Cambridge International AS & A Level Mathematics 9709: Representation of data – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 5.1, Representation of data, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). It is AS Level content, examined on Paper 5. A scientific calculator is allowed in all 9709 papers; show every step. Give non-exact answers to 3 significant figures.

They practise every 5.1 outcome. Revise first with the Probability & Statistics 1 study guide and revision notes. The topic-level practice questions cover the rest of Paper 5.

Questions

1. A café records the waiting time, to the nearest second, of 250 customers.

(a) State, with a reason, whether a stem-and-leaf diagram or a histogram is more suitable. [1] (b) Give one advantage of a box-and-whisker plot over a histogram when comparing these times with those of a second café. [1] (c) Give one advantage of a stem-and-leaf diagram over a box-and-whisker plot. [1]

2. The times, in seconds, taken by two groups of 15 people to solve a puzzle are shown.

        Group P |   | Group Q
          7 4 1 | 2 |
    8 6 5 3 2 0 | 3 | 0 7
        7 4 3 1 | 4 | 0 1 2 3 5 6 7 8 9
            8 2 | 5 | 1 3 8
                | 6 | 4
Key: 4 | 3 | 5 means 34 seconds for Group P and 35 seconds for Group Q

(a) Find the median and the interquartile range for each group. Take each quartile as the median of one half of the data, excluding the overall median. [4] (b) Make two comparisons between the times of the two groups. [2] (c) State the values you would plot to draw box-and-whisker plots for both groups on one diagram. [2]

3. The masses, in grams, of 60 apples are summarised in a box-and-whisker plot. The lowest value is 96, the lower quartile is 112, the median is 120, the upper quartile is 141 and the highest value is 178.

(a) Find the range and the interquartile range. [2] (b) Describe the skewness of the distribution, giving a reason. [2] (c) Estimate the number of apples with mass greater than 141 g. [1] (d) State one piece of information lost by showing the data only as a box-and-whisker plot. [1]

4. In a histogram of the heights of some plants, the bar for the class 20 ≤ h < 25 cm is 3 cm wide and 7.2 cm high. It represents 18 plants. The bar for the class 25 ≤ h < 40 cm is 4.8 cm high.

(a) Find the width of the bar for 25 ≤ h < 40. [1] (b) Find the number of plants in the class 25 ≤ h < 40. [3]

5. For 20 values, Σx = 642 and Σx² = 21 238. It was then found that one value had been recorded as 13 when it should have been 31. Find the correct mean and standard deviation. [5]

6. For 25 values of x, Σ(x − a) = −40 and Σ(x − a)² = 464, where a is a constant. The mean of x is 46.4.

(a) Find the value of a. [2] (b) Find the standard deviation of x. [2] (c) Find Σx². [2]

7. 18 girls have heights with mean 152 cm and standard deviation 5 cm; 12 boys have mean 160 cm and standard deviation 6 cm. Find the mean and standard deviation of the heights of all 30 children. [6]

8. The numbers of minutes late of nine trains are

0, 0, 0, 1, 2, 3, 4, 6, 47

(a) Find the mean, the median and the mode. [3] (b) State, with a reason, which average best describes a typical train. [1] (c) Find the standard deviation. [2] (d) State, with a reason, whether the interquartile range or the standard deviation is the better measure of spread for these data. [1]

9. The times, t seconds, taken to answer 100 telephone calls are shown.

Time (s)	0 ≤ t < 20	20 ≤ t < 30	30 ≤ t < 40	40 ≤ t < 60	60 ≤ t < 100
Frequency	12	21	27	24	16

(a) Calculate the frequency densities you would use to draw a histogram. [2] (b) Write down the points you would plot for a cumulative frequency graph. [1] (c) Assume the graph is drawn by joining the points with straight lines. Estimate the median and the interquartile range. [4] (d) Estimate the number of calls that took more than 50 seconds. [2] (e) Estimate the 90th percentile. [1]

10. Use the table in Question 9.

(a) Calculate estimates of the mean and the standard deviation of the times. [5] (b) Explain why your answers to (a) are only estimates. [1] (c) Compare your mean with the median from Question 9(c), and state what this suggests about the distribution. [2]

11. A data set of 10 values has mean 14 and standard deviation 3. Two more values, 22 and k, are added. The mean of all 12 values is 15.

(a) Find k. [3] (b) Find the standard deviation of all 12 values. [3]

Answers

1. (a) A histogram, because 250 values are too many to list as leaves [1]. (b) Two box plots on one scale show medians, quartiles and spread side by side [1]. (c) It keeps every raw value, so you can still find the mode or the mean [1]. Examiner insight: “it is clearer” earns nothing; each reason must fit the situation.

2. (a) n = 15: the median is the 8th value; each half has 7 values, so the quartiles are the 4th and 12th [1]. P: median 36 s [1], Q₁ = 30, Q₃ = 44, IQR 14 s. Q: median 46 s [1], Q₁ = 41, Q₃ = 51, IQR 10 s [1]. (b) Group Q took longer on average (median 46 > 36) [1]. Group P’s times were more spread out (IQR 14 > 10) [1]. (c) One labelled linear scale from 20 to 65 seconds [1]. P: 21, 30, 36, 44, 58. Q: 30, 41, 46, 51, 64 [1]. Examiner insight: in (b) one comment must be on centre and one on spread, each in context; two comments on centre score one mark only.

3. (a) Range = 178 − 96 = 82 g [1]. IQR = 141 − 112 = 29 g [1]. (b) Q₃ − median = 21 is larger than median − Q₁ = 8 [1], so the distribution has positive skew [1]. (c) A quarter of the apples lie above Q₃: 60 ÷ 4 = 15 [1]. (d) The individual masses, so the mode or exact mean cannot be found [1]. Examiner insight: a direction of skew without comparing the two halves of the box loses the reason mark.

4. (a) 5 cm of height → 3 cm of width, so 15 cm → 9 cm [1]. (b) Area of the first bar = 3 × 7.2 = 21.6 cm² for 18 plants, so 1 plant = 1.2 cm² [1]. Area of the second bar = 9 × 4.8 = 43.2 cm² [1]. 43.2 ÷ 1.2 = 36 plants [1]. Examiner insight: using bar heights alone (4.8 ÷ 7.2 × 18 = 12) scores zero; frequency is proportional to area.

5. Correct Σx = 642 − 13 + 31 = 660 [1]. Correct Σx² = 21 238 − 169 + 961 = 22 030 [1]. Mean = 660 ÷ 20 = 33 [1]. Variance = 22 030 ÷ 20 − 33² = 1101.5 − 1089 = 12.5 [1]. Standard deviation = √12.5 = 3.54 (3 s.f.) [1]. Examiner insight: Σx² must change by 31² − 13², not by 18²; that error loses the last three marks.

6. (a) Mean of (x − a) = −40 ÷ 25 = −1.6 [1], so 46.4 = a − 1.6 and a = 48 [1]. (b) Variance = 464 ÷ 25 − (−1.6)² = 18.56 − 2.56 = 16 [1], so standard deviation = 4 [1]. (c) Σx² = 25 × (16 + 46.4²) [1] = 25 × 2168.96 = 54 224 [1]. Examiner insight: coding by subtracting a does not change the standard deviation, so do not add a back on in (b).

7. Girls: Σx = 18 × 152 = 2736. Boys: Σx = 12 × 160 = 1920 [1]. Girls: Σx² = 18 × (5² + 152²) = 416 322 [1]. Boys: Σx² = 12 × (6² + 160²) = 307 632 [1]. Combined mean = 4656 ÷ 30 = 155.2 cm [1]. Variance = 723 954 ÷ 30 − 155.2² = 24 131.8 − 24 087.04 = 44.76 [1]. Standard deviation = 6.69 cm (3 s.f.) [1]. Examiner insight: averaging the two standard deviations earns no method mark.

8. (a) Mean = 63 ÷ 9 = 7 minutes [1]. Median (5th value) = 2 minutes [1]. Mode = 0 minutes [1]. (b) The median, because the mean is pulled up by the extreme value 47 [1]. (c) Σx² = 2275 [1]. Standard deviation = √(2275 ÷ 9 − 7²) = √203.78 = 14.3 minutes [1]. (d) The IQR, because it is not affected by the extreme value 47 [1]. Examiner insight: the reason must name the extreme value; “the median is more accurate” earns nothing.

9. (a) Frequency density = frequency ÷ class width [1]: 0.6, 2.1, 2.7, 1.2, 0.4 [1]. (b) (0, 0), (20, 12), (30, 33), (40, 60), (60, 84), (100, 100), plotted at upper class boundaries [1]. (c) Median at cf 50: 30 + (17 ÷ 27) × 10 = 36.3 s [1]. Q₁ at cf 25: 20 + (13 ÷ 21) × 10 = 26.2 [1]. Q₃ at cf 75: 40 + (15 ÷ 24) × 20 = 52.5 [1]. IQR = 52.5 − 26.2 = 26.3 s [1]. (d) cf at 50 s = 60 + (10 ÷ 20) × 24 = 72 [1]. More than 50 s: 100 − 72 = 28 calls [1]. (e) cf 90: 60 + (6 ÷ 16) × 40 = 75 s [1]. Examiner insight: plotting at mid-points instead of upper class boundaries makes every later reading wrong and loses the accuracy marks.

10. (a) Mid-points 10, 25, 35, 50, 80 [1]. Σxf = 4070, so mean = 4070 ÷ 100 = 40.7 s [1]. Σx²f = 209 800 [1]. Variance = 209 800 ÷ 100 − 40.7² = 2098 − 1656.49 = 441.51 [1]. Standard deviation = 21.0 s (3 s.f.) [1]. (b) The exact times are unknown, so each value is replaced by its class mid-point [1]. (c) The mean (40.7 s) is greater than the median (36.3 s) [1], which suggests positive skew [1]. Examiner insight: using class widths or upper boundaries instead of mid-points loses the first mark and all accuracy marks.

11. (a) Original Σx = 10 × 14 = 140 [1]. New Σx = 12 × 15 = 180 [1]. k = 180 − 140 − 22 = 18 [1]. (b) Original Σx² = 10 × (3² + 14²) = 2050 [1]. New Σx² = 2050 + 22² + 18² = 2858 [1]. Standard deviation = √(2858 ÷ 12 − 15²) = √13.17 = 3.63 (3 s.f.) [1]. Examiner insight: using 10 × 14² as the original Σx² ignores the variance and loses the method mark.

Where marks are usually lost

• Bar height used as frequency on a histogram with unequal class widths.
• Cumulative frequency plotted at class mid-points or lower boundaries.
• Left-hand leaves of a back-to-back diagram read in the wrong order.
• Skewness stated without comparing the two halves of the box.
• Comparisons made without context.
• Σx² not updated when a value is corrected.
• Two means averaged without weighting by the group sizes.
• Variance given instead of standard deviation.

Next steps

• Probability & Statistics 1 revision notes
• Probability & Statistics 1 study guide
• Probability & Statistics 1 practice questions for the rest of Paper 5
• 9709 course hub and printable checklist
• 9709 AS diagnostic, the 9709 self-check bank and all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027, Version 4, Cambridge Assessment International Education (part of Cambridge University Press & Assessment). Section 5.1, Representation of data (Probability & Statistics 1, for Paper 5).