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S4

Interpret, analyse and compare data-set distributions via graphs (incl. box plots), central tendency (median, mean, mode, modal class) and spread (range, outliers, quartiles, inter-quartile range)

Averages and spread

Worked answers, methods and verified real exam appearances for S4 on Edexcel GCSE Maths 1MA1.

Definition

Specification S4: interpret, analyse and compare the distributions of data sets from univariate empirical distributions through appropriate graphical representation involving discrete, continuous and grouped data, including box plots, and through appropriate measures of central tendency (median, mean, mode and modal class) and spread (range, including consideration of outliers, quartiles and inter-quartile range).

Explanation

  • A comparison scores when it names a measure of centre and a measure of spread, quotes the figures, and says what they mean in context. From a stem-and-leaf diagram, count to the middle value for the median and subtract the smallest from the largest for the range.
  • Higher tier: on a cumulative frequency graph read the median at half the total and the quartiles at a quarter and three quarters; the interquartile range Q3−Q1Q_3-Q_1 measures the middle half and ignores outliers.
  • A box plot shows both directly.
  • Estimate a mean from a grouped table with midpoints, ∑fx÷∑f\sum fx\div\sum f; with two known group means, multiply each by its size to recover totals.
  • Compare median with median and range with range.

Worked example

1

For a group of Year 9 girls, the least height is 152152 cm, the median height is 166166 cm and the greatest height is 176176 cm. The heights, in cm, of 1515 Year 9 boys are 156, 158, 161, 163, 165, 166, 169, 171, 172, 174, 175, 178, 179, 181, 183156,\ 158,\ 161,\ 163,\ 165,\ 166,\ 169,\ 171,\ 172,\ 174,\ 175,\ 178,\ 179,\ 181,\ 183. Compare the distribution of the girls' heights with the distribution of the boys' heights.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.The boys' median is the 88th of the 1515 ordered values: 171171 cm.
  2. 2.The boys' range is 183−156=27183-156=27 cm; the girls' range is 176−152=24176-152=24 cm.
  3. 3.Compare centres: the boys' median height of 171171 cm is greater than the girls' 166166 cm, so the boys are typically taller.
  4. 4.Compare spreads: the boys' range of 2727 cm is greater than the girls' 2424 cm, so the boys' heights are more spread out.

Answer: Boys' median 171171 cm and range 2727 cm; the boys are typically taller than the girls and their heights are more varied.

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for identification of the median (171) or the range (27) of the boys, or the range (24) of the girlsB1
for a correct comparison of medians and a correct comparison of ranges supported by correct figures, eg the median height for boys (171) is greater than the median for girls (166) and the range for boys (27) is greater than the range for girls (24)C2At least one comparison must be in context. (C1 for a correct comparison of medians or of ranges that could ft their figures)

Common mistakes

  • Don't fall into the trap of giving comparison figures without stating what they mean in context.
  • Don't fall into the trap of finding the median before putting raw data in order.

Mark-scheme wording to use

  • “At least one comparison must be in context referring to height or quoting cm. Simply quoting values for median, range is insufficient; they must be compared” — Pearson 1MA1, June 2020 Paper 1H, Q7
  • “B1 for correctly identifying the median of (School) A as 57 C1 (ft, dep on value for the median being stated) for making a correct comparison of median eg ‘the median of (School) A is higher than the median of (School) B’” — Pearson 1MA1, November 2024 Paper 3H, Q6
  • “for a first step in the process … for a complete process eg (50 × 167.6 - 20 × 182) ÷ 30” — Pearson 1MA1, November 2022 Paper 3H, Q8

What the examiner wants

Stated values and a direction. The November 2024 Paper 3H report says the median of 5757 had to be clearly stated, not just circled on the stem-and-leaf diagram, and that a comparison must say which group is greater; simply giving the difference between the values was not enough.

Exam tip

A full comparison usually needs both a typical-value statement and a spread statement.

Worked practice

Q1
Tier 1 · Easy

1

For the data 4,6,6,9,104,6,6,9,10, work out the mean and the range.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • Mean =7=7; range =6=6.
2The total is 4+6+6+9+10=354+6+6+9+10=35, so the mean is 35/5=735/5=7. The range is 10−4=610-4=6.
Q2
Tier 2 · Standard

2

Data set A has median 2424 and range 3333. Data set B has median 2525 and range 2929. Compare the two distributions.

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • B has the slightly higher median: 2525 compared with 2424.
  • B has the smaller range: 2929 compared with 3333, so its values are less spread out overall.
2Compare the typical values using the medians, then compare spread using the ranges. B has both the larger centre and the smaller spread.
Q3
Tier 3 · Hard

3

The data are 12,13,13,14,15,15,16,5212,13,13,14,15,15,16,52. Work out the mean, median and range. Decide which of the mean or median better describes a typical value for these data. Give a reason.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • Mean =18.75=18.75, median =14.5=14.5, range =40=40.
  • The median is more representative because 5252 is an outlier that pulls up the mean.
4The total is 150150, giving mean 150/8=18.75150/8=18.75. The median is (14+15)/2=14.5(14+15)/2=14.5 and the range is 52−12=4052-12=40. The isolated value 5252 has a strong effect on the mean but not on the median, so the median better represents the main cluster.
Q4
Tier 1 · Easy

4

For the data 2,4,4,6,9,10,122,4,4,6,9,10,12, write down the mode and the median.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • Mode =4=4; median =6=6.
244 occurs most often, so it is the mode. The data are ordered and the fourth of the seven values is 66, so the median is 66.
Q5
Tier 2 · Standard

5

Five readings have mean 1818. A sixth reading of 2424 is added. Work out the new mean and explain why it is greater than 1818.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • New mean =19=19.
  • The added reading is greater than the original mean, so it raises the mean.
3The original total is 5×18=905\times18=90. The new total is 90+24=11490+24=114, so the new mean is 114/6=19114/6=19. Since 2424 is above the original mean, the mean increases.
Q6
Tier 3 · Hard

6

The ordered data are 6,9,11,x,18,206,9,11,x,18,20. Their mean is 1313. Work out xx, the median and the range.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • x=14x=14; median =12.5=12.5; range =14=14.
4The total of all six values is 6×13=786\times13=78. The known values total 6+9+11+18+20=646+9+11+18+20=64, so x=78−64=14x=78-64=14. The median is the mean of the third and fourth values: (11+14)/2=12.5(11+14)/2=12.5. The range is 20−6=1420-6=14.
Q7
Tier 2 · Standard

7

A data set has mean 1212, median 1111 and range 99. Every value in the data set is doubled and then 33 is added. Work out the mean, median and range of the new data set.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • Mean =27=27; median =25=25; range =18=18.
3The mean and median both undergo the same transformation as every value: 2×12+3=272\times12+3=27 and 2×11+3=252\times11+3=25. Doubling every value doubles the range, while adding the same amount to every value does not change the range, so the new range is 2×9=182\times9=18.
Q8
Tier 3 · Hard

8

Seven integer values are written in order of size: 3,7,a,11,b,16,203,7,a,11,b,16,20. Their mean is 1111 and their only mode is 77. Work out aa, bb and the range.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • a=7a=7, b=13b=13 and the range is 1717.
4For 77 to be the only mode, it must occur again, so a=7a=7. A mean of 1111 for seven values gives a total of 7777. The known values, including a=7a=7, total 6464, so b=77−64=13b=77-64=13. This is consistent with the stated order. The range is 20−3=1720-3=17.
Q9
Tier 3 · Hard

9

The values 44, 77, 1010 and 1313 have frequencies 22, 55, kk and 11 respectively. The mean is 88. Work out kk, the median, the mode and the range.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • k=4k=4, median =7=7, mode =7=7 and range =9=9.
4The total frequency is 8+k8+k and the total of the values is 2(4)+5(7)+10k+13=56+10k2(4)+5(7)+10k+13=56+10k. Therefore 56+10k=8(8+k)56+10k=8(8+k), so 2k=82k=8 and k=4k=4. There are 1212 values; the sixth and seventh are both 77, so the median is 77. The greatest frequency is 55, so the mode is 77, and the range is 13−4=913-4=9.
Q10
Tier 3 · Hard

10

Higher only: Nine values are written in order: 4,7,9,12,15,18,x,26,314,7,9,12,15,18,x,26,31. The upper quartile is 2424. The median is excluded when the lower and upper halves are formed. Work out xx, the lower quartile and the interquartile range.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • x=22x=22, lower quartile =8=8 and interquartile range =16=16.
4After excluding the median 1515, the upper half is 18,x,26,3118,x,26,31. Its median is (x+26)/2=24(x+26)/2=24, so x=22x=22, which is consistent with the stated order. The lower half is 4,7,9,124,7,9,12, so the lower quartile is (7+9)/2=8(7+9)/2=8. The interquartile range is 24−8=1624-8=16.

Verified exam appearances

SeriesPaperQuestionMarksCalculatorTierLinks
2019-112FQ96AllowedFoundationQPMS
2023-113FQ224AllowedFoundationQPMS
2019-113HQ43AllowedHigherQPMS
2019-061FQ73Non-calculatorFoundationQPMS
2019-061HQ114Non-calculatorHigherQPMS
2024-062HQ117AllowedHigherQPMS
2024-112FQ75AllowedFoundationQPMS
2024-063HQ115AllowedHigherQPMS
2023-111FQ214Non-calculatorFoundationQPMS
2022-061HQ83Non-calculatorHigherQPMS
2019-112HQ113AllowedHigherQPMS
2024-113FQ154AllowedFoundationQPMS
2022-111FQ73Non-calculatorFoundationQPMS
2023-111FQ86Non-calculatorFoundationQPMS
2023-113HQ34AllowedHigherQPMS
2022-062HQ105AllowedHigherQPMS
2022-113HQ83AllowedHigherQPMS
2024-111HQ96Non-calculatorHigherQPMS
2023-063FQ96AllowedFoundationQPMS
2019-063HQ33AllowedHigherQPMS
2022-112FQ155AllowedFoundationQPMS
2024-063FQ72AllowedFoundationQPMS
2022-062FQ154AllowedFoundationQPMS
2022-061HQ106Non-calculatorHigherQPMS
2023-111FQ11Non-calculatorFoundationQPMS
2021-112HQ94AllowedHigherQPMS
2023-111HQ44Non-calculatorHigherQPMS
2019-063FQ263AllowedFoundationQPMS
2019-113HQ123AllowedHigherQPMS
2023-061FQ122Non-calculatorFoundationQPMS
2022-062FQ31AllowedFoundationQPMS
2023-062FQ174AllowedFoundationQPMS
2022-112FQ173AllowedFoundationQPMS
2023-112HQ153AllowedHigherQPMS
2019-113FQ132AllowedFoundationQPMS
2024-113FQ224AllowedFoundationQPMS
2019-111HQ106Non-calculatorHigherQPMS
2019-111HQ63Non-calculatorHigherQPMS
2019-113FQ253AllowedFoundationQPMS
2024-061HQ83Non-calculatorHigherQPMS
2024-061FQ154Non-calculatorFoundationQPMS
2022-062FQ183AllowedFoundationQPMS
2024-113HQ64AllowedHigherQPMS
2021-111HQ123Non-calculatorHigherQPMS
2022-113FQ83AllowedFoundationQPMS
2023-063FQ124AllowedFoundationQPMS
2022-063FQ144AllowedFoundationQPMS
2022-112HQ113AllowedHigherQPMS
2019-111FQ253Non-calculatorFoundationQPMS
2022-113FQ283AllowedFoundationQPMS
2019-063FQ183AllowedFoundationQPMS
2024-062FQ284AllowedFoundationQPMS

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