Cambridge O Level Statistics: Formation of Frequency Distributions (4040) – 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 3.1 of the Cambridge O Level Statistics (4040) syllabus for examination in 2025, 2026 and 2027: class limits, boundaries and mid-points, and class intervals for grouped frequency distributions. Recall the methods from the study guide or the revision notes. See also the Cambridge O Level Statistics hub, the printable checklist and the free diagnostics. Show your working. Give answers exactly unless a question says otherwise.

Questions

1. A shop records the number of emails it receives each day, grouped as 1-5, 6-10, 11-15 and 16-20.

(a) State the lower and upper boundaries of the class 6-10. [2] (b) State the width of this class. [1] (c) Find the mid-point of this class. [1]

2. Times to complete a task are recorded to 1 decimal place of a minute and grouped as 4.0-4.4, 4.5-4.9 and 5.0-5.4. For the class 4.5-4.9, find the boundaries, the width and the mid-point. [4]

3. The masses m kg of some parcels are grouped as 60 ≤ m < 70, 70 ≤ m < 85, 85 ≤ m < 100 and 100 ≤ m < 130.

(a) Explain why you do not need to adjust these classes to find boundaries. [1] (b) Find the width and the mid-point of each class. [4]

4. The ages of people at a club are recorded in completed years and grouped as 20-24, 25-29 and 30-39.

(a) Marta is 24 years 11 months old. State her class. [1] (b) Write the exact range of ages covered by 20-24 and state its width. [2] (c) Find the mid-point of the class 30-39. [1]

5. A student groups some whole-number test scores as 0-10, 10-20, 20-30 and 40-50.

(a) Give two faults in these classes. [2] (b) Write a corrected set of classes, with equal widths, that covers scores from 0 to 59. [2]

6. Data are recorded as whole numbers.

(a) A class has boundaries 24.5 and 34.5. State its limits, width and mid-point. [3] (b) Another class has a mid-point of 47 and a width of 5. Find its lower and upper limits. [3]

7. The times, in minutes, that 24 customers waited are: 23, 41, 17, 35, 28, 52, 19, 33, 44, 26, 38, 15, 47, 31, 22, 29, 36, 12, 40, 25, 34, 55, 21, 30.

(a) Copy and complete a tally and frequency table using five classes of width 10 beginning at 10-19. [3] (b) State the boundaries of the class 30-39 and its mid-point. [2] (c) Show that your table accounts for every one of the customers. [1]

8. A teacher has 60 scores between 3 and 98 and wants to group them. She considers class widths of 1, 10 and 50. Evaluate each choice and recommend one. [4]

9. A clinic records waiting times in minutes.

Time t (minutes)	0 ≤ t < 5	5 ≤ t < 15	15 ≤ t < 30	30 ≤ t < 60
Frequency	6	14	12	8

(a) State the width of each class. [2] (b) Find the mid-point of each class. [2] (c) Estimate the mean waiting time to 3 significant figures. [3] (d) Give a reason why the mean found in (c) is only approximate. [1]

10. Heights are measured and recorded to the nearest centimetre, then grouped as 150-154, 155-159 and 160-164.

(a) Find the boundaries of 155-159. [2] (b) A girl’s true height is 154.6 cm. State the class she is placed in and explain. [2] (c) State the smallest and largest true heights that can be placed in 155-159, using inequalities. [2]

11. Rainfall is recorded to 1 decimal place of a millimetre. A student works on the class 5.0-9.9 and writes: “The next class starts at 10.0, so the boundaries are 4.9 and 10.0. The width is 5.1 and the mid-point is (5.0 + 10.0) ÷ 2 = 7.5.”

(a) Identify the three errors. [3] (b) Give the correct boundaries, width and mid-point. [3]

Answers

1. (a) The gap between 5 and 6 is 1, so the half-gap is 0.5 [1]. Boundaries 5.5 and 10.5 [1]. (b) 10.5 − 5.5 = 5 [1]. (c) (5.5 + 10.5) ÷ 2 = 8 [1]. Examiner insight: A width of 4 from 10 − 6 is a common slip. Working from the boundaries, 10.5 − 5.5, avoids it.

2. The gap is 0.1, so the half-gap is 0.05 [1]. Boundaries 4.45 and 4.95 [1]. Width 4.95 − 4.45 = 0.5 [1]. Mid-point (4.45 + 4.95) ÷ 2 = 4.7 [1]. Examiner insight: Using 0.5 as the half-gap for decimal data gives boundaries that cut into the next class, so check that your upper boundary is lower than the next class’s lower limit.

3. (a) The classes are written with inequalities, so the numbers printed are already the boundaries [1]. (b) Widths: 70 − 60 = 10 [1], 85 − 70 = 15, 100 − 85 = 15, 130 − 100 = 30 [1]. Mid-points: 65, 77.5, 92.5, 115 [2]. Examiner insight: When a question gives four classes and asks for two measures, a small table keeps each value easy to find.

4. (a) She has not yet had her 25th birthday, so her class is 20-24 [1]. (b) The class holds exact ages from 20 up to but not including 25, so 20 ≤ age < 25 [1]. Width 5 [1]. (c) The class covers exact ages from 30 to 40, so the mid-point is (30 + 40) ÷ 2 = 35 [1]. Examiner insight: For ages in completed years, the exact range ends at the next whole number, so a mid-point of 34.5 taken from the printed limits of 30 and 39 does not match the ages being described.

5. (a) The classes overlap, because 10 and 20 each appear in two classes [1]. The range 30 to 40 has no class, so a score in it cannot be placed [1]. (b) 0-9, 10-19 and so on in steps of ten, ending with 50-59 [1]. Each class has width 10, no value is in two classes, and nothing is missing [1]. Examiner insight: A corrected set must show no overlap, no gap and equal widths, and a one-line check of each condition makes it easy to credit.

6. (a) Limits 25 and 34 [1]. Width 34.5 − 24.5 = 10 [1]. Mid-point (24.5 + 34.5) ÷ 2 = 29.5 [1]. (b) Half the width is 2.5, so the boundaries are 47 − 2.5 = 44.5 and 47 + 2.5 = 49.5 [1]. Moving each inwards by 0.5 gives the limits [1]: 45 and 49 [1]. Examiner insight: Working backwards from a mid-point is safer through the boundaries first, because a slip in a jump straight to the limits is hard to trace.

7. (a) Tallies give 10-19: 4 (12, 15, 17, 19) and 20-29: 7 [1]; 30-39: 7 and 40-49: 4 [1]; 50-59: 2 [1]. Frequencies 4, 7, 7, 4, 2. (b) Boundaries 29.5 and 39.5 [1]. Mid-point 34.5 [1]. (c) 4 + 7 + 7 + 4 + 2 = 24, which equals the 24 customers [1]. Examiner insight: Adding up the frequencies is the standard check for a tally, and writing the sum out shows it was done.

8. Width 1 would need 96 classes for only 60 values, so most rows would be empty or hold one value and the shape would not show [1]. Width 50 gives only 2 classes, which hides the shape of the data [1]. Width 10 gives 10 classes, with an average of about 6 values per class, which is enough to show the shape [1]. Recommend width 10, such as 0-9, 10-19 and so on up to 90-99 [1]. Examiner insight: An evaluation scores best when it uses the numbers in the question, here 60 values and a range of 95, and not a general remark such as “too many”.

9. (a) Widths 5, 10, 15, 30 [2]. (b) Mid-points 2.5, 10, 22.5, 45 [2]. (c) Σfx = 6 × 2.5 + 14 × 10 + 12 × 22.5 + 8 × 45 = 15 + 140 + 270 + 360 = 785 [1]. Σf = 40 [1]. Mean = 785 ÷ 40 = 19.625 = 19.6 minutes [1]. (d) The actual times are not known, and each value is taken to be at the mid-point of its class [1]. Examiner insight: Show the products f × mid-point in a row or column, so a slip in one product can be separated from the method.

10. (a) The gap is 1, so the half-gap is 0.5 [1]. Boundaries 154.5 and 159.5 [1]. (b) 154.6 rounds to 155 [1], so she is in 155-159, not 150-154 [1]. (c) 154.5 ≤ h < 159.5 [2]. Examiner insight: Inequality signs are better than the words “154.5 to 159.5”, because the signs show which end is included.

11. (a) The gap is 0.1, so the half-gap is 0.05, not 0.1 [1]. The width should be 5.0, not 5.1, because the boundaries were wrong [1]. The mid-point used the limit 10.0 instead of the upper limit 9.9 or the boundary 9.95 [1]. (b) Boundaries 4.95 and 9.95 [1]. Width 9.95 − 4.95 = 5.0 [1]. Mid-point (4.95 + 9.95) ÷ 2 = 7.45 [1]. Examiner insight: When asked to identify errors, name each one and say what should have been done, since a bare corrected answer does not show that the fault was found.

Where marks are usually lost

• Taking the width as upper limit minus lower limit, giving 4 for 6-10.
• Using a half-gap of 0.5 for data recorded to 1 decimal place.
• Writing boundaries as whole numbers, such as 5 and 10, instead of 5.5 and 10.5.
• Adjusting a class written with inequalities, when its printed numbers are already boundaries.
• Using the printed limits for the mid-point of an age class recorded in completed years.
• Giving a range of true values without inequality signs.
• Commenting on class choices with no figures from the question.
• Not checking that the frequency column totals the size of the sample.

Next steps

• Review the methods in the revision notes and the study guide.
• Visit the Cambridge O Level Statistics hub and tick off the checklist.
• Take the free 10-minute diagnostics to find weak spots.
• Ask for a free trial class for help with any question you could not finish.

Frequently asked questions

What is the difference between class limits and class boundaries? Class limits are the numbers printed for a class, such as 6-10. Class boundaries are found by moving half the gap between neighbouring classes outwards, so for whole-number data the class 6-10 has boundaries 5.5 and 10.5.

How do you find the width of a class interval? Subtract the lower boundary from the upper boundary, not the lower limit from the upper limit. Taking upper limit minus lower limit is one of the most common ways to lose a mark on this topic.

What half-gap do you use for data recorded to 1 decimal place? The gap between neighbouring classes is 0.1, so the half-gap is 0.05. Using 0.5 gives boundaries that cut into the next class.

Do you adjust classes written with inequalities to find boundaries? No. When classes are written as inequalities, such as 60 ≤ m < 70, the printed numbers are already the boundaries, so no adjustment is needed.

Which part of the Cambridge O Level Statistics 4040 syllabus do these questions cover? They cover section 3.1 of the 4040 syllabus for examination in 2025, 2026 and 2027: class limits, boundaries, mid-points and class intervals for grouped frequency distributions. The questions are original, not past-paper questions.

Official syllabus

Cambridge O Level Statistics (4040), syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment (Cambridge International). Section 3.1: class measures for grouped frequency distributions, including class limits, boundaries and mid-points, and class intervals.