GCSE Histograms and cumulative frequency Practice Questions

Free GCSE Histograms and cumulative frequency practice questions with full step-by-step worked solutions. Covers frequency density, class width, grouped data, area of a bar. Practise exam-style problems and check your method.

frequency densityclass widthgrouped dataarea of a barreading a histogramcumulative frequency
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
The table shows the waiting time, in minutes, of 160 customers at a post office. 0<t≤100 < t \le 10: 1010; 10<t≤2010 < t \le 20: 3030; 20<t≤3020 < t \le 30: 4040; 30<t≤4030 < t \le 40: 4040; 40<t≤5040 < t \le 50: 2020; 50<t≤7050 < t \le 70: 2020. Work out the frequency density for the class 30<t≤4030 < t \le 40.
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Worked solution

  1. Recall the formula for frequency density

    frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

    In a histogram the height of a bar is the frequency density, not the frequency. It is the frequency shared out over the width of the class.

  2. Work out the class width

    class width=40−30=10\text{class width} = 40 - 30 = 10

    The class 30<t≤4030 < t \le 40 runs from 3030 to 4040, so it is 1010 minutes wide.

  3. Divide the frequency by the class width

    frequency density=4010=4\text{frequency density} = \frac{40}{10} = 4

    There are 4040 customers in a class of width 1010, so the frequency density is 44.

Answer
frequency density=4\text{frequency density} = 4
Question 2
2 markseasy
The cumulative frequency graph shows the mass, in grams, of 140 letters posted in one day. Use the graph to estimate the number of letters with m≤60m \le 60.
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Worked solution

  1. Say what the vertical axis means

    cumulative frequency=running total\text{cumulative frequency} = \text{running total}

    A cumulative frequency of cc at m=xm = x means about cc letters have a value of xx or less.

  2. Read the cumulative frequency at 60

    m≤60:cumulative frequency=120m \le 60: \quad \text{cumulative frequency} = 120

    Going up from 6060 on the horizontal axis to the curve and across gives 120120.

  3. Subtract to get only the range asked for

    answer=120\text{answer} = 120

    The reading is already the answer.

Answer
estimate=120 letters\text{estimate} = 120 \text{ letters}
Question 3
2 marksintermediate
The cumulative frequency graph shows the daily rainfall, in mm, of 200 days at a weather station. Use the graph to estimate the number of days with 10<r≤3010 < r \le 30.
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Worked solution

  1. Say what the vertical axis means

    cumulative frequency=running total\text{cumulative frequency} = \text{running total}

    A cumulative frequency of cc at r=xr = x means about cc days have a value of xx or less.

  2. Read the cumulative frequency at 10

    r≤10:cumulative frequency=50r \le 10: \quad \text{cumulative frequency} = 50

    Going up from 1010 on the horizontal axis to the curve and across gives 5050.

  3. Read the cumulative frequency at 30

    r≤30:cumulative frequency=150r \le 30: \quad \text{cumulative frequency} = 150

    Going up from 3030 to the curve and across gives 150150 — that is everyone with r≤30r \le 30, including those already counted at 1010.

  4. Subtract to get only the range asked for

    150−50=100150 - 50 = 100

    Everybody with r≤10r \le 10 has to be taken away from everybody with r≤30r \le 30, leaving those in between.

  5. Note the mistake to avoid

    plot at the upper bound, not the midpoint\text{plot at the upper bound, not the midpoint}

    Cumulative frequency is plotted at the UPPER class boundary. Plotting it at the midpoint is the classic error — at the midpoint you have not yet counted the whole class.

  6. State the answer

    estimate=100 days\text{estimate} = 100 \text{ days}

    About 100100 days are in the range. It is an estimate because the data is grouped.

Answer
estimate=100 days\text{estimate} = 100 \text{ days}
Question 4
3 markshard
The histogram shows the mass, in grams, of some letters posted in one day. The classes do not all have the same width. Why must frequency density, and not frequency, be plotted on the vertical axis?
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Worked solution

  1. Write down the class widths

    widths:20−0=20,40−20=20,50−40=10,60−50=10,80−60=20,120−80=40\text{widths}: \quad 20 - 0 = 20, \quad 40 - 20 = 20, \quad 50 - 40 = 10, \quad 60 - 50 = 10, \quad 80 - 60 = 20, \quad 120 - 80 = 40

    The widths run from 1010 up to 4040, so they are not all the same.

  2. Imagine plotting frequency instead

    bar=frequency×class width\text{bar} = \text{frequency} \times \text{class width}

    If the HEIGHT of a bar were the frequency, then a class of width 4040 would get 40÷10=440 \div 10 = 4 times the area of a class of width 1010 with the same frequency. The picture would exaggerate the wide classes.

  3. Write down what frequency density does instead

    frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

    Dividing by the class width shares the frequency out over the class, so a wide class is not given extra height for being wide.

  4. Check what the area of a bar then becomes

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

    The class width cancels, so the AREA of each bar is its frequency. Areas can now be compared fairly whatever the widths are.

  5. Check that frequency density is not just the frequency

    heights:0.5,2.5,5,1,0.4,0.3\text{heights}: \quad 0.5, \quad 2.5, \quad 5, \quad 1, \quad 0.4, \quad 0.3

    The heights are not the frequencies, and they are not all whole numbers either, so any option claiming that is false.

  6. Check that the frequency densities do not add to 1

    0.5+2.5+5+1+0.4+0.3=9.7≠10.5 + 2.5 + 5 + 1 + 0.4 + 0.3 = 9.7 \ne 1

    Frequency density is not a probability. There is no reason for the heights to add up to anything in particular.

  7. Check the widths really are unequal

    10≠4010 \ne 40

    The class 40<m≤5040 < m \le 50 is 1010 wide and 80<m≤12080 < m \le 120 is 4040 wide, so any option that says the widths are equal is false.

  8. 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.

  9. Say what would go wrong on the picture

    80<m≤120:frequency=12,height=0.380 < m \le 120: \quad \text{frequency} = 12, \quad \text{height} = 0.3

    The class 80<m≤12080 < m \le 120 holds 1212 letters, but its bar is only 0.30.3 tall, because those letters are spread over 4040 grams. That is exactly the point of frequency density.

  10. State the answer

    unequal widths⇒plot frequency density so that area=frequency\text{unequal widths} \Rightarrow \text{plot frequency density so that area} = \text{frequency}

    Frequency density is used so that the area of each bar is its frequency, which makes classes of different widths comparable.

Answer
unequal widths⇒area=frequency\text{unequal widths} \Rightarrow \text{area} = \text{frequency}
Question 5
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.
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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}

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