Hard GCSE Quartiles and box plots Questions

Challenging, exam-style GCSE Quartiles and box plots questions with worked solutions. Stretch yourself on the hardest quartiles, ordered data, interquartile range, median problems.

quartilesordered datainterquartile rangemedianlower quartileupper quartile
GCSE Higher34 questionsStep-by-step solutions
Question 1
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}
Question 2
5 markschallenging
For a set of data the smallest value is 88 and the range is 3434. The median is 2222. The upper quartile is 55 more than the median and the lower quartile is 66 less than the median. Work out the largest value.
Show worked solution

Worked solution

  1. Write down what the question gives you

    smallest=8,range=34,median=22\text{smallest} = 8, \quad \text{range} = 34, \quad \text{median} = 22

    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+5=22+5=27\text{UQ} = \text{median} + 5 = 22 + 5 = 27

    The upper quartile is 55 more than the median, so it is 22+5=2722 + 5 = 27 pounds.

  5. Work out the lower quartile

    LQ=median−6=22−6=16\text{LQ} = \text{median} - 6 = 22 - 6 = 16

    The lower quartile is 66 less than the median, so it is 22−6=1622 - 6 = 16 pounds.

  6. Work out the interquartile range

    IQR=27−16=11\text{IQR} = 27 - 16 = 11

    The interquartile range is the upper quartile minus the lower quartile: 27−16=1127 - 16 = 11 pounds. Notice it is also just 5+65 + 6 — the two distances from the median added together.

  7. Work out the largest value

    largest=smallest+range=8+34=42\text{largest} = \text{smallest} + \text{range} = 8 + 34 = 42

    Since range == largest −- smallest, the largest value is 8+34=428 + 34 = 42 pounds.

  8. Write out the five-number summary

    8, 16, 22, 27, 428, \ 16, \ 22, \ 27, \ 42

    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

    8≤16≤22≤27≤428 \le 16 \le 22 \le 27 \le 42

    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:16→27,whiskers:8 and 42\text{box}: 16 \rightarrow 27, \quad \text{whiskers}: 8 \text{ and } 42

    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

    34−11=2334 - 11 = 23

    The range is 3434 and the interquartile range is 1111, so the range exceeds the interquartile range by 2323 pounds.

  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

    34 vs 1134 \text{ vs } 11

    The range, 3434, stretches all the way to the extreme values, so a single freak reading changes it. The interquartile range, 1111, 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 55 above the median but the lower quartile is 66 below it.

  15. State the answer

    largest=42\text{largest} = 42

    So the largest value is 4242 pounds.

Answer
largest=42 pounds\text{largest} = 42 \text{ pounds}
Question 3
6 markschallenging
For a set of data the smallest value is 1616 and the range is 3737. The median is 3737. The upper quartile is 99 more than the median and the lower quartile is 1212 less than the median. Work out the interquartile range.
Show worked solution

Worked solution

  1. Write down what the question gives you

    smallest=16,range=37,median=37\text{smallest} = 16, \quad \text{range} = 37, \quad \text{median} = 37

    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=37+9=46\text{UQ} = \text{median} + 9 = 37 + 9 = 46

    The upper quartile is 99 more than the median, so it is 37+9=4637 + 9 = 46 points.

  5. Work out the lower quartile

    LQ=median−12=37−12=25\text{LQ} = \text{median} - 12 = 37 - 12 = 25

    The lower quartile is 1212 less than the median, so it is 37−12=2537 - 12 = 25 points.

  6. Work out the interquartile range

    IQR=46−25=21\text{IQR} = 46 - 25 = 21

    The interquartile range is the upper quartile minus the lower quartile: 46−25=2146 - 25 = 21 points. Notice it is also just 9+129 + 12 — the two distances from the median added together.

  7. Work out the largest value

    largest=smallest+range=16+37=53\text{largest} = \text{smallest} + \text{range} = 16 + 37 = 53

    Since range == largest −- smallest, the largest value is 16+37=5316 + 37 = 53 points.

  8. Write out the five-number summary

    16, 25, 37, 46, 5316, \ 25, \ 37, \ 46, \ 53

    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

    16≤25≤37≤46≤5316 \le 25 \le 37 \le 46 \le 53

    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:25→46,whiskers:16 and 53\text{box}: 25 \rightarrow 46, \quad \text{whiskers}: 16 \text{ and } 53

    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

    37−21=1637 - 21 = 16

    The range is 3737 and the interquartile range is 2121, so the range exceeds the interquartile range by 1616 points.

  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

    37 vs 2137 \text{ vs } 21

    The range, 3737, stretches all the way to the extreme values, so a single freak reading changes it. The interquartile range, 2121, 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 1212 below it.

  15. State the answer

    IQR=21\text{IQR} = 21

    So the interquartile range is 2121 points.

Answer
IQR=21 points\text{IQR} = 21 \text{ points}
Question 4
5 markschallenging
For a set of data the smallest value is 1717 and the range is 2727. The median is 3030. The upper quartile is 55 more than the median and the lower quartile is 33 less than the median. Work out the lower quartile.
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Worked solution

  1. Write down what the question gives you

    smallest=17,range=27,median=30\text{smallest} = 17, \quad \text{range} = 27, \quad \text{median} = 30

    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+5=30+5=35\text{UQ} = \text{median} + 5 = 30 + 5 = 35

    The upper quartile is 55 more than the median, so it is 30+5=3530 + 5 = 35 millimetres.

  5. Work out the lower quartile

    LQ=median−3=30−3=27\text{LQ} = \text{median} - 3 = 30 - 3 = 27

    The lower quartile is 33 less than the median, so it is 30−3=2730 - 3 = 27 millimetres.

  6. Work out the interquartile range

    IQR=35−27=8\text{IQR} = 35 - 27 = 8

    The interquartile range is the upper quartile minus the lower quartile: 35−27=835 - 27 = 8 millimetres. Notice it is also just 5+35 + 3 — the two distances from the median added together.

  7. Work out the largest value

    largest=smallest+range=17+27=44\text{largest} = \text{smallest} + \text{range} = 17 + 27 = 44

    Since range == largest −- smallest, the largest value is 17+27=4417 + 27 = 44 millimetres.

  8. Write out the five-number summary

    17, 27, 30, 35, 4417, \ 27, \ 30, \ 35, \ 44

    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

    17≤27≤30≤35≤4417 \le 27 \le 30 \le 35 \le 44

    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:27→35,whiskers:17 and 44\text{box}: 27 \rightarrow 35, \quad \text{whiskers}: 17 \text{ and } 44

    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

    27−8=1927 - 8 = 19

    The range is 2727 and the interquartile range is 88, so the range exceeds the interquartile range by 1919 millimetres.

  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

    27 vs 827 \text{ vs } 8

    The range, 2727, stretches all the way to the extreme values, so a single freak reading changes it. The interquartile range, 88, 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 55 above the median but the lower quartile is 33 below it.

  15. State the answer

    LQ=27\text{LQ} = 27

    So the lower quartile is 2727 millimetres.

Answer
LQ=27 millimetres\text{LQ} = 27 \text{ millimetres}
Question 5
5 markschallenging
For a set of data the smallest value is 44 and the range is 2929. The median is 1616. The upper quartile is 77 more than the median and the lower quartile is 55 less than the median. Work out the upper quartile.
Show worked solution

Worked solution

  1. Write down what the question gives you

    smallest=4,range=29,median=16\text{smallest} = 4, \quad \text{range} = 29, \quad \text{median} = 16

    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+7=16+7=23\text{UQ} = \text{median} + 7 = 16 + 7 = 23

    The upper quartile is 77 more than the median, so it is 16+7=2316 + 7 = 23 minutes.

  5. Work out the lower quartile

    LQ=median−5=16−5=11\text{LQ} = \text{median} - 5 = 16 - 5 = 11

    The lower quartile is 55 less than the median, so it is 16−5=1116 - 5 = 11 minutes.

  6. Work out the interquartile range

    IQR=23−11=12\text{IQR} = 23 - 11 = 12

    The interquartile range is the upper quartile minus the lower quartile: 23−11=1223 - 11 = 12 minutes. Notice it is also just 7+57 + 5 — the two distances from the median added together.

  7. Work out the largest value

    largest=smallest+range=4+29=33\text{largest} = \text{smallest} + \text{range} = 4 + 29 = 33

    Since range == largest −- smallest, the largest value is 4+29=334 + 29 = 33 minutes.

  8. Write out the five-number summary

    4, 11, 16, 23, 334, \ 11, \ 16, \ 23, \ 33

    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

    4≤11≤16≤23≤334 \le 11 \le 16 \le 23 \le 33

    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:11→23,whiskers:4 and 33\text{box}: 11 \rightarrow 23, \quad \text{whiskers}: 4 \text{ and } 33

    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

    29−12=1729 - 12 = 17

    The range is 2929 and the interquartile range is 1212, so the range exceeds the interquartile range by 1717 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

    29 vs 1229 \text{ vs } 12

    The range, 2929, stretches all the way to the extreme values, so a single freak reading changes it. The interquartile range, 1212, 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 77 above the median but the lower quartile is 55 below it.

  15. State the answer

    UQ=23\text{UQ} = 23

    So the upper quartile is 2323 minutes.

Answer
UQ=23 minutes\text{UQ} = 23 \text{ minutes}

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