Quartiles and box plots Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Quartiles and box plots questions. See exactly how to solve problems on quartiles, ordered data, lower quartile, median.

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.

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
1 markeasy
Here are the heights, in centimetres, of 1111 plants. 9,11,15,19,22,25,29,31,37,42,449, 11, 15, 19, 22, 25, 29, 31, 37, 42, 44 Work out the median.

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 median

    n+12=122=6\frac{n+1}{2} = \frac{12}{2} = 6

    With n=11n = 11, n+12=6\frac{n+1}{2} = 6. Position 66 is a whole number, so the quartile is simply the 66th value in the ordered list.

  3. Read off the median

    Q2=value at position 6=25Q_2 = \text{value at position } 6 = 25

    The median is 2525 centimetres: half of the plants are below this value and half are above it.

Answer
median=25 centimetres\text{median} = 25 \text{ centimetres}
Question 3
1 markeasy
Here are the masses, in grams, of 1515 apples. 17,23,26,31,36,43,47,55,61,65,70,73,75,79,8717, 23, 26, 31, 36, 43, 47, 55, 61, 65, 70, 73, 75, 79, 87 Work out the upper quartile.

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 upper quartile

    3(n+1)4=484=12\frac{3(n+1)}{4} = \frac{48}{4} = 12

    With n=15n = 15, 3(n+1)4=12\frac{3(n+1)}{4} = 12. Position 1212 is a whole number, so the quartile is simply the 1212th value in the ordered list.

  3. Read off the upper quartile

    Q3=value at position 12=73Q_3 = \text{value at position } 12 = 73

    The upper quartile is 7373 grams. Three quarters of the data lies below it, so only a quarter lies above it.

Answer
UQ=73 grams\text{UQ} = 73 \text{ grams}
Question 4
2 markseasy
Here are the marks, in marks, of 77 students. 16,20,23,27,31,33,3416, 20, 23, 27, 31, 33, 34 Work out the interquartile range.

Worked solution

  1. Work out the positions of the two quartiles

    n+14=2,3(n+1)4=6\frac{n+1}{4} = 2, \quad \frac{3(n+1)}{4} = 6

    With n=7n = 7 the lower quartile sits at position 22 and the upper quartile at position 66 in the ordered list.

  2. Read off the lower and upper quartiles

    Q1=20,Q3=33Q_1 = 20, \quad Q_3 = 33

    Counting to position 22 gives Q1=20Q_1 = 20 and counting to position 66 gives Q3=33Q_3 = 33.

  3. Work out the interquartile range

    IQR=Q3Q1=3320=13\text{IQR} = Q_3 - Q_1 = 33 - 20 = 13

    The interquartile range is the upper quartile minus the lower quartile: 3320=1333 - 20 = 13 marks. It is the width of the middle half of the data.

Answer
IQR=13 marks\text{IQR} = 13 \text{ marks}
Question 5
1 markeasy
Here are the ages, in years, of 99 members. 21,27,33,36,41,43,47,49,5021, 27, 33, 36, 41, 43, 47, 49, 50 Work out the range.

Worked solution

  1. Put the values in order of size, smallest first

    21,27,33,36,41,43,47,49,5021, 27, 33, 36, 41, 43, 47, 49, 50

    Quartiles are found by counting through the data in order, so the very first thing to do is sort it. Everything after this depends on the list being in order.

  2. Write down the smallest and the largest value

    smallest=21,largest=50\text{smallest} = 21, \quad \text{largest} = 50

    Once the list is in order the smallest value is the first, 2121, and the largest is the last, 5050.

  3. Work out the range

    range=5021=29\text{range} = 50 - 21 = 29

    The range is largest minus smallest: 5021=2950 - 21 = 29 years. It uses only the two most extreme values.

Answer
range=29 years\text{range} = 29 \text{ years}

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