Harder rounding: nearest 0.1 min and nearest 50 years

The same ±half-unit rule scales to the rounding unit. (a) Times to the nearest $0.1$ min labelled $2.6\text{–}2.8$ mean $2.55\le t<2.85$ — width $0.3$ , not 0.2. (b) Ages to the nearest 50 years labelled $50\text{–}150$ mean $25\le A<175$ — width 150 (each boundary shifts $\pm25$ ).

Answer

\text{nearest }0.1\Rightarrow \pm0.05;\quad \text{nearest }50\Rightarrow \pm25

Common mistake

Using the label width (14–15 → width 1) instead of true boundaries (width 2), plotting raw frequency as the height, or labelling the axis “frequency”. And “tallest bar = most data” is false — a wider, shorter class can hold more (area, not height).

Reading a count back off

Frequency is area, so read counts back by finding an area.
For part of a class, take the same fraction of the bar area. This assumes the values are spread evenly inside that class.
On a drawn histogram, you can also count blocks of grid squares and divide by how many blocks make one item.

36 athletes again. Two squares (blocks) make one athlete, so any region can be counted by blocks ÷ 2 — the dashed line marks 13.0 min.

Worked example

Count athletes back off the histogram

Use the 36-athlete histogram (densities 4, 7, 6 over widths 1, 2, 3). Estimate (a) those who took less than 13.0 min, (b) those between 14.5 and 17.5 min, (c) the time band of the slowest 3 athletes.

(a) $12.5\le t<13.5$ holds 4; $<13.0$ is half of it: $\tfrac12\times4=2$ athletes.
(b) From 14.5 to 15.5 is $\tfrac12$ of the width-2 class: $\tfrac12\times14=7$ ; from 15.5 to 17.5 is $\tfrac23$ of the width-3 class: $\tfrac23\times18=12$ . Total $7+12=19$ .
(c) The slowest 3 are the top of the last class. $3$ of its 18 over width 3, working in from 18.5: they lie between $18.0$ and $18.5$ min.

Answer

(a)\ 2,\quad (b)\ 19,\quad (c)\ 18.0\text{ to }18.5\text{ min}

Worked example

Estimate a part-class count across two classes

Back to the 100 children ( $40\text{–}50$ holds 40, $50\text{–}70$ holds 60). Estimate the number with masses between 45 and 63 kg.

Split at the boundary: 45–50, then 50–63.
45–50 is half of the first class: $\tfrac{5}{10}\times40=20$ .
50–63 is $\tfrac{13}{20}$ of the second (= width × density = $13\times3$ ): $\tfrac{13}{20}\times60=39$ .
Add: $20+39=59$ .

Answer

\approx 59 \text{ children}

Counts come back as AREA; for part of a class take the proportional area. Split at any class boundary first, then add the pieces.

Examiner note

Show the frequency-density row (freq ÷ width) explicitly — that line earns the method mark; an estimate should show the proportional-area fraction (e.g.

\tfrac{13}{20}\times60

) before the number.

PreviousTypes of data & stem-and-leaf NextCumulative frequency graphs

Histograms

A histogram is not just a bar chart for grouped data.
In a histogram, area is frequency.
The height is frequency density: how crowded the class is, not just how many values it holds.

Area = frequency

When classes have different widths, frequency alone is not enough.
A tall narrow bar and a short wide bar can hold the same frequency because frequency is area.
So the height must be frequency density, and histogram bars must touch.

\text{frequency density}=\dfrac{\text{frequency}}{\text{class width}}\qquad\Longrightarrow\qquad \text{frequency}=\text{width}\times\text{density}=\text{area}

If labels are rounded, convert each class to true boundaries (± half the rounding unit).
width = upper boundary − lower boundary (not the label span).
frequency density = frequency ÷ width.
Plot density (label the axis “frequency density”); bars touch.
Reverse: a part-class count is its proportional area.

100 children. The second bar is SHORTER (density 3 vs 4) yet holds MORE children (area 60 vs 40) — proof that height is not the count.

Worked example

Start simple: equal labels, unequal widths

100 children: $40\le m<50$ has 40; $50\le m<70$ has 60. Widths $10,\ 20$ ; densities $40\div10=4$ and $60\div20=3$ ; areas $40,\ 60$ , total 100. The wider class is shorter yet holds more.

Answer

\text{densities } 4 \text{ and } 3

Plot frequency density (= frequency ÷ width), never raw frequency. Each bar's area IS its frequency.

Worked example

Find a missing frequency, then the densities

690 medical samples: $4\le m<12$ (224), $12\le m<24$ (396), $24\le m<28$ ( $p$ ). Find $p$ and the three frequency densities.

Frequencies add to the total: $p=690-224-396=70$ .
These labels are already true boundaries (the $\le\,/\,<$ form), so widths are $8,\ 12,\ 4$ .
Densities: $224\div8=28$ , $396\div12=33$ , $70\div4=17.5$ .

Answer

p=70;\quad \text{densities } 28,\ 33,\ 17.5

True boundaries: the classic trap

Rounded labels hide the gaps. Times to the nearest minute labelled $14\text{–}15$ mean $13.5\le t<15.5$ — width $2$ , not 1. Always rebuild the boundaries before any width.

Worked example

Rounded to the nearest minute

Answer

\text{densities } 4,\ 7,\ 6

Rounded labels hide half a unit each side: use TRUE boundaries for the width, then density = frequency ÷ width.

Worked example

Harder rounding: nearest 0.1 min and nearest 50 years

Answer

\text{nearest }0.1\Rightarrow \pm0.05;\quad \text{nearest }50\Rightarrow \pm25

Common mistake

Reading a count back off

Frequency is area, so read counts back by finding an area.
For part of a class, take the same fraction of the bar area. This assumes the values are spread evenly inside that class.
On a drawn histogram, you can also count blocks of grid squares and divide by how many blocks make one item.

36 athletes again. Two squares (blocks) make one athlete, so any region can be counted by blocks ÷ 2 — the dashed line marks 13.0 min.

Worked example

Count athletes back off the histogram

Use the 36-athlete histogram (densities 4, 7, 6 over widths 1, 2, 3). Estimate (a) those who took less than 13.0 min, (b) those between 14.5 and 17.5 min, (c) the time band of the slowest 3 athletes.

(a) $12.5\le t<13.5$ holds 4; $<13.0$ is half of it: $\tfrac12\times4=2$ athletes.
(b) From 14.5 to 15.5 is $\tfrac12$ of the width-2 class: $\tfrac12\times14=7$ ; from 15.5 to 17.5 is $\tfrac23$ of the width-3 class: $\tfrac23\times18=12$ . Total $7+12=19$ .
(c) The slowest 3 are the top of the last class. $3$ of its 18 over width 3, working in from 18.5: they lie between $18.0$ and $18.5$ min.

Answer

(a)\ 2,\quad (b)\ 19,\quad (c)\ 18.0\text{ to }18.5\text{ min}

Worked example

Estimate a part-class count across two classes

Back to the 100 children ( $40\text{–}50$ holds 40, $50\text{–}70$ holds 60). Estimate the number with masses between 45 and 63 kg.

Split at the boundary: 45–50, then 50–63.
45–50 is half of the first class: $\tfrac{5}{10}\times40=20$ .
50–63 is $\tfrac{13}{20}$ of the second (= width × density = $13\times3$ ): $\tfrac{13}{20}\times60=39$ .
Add: $20+39=59$ .

Answer

\approx 59 \text{ children}

Counts come back as AREA; for part of a class take the proportional area. Split at any class boundary first, then add the pieces.

Examiner note

Show the frequency-density row (freq ÷ width) explicitly — that line earns the method mark; an estimate should show the proportional-area fraction (e.g.

\tfrac{13}{20}\times60

) before the number.