Cambridge O Level Statistics: Effects of a constant and linear transformation to a given mean and SD (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 sections 6.1 and 6.2 of the Cambridge O Level Statistics (4040) syllabus for examination in 2025, 2026 and 2027: the effect on the mean and standard deviation of adding a constant and of multiplying by a constant, and linear transformation of data to a given mean and standard deviation. Standard deviations use the population form, dividing by n. Use the study guide or the revision notes if you need to recall a method. See also the Cambridge O Level Statistics hub, the printable checklist and the free 10-minute diagnostics. Show your working. Give answers to 3 significant figures unless a question says otherwise.

Questions

1. A set of data has mean 24 and standard deviation 5.

(a) Each value is increased by 10. Write down the new mean and the new standard deviation. [2] (b) The original values are each multiplied by 3. Write down the new mean and the new standard deviation. [2]

2. The masses of parcels have mean 3.2 kg and standard deviation 0.4 kg. The masses are converted to grams.

(a) Find the mean and the standard deviation in grams. [2] (b) A student says the variance in grams is 1000 times the variance in kg. Find the variance in grams and explain whether the student is correct. [2]

3. Five values of x are 2, 4, 6, 8 and 10.

(a) Calculate the mean and the standard deviation of x. [3] (b) Each value is transformed using y = 2x − 3. Use the rules for transformations to find the mean and standard deviation of y. [2] (c) List the values of y and calculate their standard deviation directly to confirm your answer to (b). [2]

4. A courier records the delivery times, x minutes, of 25 parcels. Σx = 1500 and Σx² = 92 500.

(a) Calculate the mean and the standard deviation of x. [3] (b) The cost of a delivery in pesos is y = 20 + 5x. Find the mean and the standard deviation of the cost. [2] (c) Find the variance of the cost. [1]

5. Some measurements, x, are coded using y = (x − 200) / 10. The coded values have mean 4.6 and standard deviation 2.1.

(a) Write x in terms of y. [1] (b) Find the mean and standard deviation of x. [2] (c) Find the variance of x. [1]

6. A set of scores has mean 38 and standard deviation 6. The scores are to be transformed using y = a + bx so that the new mean is 100 and the new standard deviation is 15. Find a and b. [4]

7. The temperatures x of a set of readings have mean 40 and standard deviation 8. New values are found using y = 100 − 2x.

(a) Find the mean and the standard deviation of y. [2] (b) State the variance of y and explain why the standard deviation of y is not negative. [2]

8. Paper A has raw marks with mean 52 and standard deviation 14. Paper B has raw marks with mean 64 and standard deviation 7. The Paper A marks are to be scaled using y = a + bx so that they have the same mean and standard deviation as Paper B.

(a) Find a and b. [4] (b) A candidate scored 80 on Paper A. Find the scaled mark. [2]

9. (a) Data x are transformed using y = 4x − 7. The y values have mean 29 and standard deviation 10. Find the mean and standard deviation of x. [3] (b) A second set of data w is transformed using v = 15 − 3w. The v values have mean 3 and standard deviation 12. Find the mean and standard deviation of w. [2]

10. A quiz awards 10, 20, 30, 40 or 50 points. The frequencies for 20 players are shown.

Points x	10	20	30	40	50
Frequency	1	3	5	5	6

(a) Calculate the mean number of points. [2] (b) Calculate the standard deviation. [3] (c) The points are to be rescaled using y = a + bx so that the mean is 60 and the standard deviation is 9. Find a and b. [3] (d) Find the rescaled values for players who scored 10 and 50 points. [2]

11. Data x have mean x̄ and standard deviation s. New data are given by y = a + bx, where a and b are constants and b ≠ 0.

(a) Show that the mean of y is a + b x̄. [2] (b) Show that the standard deviation of y is |b| s. [3] (c) State which of a and b has no effect on the standard deviation. [1]

Answers

1. (a) Mean = 24 + 10 = 34 [1]; standard deviation is unchanged at 5 [1]. (b) Mean = 3 × 24 = 72 [1]; standard deviation = 3 × 5 = 15 [1]. Examiner insight: In part (a), state clearly that the standard deviation stays 5. Writing 15 or 10 suggests the two rules have been swapped, so check each rule against its example before you write the answer.

2. (a) Multiply by 1000: mean = 3200 g [1]; standard deviation = 400 g [1]. (b) Variance in kg² = 0.4² = 0.16, so variance in g² = 400² = 160 000 [1]. The variance is multiplied by 1000² = 1 000 000, not 1000, so the student is wrong [1]. Examiner insight: Give a reason that uses the square of the multiplier. A bare “no” with the figure 160 000 and no comparison with 1000 leaves the explanation incomplete.

3. (a) Mean = 30 / 5 = 6 [1]. Squared deviations: 16, 4, 0, 4, 16, total 40, so variance = 40 / 5 = 8 [1]. Standard deviation = √8 = 2.83 [1]. (b) Mean of y = 2 × 6 − 3 = 9 [1]; standard deviation of y = 2 × 2.83 = 5.66 [1]. (c) y values: 1, 5, 9, 13, 17. Deviations from 9: −8, −4, 0, 4, 8, squares total 64 + 16 + 0 + 16 + 64 = 160 [1]. Variance = 32, standard deviation = √32 = 5.66, which agrees [1]. Examiner insight: A “confirm” part is answered by the working you show. Listing the y values and the deviations is what makes the check visible; stating 5.66 alone is not a check.

4. (a) Mean = 1500 / 25 = 60 [1]. Variance = 92 500 / 25 − 60² = 3700 − 3600 = 100 [1]. Standard deviation = 10 [1]. (b) Mean cost = 20 + 5 × 60 = 320 [1]; standard deviation = 5 × 10 = 50 [1]. (c) Variance = 50² = 2500 [1]. Examiner insight: Subtract the square of the mean from Σx² / n, not from Σx². Writing the line 3700 − 3600 makes your method clear even if you slip afterwards.

5. (a) y = (x − 200) / 10, so x = 200 + 10y [1]. (b) Mean of x = 200 + 10 × 4.6 = 246 [1]; standard deviation = 10 × 2.1 = 21 [1]. (c) Variance = 21² = 441 [1]. Examiner insight: Rearrange the coding before you apply the rules. Using x = 10y − 200 or adding 200 to the standard deviation are the usual slips.

6. b = 15 / 6 = 2.5 [1]. Mean condition: 100 = a + 2.5 × 38 [1]. a = 100 − 95 = 5 [1]. y = 5 + 2.5x, so a = 5 and b = 2.5 [1]. Examiner insight: Find b from the standard deviations first. If you start with the means you have two unknowns and no clear way to separate them.

7. (a) Mean of y = 100 − 2 × 40 = 20 [1]; standard deviation = |−2| × 8 = 16 [1]. (b) Variance = 16² = 256 [1]. A standard deviation is a square root measuring spread, so it cannot be negative, and the sign of b is removed [1]. Examiner insight: Writing −16 for the standard deviation is an error even when the mean is right. Say that the sign of b affects the mean but not the spread.

8. (a) b = 7 / 14 = 0.5 [1]. Using the means: 64 = a + 0.5 × 52 [1]. a = 64 − 26 = 38 [1]. a = 38, b = 0.5 [1]. (b) Scaled mark = 38 + 0.5 × 80 [1] = 78 [1]. Examiner insight: Check your answer by substituting the old mean, 52, into your rule. It must return 64. That one line catches arithmetic slips before you use the rule in part (b).

9. (a) Standard deviation of x = 10 / 4 = 2.5 [1]. Mean: 29 = 4 × (mean of x) − 7 [1], so the mean of x = 36 / 4 = 9 [1]. (b) Standard deviation of w = 12 / 3 = 4 [1]; 3 = 15 − 3 × (mean of w), so the mean of w = 4 [1]. Examiner insight: Reverse the rule in the right order: undo a first for the mean, and divide by |b| for the standard deviation. In (b) the divisor is 3, not −3.

10. (a) Σfx = 10 + 60 + 150 + 200 + 300 = 720 [1]. Mean = 720 / 20 = 36 [1]. (b) Σfx² = 100 + 1200 + 4500 + 8000 + 15 000 = 28 800 [1]. Variance = 28 800 / 20 − 36² = 1440 − 1296 = 144 [1]. Standard deviation = 12 [1]. (c) b = 9 / 12 = 0.75 [1]. 60 = a + 0.75 × 36, so a = 60 − 27 = 33 [1]. y = 33 + 0.75x, so a = 33 and b = 0.75 [1]. (d) For 10 points: 33 + 7.5 = 40.5 [1]. For 50 points: 33 + 37.5 = 70.5 [1]. Examiner insight: In a frequency table, multiply each x² by its frequency. Using Σfx² = Σ(fx)² is a common error that gives a very different variance.

11. (a) Σy = Σ(a + bx) = na + bΣx [1]. Dividing by n gives ȳ = a + b x̄ [1]. (b) Each deviation is y − ȳ = (a + bx) − (a + b x̄) = b(x − x̄) [1]. So Σ(y − ȳ)² / n = b² × Σ(x − x̄)² / n = b² s² [1]. The standard deviation is √(b² s²) = |b| s [1]. (c) The constant a has no effect on the standard deviation [1]. Examiner insight: In a “show that” question, every line of the chain must appear. Going straight from y − ȳ to the result without the line b(x − x̄) leaves a gap in the argument.

Where marks are usually lost

• Multiplying the standard deviation by b² instead of b.
• Leaving the variance multiplied by b instead of b².
• Adding the constant a to the standard deviation.
• Finding a before b in a 6.2 question.
• Using a negative standard deviation when b is negative.
• Rearranging a coded relationship the wrong way, such as writing x = 10y − 200 instead of x = 200 + 10y.
• In a frequency table, forgetting to multiply x² by the frequency when finding the variance.
• Giving the answer with no working. The syllabus says essential working must be shown for full marks to be awarded.
• Leaving out the final check that the old mean maps to the target mean.

Next steps

• Review the revision notes and the study guide.
• Go to the Cambridge O Level Statistics hub and tick off topics on the printable checklist.
• Take one of the free 10-minute diagnostics to find your next weak topic.
• Book a free trial class if you would like help with this topic.

Official syllabus

Cambridge International, Cambridge O Level Statistics 4040, syllabus for examination in 2025, 2026 and 2027, sections 6.1 and 6.2.