Histograms GCSE Questions, Worked Examples and Answers

$27-15=\boxed{12}$

$24\div8=\boxed{3}$

$12\times2.5=\boxed{30}$

$15\div5=3$

$20\div10=2$

$30\div20=1.5$

$\boxed{3,\ 2,\ 1.5}$

Use the class boundaries for bar widths and the calculated densities for heights: $3$, $2$ and $1.5$. The bars touch because the variable is continuous.

Frequency is width × density.

$5(2)=10$

$10(3)=30$

$20(1)=20$

$15(2)=30$

$\boxed{10,\ 30,\ 20,\ 30}$

$10+30+20+30=\boxed{90}$

$\text{frequency A}=5\times4=20$

$\text{frequency B}=20\times1.5=30$

$\boxed{\text{Bar B represents the greater frequency.}}$

$\text{missing frequency}=90-(18+27+25)$

$=90-70=20$

$\text{density}=20\div10=\boxed{2}$

The known density is

$12\div8=1.5$

So $3$ cm represents $1.5$ density units, or $2$ cm per density unit.

The second density is

$5\div2=2.5$

$\text{frequency}=6\times2.5=\boxed{15}$

The selected width is $10$ out of the full width $30$.

$\frac{10}{30}\times42=14$

$\boxed{14\text{ values (estimated)}}$

The frequencies are the bar areas.

$10(1)=10,\qquad10(3)=30,\qquad20(1)=20$

The class midpoints are $5$, $15$ and $30$.

$\text{estimated mean}=\frac{5(10)+15(30)+30(20)}{60}$

$=\frac{1100}{60}=18.333\ldots$

$\boxed{18.3\text{ (estimated)}}$

There are $60$ values, so use the $30$th value.

The first class contains $10$, leaving $20$ values to move through the second class of frequency $30$.

$10+\frac{20}{30}\times10=16.666\ldots$

$\boxed{16.7\text{ (estimated)}}$

Height represents frequency density, not frequency. Frequency depends on both width and height:

$\text{frequency}=\text{class width}\times\text{frequency density}$

A shorter but much wider bar can therefore contain more data.

For $0<x\leq10$,

$\text{frequency}=10\times1.8=18$

For $10<x\leq25$, the width is $15$.

$\text{density}=36\div15=2.4$

For $25<x\leq45$, the width is $20$.

$\text{frequency}=20\times1.3=26$

$\text{total}=18+36+26=\boxed{80}$

The missing entries are $\boxed{18,\ 2.4,\ 26}$.