GCSE Quartiles and box plots Practice Questions

Free GCSE Quartiles and box plots practice questions with full step-by-step worked solutions. Covers quartiles, ordered data, lower quartile, median. Practise exam-style problems and check your method.

quartilesordered datalower quartilemedianupper quartileinterquartile range
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Here are the times, in seconds, of 1111 runners. 6,9,13,15,18,22,26,30,31,34,386, 9, 13, 15, 18, 22, 26, 30, 31, 34, 38 Work out the lower quartile.
Show worked solution

Worked solution

  1. Write down the rule for the quartile positions

    LQ at position n+14,median at position n+12,UQ at position 3(n+1)4\text{LQ at position } \frac{n+1}{4}, \quad \text{median at position } \frac{n+1}{2}, \quad \text{UQ at position } \frac{3(n+1)}{4}

    For a list of raw data the lower quartile sits at position n+14\frac{n+1}{4}, the median at position n+12\frac{n+1}{2} and the upper quartile at position 3(n+1)4\frac{3(n+1)}{4}. These are POSITIONS in the ordered list, not values.

  2. Work out the position of the lower quartile

    n+14=124=3\frac{n+1}{4} = \frac{12}{4} = 3

    With n=11n = 11, n+14=3\frac{n+1}{4} = 3. Position 33 is a whole number, so the quartile is simply the 33th value in the ordered list.

  3. Read off the lower quartile

    Q1=value at position 3=13Q_1 = \text{value at position } 3 = 13

    Counting to position 33 in the ordered list gives a lower quartile of 1313 seconds. A quarter of the data lies below it.

Answer
LQ=13 seconds\text{LQ} = 13 \text{ seconds}
Question 2
2 markseasy
The box plots show the waiting times, in minutes, of the patients in Kelby and in Lyndon. Compare the two distributions. Which statement is correct?
Show worked solution

Worked solution

  1. Read the median off each box plot

    medianKelby=20,medianLyndon=22\text{median}_{\text{Kelby}} = 20, \quad \text{median}_{\text{Lyndon}} = 22

    The median is the line inside each box: 2020 for Kelby and 2222 for Lyndon.

  2. Work out each interquartile range

    IQRKelby=25−18=7,IQRLyndon=27−17=10\text{IQR}_{\text{Kelby}} = 25 - 18 = 7, \quad \text{IQR}_{\text{Lyndon}} = 27 - 17 = 10

    The interquartile range is the width of the box: 77 for Kelby and 1010 for Lyndon.

  3. State the answer

    median 20<22 and IQR 7<10\text{median } 20 < 22 \text{ and IQR } 7 < 10

    So the correct comparison is: The median for Kelby (2020) is lower than the median for Lyndon (2222), so Kelby generally had lower waiting times. The interquartile range for Kelby (77) is smaller than the interquartile range for Lyndon (1010), so the waiting times for Kelby are more consistent.

Answer
median 20<22 and IQR 7<10\text{median } 20 < 22 \text{ and IQR } 7 < 10
Question 3
2 marksintermediate
The box plots show the scores, in points, of the players in Oakley and in Preston. Compare the two distributions. Which statement is correct?
Show worked solution

Worked solution

  1. Read the median off each box plot

    medianOakley=17,medianPreston=14\text{median}_{\text{Oakley}} = 17, \quad \text{median}_{\text{Preston}} = 14

    The median is the line inside each box: 1717 for Oakley and 1414 for Preston.

  2. Compare the medians

    17>1417 > 14

    The median for Oakley is higher than the median for Preston, so on average the scores for Oakley were higher. This is the comparison of CENTRE — one of the two things the question is marked on.

  3. Read the quartiles off each box plot

    Oakley:Q1=15, Q3=22;Preston:Q1=11, Q3=20\text{Oakley}: Q_1 = 15, \ Q_3 = 22; \quad \text{Preston}: Q_1 = 11, \ Q_3 = 20

    The quartiles are the two edges of each box — the left edge is the lower quartile and the right edge is the upper quartile.

  4. Work out each interquartile range

    IQROakley=22−15=7,IQRPreston=20−11=9\text{IQR}_{\text{Oakley}} = 22 - 15 = 7, \quad \text{IQR}_{\text{Preston}} = 20 - 11 = 9

    The interquartile range is the width of the box: 77 for Oakley and 99 for Preston.

  5. Compare the interquartile ranges

    7<97 < 9

    The interquartile range for Oakley is smaller than the interquartile range for Preston. This is the comparison of SPREAD — the second thing the question is marked on.

  6. State the answer

    median 17>14 and IQR 7<9\text{median } 17 > 14 \text{ and IQR } 7 < 9

    So the correct comparison is: The median for Oakley (1717) is higher than the median for Preston (1414), so Oakley generally had higher scores. The interquartile range for Oakley (77) is smaller than the interquartile range for Preston (99), so the scores for Oakley are more consistent.

Answer
median 17>14 and IQR 7<9\text{median } 17 > 14 \text{ and IQR } 7 < 9
Question 4
4 markshard
For a set of data the smallest value is 66 and the range is 3030. The median is 1515. The upper quartile is 88 more than the median and the lower quartile is 66 less than the median. Work out the interquartile range.
Show worked solution

Worked solution

  1. Write down what the question gives you

    smallest=6,range=30,median=15\text{smallest} = 6, \quad \text{range} = 30, \quad \text{median} = 15

    A box plot needs five numbers. The question gives two of them directly and describes the other three, so each one has to be built up in turn.

  2. Recall what the interquartile range is

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    The interquartile range is the upper quartile minus the lower quartile — the width of the box on the box plot.

  3. Work out the upper quartile

    UQ=median+8=15+8=23\text{UQ} = \text{median} + 8 = 15 + 8 = 23

    The upper quartile is 88 more than the median, so it is 15+8=2315 + 8 = 23 years.

  4. Work out the lower quartile

    LQ=median−6=15−6=9\text{LQ} = \text{median} - 6 = 15 - 6 = 9

    The lower quartile is 66 less than the median, so it is 15−6=915 - 6 = 9 years.

  5. Work out the interquartile range

    IQR=23−9=14\text{IQR} = 23 - 9 = 14

    The interquartile range is the upper quartile minus the lower quartile: 23−9=1423 - 9 = 14 years. Notice it is also just 8+68 + 6 — the two distances from the median added together.

  6. Recall what the range is

    range=largest−smallest\text{range} = \text{largest} - \text{smallest}

    The range is the distance from the smallest value to the largest, so it can be rearranged to give the largest value once the smallest and the range are known.

  7. Work out the largest value

    largest=smallest+range=6+30=36\text{largest} = \text{smallest} + \text{range} = 6 + 30 = 36

    Since range == largest −- smallest, the largest value is 6+30=366 + 30 = 36 years.

  8. Check the five numbers are in order

    6≤9≤15≤23≤366 \le 9 \le 15 \le 23 \le 36

    The five-number summary must never decrease. It does not, so the numbers are consistent with a real box plot.

  9. Note the mistake to avoid

    IQR≠12×range\text{IQR} \ne \frac{1}{2} \times \text{range}

    The interquartile range is not half the range, and it is not symmetric about the median either — here the upper quartile is 88 above the median but the lower quartile is 66 below it.

  10. State the answer

    IQR=14\text{IQR} = 14

    So the interquartile range is 1414 years.

Answer
IQR=14 years\text{IQR} = 14 \text{ years}
Question 5
6 markschallenging
For a set of data the smallest value is 77 and the range is 3838. The median is 2828. The upper quartile is 99 more than the median and the lower quartile is 1111 less than the median. Work out the difference between the range and the interquartile range.
Show worked solution

Worked solution

  1. Write down what the question gives you

    smallest=7,range=38,median=28\text{smallest} = 7, \quad \text{range} = 38, \quad \text{median} = 28

    A box plot needs five numbers. The question gives two of them directly and describes the other three, so each one has to be built up in turn.

  2. Recall what the range is

    range=largest−smallest\text{range} = \text{largest} - \text{smallest}

    The range is the distance from the smallest value to the largest, so it can be rearranged to give the largest value once the smallest and the range are known.

  3. Recall what the interquartile range is

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    The interquartile range is the upper quartile minus the lower quartile — the width of the box on the box plot.

  4. Work out the upper quartile

    UQ=median+9=28+9=37\text{UQ} = \text{median} + 9 = 28 + 9 = 37

    The upper quartile is 99 more than the median, so it is 28+9=3728 + 9 = 37 minutes.

  5. Work out the lower quartile

    LQ=median−11=28−11=17\text{LQ} = \text{median} - 11 = 28 - 11 = 17

    The lower quartile is 1111 less than the median, so it is 28−11=1728 - 11 = 17 minutes.

  6. Work out the interquartile range

    IQR=37−17=20\text{IQR} = 37 - 17 = 20

    The interquartile range is the upper quartile minus the lower quartile: 37−17=2037 - 17 = 20 minutes. Notice it is also just 9+119 + 11 — the two distances from the median added together.

  7. Work out the largest value

    largest=smallest+range=7+38=45\text{largest} = \text{smallest} + \text{range} = 7 + 38 = 45

    Since range == largest −- smallest, the largest value is 7+38=457 + 38 = 45 minutes.

  8. Write out the five-number summary

    7, 17, 28, 37, 457, \ 17, \ 28, \ 37, \ 45

    Smallest, lower quartile, median, upper quartile, largest: the complete five-number summary the box plot is drawn from.

  9. Check the five numbers are in order

    7≤17≤28≤37≤457 \le 17 \le 28 \le 37 \le 45

    The five-number summary must never decrease. It does not, so the numbers are consistent with a real box plot.

  10. Draw the box plot

    box:17→37,whiskers:7 and 45\text{box}: 17 \rightarrow 37, \quad \text{whiskers}: 7 \text{ and } 45

    With all five numbers known the box plot can be drawn: box from the lower to the upper quartile, median line inside it, whiskers out to the extremes.

  11. Work out the difference between the range and the interquartile range

    38−20=1838 - 20 = 18

    The range is 3838 and the interquartile range is 2020, so the range exceeds the interquartile range by 1818 minutes.

  12. Say what the range and the interquartile range each measure

    range=all of the data,IQR=middle 50%\text{range} = \text{all of the data}, \quad \text{IQR} = \text{middle } 50\%

    The range measures the total spread, from the smallest value to the largest. The interquartile range measures only the spread of the middle half, so the two answer different questions.

  13. Explain why the interquartile range ignores the extremes

    38 vs 2038 \text{ vs } 20

    The range, 3838, stretches all the way to the extreme values, so a single freak reading changes it. The interquartile range, 2020, throws away the top and bottom quarters, so it is not affected by extremes at all.

  14. Note the mistake to avoid

    IQR≠12×range\text{IQR} \ne \frac{1}{2} \times \text{range}

    The interquartile range is not half the range, and it is not symmetric about the median either — here the upper quartile is 99 above the median but the lower quartile is 1111 below it.

  15. State the answer

    range−IQR=18\text{range} - \text{IQR} = 18

    So the difference between the range and the interquartile range is 1818 minutes.

Answer
range−IQR=18 minutes\text{range} - \text{IQR} = 18 \text{ minutes}

Unlock 65 more Quartiles and box plots questions

Create a free account to work through every GCSE Quartiles and box plots question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Quartiles and box plots practice

Hard QuestionsWorked Solutions

Related Statistics topics

Averages and spreadCharts and tablesComparing distributionsHistograms and cumulative frequencySamplingScatter graphs and correlation