Hard GCSE Histograms and cumulative frequency Questions

Challenging, exam-style GCSE Histograms and cumulative frequency questions with worked solutions. Stretch yourself on the hardest reading a histogram, area of a bar, proportion of a class, estimating from grouped data problems.

reading a histogramarea of a barproportion of a classestimating from grouped datahistogram with no vertical scalefrequency density
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
The histogram shows the height, in cm, of some plants in a greenhouse. The vertical axis has no numbers on it. There are 3030 plants in the class 15<h≤2015 < h \le 20. Work out the total number of plants.
Show worked solution

Worked solution

  1. Say what can still be read from the histogram

    no scale⇒compare areas\text{no scale} \Rightarrow \text{compare areas}

    With no numbers on the vertical axis, heights can still be compared against the gridlines. Call one gridline one unit of height.

  2. Recall that the frequency is the area of the bar

    frequency=frequency density×class width=area of the bar\text{frequency} = \text{frequency density} \times \text{class width} = \text{area of the bar}

    Frequency is proportional to area, so one known frequency fixes the scale of the whole histogram.

  3. Write down the class widths

    widths:10−0=10,15−10=5,20−15=5,25−20=5,30−25=5,40−30=10\text{widths}: \quad 10 - 0 = 10, \quad 15 - 10 = 5, \quad 20 - 15 = 5, \quad 25 - 20 = 5, \quad 30 - 25 = 5, \quad 40 - 30 = 10

    The classes are NOT all the same width, so equal heights do not mean equal frequencies.

  4. Read the height of each bar in units

    heights:3.5u,4u,6u,5u,3u,1.5u\text{heights}: \quad 3.5u, \quad 4u, \quad 6u, \quad 5u, \quad 3u, \quad 1.5u

    These are the heights measured against the gridlines, from left to right.

  5. Use the given frequency to find what one unit of area is worth

    6×5=30u=30⇒1u=16 \times 5 = 30u = 30 \Rightarrow 1u = 1

    The bar for 15<h≤2015 < h \le 20 has area 3030 units and represents 3030 plants, so one unit of area is 11 plants.

  6. Work out the frequency for the class 0 to 10

    0<h≤10:3.5×10=35u⇒350 < h \le 10: \quad 3.5 \times 10 = 35u \Rightarrow 35

    Area 3535 units, and one unit is worth 11 plants, so this class holds 3535 plants.

  7. Work out the frequency for the class 10 to 15

    10<h≤15:4×5=20u⇒2010 < h \le 15: \quad 4 \times 5 = 20u \Rightarrow 20

    Area 2020 units, and one unit is worth 11 plants, so this class holds 2020 plants.

  8. Work out the frequency for the class 15 to 20

    15<h≤20:6×5=30u⇒3015 < h \le 20: \quad 6 \times 5 = 30u \Rightarrow 30

    Area 3030 units, and one unit is worth 11 plants, so this class holds 3030 plants.

  9. Work out the frequency for the class 20 to 25

    20<h≤25:5×5=25u⇒2520 < h \le 25: \quad 5 \times 5 = 25u \Rightarrow 25

    Area 2525 units, and one unit is worth 11 plants, so this class holds 2525 plants.

  10. Work out the frequency for the class 25 to 30

    25<h≤30:3×5=15u⇒1525 < h \le 30: \quad 3 \times 5 = 15u \Rightarrow 15

    Area 1515 units, and one unit is worth 11 plants, so this class holds 1515 plants.

  11. Work out the frequency for the class 30 to 40

    30<h≤40:1.5×10=15u⇒1530 < h \le 40: \quad 1.5 \times 10 = 15u \Rightarrow 15

    Area 1515 units, and one unit is worth 11 plants, so this class holds 1515 plants.

  12. Add the frequencies

    35+20+30+25+15+15=14035 + 20 + 30 + 25 + 15 + 15 = 140

    Adding the six areas (converted to frequencies) gives the total.

  13. Note the mistake to avoid

    height≠frequency\text{height} \ne \text{frequency}

    The height of a bar is the frequency DENSITY, not the frequency. Reading the height as a frequency is the single most common error in this topic: you must multiply by the class width.

  14. Check the answer against the given class

    30140=314\frac{30}{140} = \frac{3}{14}

    The class 15<h≤2015 < h \le 20 was given as 3030 plants out of 140140. Its bar takes up the same share of the total AREA of the histogram, which is the consistency check that the scale was right.

  15. State the answer

    total=140 plants\text{total} = 140 \text{ plants}

    There are 140140 plants altogether.

Answer
140 plants140 \text{ plants}
Question 2
6 markschallenging
The histogram shows the waiting time, in minutes, of some customers at a post office. The vertical axis has no numbers on it. There are 4040 customers in the class 20<t≤3020 < t \le 30. Work out the total number of customers.
Show worked solution

Worked solution

  1. Say what can still be read from the histogram

    no scale⇒compare areas\text{no scale} \Rightarrow \text{compare areas}

    With no numbers on the vertical axis, heights can still be compared against the gridlines. Call one gridline one unit of height.

  2. Recall that the frequency is the area of the bar

    frequency=frequency density×class width=area of the bar\text{frequency} = \text{frequency density} \times \text{class width} = \text{area of the bar}

    Frequency is proportional to area, so one known frequency fixes the scale of the whole histogram.

  3. Write down the class widths

    widths:10−0=10,20−10=10,30−20=10,40−30=10,50−40=10,70−50=20\text{widths}: \quad 10 - 0 = 10, \quad 20 - 10 = 10, \quad 30 - 20 = 10, \quad 40 - 30 = 10, \quad 50 - 40 = 10, \quad 70 - 50 = 20

    The classes are NOT all the same width, so equal heights do not mean equal frequencies.

  4. Read the height of each bar in units

    heights:1u,3u,4u,4u,2u,1u\text{heights}: \quad 1u, \quad 3u, \quad 4u, \quad 4u, \quad 2u, \quad 1u

    These are the heights measured against the gridlines, from left to right.

  5. Use the given frequency to find what one unit of area is worth

    4×10=40u=40⇒1u=14 \times 10 = 40u = 40 \Rightarrow 1u = 1

    The bar for 20<t≤3020 < t \le 30 has area 4040 units and represents 4040 customers, so one unit of area is 11 customers.

  6. Work out the frequency for the class 0 to 10

    0<t≤10:1×10=10u⇒100 < t \le 10: \quad 1 \times 10 = 10u \Rightarrow 10

    Area 1010 units, and one unit is worth 11 customers, so this class holds 1010 customers.

  7. Work out the frequency for the class 10 to 20

    10<t≤20:3×10=30u⇒3010 < t \le 20: \quad 3 \times 10 = 30u \Rightarrow 30

    Area 3030 units, and one unit is worth 11 customers, so this class holds 3030 customers.

  8. Work out the frequency for the class 20 to 30

    20<t≤30:4×10=40u⇒4020 < t \le 30: \quad 4 \times 10 = 40u \Rightarrow 40

    Area 4040 units, and one unit is worth 11 customers, so this class holds 4040 customers.

  9. Work out the frequency for the class 30 to 40

    30<t≤40:4×10=40u⇒4030 < t \le 40: \quad 4 \times 10 = 40u \Rightarrow 40

    Area 4040 units, and one unit is worth 11 customers, so this class holds 4040 customers.

  10. Work out the frequency for the class 40 to 50

    40<t≤50:2×10=20u⇒2040 < t \le 50: \quad 2 \times 10 = 20u \Rightarrow 20

    Area 2020 units, and one unit is worth 11 customers, so this class holds 2020 customers.

  11. Work out the frequency for the class 50 to 70

    50<t≤70:1×20=20u⇒2050 < t \le 70: \quad 1 \times 20 = 20u \Rightarrow 20

    Area 2020 units, and one unit is worth 11 customers, so this class holds 2020 customers.

  12. Add the frequencies

    10+30+40+40+20+20=16010 + 30 + 40 + 40 + 20 + 20 = 160

    Adding the six areas (converted to frequencies) gives the total.

  13. Note the mistake to avoid

    height≠frequency\text{height} \ne \text{frequency}

    The height of a bar is the frequency DENSITY, not the frequency. Reading the height as a frequency is the single most common error in this topic: you must multiply by the class width.

  14. Check the answer against the given class

    40160=14\frac{40}{160} = \frac{1}{4}

    The class 20<t≤3020 < t \le 30 was given as 4040 customers out of 160160. Its bar takes up the same share of the total AREA of the histogram, which is the consistency check that the scale was right.

  15. State the answer

    total=160 customers\text{total} = 160 \text{ customers}

    There are 160160 customers altogether.

Answer
160 customers160 \text{ customers}
Question 3
6 markschallenging
The histogram shows the daily rainfall, in mm, of 200 days at a weather station. Estimate the percentage of days for which 20<r≤4520 < r \le 45.
Show worked solution

Worked solution

  1. Say what a percentage needs

    number with 20<r≤45200×100\frac{\text{number with } 20 < r \le 45}{200} \times 100

    A percentage needs two numbers: how many days are in the range, and how many there are altogether (200200).

  2. Recall that the frequency is the area of the bar

    frequency=frequency density×class width=area of the bar\text{frequency} = \text{frequency density} \times \text{class width} = \text{area of the bar}

    The number in the range is the AREA of the shaded region, so that area has to be found first.

  3. Write down the class widths

    widths:5−0=5,10−5=5,20−10=10,30−20=10,40−30=10,60−40=20\text{widths}: \quad 5 - 0 = 5, \quad 10 - 5 = 5, \quad 20 - 10 = 10, \quad 30 - 20 = 10, \quad 40 - 30 = 10, \quad 60 - 40 = 20

    The classes are not all the same width, so each frequency must be found as an area.

  4. Read the frequency density of each bar the range touches

    heights:5,3,1\text{heights}: \quad 5, \quad 3, \quad 1

    These are the bar heights, read off the vertical axis.

  5. Split the shaded region into one rectangle per bar

    5×(30−20)=5×10=503×(40−30)=3×10=301×(45−40)=1×5=5\begin{aligned} 5 \times (30 - 20) = 5 \times 10 = 50 \\ 3 \times (40 - 30) = 3 \times 10 = 30 \\ 1 \times (45 - 40) = 1 \times 5 = 5 \end{aligned}

    Each piece is the height of a bar times the width of the part of that class inside the range.

  6. Work out the part of the bar 20 to 30 that lies inside the range

    20 to 30:5×10=5020 \text{ to } 30: \quad 5 \times 10 = 50

    The whole bar 20<r≤3020 < r \le 30 lies inside the range, so all of its area counts: 5050 days.

  7. Work out the part of the bar 30 to 40 that lies inside the range

    30 to 40:3×10=3030 \text{ to } 40: \quad 3 \times 10 = 30

    The whole bar 30<r≤4030 < r \le 40 lies inside the range, so all of its area counts: 3030 days.

  8. Work out the part of the bar 40 to 60 that lies inside the range

    40 to 45:1×5=540 \text{ to } 45: \quad 1 \times 5 = 5

    The range covers only 4040 to 4545 of the class 40<r≤6040 < r \le 60, a width of 55. The bar has height 11, so the area of that piece is 1×5=51 \times 5 = 5 — the estimated number of days in it. This proportional part of the bar is the whole idea of the question.

  9. Add the areas to get the number in the range

    50+30+5=8550 + 30 + 5 = 85

    About 8585 days lie in the range.

  10. Write the number as a fraction of the total

    85200\frac{85}{200}

    There are 200200 days altogether, so this fraction of them lies in the range.

  11. Multiply by 100 to turn the fraction into a percentage

    85200×100=42.5\frac{85}{200} \times 100 = 42.5

    That gives 42.5%42.5\%.

  12. Say why the answer is an estimate

    grouped data⇒estimate\text{grouped data} \Rightarrow \text{estimate}

    The data is grouped, so the individual values are not known. The method assumes the values are spread evenly across each class, which is why the answer is an estimate and not an exact count.

  13. Note the mistake to avoid

    height≠frequency\text{height} \ne \text{frequency}

    The height of a bar is the frequency DENSITY, not the frequency. Reading the height as a frequency is the single most common error in this topic: you must multiply by the class width.

  14. Check the answer is sensible

    0<42.5<1000 < 42.5 < 100

    A percentage of a total has to lie between 00 and 100100, and it does.

  15. State the answer

    42.5%42.5\%

    About 42.5%42.5\% of the days have 20<r≤4520 < r \le 45.

Answer
42.5%42.5\%
Question 4
6 markschallenging
The histogram shows the waiting time, in minutes, of 160 customers at a post office. Estimate the percentage of customers for which 25<t≤4525 < t \le 45.
Show worked solution

Worked solution

  1. Say what a percentage needs

    number with 25<t≤45160×100\frac{\text{number with } 25 < t \le 45}{160} \times 100

    A percentage needs two numbers: how many customers are in the range, and how many there are altogether (160160).

  2. Recall that the frequency is the area of the bar

    frequency=frequency density×class width=area of the bar\text{frequency} = \text{frequency density} \times \text{class width} = \text{area of the bar}

    The number in the range is the AREA of the shaded region, so that area has to be found first.

  3. Write down the class widths

    widths:10−0=10,20−10=10,30−20=10,40−30=10,50−40=10,70−50=20\text{widths}: \quad 10 - 0 = 10, \quad 20 - 10 = 10, \quad 30 - 20 = 10, \quad 40 - 30 = 10, \quad 50 - 40 = 10, \quad 70 - 50 = 20

    The classes are not all the same width, so each frequency must be found as an area.

  4. Read the frequency density of each bar the range touches

    heights:4,4,2\text{heights}: \quad 4, \quad 4, \quad 2

    These are the bar heights, read off the vertical axis.

  5. Split the shaded region into one rectangle per bar

    4×(30−25)=4×5=204×(40−30)=4×10=402×(45−40)=2×5=10\begin{aligned} 4 \times (30 - 25) = 4 \times 5 = 20 \\ 4 \times (40 - 30) = 4 \times 10 = 40 \\ 2 \times (45 - 40) = 2 \times 5 = 10 \end{aligned}

    Each piece is the height of a bar times the width of the part of that class inside the range.

  6. Work out the part of the bar 20 to 30 that lies inside the range

    25 to 30:4×5=2025 \text{ to } 30: \quad 4 \times 5 = 20

    The range covers only 2525 to 3030 of the class 20<t≤3020 < t \le 30, a width of 55. The bar has height 44, so the area of that piece is 4×5=204 \times 5 = 20 — the estimated number of customers in it. This proportional part of the bar is the whole idea of the question.

  7. Work out the part of the bar 30 to 40 that lies inside the range

    30 to 40:4×10=4030 \text{ to } 40: \quad 4 \times 10 = 40

    The whole bar 30<t≤4030 < t \le 40 lies inside the range, so all of its area counts: 4040 customers.

  8. Work out the part of the bar 40 to 50 that lies inside the range

    40 to 45:2×5=1040 \text{ to } 45: \quad 2 \times 5 = 10

    The range covers only 4040 to 4545 of the class 40<t≤5040 < t \le 50, a width of 55. The bar has height 22, so the area of that piece is 2×5=102 \times 5 = 10 — the estimated number of customers in it. This proportional part of the bar is the whole idea of the question.

  9. Add the areas to get the number in the range

    20+40+10=7020 + 40 + 10 = 70

    About 7070 customers lie in the range.

  10. Write the number as a fraction of the total

    70160\frac{70}{160}

    There are 160160 customers altogether, so this fraction of them lies in the range.

  11. Multiply by 100 to turn the fraction into a percentage

    70160×100=43.75\frac{70}{160} \times 100 = 43.75

    That gives 43.75%43.75\%.

  12. Say why the answer is an estimate

    grouped data⇒estimate\text{grouped data} \Rightarrow \text{estimate}

    The data is grouped, so the individual values are not known. The method assumes the values are spread evenly across each class, which is why the answer is an estimate and not an exact count.

  13. Note the mistake to avoid

    height≠frequency\text{height} \ne \text{frequency}

    The height of a bar is the frequency DENSITY, not the frequency. Reading the height as a frequency is the single most common error in this topic: you must multiply by the class width.

  14. Check the answer is sensible

    0<43.75<1000 < 43.75 < 100

    A percentage of a total has to lie between 00 and 100100, and it does.

  15. State the answer

    43.75%43.75\%

    About 43.75%43.75\% of the customers have 25<t≤4525 < t \le 45.

Answer
43.75%43.75\%
Question 5
5 markschallenging
The histogram shows the height, in cm, of some plants in a greenhouse. Estimate the number of plants for which 5<h≤355 < h \le 35.
Show worked solution

Worked solution

  1. Say what is being asked for

    5<h≤355 < h \le 35

    The range from 55 to 3535 does not line up with the class boundaries, so part of at least one bar has to be taken. The shaded region is what has to be measured.

  2. Recall that the frequency is the area of the bar

    frequency=frequency density×class width=area of the bar\text{frequency} = \text{frequency density} \times \text{class width} = \text{area of the bar}

    The number of plants in a range is the AREA under the histogram over that range — never the height.

  3. Split the shaded region into one rectangle per bar

    3.5×(10−5)=3.5×5=17.54×(15−10)=4×5=206×(20−15)=6×5=305×(25−20)=5×5=253×(30−25)=3×5=151.5×(35−30)=1.5×5=7.5\begin{aligned} 3.5 \times (10 - 5) = 3.5 \times 5 = 17.5 \\ 4 \times (15 - 10) = 4 \times 5 = 20 \\ 6 \times (20 - 15) = 6 \times 5 = 30 \\ 5 \times (25 - 20) = 5 \times 5 = 25 \\ 3 \times (30 - 25) = 3 \times 5 = 15 \\ 1.5 \times (35 - 30) = 1.5 \times 5 = 7.5 \end{aligned}

    Each piece is a rectangle: the height of the bar times the width of the piece of that class that lies inside the range.

  4. Work out the part of the bar 0 to 10 that lies inside the range

    5 to 10:3.5×5=17.55 \text{ to } 10: \quad 3.5 \times 5 = 17.5

    The range covers only 55 to 1010 of the class 0<h≤100 < h \le 10, a width of 55. The bar has height 3.53.5, so the area of that piece is 3.5×5=17.53.5 \times 5 = 17.5 — the estimated number of plants in it. This proportional part of the bar is the whole idea of the question.

  5. Work out the part of the bar 10 to 15 that lies inside the range

    10 to 15:4×5=2010 \text{ to } 15: \quad 4 \times 5 = 20

    The whole bar 10<h≤1510 < h \le 15 lies inside the range, so all of its area counts: 2020 plants.

  6. Work out the part of the bar 15 to 20 that lies inside the range

    15 to 20:6×5=3015 \text{ to } 20: \quad 6 \times 5 = 30

    The whole bar 15<h≤2015 < h \le 20 lies inside the range, so all of its area counts: 3030 plants.

  7. Work out the part of the bar 20 to 25 that lies inside the range

    20 to 25:5×5=2520 \text{ to } 25: \quad 5 \times 5 = 25

    The whole bar 20<h≤2520 < h \le 25 lies inside the range, so all of its area counts: 2525 plants.

  8. Work out the part of the bar 25 to 30 that lies inside the range

    25 to 30:3×5=1525 \text{ to } 30: \quad 3 \times 5 = 15

    The whole bar 25<h≤3025 < h \le 30 lies inside the range, so all of its area counts: 1515 plants.

  9. Work out the part of the bar 30 to 40 that lies inside the range

    30 to 35:1.5×5=7.530 \text{ to } 35: \quad 1.5 \times 5 = 7.5

    The range covers only 3030 to 3535 of the class 30<h≤4030 < h \le 40, a width of 55. The bar has height 1.51.5, so the area of that piece is 1.5×5=7.51.5 \times 5 = 7.5 — the estimated number of plants in it. This proportional part of the bar is the whole idea of the question.

  10. Add the areas of the pieces

    17.5+20+30+25+15+7.5=11517.5 + 20 + 30 + 25 + 15 + 7.5 = 115

    The total shaded area is 115115, so about 115115 plants lie in the range.

  11. Express the estimate as a fraction of the whole data set

    115140\frac{115}{140}

    The estimate is 115115 out of 140140 plants — a useful sanity check that the answer is the right size.

  12. Say why the answer is an estimate

    grouped data⇒estimate\text{grouped data} \Rightarrow \text{estimate}

    The data is grouped, so the individual values are not known. The method assumes the values are spread evenly across each class, which is why the answer is an estimate and not an exact count.

  13. Note the mistake to avoid

    height≠frequency\text{height} \ne \text{frequency}

    The height of a bar is the frequency DENSITY, not the frequency. Reading the height as a frequency is the single most common error in this topic: you must multiply by the class width.

  14. Check the answer is sensible

    0<115<1400 < 115 < 140

    The estimate must lie between 00 and the total number of plants, 140140, and it does.

  15. State the answer

    estimate=115 plants\text{estimate} = 115 \text{ plants}

    About 115115 plants have 5<h≤355 < h \le 35.

Answer
estimate=115 plants\text{estimate} = 115 \text{ plants}

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