Challenge · Histograms

Calculator allowed13 marks

These problems use frequency density both ways (FD $=$ frequency ÷ width, and frequency $=$ FD × width $=$ area) and estimating within a class. A worked example shows how to start; hints are at the foot. Calculator allowed.

Worked example: how to start
On a histogram, the bar for the class $10
Class width $=30-10=20$.
Frequency $=$ FD $\times$ width $=2.4\times20$.
$=48$ values.
1A histogram bar for the class $15[2 marks]
Answer:
2A class $10[2 marks]
Answer:
3A histogram has two bars: class $0[3 marks]
Answer:
4The class $10[3 marks]
Answer:
5Class A is $20[3 marks]
Answer:
★Extension: Two classes have the SAME frequency. One has width $8$ and frequency density $3$. The other has width $12$. Work out its frequency density.
Stuck? Hints (don't peek unless you need to)1. Frequency $=$ FD $\times$ class width.2. Frequency density $=$ frequency ÷ class width.3. Work out each frequency (FD × width), then add.4. Find the frequency of the whole class, then take the fraction $\dfrac{5}{20}$ of it (or use FD × the smaller width).5. Work out the class B frequency (FD × width), then compare with $60$.

Solutions & mark scheme · Histograms

Total: 13 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1A histogram bar for the class $15[2]
Model solution
Width $=35-15=20$.
Frequency $=2.3\times20=46$.
Answer: $46$
Marks
✔1Width $=20$
✔1$2.3\times20=46$
2A class $10[2]
Model solution
Width $=16-10=6$.
$\text{FD}=\dfrac{21}{6}=3.5$.
Answer: $3.5$
Marks
✔1Width $=6$
✔1$21\div6=3.5$
3A histogram has two bars: class $0[3]
Model solution
First class: $1.5\times20=30$.
Second class: $0.8\times30=24$.
Total $=30+24=54$.
Answer: $54$
Marks
✔1First frequency $=30$
✔1Second frequency $=24$
✔1Total $=54$
4The class $10[3]
Model solution
The part $10$ to $15$ has width $5$.
Estimate $=$ FD $\times$ width $=2.4\times5$.
$=12$ values.
Answer: $12$
Marks
✔1Uses the part-width $5$
✔1$2.4\times5$
✔1$=12$
5Class A is $20[3]
Model solution
Class B frequency $=4\times(50-40)=4\times10=40$.
Class A has $60$, class B has $40$.
Class A has more, by $60-40=20$.
Answer: $\text{A, by }20$
Marks
✔1Class B frequency $=40$
✔1Compares $60$ and $40$
✔1A, by $20$
★Extension
First frequency $=3\times8=24$.
Same frequency, so the second has frequency $24$ over width $12$.
$\text{FD}=\dfrac{24}{12}=2$.
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