Pearson Edexcel International Advanced Level Mathematics: Models and data (YMA01) – 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 Unit S1: Statistics 1, sections 1.1 and 2.1 to 2.4, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). The specification states that S1 is an optional unit for IAS and IAL Mathematics and Further Mathematics. For ungrouped quartiles, round a non-whole position (n/4, n/2, 3n/4) up, and take the mean of two values when it is whole. For grouped data, interpolate. Give answers to 3 significant figures unless told otherwise.

Study first with the study guide and the revision notes. Also see the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A bus company models the journey time between two stops as exactly 12 minutes every day.

(a) Give one reason why this model is unrealistic. [1] (b) A sample of 30 journeys has mean 12.6 minutes and standard deviation 3.1 minutes. Comment on the model in the light of these data. [2]

2. The waiting times, t minutes, of 90 patients are summarised below. A histogram is drawn with frequency density on the vertical axis. The bar for the class 10 ≤ t < 20 has height 3.0.

Time t	0–5	5–10	10–20	20–40	40–60
Frequency	12	21	k	18	9

(a) Find k, and verify that the total is 90. [2] (b) Find the heights of the bars for 0–5 and 20–40. [2] (c) Estimate the number of patients who waited between 7 and 12 minutes. [2]

3. Eleven ordered readings are 14, 17, 19, 19, 22, 25, 26, 30, 33, 38, 52.

(a) Find the lower quartile, the median and the upper quartile. [3] (b) An outlier is a value more than 1.5 × IQR above Q3 or below Q1. Decide whether 52 is an outlier. [2]

4. The numbers of customers served per hour by cashiers A and B are shown.

                   |  1 | 9
               5 3 |  2 | 2 6 8
           8 6 4 1 |  3 | 1 3 5 7
           7 4 3 1 |  4 | 1 8
               8 2 |  5 | 5
Key: 7 | 4 means 47 for A; 4 | 8 means 48 for B

(a) Find the median and interquartile range for A. [3] (b) For B the median is 33 and the interquartile range is 15. Compare the two cashiers. [2]

5. The masses x kg of loads are coded as y = (x − 1200)/50. The coded values have mean 3.4 and standard deviation 1.2. Find the mean and standard deviation of x. [3]

6. For 13 readings, Σx = 247 and Σx² = 4801.

(a) Find the mean and standard deviation. [3] (b) The reading 19 is removed. Find the new mean and standard deviation. [3]

7. The lifetimes of 50 batteries are recorded in hours.

Lifetime	20–30	30–35	35–40	40–50	50–70
Frequency	4	9	17	14	6

(a) Estimate the mean. [2] (b) Use interpolation to estimate the median. [2] (c) Estimate the interquartile range. [3] (d) Describe the skewness, giving a reason. [1]

8. In a test, the quartiles are Q1 = 33, Q2 = 41 and Q3 = 47. The three lowest marks are 11, 26 and 29, and the three highest are 52, 57 and 71. A mark is an outlier if it lies more than 1.5 × IQR above Q3 or below Q1.

(a) Identify all outliers. [3] (b) State the values at the ends of the whiskers on a box plot. [2] (c) Describe the skewness shown by the quartiles. [1]

9. For 20 marks x, Σ(x − 60) = −30 and Σ(x − 60)² = 1325.

(a) Find the mean of x. [2] (b) Find the standard deviation of x. [2] (c) Every mark is changed to 1.2x + 4. Find the new mean and standard deviation. [2]

10. Two machines fill crates with fruit (mass in kg). For machine A, n = 40, Σx = 1000 and Σx² = 25 230. For machine B, n = 50, Σx = 1230 and Σx² = 31 520.

(a) Find the mean and standard deviation for each machine. [4] (b) Find the mean and standard deviation of all 90 crates. [3] (c) Compare the machines. [2]

11. The lengths of 51 calls, in minutes, are shown on a histogram. The frequency densities are 2.5 for 0–4, 6.0 for 4–6, 3.5 for 6–10, 1.2 for 10–20 and 0.15 for 20–40.

(a) Find the frequency of each class. [2] (b) Estimate the median. [2] (c) Estimate the mean. [2] (d) Describe the skewness, with a reason. [1] (e) Estimate the standard deviation. [3]

Answers

1. (a) Traffic and the number of passengers vary, so the time cannot be constant. [1] (b) A standard deviation of 3.1 shows that the times vary quite widely, so a fixed 12 minutes fits poorly. [1] The mean 12.6 is near 12, so the model may work for a typical day; refine it with a spread of times. [1] [3] Examiner insight: When asked to comment, quote the numbers given and say what they mean for the model.

2. (a) k = 3.0 × 10 = 30. [1] Total 12 + 21 + 30 + 18 + 9 = 90. [1] [2] (b) 0–5: 12 ÷ 5 = 2.4. [1] 20–40: 18 ÷ 20 = 0.9. [1] [2] (c) The 5–10 bar has density 21 ÷ 5 = 4.2, so 7 to 10 gives 3 × 4.2 = 12.6. [1] Add 2 × 3.0 = 6 for 10 to 12: 18.6. [1] [2] Examiner insight: A histogram estimate assumes values are spread evenly within each class, so say so.

3. (a) n/4 = 2.75, round up to the 3rd value: Q1 = 19. [1] n/2 = 5.5, so the 6th value: median = 25. [1] 3n/4 = 8.25, so the 9th value: Q3 = 33. [1] [3] (b) IQR = 33 − 19 = 14, so the upper limit is 33 + 21 = 54. [1] Since 52 < 54, 52 is not an outlier. [1] [2] Examiner insight: Show the limit as well as the verdict: comparing 52 with 54 shows the rule was applied.

4. (a) A has n = 12. The median lies between the 6th and 7th values: (38 + 41)/2 = 39.5. [1] Q1 is the mean of the 3rd and 4th: (31 + 34)/2 = 32.5, and Q3 the mean of the 9th and 10th: (44 + 47)/2 = 45.5. [1] IQR = 45.5 − 32.5 = 13. [1] [3] (b) B’s median is lower (33 against 39.5), so A typically serves more customers per hour. [1] B’s IQR is larger (15 against 13), so B is less consistent. [1] [2] Examiner insight: Count the leaves to confirm n before locating positions.

5. x = 50y + 1200. [1] Mean = 50 × 3.4 + 1200 = 1370. [1] Standard deviation = 50 × 1.2 = 60. [1] [3] Examiner insight: The constant 1200 shifts the mean but not the standard deviation.

6. (a) Mean = 247 ÷ 13 = 19. [1] Variance = 4801/13 − 19² = 369.31 − 361 = 8.31. [1] Standard deviation = 2.88. [1] [3] (b) New n = 12, Σx = 228, Σx² = 4801 − 361 = 4440. [1] Mean = 228 ÷ 12 = 19. [1] Variance = 4440/12 − 361 = 9, so standard deviation = 3. [1] [3] Examiner insight: Removing a value equal to the mean leaves the mean unchanged but raises the standard deviation.

7. (a) Mid-points 25, 32.5, 37.5, 45, 60. Σfx = 100 + 292.5 + 637.5 + 630 + 360 = 2020. [1] Mean = 2020 ÷ 50 = 40.4 hours. [1] [2] (b) Cumulative frequencies 4, 13, 30, 44, 50. Position 25 is in 35–40: 35 + 5 × (25 − 13)/17 = 38.5. [1][1] [2] (c) Q1 at position 12.5 in 30–35: 30 + 5 × (12.5 − 4)/9 = 34.72. [1] Q3 at 37.5 in 40–50: 40 + 10 × (37.5 − 30)/14 = 45.36. [1] IQR = 10.6 hours. [1] [3] (d) The mean 40.4 is above the median 38.5, so the skew is positive. [1] [1] Examiner insight: Keep unrounded values until the end: subtracting the rounded quartiles gives 10.7.

8. (a) IQR = 47 − 33 = 14, so the limits are 33 − 21 = 12 and 47 + 21 = 68. [1] 11 is below 12, so 11 is an outlier. [1] 71 is above 68, so 71 is an outlier, and no other value is. [1] [3] (b) The lower whisker ends at 26 and the upper whisker at 57, the extreme values inside the limits. [1][1] [2] (c) Q2 − Q1 = 8 and Q3 − Q2 = 6, so the skew is slightly negative. [1] [1] Examiner insight: Whiskers end at the last value that is not an outlier.

9. (a) Mean of (x − 60) = −30 ÷ 20 = −1.5. [1] Mean of x = 58.5. [1] [2] (b) Variance = 1325/20 − (−1.5)² = 66.25 − 2.25 = 64. [1] Standard deviation = 8. [1] [2] (c) New mean = 1.2 × 58.5 + 4 = 74.2. [1] New standard deviation = 1.2 × 8 = 9.6; adding 4 does not change the spread. [1] [2] Examiner insight: Square the mean of the coded values, not the mean of x.

10. (a) A: mean 1000/40 = 25.0, variance 25 230/40 − 625 = 5.75, standard deviation 2.40. [1][1] B: mean 1230/50 = 24.6, variance 31 520/50 − 24.6² = 630.4 − 605.16 = 25.24, standard deviation 5.02. [1][1] [4] (b) Combined n = 90, Σx = 2230, mean = 24.8. [1] Σx² = 56 750, variance = 56 750/90 − (2230/90)² = 630.56 − 613.94 = 16.62. [1] Standard deviation = 4.08. [1] [3] (c) The means are close (25.0 and 24.6), so the machines fill similar average amounts. [1] A has the much smaller standard deviation (2.40 against 5.02), so A is more consistent. [1] [2] Examiner insight: To combine groups, add the sums and the sums of squares; do not average the standard deviations.

11. (a) Frequency = density × width: 10, 12, 14, 12, 3. [1] Total = 51. [1] [2] (b) Position 25.5; cumulative 10, 22, 36. It lies in 6–10: 6 + 4 × (25.5 − 22)/14 = 7.0 minutes. [1][1] [2] (c) Mid-points 2, 5, 8, 15, 30. Σfx = 20 + 60 + 112 + 180 + 90 = 462. [1] Mean = 462 ÷ 51 = 9.06 minutes. [1] [2] (d) The mean 9.06 is greater than the median 7.0, so the skew is positive. [1] [1] (e) Σfx² = 40 + 300 + 896 + 2700 + 2700 = 6636. [1] Variance = 6636/51 − (462/51)² = 130.12 − 82.06 = 48.06. [1] Standard deviation = 6.93 minutes. [1] [3] Examiner insight: Frequencies, not densities, go into Σfx and Σfx².

Where marks are usually lost

• Using frequency density as if it were frequency in a mean or variance calculation.
• Forgetting to square the mean when finding variance, or leaving variance as the final answer.
• Rounding quartiles before finding the interquartile range.
• Ending a whisker at an outlier, or not showing the outlier limits.
• Decoding the mean of a coded set correctly but giving the standard deviation without multiplying by b.
• Comparing two data sets without numbers, or without both location and spread.
• Averaging standard deviations when combining two groups.

Next steps

• Re-read the revision notes and study guide.
• Tick off topics on the printable checklist and visit the course hub.
• Take the free 10-minute diagnostics to find weak spots.
• Book a free trial class for live help.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, published by Pearson Education Limited: Unit S1: Statistics 1, sections 1.1 and 2.1 to 2.4.