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AS & A Level Mathematics (9709)

Representation of data Banker

38 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 51 · Q3 (a, b, c)9 marksComparing distributions · Stem-and-leaf / quartiles
  • 2026 May/June · Paper 52 · Q6 (a, b, c)8 marksStem-and-leaf / quartiles · Summary statistics (coded sums)
  • 2026 May/June · Paper 53 · Q3 (a, b, c)8 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2026 May/June · Paper 55 · Q1 (a, b)4 marksSummary statistics (coded sums)
  • 2026 May/June · Paper 55 · Q2 (a, b)4 marksHistograms / cumulative frequency

2025

  • 2025 Oct/Nov · Paper 51 · Q3 (a, b)8 marksStem-and-leaf / quartiles · Summary statistics (coded sums)
  • 2025 Oct/Nov · Paper 52 · Q5 (a, b)7 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2025 Oct/Nov · Paper 53 · Q6 (a, b, c, d)10 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles
  • 2025 Oct/Nov · Paper 55 · Q4 (a, b, c, d)11 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles
  • 2025 May/June · Paper 51 · Q3 (a, b, c, d)12 marksStem-and-leaf / quartiles · Summary statistics (coded sums)
  • 2025 May/June · Paper 52 · Q5 (a, b, c)10 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2025 May/June · Paper 53 · Q1 (a, b)4 marksSummary statistics (coded sums)
  • 2025 May/June · Paper 53 · Q4 (a, b)5 marksBox-and-whisker plots · Histograms / cumulative frequency
  • 2025 May/June · Paper 55 · Q5 (a, b, c)10 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles

2024

  • 2024 Oct/Nov · Paper 51 · Q3 (a, b, c)8 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2024 Oct/Nov · Paper 52 · Q6 (a, b, c)10 marksStem-and-leaf / quartiles · Summary statistics (coded sums)
  • 2024 Oct/Nov · Paper 53 · Q4 (a, b, c)11 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2024 May/June · Paper 51 · Q1 (a, b)4 marksOfficial paper ↗Summary statistics (coded sums)
  • 2024 May/June · Paper 51 · Q3 (a, b)6 marksOfficial paper ↗Histograms / cumulative frequency
  • 2024 May/June · Paper 52 · Q4 (a, b, c)7 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles
  • 2024 May/June · Paper 53 · Q4 (a, b, c)8 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles

2023

  • 2023 Oct/Nov · Paper 51 · Q1 (a, b)4 marksHistograms / cumulative frequency
  • 2023 Oct/Nov · Paper 51 · Q4 (a, b)9 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2023 Oct/Nov · Paper 52 · Q4 (a, b, c)8 marksComparing distributions · Stem-and-leaf / quartiles
  • 2023 Oct/Nov · Paper 53 · Q4 (a, b, c)10 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2023 May/June · Paper 51 · Q1 (a, b)4 marksSummary statistics (coded sums)
  • 2023 May/June · Paper 51 · Q5 (a, b, c)7 marksHistograms / cumulative frequency
  • 2023 May/June · Paper 52 · Q3 (a, b, c)7 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles
  • 2023 May/June · Paper 53 · Q4 (a, b, c)8 marksComparing distributions · Stem-and-leaf / quartiles · Summary statistics (coded sums)

2022

  • 2022 Oct/Nov · Paper 51 · Q3 (a, b, c)9 marksComparing distributions · Stem-and-leaf / quartiles
  • 2022 Oct/Nov · Paper 52 · Q4 (a, b)7 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2022 Oct/Nov · Paper 53 · Q13 marksSummary statistics (coded sums)
  • 2022 Oct/Nov · Paper 53 · Q3 (a, b, c)7 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2022 May/June · Paper 51 · Q3 (a, b, c, d)9 marksHistograms / cumulative frequency · Mean & sd from grouped data
  • 2022 May/June · Paper 52 · Q13 marksSummary statistics (coded sums)
  • 2022 May/June · Paper 52 · Q3 (a, b, c)7 marksBox-and-whisker plots · Comparing distributions · Stem-and-leaf / quartiles
  • 2022 May/June · Paper 53 · Q1 (a, b)3 marksHistograms / cumulative frequency
  • 2022 May/June · Paper 53 · Q2 (a, b, c)3 marksStem-and-leaf / quartiles · Summary statistics (coded sums)

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Stem-and-leaf / quartiles · Core · Paper 5 style · 5 marks

The times, in minutes, taken by 15 students in Karachi to travel to college one morning are shown in the stem-and-leaf diagram.1 | 2 5 82 | 1 2 4 5 7 83 | 1 3 6 84 | 25 | 5Key: 2 | 4 means 24 minutes

  1. (a) Find the median and the interquartile range of the times. [3]
  2. (b) Find the mean time, correct to 3 significant figures. [1]
  3. (c) Give a reason why the median may be a better average than the mean for these data. [1]
Show answer
  1. (a) Median 27 minutes; LQ 21, UQ 36, IQR 15 minutes
    B1 median 27B1 quartiles 21 and 36B1 IQR 15
  2. (b) 28.5 minutes
    B1 427/15 = 28.5
  3. (c) The value 55 is much larger than the rest (data positively skewed), so it pulls the mean up; the median is not affected by extreme values
    B1
Q2Stem-and-leaf / quartiles · Core · Paper 5 style · 6 marks

The marks out of 50 scored in a quiz by 11 students in class P and 11 students in class Q are shown in the back-to-back stem-and-leaf diagram. Leaves for class P are on the left.8 | 1 |9 5 2 | 2 | 1 67 4 3 1 | 3 | 0 2 5 8 87 4 0 | 4 | 1 3 5 9Key: 5 | 2 | 6 means 25 marks for a student in class P and 26 marks for a student in class Q

  1. (a) Find the median and the interquartile range for each class. [4]
  2. (b) Make two comparisons between the marks of the two classes. [2]
Show answer
  1. (a) Class P: median 33, IQR 40 − 25 = 15. Class Q: median 38, IQR 43 − 30 = 13
    B1 medians 33 and 38B1 quartiles for P: 25, 40B1 quartiles for Q: 30, 43B1 IQRs 15 and 13
  2. (b) Class Q scored higher on average (median 38 > 33); Class Q's marks were slightly less spread (IQR 13 < 15)
    B1 average comparison in contextB1 spread comparison in context
Q3Stem-and-leaf / quartiles · Core · Paper 5 style · 5 marks

The battery life, in minutes, of 19 phones of one model was measured during a video-streaming test.31 | 232 | 533 | 834 | 1 635 | 2 5 936 | 1 4 837 | 2 738 | 339 | 0 840 | 541 | 42 | 143 | 6Key: 35 | 9 means 359 minutes

  1. (a) Find the range and the median of the battery lives. [2]
  2. (b) Find the interquartile range. [2]
  3. (c) Find the percentage of these phones with a battery life of more than 380 minutes, correct to 1 decimal place. [1]
Show answer
  1. (a) Range 124 minutes; median 364 minutes
    B1 436 − 312 = 124B1 364
  2. (b) 390 − 346 = 44 minutes
    B1 quartiles 346, 390B1 44
  3. (c) 6/19 = 31.6%
    B1
Q4Stem-and-leaf / quartiles · Standard · Paper 5 style · 7 marks

A botanist in Murree measured the heights, in cm, of 19 young pine seedlings after one year. The results are shown below.3 | 44 | 1 5 7 85 | 2 3 5 6 8 96 | 1 2 4 6 77 | 1 48 | 9 | 6Key: 5 | 3 means 53 cm

  1. (a) Find the median and the quartiles of the heights. [3]
  2. (b) An outlier is defined as any value more than 1.5 × IQR above the upper quartile or below the lower quartile. Determine whether there are any outliers, showing your working. [3]
  3. (c) Suggest one reason why the researcher might keep the value 96 in the data. [1]
Show answer
  1. (a) Median 58 cm, LQ 48 cm, UQ 66 cm
    B1 B1 B1
  2. (b) IQR = 18; 1.5 × 18 = 27; limits 21 and 93; 96 > 93 so 96 cm is an outlier; no low outliers
    M1 1.5 × 18 = 27A1 limits 21 and 93A1 96 is the only outlier
  3. (c) It may be a genuine measurement of an unusually tall plant, not a mistake
    B1
Q5Stem-and-leaf / quartiles · Standard · Paper 5 style · 5 marks
  1. (a) The stem-and-leaf diagram shows the masses, in grams, of 15 guavas from a market stall, where x is an unknown digit.3 | 2 5 84 | 0 1 3 x 6 7 95 | 0 2 4 7 8Key: 4 | 1 means 41 gGiven that the mean mass is 45.8 g, find the value of x. [3]
  2. (b) Find the interquartile range of the masses. [2]
Show answer
  1. (a) x = 5
    M1 sum of the other 14 values = 642M1 642 + 40 + x = 15 × 45.8 = 687A1 x = 5
  2. (b) 52 − 40 = 12 g
    B1 LQ 40, UQ 52B1 12
Q6Stem-and-leaf / quartiles · Standard · Paper 5 style · 6 marks

The rainfall, in mm, recorded on 19 days in August at a weather station in Abbottabad was:23 8 14 31 17 26 12 9 19 35 22 16 28 11 40 25 18 13 21

  1. (a) Represent the data in a stem-and-leaf diagram with stems 0, 1, 2, 3, 4 and a key. [3]
  2. (b) Find the median and the interquartile range. [3]
Show answer
  1. (a) 0 | 8 91 | 1 2 3 4 6 7 8 92 | 1 2 3 5 6 83 | 1 54 | 0Key: 1 | 6 means 16 mm
    B1 correct stemsB1 leaves correct and orderedB1 key
  2. (b) Median 19 mm; IQR 26 − 13 = 13 mm
    B1 median 19B1 quartiles 13 and 26B1 IQR 13
Q7Stem-and-leaf / quartiles · Challenge · Paper 5 style · 7 marks

The resting pulse rates, in beats per minute, of 15 athletes and 15 non-athletes at a sports club in Faisalabad are shown. Leaves for athletes are on the left.8 | 4 |8 8 6 5 3 2 | 5 |9 6 4 3 1 0 | 6 | 2 5 84 1 | 7 | 0 1 2 4 5 6 8 9| 8 | 1 4 8| 9 | 3Key: 8 | 5 | 2 means 58 for an athlete and 52 for a non-athlete

  1. (a) Find the median and the quartiles for each group. [4]
  2. (b) Compare the resting pulse rates of the two groups in two different ways. [2]
  3. (c) State one advantage of a back-to-back stem-and-leaf diagram over two box-and-whisker plots for these data. [1]
Show answer
  1. (a) Athletes: LQ 55, median 60, UQ 66. Non-athletes: LQ 70, median 75, UQ 81
    B1 both mediansB1 B1 quartiles for each groupB1 all correct
  2. (b) Athletes have lower pulse rates on average (median 60 vs 75); the spreads are similar (both IQRs 11), although the non-athletes have a larger range (31 vs 26)
    B1 averageB1 spread (both IQRs are 11; ranges 26 and 31)
  3. (c) It shows every individual value (the raw data are kept), so the shape of each distribution can be seen in detail
    B1
Q8Stem-and-leaf / quartiles · Challenge · Paper 5 style · 5 marks
  1. (a) The ages, in years, of 11 members of a cricket club committee are, in order,22 25 28 31 33 3a 41 4b 50 54 59where a and b are unknown digits. The median age is 37 years and the mean age is 39 years. Find a and b. [3]
  2. (b) Find the interquartile range of the ages. [2]
Show answer
  1. (a) a = 7, b = 9
    B1 median is the 6th value, so a = 7M1 total 11 × 39 = 429A1 4b = 429 − 380 = 49, so b = 9
  2. (b) 50 − 28 = 22 years
    B1 LQ 28, UQ 50B1 22
Q9Histograms / cumulative frequency · Core · Paper 5 style · 5 marks

The times, t seconds, taken by 100 people to solve a puzzle at a Lahore science fair are summarised below.Time t (s): 0 ≤ t < 20 | 20 ≤ t < 30 | 30 ≤ t < 40 | 40 ≤ t < 60 | 60 ≤ t < 100Frequency: 14 | 22 | 30 | 24 | 10

  1. (a) Calculate the frequency density for each class, ready to draw a histogram. [3]
  2. (b) Estimate the number of people who took more than 50 seconds. [2]
Show answer
  1. (a) 0.7, 2.2, 3.0, 1.2, 0.25 (per second)
    M1 frequency ÷ class widthA1 at least three correctA1 all five correct
  2. (b) 22
    M1 ½ × 24 + 10A1 22
Q10Histograms / cumulative frequency · Core · Paper 5 style · 5 marks

The lengths, l mm, of 141 leaves from a neem tree are summarised below.20 ≤ l < 25: 1225 ≤ l < 30: 2830 ≤ l < 35: 4535 ≤ l < 40: 1640 ≤ l < 50: 3250 ≤ l < 60: 8

  1. (a) Calculate the frequency densities needed to draw a histogram of these data. [3]
  2. (b) Estimate the number of leaves with lengths between 33 mm and 45 mm. [2]
Show answer
  1. (a) 2.4, 5.6, 9, 3.2, 3.2, 0.8 (leaves per mm)
    M1 divide by class widths 5, 5, 5, 5, 10, 10A1 at least four correctA1 all correct
  2. (b) 50
    M1 (2/5) × 45 + 16 + (5/10) × 32A1 18 + 16 + 16 = 50
Q11Histograms / cumulative frequency · Standard · Paper 5 style · 6 marks

A histogram shows the masses, m kg, of parcels handled by a courier office in Peshawar in one day. The classes and bar heights are:0 ≤ m < 2: height 3 cm2 ≤ m < 3: height 8 cm3 ≤ m < 5: height 5 cm5 ≤ m < 10: height 1.2 cmThe class 2 ≤ m < 3 contains 40 parcels.

  1. (a) Find the frequency of each class and the total number of parcels. [4]
  2. (b) Estimate the number of parcels with masses between 2.5 kg and 4 kg. [2]
Show answer
  1. (a) 30, 40, 50, 30; total 150
    B1 one unit of area (1 cm × 1 kg) represents 5 parcelsM1 frequency = 5 × height × widthA1 30, 50, 30A1 total 150
  2. (b) 45
    M1 ½ × 40 + ½ × 50A1 45
Q12Histograms / cumulative frequency · Standard · Paper 5 style · 7 marks

The daily electricity use, x kWh, of 200 households in Lahore is summarised in the cumulative frequency table.x ≤ 5: 12x ≤ 10: 46x ≤ 15: 108x ≤ 20: 160x ≤ 25: 188x ≤ 30: 200(No household used 0 kWh or less.)

  1. (a) Estimate the median daily use, assuming the points of the cumulative frequency graph are joined with straight lines. [2]
  2. (b) Estimate the interquartile range. [3]
  3. (c) Estimate the number of households that used more than 22 kWh. [2]
Show answer
  1. (a) 14.4 kWh
    M1 100th value: 10 + (54/62) × 5A1 14.4 (14.35…)
  2. (b) 8.72 kWh (UQ 19.04, LQ 10.32)
    M1 50th and 150th valuesA1 LQ 10.32, UQ 19.04A1 IQR 8.72 (accept 8.6–8.8 from a graph)
  3. (c) 28.8, about 29
    M1 cf at 22: 160 + (2/5) × 28 = 171.2A1 200 − 171.2 ≈ 29
Q13Histograms / cumulative frequency · Standard · Paper 5 style · 5 marks

The ages, a years, of 80 visitors to a museum in Lahore one morning are shown in a histogram. The frequency densities (heights) of some bars and the frequencies of some classes are known:0 ≤ a < 10: frequency 810 ≤ a < 20: bar height 1.620 ≤ a < 40: frequency 24, bar height 1.240 ≤ a < 70: frequency 1870 ≤ a < 90: frequency unknown

  1. (a) Find the frequency of the class 10 ≤ a < 20 and the frequency density of the class 70 ≤ a < 90. [4]
  2. (b) State which class contains the median age. [1]
Show answer
  1. (a) 16; 0.7
    B1 scale: 24 ÷ 20 = 1.2 matches, so height = frequency densityB1 1.6 × 10 = 16M1 80 − 8 − 16 − 24 − 18 = 14A1 14 ÷ 20 = 0.7
  2. (b) 20 ≤ a < 40
    B1 the 40th/41st values lie in 20–40 (cumulative 24 then 48)
Q14Histograms / cumulative frequency · Challenge · Paper 5 style · 7 marks

The typing speeds, w words per minute, of 120 applicants for an office job in Rawalpindi are summarised in a cumulative frequency table. The slowest applicant typed more than 10 words per minute. Assume the points of the cumulative frequency graph are joined by straight lines.w ≤ 20: 6w ≤ 30: 20w ≤ 40: 52w ≤ 50: 90w ≤ 60: 110w ≤ 70: 120

  1. (a) Estimate the number of applicants with speeds of more than 45 words per minute. [2]
  2. (b) The fastest 20% of applicants are invited to an interview. Estimate the least speed needed for an interview. [3]
  3. (c) Estimate the 10th percentile of the speeds. [2]
Show answer
  1. (a) 49
    M1 cf at 45 = 52 + ½ × 38 = 71A1 120 − 71 = 49
  2. (b) 53 words per minute
    B1 20% of 120 = 24, so the 96th valueM1 50 + (6/20) × 10A1 53
  3. (c) 24.3 words per minute
    M1 12th value: 20 + (6/14) × 10A1 24.3
Q15Histograms / cumulative frequency · Challenge · Paper 5 style · 7 marks

100 students at a college in Multan recorded the number of text messages they sent on one day.Number of messages: 1–5 | 6–10 | 11–20 | 21–40Number of students: 15 | 25 | 40 | 20

  1. (a) State the class boundaries of the class 11–20 and find the frequency densities for all four classes. [3]
  2. (b) Estimate the number of students who sent more than 30 messages. [2]
  3. (c) Calculate an estimate of the mean number of messages. [2]
Show answer
  1. (a) 10.5 and 20.5; densities 3, 5, 4, 1
    B1 10.5, 20.5M1 widths 5, 5, 10, 20A1 3, 5, 4, 1
  2. (b) 10
    M1 (40.5 − 30.5)/20 × 20A1 10
  3. (c) 14.75
    M1 midpoints 3, 8, 15.5, 30.5A1 1475/100 = 14.75
Q16Summary statistics (coded sums) · Core · Paper 5 style · 3 marks
  1. (a) The lengths, x cm, of 20 fish caught at Keenjhar Lake are summarised by Σ(x − 50) = 84 and Σ(x − 50)² = 2400. Find the mean and the standard deviation of the lengths. [3]
Show answer
  1. (a) Mean 54.2 cm; standard deviation 10.1 cm
    B1 mean = 50 + 84/20 = 54.2M1 variance = 2400/20 − 4.2² = 102.36A1 sd = 10.1
Q17Summary statistics (coded sums) · Standard · Paper 5 style · 6 marks

A sample of 12 mangoes from orchard A in Multan has mean mass 250 g and standard deviation 8 g. A sample of 18 mangoes from orchard B has mean mass 238 g and standard deviation 6 g.

  1. (a) Find Σx and Σx² for each orchard. [2]
  2. (b) Find the mean and the standard deviation of the masses of all 30 mangoes. [4]
Show answer
  1. (a) Orchard A: Σx = 3000, Σx² = 750 768; Orchard B: Σx = 4284, Σx² = 1 020 240
    B1 Σx valuesB1 Σx² = n(σ² + x̄²) for both
  2. (b) Mean 242.8 g; standard deviation 9.04 g
    M1 (3000 + 4284)/30A1 242.8M1 (750 768 + 1 020 240)/30 − 242.8²A1 9.04
Q18Summary statistics (coded sums) · Standard · Paper 5 style · 5 marks
  1. (a) For a set of 10 values of x, Σx = 360 and Σx² = 13 500. Find Σ(x − 30)². [2]
  2. (b) Find the standard deviation of x, correct to 3 significant figures, and state the standard deviation of the values 5x + 2. [3]
Show answer
  1. (a) 900
    M1 Σx² − 60Σx + 900 × 10A1 13 500 − 21 600 + 9000 = 900
  2. (b) 7.35; 36.7
    M1 13 500/10 − 36² = 54A1 √54 = 7.35B1 5 × 7.35 = 36.7
Q19Summary statistics (coded sums) · Challenge · Paper 5 style · 6 marks
  1. (a) A set of n values of x has Σ(x − 10) = 48 and Σ(x − 10)² = 400. The mean of x is 12.4. Find n. [2]
  2. (b) Find the standard deviation of x. [2]
  3. (c) Find Σx². [2]
Show answer
  1. (a) n = 20
    M1 48/n = 2.4A1 n = 20
  2. (b) 3.77
    M1 400/20 − 2.4² = 14.24A1 3.77
  3. (c) 3360
    M1 Σx² = n(σ² + x̄²) = 20(14.24 + 153.76)A1 3360
Q20Summary statistics (coded sums) · Challenge · Paper 5 style · 6 marks

The times, x seconds, taken by 25 runners at a school sports day in Quetta to run 200 m were recorded. The mean was 40.8 s and the standard deviation was 5.2 s. It was later found that one time had been recorded as 52 s when it should have been 42 s.

  1. (a) Find Σx and Σx² for the 25 recorded values. [2]
  2. (b) Find the correct mean and the correct standard deviation. [4]
Show answer
  1. (a) Σx = 1020; Σx² = 42 292
    B1 25 × 40.8B1 25(5.2² + 40.8²)
  2. (b) Mean 40.4 s; standard deviation 4.68 s
    M1 Σx = 1020 − 52 + 42 = 1010A1 mean 40.4M1 Σx² = 42 292 − 52² + 42² = 41 352, variance 41 352/25 − 40.4²A1 4.68
Q21Mean & sd from grouped data · Core · Paper 5 style · 5 marks

40 students in Hyderabad recorded the time, t hours, they spent on homework in one week.Time t: 0 ≤ t < 2 | 2 ≤ t < 4 | 4 ≤ t < 6 | 6 ≤ t < 10Frequency: 6 | 14 | 12 | 8

  1. (a) Calculate estimates of the mean and the standard deviation of the time spent. [4]
  2. (b) Explain why your answers are only estimates. [1]
Show answer
  1. (a) Mean 4.3 hours; standard deviation 2.26 hours
    B1 midpoints 1, 3, 5, 8M1 Σfx/Σf = 172/40M1 √(Σfx²/Σf − mean²) = √(944/40 − 4.3²) = √5.11A1 4.3 and 2.26
  2. (b) The exact times are not known; each class is represented by its midpoint
    B1
Q22Mean & sd from grouped data · Standard · Paper 5 style · 6 marks

The speeds, v km/h, of 100 cars on the M2 motorway were recorded by a speed camera.Speed v: 60 ≤ v < 80 | 80 ≤ v < 100 | 100 ≤ v < 110 | 110 ≤ v < 120 | 120 ≤ v < 140Frequency: 8 | 22 | 35 | 25 | 10

  1. (a) Calculate estimates of the mean and the standard deviation of the speeds. [4]
  2. (b) The speed limit on this part of the motorway is 120 km/h. Estimate the percentage of these cars exceeding the limit, and comment on whether the mean or the percentage is more useful to the police. [2]
Show answer
  1. (a) Mean 103.9 km/h; standard deviation 15.3 km/h
    B1 midpoints 70, 90, 105, 115, 130M1 mean 10 390/100M1 variance = Σfx²/100 − 103.9²A1 103.9 and 15.3
  2. (b) 10%; the percentage, because the mean (below the limit) hides the cars that are speeding
    B1 10%B1 sensible comment
Q23Mean & sd from grouped data · Standard · Paper 5 style · 5 marks

The masses of 100 eggs from a poultry farm near Sheikhupura were measured to the nearest gram.Mass (g): 45–49 | 50–54 | 55–59 | 60–64 | 65–69Frequency: 9 | 23 | 38 | 21 | 9

  1. (a) Write down the midpoints of the classes, explaining why the first midpoint is 47. [2]
  2. (b) Calculate estimates of the mean and standard deviation of the masses. [3]
Show answer
  1. (a) 47, 52, 57, 62, 67; the class 45–49 has boundaries 44.5 and 49.5, midpoint 47
    B1 midpointsB1 reason
  2. (b) Mean 56.9 g; standard deviation 5.38 g
    M1 Σfx/100 = 5690/100M1 variance = Σfx²/100 − 56.9²A1 56.9 and 5.38
Q24Mean & sd from grouped data · Challenge · Paper 5 style · 6 marks

The distances, d km, travelled to work by the employees of a company in Islamabad are summarised below.Distance d: 10 ≤ d < 20 | 20 ≤ d < 30 | 30 ≤ d < 40 | 40 ≤ d < 60Frequency: 5 | k | 20 | 12

  1. (a) Given that an estimate of the mean distance is 32.5 km, find k. [3]
  2. (b) Calculate an estimate of the standard deviation of the distances. [3]
Show answer
  1. (a) k = 23
    M1 (75 + 25k + 700 + 600)/(37 + k) = 32.5M1 1375 + 25k = 1202.5 + 32.5kA1 k = 23
  2. (b) 10.5 km
    M1 Σfx² = 5(225) + 23(625) + 20(1225) + 12(2500) = 70 000M1 √(70 000/60 − 32.5²)A1 10.5
Q25Mean & sd from grouped data · Challenge · Paper 5 style · 7 marks

A call centre in Karachi recorded the durations, t minutes, of 100 calls.Duration t: 0 ≤ t < 2 | 2 ≤ t < 5 | 5 ≤ t < 10 | 10 ≤ t < 20 | 20 ≤ t < 30Frequency: 18 | 30 | 26 | 16 | 10

  1. (a) Calculate estimates of the mean and the standard deviation of the call durations. [4]
  2. (b) Estimate the number of calls that lasted longer than the mean plus one standard deviation. [2]
  3. (c) State, with a reason, whether the median would be less than or greater than the mean. [1]
Show answer
  1. (a) Mean 8.08 minutes; standard deviation 7.19 minutes
    B1 midpoints 1, 3.5, 7.5, 15, 25M1 mean 808/100M1 √(Σfx²/100 − 8.08²)A1 8.08 and 7.19
  2. (b) About 18 (17.6)
    M1 mean + sd ≈ 15.27; (20 − 15.27)/10 × 16 + 10A1 ≈ 17.6, so about 18 (accept 17–18)
  3. (c) Less than the mean: the distribution is positively skewed (long tail of long calls); the 50th call lies in the class 5–10 (cumulative 48 then 74), median ≈ 5 + (2/26) × 5 ≈ 5.4 < 8.08
    B1
Q26Box-and-whisker plots · Core · Paper 5 style · 5 marks

The daily maximum temperatures, in °C, in Jacobabad over 31 days in June have the following summary: lowest 39, lower quartile 43, median 44, upper quartile 47, highest 54.

  1. (a) Find the interquartile range and determine whether the highest temperature is an outlier, where an outlier is a value more than 1.5 × IQR above the upper quartile. [3]
  2. (b) Describe the shape of the distribution, giving a reason. [2]
Show answer
  1. (a) IQR = 4; 47 + 6 = 53; 54 > 53, so it is an outlier
    B1 IQR 4M1 47 + 1.5 × 4A1 54 > 53, outlier
  2. (b) Positively skewed: UQ − median (3) > median − LQ (1) and the upper whisker is longer
    B1 positive skewB1 reason
Q27Box-and-whisker plots · Standard · Paper 5 style · 6 marks

The numbers of customers served in each of 15 half-hour periods at a bank counter in Gujranwala were:0 | 5 81 | 2 4 5 7 82 | 0 1 3 4 6 93 | 34 | 7Key: 2 | 3 means 23 customers

  1. (a) Find the five values needed to draw a box-and-whisker plot of the data. [3]
  2. (b) Show that 47 is an outlier using the 1.5 × IQR rule, and state the value at which the upper whisker would end if outliers are plotted separately. [3]
Show answer
  1. (a) Lowest 5, LQ 14, median 20, UQ 26, highest 47
    B1 medianB1 quartilesB1 extremes
  2. (b) IQR 12, 26 + 18 = 44, 47 > 44 so outlier; whisker ends at 33
    M1 26 + 1.5 × 12A1 44 < 47B1 33
Q28Box-and-whisker plots · Challenge · Paper 5 style · 5 marks

Box-and-whisker plots summarise the waiting times, in minutes, at two clinics in Rawalpindi on one day.Clinic A: lowest 4, lower quartile 9, median 13, upper quartile 20, highest 35Clinic B: lowest 6, lower quartile 10, median 12, upper quartile 14, highest 18

  1. (a) Compare the waiting times at the two clinics in two ways. [2]
  2. (b) Clinic A saw 120 patients that day. Estimate how many of them waited between 9 and 20 minutes. [1]
  3. (c) A manager says: 'At clinic B no patient waited more than 18 minutes, but at clinic A at least a quarter of patients waited longer than any patient at clinic B.' Use the box plots to decide whether this statement is justified. [2]
Show answer
  1. (a) Clinic A has a slightly higher median (13 vs 12); Clinic A is much more spread out (IQR 11 vs 4, range 31 vs 12)
    B1 averageB1 spread
  2. (b) 60
    B1 half of the patients lie between the quartiles
  3. (c) Justified: clinic B's maximum is 18, and clinic A's upper quartile is 20 > 18, so at least 25% of clinic A patients waited at least 20 minutes, longer than any clinic B patient
    M1 compare UQ of A with max of BA1 conclusion
Q29Comparing distributions · Standard · Paper 5 style · 5 marks

Two varieties of wheat were grown on 40 one-hectare plots each in Punjab. The yields, in tonnes per hectare, are summarised:Variety X: mean 3.42, standard deviation 0.31Variety Y: mean 3.58, standard deviation 0.62

  1. (a) Compare the yields of the two varieties in two ways. [2]
  2. (b) A farmer wants a reliable harvest every year. Suggest which variety the farmer should choose, giving a reason. [1]
  3. (c) Estimate the total yield from 25 hectares of variety Y. Also state the mean and the standard deviation of the yields of variety Y in kilograms per hectare. [2]
Show answer
  1. (a) Variety Y has the higher mean yield (3.58 > 3.42 t/ha); variety X's yields are more consistent (sd 0.31 < 0.62)
    B1 average in contextB1 spread in context
  2. (b) Variety X: its standard deviation is half that of Y, so yields vary less, even though its mean is slightly lower
    B1
  3. (c) 89.5 tonnes; mean 3580 kg/ha, standard deviation 620 kg/ha
    B1 25 × 3.58 = 89.5 tonnesB1 3580 and 620 (both multiplied by 1000)
Q30Comparing distributions · Challenge · Paper 5 style · 5 marks

The monthly incomes, in thousands of rupees, of 9 workers at factory P in Sialkot are:32 35 35 38 40 41 44 46 180

  1. (a) Find the mean and the median of the monthly incomes of the workers at factory P. [2]
  2. (b) Explain which average better represents a typical worker at factory P. [1]
  3. (c) At factory Q, the monthly incomes of 9 workers have median 42 thousand rupees and mean 43 thousand rupees. A worker claims that the workers at factory P earn more because 54.6 > 43. Comment on this claim. [2]
Show answer
  1. (a) Mean 54.6 thousand rupees (491/9); median 40 thousand rupees
    B1 491/9 = 54.6B1 40
  2. (b) The median: the mean is pulled up by the single very high income of 180
    B1
  3. (c) Not justified: the high mean at P is due to one outlier (180); the median at Q (42) is higher than at P (40), so a typical worker at Q earns slightly more
    M1 compare mediansA1 conclusion with reason

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Stem-and-leaf / quartiles Banker16 questions · 9/9 sessions
Histograms / cumulative frequency Banker16 questions · 9/9 sessions
Summary statistics (coded sums) Banker12 questions · 8/9 sessions
Mean & sd from grouped data 10 questions · 7/9 sessions
Box-and-whisker plots 8 questions · 5/9 sessions
Comparing distributions Banker11 questions · 8/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Stem-and-leaf / quartiles Banker789/9
Histograms / cumulative frequency Banker699/9
Summary statistics (coded sums) Banker428/9
Mean & sd from grouped data 427/9
Box-and-whisker plots 265/9
Comparing distributions Banker158/9

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