OCR A Level Mathematics A: Sampling and interpreting single-variable data (H240) – 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 follow sections 2.01a to 2.01d and 2.02a, 2.02b and 2.02f to 2.02j of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026). All of the content is Stage 1, shared with AS Level. Give estimates as estimates and state any definition of outlier you use.

Read the study guide first if a topic is new, and use the revision notes for a quick recap. The OCR A Level Mathematics hub and printable checklist show the rest of the course, and the free 10-minute diagnostics help you pick what to practise next.

Questions

1. A council wants to know how the 18,000 working residents of a town travel to work. It posts a questionnaire to 500 addresses.

(a) State the population and the sample. [2] (b) Give a reason why the council might prefer a sample to a census. [1]

2. A club has 260 members on a numbered list, 001 to 260. Describe how to select a simple random sample of 5 members using a random number generator. Say what you do with the generated numbers 144, 301 and 144. [4]

3. A shop has takings records for 364 consecutive days, listed in date order starting on a Monday. The manager takes every 7th day from a random start to estimate the average daily takings.

(a) Explain why this sample is likely to give a biased estimate. [3] (b) Suggest a better method. [1]

4. An interviewer is told to question 20 men and 20 women outside a shopping centre between 10 am and noon on a weekday about local parking charges.

(a) Name the sampling method. [1] (b) Give two reasons why conclusions about all residents may be unreliable. [2] (c) Suggest one improvement. [1]

5. A histogram shows the journey times of 100 commuters. The classes are 0 to <10, 10 to <15, 15 to <25, 25 to <45 and 45 to <65 minutes. The 0 to <10 class has frequency 8, the 10 to <15 bar has height 3.2, the 15 to <25 class has frequency 34, the 25 to <45 class has frequency 30 and the 45 to <65 class has frequency 12.

(a) Find the frequency of the 10 to <15 class and the height of the 25 to <45 bar. [2] (b) Estimate how many journeys took between 12 and 20 minutes. [2] (c) A journey is chosen at random from the 100. Find the probability that it took under 15 minutes. [2]

6. The number of emails received on 15 days is, in order: 3, 5, 6, 6, 7, 8, 9, 10, 10, 11, 12, 13, 14, 15, 21. You are given that Σx = 150 and Σx² = 1796.

(a) Find the median, the quartiles and the interquartile range. [3] (b) Decide whether 21 is an outlier using the 1.5 × IQR definition. [2] (c) Decide whether 21 is an outlier using the 2 standard deviation definition. [3]

7. For 30 readings, Σx = 1170 and Σx² = 48060.

(a) Find the mean and standard deviation. [3] (b) A second batch of 30 readings has mean 39 and standard deviation 4. Compare the two batches. [2]

8. The masses of 50 parcels are grouped: 0 to <20 kg (6 parcels), 20 to <30 (14), 30 to <40 (18), 40 to <60 (10), 60 to <100 (2).

(a) Estimate the mean and standard deviation. [4] (b) Estimate the median. [2] (c) Comment on the shape of the distribution. [1]

9. In question 7, one reading was recorded as 62 but should have been 26.

(a) Find the corrected mean and standard deviation. [4] (b) Another reading is missing. Explain how to treat it. [2]

10. (a) State one advantage and one disadvantage of a box-and-whisker diagram compared with a histogram. [2] (b) A bar chart of waiting times uses unequal class widths, with frequency as the bar height. Explain why this misleads and what should be plotted instead. [2]

11. A library records its visitors on 40 days, taken as the first 40 days of January. The lower quartile is 62, the upper quartile is 90, the mean is 78 and the standard deviation is 14. The largest value is 140 and the second largest is 112.

(a) Criticise the sampling method for estimating typical daily visitors over a year. [2] (b) Show that 140 is an outlier by both definitions. [3] (c) Test 112 using both definitions and comment. [3] (d) The 140 was caused by a one-off event and is removed. Find the new mean. [2]

Answers

1. (a) Population: all 18,000 working residents of the town [1]. Sample: the 500 addresses surveyed (or the people replying) [1]. (b) A census would take much longer and cost far more [1]. [3] Examiner insight: name the sample in context rather than writing “the people asked”; a vague label may not show you have identified it.

2. Number the members 001 to 260 as the sampling frame [1]. Generate random three-digit numbers [1]. Reject 301 because it is outside the range [1]. Reject the second 144 because it repeats, and keep generating until 5 different members are chosen [1]. [4] Examiner insight: “pick at random” on its own describes nothing; give the frame, the numbering and the rule for rejects.

3. (a) Every 7th day gives the same day of the week each time [1]. Takings differ by weekday, for example weekends against Mondays [1]. So the sample represents one weekday only and the estimate is biased [1]. (b) Select days by simple random sampling from the numbered list [1]. [4] Examiner insight: tie the bias to this data (the weekly cycle), not just “systematic sampling is biased”.

4. (a) Quota sampling [1]. (b) It is not random because the interviewer chooses who to ask [1]. Only people at that place on a weekday morning are reached, so most workers are missed [1]. (c) Choose residents by simple random sampling from a list of addresses [1]. [4] Examiner insight: give two distinct points for two marks; two wordings of “not random” read as one point.

5. (a) Frequency = 3.2 × 5 = 16 [1]. Height = 30 ÷ 20 = 1.5 [1]. (b) 12 to 15: 3 × 3.2 = 9.6 [1]. 15 to 20: 5 × 3.4 = 17, so total ≈ 26.6 (about 27 journeys), an estimate [1]. (c) (8 + 16)/100 [1] = 0.24 [1]. [6] Examiner insight: for a part of a bar, multiply the density by the width of the part you want; using the whole class frequency would overstate the estimate.

6. (a) Median is the 8th value: 10 [1]. Q1 is the 4th value: 6, Q3 the 12th: 13 [1]. IQR = 13 − 6 = 7 [1]. (b) Upper fence = 13 + 1.5 × 7 = 23.5 [1]. 21 < 23.5, so not an outlier [1]. (c) Mean = 150/15 = 10 [1]. Variance = 1796/15 − 10² ≈ 19.73, so sd ≈ 4.44 [1]. 10 + 2 × 4.44 ≈ 18.9 < 21, so 21 is an outlier by this definition [1]. [8] Examiner insight: the two definitions can disagree, so say which one you are applying and compare with the nearer quartile.

7. (a) Mean = 1170/30 = 39 [1]. Variance = 48060/30 − 39² = 1602 − 1521 = 81 [1]. sd = 9 [1]. (b) The averages are the same [1]. The second batch has the smaller standard deviation, so it is more consistent [1]. [5] Examiner insight: a comparison needs a comment on average and on spread, written about the readings and not only about the numbers.

8. (a) Midpoints 10, 25, 35, 50, 80 [1]. Σfx = 1700, so mean = 34 kg [1]. Σfx² = 69200 [1]. Variance = 69200/50 − 34² = 1384 − 1156 = 228, so sd ≈ 15.1 kg [1]. (b) The 25th value lies in 30 to <40, where 20 parcels are below 30 [1]. Median ≈ 30 + (5/18) × 10 ≈ 32.8 kg [1]. (c) The mean exceeds the median, and the long upper class suggests positive skew [1]. [7] Examiner insight: write that these are estimates because midpoints stand in for the values; show Σfx² so a slip is easy to trace.

9. (a) Corrected Σx = 1170 − 62 + 26 = 1134, so mean = 37.8 [1]. Corrected Σx² = 48060 − 3844 + 676 = 44892 [1]. Variance = 44892/30 − 37.8² = 1496.4 − 1428.84 = 67.56 [1]. sd ≈ 8.22 [1]. (b) Leave it out of the calculation and use n = 29, rather than entering 0 [1]. Check whether the gap hides a pattern, such as a faulty sensor [1]. [6] Examiner insight: correct both Σx and Σx² before recalculating; adjusting the mean alone leaves the spread wrong.

10. (a) Advantage: shows median, quartiles and skew clearly and compares several sets easily [1]. Disadvantage: individual values and shape detail are hidden [1]. (b) Wider classes would look larger although they may hold few items per minute [1]. Plot frequency density (frequency ÷ class width) in a histogram [1]. [4] Examiner insight: give the quantity plotted, frequency density, and the reason, because area then stands for frequency.

11. (a) The first 40 days of January are not random and are likely to be unrepresentative of the year (holidays, weather) [1]. So the estimate may be biased; a simple random sample of days across the year is better [1]. (b) IQR = 90 − 62 = 28; 90 + 1.5 × 28 = 132 < 140 [1]. 78 + 2 × 14 = 106 < 140 [1]. So 140 is an outlier by both [1]. (c) Definition 1: 112 < 132, so not an outlier [1]. Definition 2: 112 > 106, so an outlier [1]. The definitions disagree, so the conclusion depends on the definition chosen and should be stated [1]. (d) Total of the 40 values = 40 × 78 = 3120 [1]. New mean = (3120 − 140)/39 ≈ 76.4 [1]. [10] Examiner insight: keep the removal separate from the test; say why a value is removed (a known cause) rather than only that it is an outlier.

Where marks are usually lost

• Describing a sample as biased without saying who is missed in this context.
• Calling an opportunity or quota sample random.
• Using frequency, not frequency density, for the heights in a histogram with unequal widths.
• Taking the whole class frequency when only part of a class is needed.
• Using class boundaries instead of midpoints for grouped mean and standard deviation.
• Forgetting to subtract the mean squared in the variance.
• Stating grouped results as exact values.
• Applying an outlier definition from the wrong quartile or without writing the fence.
• Entering 0 for missing data.
• Comparing two sets with an average only.

Next steps

• Recap with the revision notes.
• Re-read the study guide.
• Open the OCR A Level Mathematics hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

The OCR A Level in Mathematics A (H240) specification, Version 3.1 (August 2026), published by OCR, sections 2.01 and 2.02.