Revision mat · Histograms

Calculator allowed13 marks

Question mix weighted by 64 real exam questions from 41 GCSE papers (2017–24), so the most common question types get the most space. Show your working.

Frequency density $\text{FD}=\dfrac{\text{frequency}}{\text{class width}}$Frequency $\text{frequency}=\text{FD}\times\text{class width}$Area = frequency $\text{area of a bar}=\text{frequency}$Class width $\text{width}=\text{upper}-\text{lower}$Total frequency $\textstyle\sum(\text{FD}\times\text{width})$
Interpret≈61% of real exam Qs
1The histogram shows the height of some plants.
Estimate how many had a height between 30 and 40 cm.
01230510152025303540Height, h (cm)Frequency density
Answer: (2 marks)
2The histogram shows the speed of some cars.
Estimate how many had a speed between 10 and 60 mph.
0120102030405060Speed, s (mph)Frequency density
Answer: (3 marks)
Frequency density≈27% of real exam Qs
3A histogram is to be drawn for this data.
Work out the height (frequency density) of the bar for the class $40 < w \le 50$.
Weight, w (kg)Frequency
0 < w ≤ 2030
20 < w ≤ 3025
30 < w ≤ 4020
40 < w ≤ 5030
Answer: (2 marks)
4A histogram is to be drawn for this data.
Work out the height (frequency density) of the bar for the class $30 < s \le 40$.
Speed, s (mph)Frequency
0 < s ≤ 105
10 < s ≤ 3030
30 < s ≤ 405
40 < s ≤ 6020
Answer: (2 marks)
★ Exam capstonemixed & other forms ≈13% of real exam Qs
5The histogram shows the height of 90 plants.
Calculate an estimate for the mean height.
Give your answer to 3 significant figures.
01230102030405060Height, h (cm)Frequency density
Answer: (4 marks)

Mark scheme · Histograms

Total: 13 marks

Award the marks shown for each correct step, then add up the total out of 13. A method mark counts even if the final answer is wrong.

1The histogram shows the height of some plants.
Estimate how many had a height between 30 and 40 cm.[2 marks]
Method
Read the frequency density of that bar: 0.5. Class width $= 10$.
Frequency $= 0.5 \times 10 = 5$.
Answer: $5$
Marks
✔1 markfrequency density × class width
✔1 mark5
2The histogram shows the speed of some cars.
Estimate how many had a speed between 10 and 60 mph.[3 marks]
Method
For each bar in the range, frequency = frequency density × width.
Frequencies: 40 + 5 + 30 = 75.
Answer: $75$
Marks
✔1 markfrequency density × width for each bar
✔1 markAdd the bars in the range
✔1 mark75
3A histogram is to be drawn for this data.
Work out the height (frequency density) of the bar for the class $40 < w \le 50$.
Weight, w (kg)Frequency
0 < w ≤ 2030
20 < w ≤ 3025
30 < w ≤ 4020
40 < w ≤ 5030
[2 marks]
Method
Frequency density $= \dfrac{\text{frequency}}{\text{class width}} = \dfrac{30}{10}$.
The bar height is $3$.
Answer: $3$
Marks
✔1 markfrequency ÷ class width
✔1 mark3
4A histogram is to be drawn for this data.
Work out the height (frequency density) of the bar for the class $30 < s \le 40$.
Speed, s (mph)Frequency
0 < s ≤ 105
10 < s ≤ 3030
30 < s ≤ 405
40 < s ≤ 6020
[2 marks]
Method
Frequency density $= \dfrac{\text{frequency}}{\text{class width}} = \dfrac{5}{10}$.
The bar height is $0.5$.
Answer: $0.5$
Marks
✔1 markfrequency ÷ class width
✔1 mark0.5
5The histogram shows the height of 90 plants.
Calculate an estimate for the mean height.
Give your answer to 3 significant figures.[4 marks]
Method
Frequency of each bar = frequency density × width: 20, 50, 10, 10. Total $= 90$.
Use the class midpoints: $\sum(\text{midpoint} \times \text{frequency}) = 5\times20 + 20\times50 + 35\times10 + 50\times10 = 1950$.
Mean $= 1950 \div 90 = 21.7$ (3 s.f.).
Answer: $21.7$
Marks
✔1 markfrequency of each bar from the histogram
✔1 markΣ(midpoint × frequency) = 1950
✔1 marktotal frequency = 90
✔1 mark21.7
mathedup.co.uk · sheet A8DF