Cambridge IGCSE Statistics: Transforming, combining and comparing data sets: coding, combined means and standardising (0479) – 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.

This set practises Topic 6, Transformations of data sets, from the Cambridge IGCSE Statistics (0479) syllabus for examination in 2027, outcomes 6.1 to 6.6. The standard deviation here divides by n. Read the study guide first, and use the revision notes as a quick reference. Tick off the topic on the printable checklist, see the Statistics hub, or check your level with the free 10-minute diagnostics.

Questions

1. A data set has mean 15 and standard deviation 3.2. Each value x is coded using y = 4x − 10. (a) Calculate the mean of y. [2] (b) Calculate the standard deviation of y. [2]

2. A data set has median 40 and interquartile range 12. Find the median and interquartile range of the new data set when each value x is changed to: (a) y = 0.5x + 3 [2] (b) y = 100 − x [2]

3. The values 12, 15, 17, 17, 20, 27 have mean 18 and median 17. (a) A new value, 30, is added. State whether the mean and the median increase, decrease or stay the same, and give the new mean. [3] (b) Instead, the value 27 is removed from the original six. State what happens to the mean, with a reason, and find the new mean. [2]

4. Group P has 15 values with mean 8.2 and standard deviation 1.5. Group Q has 10 values with mean 9.7 and standard deviation 2.1. Calculate the mean and the standard deviation of all 25 values. [5]

5. A group of 8 values has mean 15.5. It is combined with a second group of 12 values. The 20 values together have mean 18.2. Find the mean of the second group. [4]

6. The data set 1, 3, 4, 5, 7 is standardised to mean 0 and standard deviation 1. Calculate the five standardised values. [4]

7. Test marks have mean 48 and standard deviation 9. They are scaled to a new mean of 100 and a new standard deviation of 15. (a) Find the scaled value of a mark of 57. [2] (b) Find the original mark that becomes 85. [2]

8. In a school sports day, the high jump results have mean 1.60 m and standard deviation 0.08 m. The javelin results have mean 60 m and standard deviation 5 m. Hira jumps 1.72 m and throws 66 m. Use standardised values to decide which result is better relative to the other competitors. [4]

9. The waiting times, in minutes, at two bus stops are summarised below.

	Mean	Standard deviation	Median	IQR
Route A	8.4	1.2	8	1.5
Route B	6.9	3.4	6	4.0

(a) Using the mean and standard deviation, compare the waiting times on the two routes. [3] (b) A company says that route B “is quicker”. Give one reason from the figures that supports this and one reason that weakens it. [2]

10. A set of 40 values x is coded using y = x − 50. The coded values have Σy = 60 and Σy² = 1 090. (a) Find the mean and the standard deviation of x. [4] A second set has 60 values with mean 48.5 and standard deviation 4. (b) Find the mean of all 100 values. [2] (c) Find the standard deviation of all 100 values. [3]

11. A taxi firm records the length of 30 journeys. The times have mean 20 minutes, standard deviation 4.5 minutes, median 19 minutes and interquartile range 6 minutes. The fare in dollars is F = 0.8t + 5, where t is the journey time in minutes. (a) Find the mean and standard deviation of the fares. [3] (b) Find the median and interquartile range of the fares. [2] (c) A 31st journey takes exactly 20 minutes. State what happens to the mean journey time and show that the standard deviation decreases, giving its new value. [3] (d) A fare of 25 dollars is charged. Show that its standardised value is the same as that of the journey time that produced it. [2]

Answers

1. (a) Mean of y = 4 × 15 − 10 [1] = 50 [1] (b) Subtracting 10 does not change the spread [1], so standard deviation = 4 × 3.2 = 12.8 [1] Examiner insight: Show the rule you are using. A wrong answer with the rule written down is easier to give credit to than a bare number.

2. (a) Median = 0.5 × 40 + 3 = 23 [1]; IQR = 0.5 × 12 = 6 [1] (b) Median = 100 − 40 = 60 [1]; the multiplier is −1, so the IQR is 1 × 12 = 12 [1] Examiner insight: A spread cannot be negative. Writing an IQR of −12 for part (b) shows the sign rule has not been understood.

3. (a) New total = 108 + 30 = 138, so new mean = 138/7 = 19.7 (3 s.f.) [1]. The mean increases because 30 is above the old mean of 18 [1]. The ordered data is 12, 15, 17, 17, 20, 27, 30, so the median is the fourth value, 17: it stays the same [1] (b) Removing 27 (above the mean 18) means the mean decreases [1]. New mean = 81/5 = 16.2 [1] Examiner insight: Give the reason by comparing the new value with the existing mean. “It gets bigger because there is more data” does not explain anything.

4. Totals: 15 × 8.2 = 123 and 10 × 9.7 = 97 [1]. Combined mean = 220/25 = 8.8 [1] Σx² for P = 15 × (1.5² + 8.2²) = 15 × 69.49 = 1 042.35 and for Q = 10 × (2.1² + 9.7²) = 985 [1] Variance = (1 042.35 + 985)/25 − 8.8² = 81.094 − 77.44 = 3.654 [1] Standard deviation = √3.654 = 1.91 (3 s.f.) [1] Examiner insight: Averaging the two standard deviations gives 1.8 and is a common wrong route. Set your working out so each stage (totals, Σx², variance, root) can be seen.

5. Total of all 20 = 20 × 18.2 = 364 [1]. Total of first group = 8 × 15.5 = 124 [1]. Total of second group = 364 − 124 = 240 [1]. Mean = 240/12 = 20 [1] Examiner insight: Work with totals, not means. Subtracting 15.5 from 18.2 and stopping there does not use the group sizes.

6. Mean = (1 + 3 + 4 + 5 + 7)/5 = 4 [1]. Squared deviations: 9, 1, 0, 1, 9, which sum to 20, so variance = 4 [1] and σ = 2 [1]. Standardised values (x − 4)/2: −1.5, −0.5, 0, 0.5, 1.5 [1] Examiner insight: Check your answer: standardised data must have mean 0, and the squares of the z-values must add up to the number of values (here 5).

7. (a) z = (57 − 48)/9 = 1 [1]; scaled value = 100 + 15 × 1 = 115 [1] (b) z = (85 − 100)/15 = −1 [1]; original = 48 + 9 × (−1) = 39 [1] Examiner insight: For the reverse direction, undo the scaling first to find z, then rebuild with the original mean and standard deviation. Do not apply the forward rule again.

8. High jump: z = (1.72 − 1.60)/0.08 = 1.5 [1]. Javelin: z = (66 − 60)/5 = 1.2 [1]. The high jump has the larger standardised value [1], so it is better relative to the other competitors, because it is 1.5 standard deviations above the mean compared with 1.2 [1] Examiner insight: A conclusion must refer to the standardised values. Saying “66 is bigger than 1.72” compares units that cannot be compared.

9. (a) Route A has the longer waits generally, because its mean (8.4 minutes) is larger than route B’s (6.9 minutes) [1]. Route A is more consistent, because its standard deviation (1.2) is smaller than B’s (3.4) [1]. Both statements refer to the context, waiting times at the stops [1] (b) Supports: B has the lower mean and median (6.9 and 6), so waits are shorter on average [1]. Weakens: B’s spread is much larger (standard deviation 3.4, IQR 4.0), so some waits will be far longer than average, and waits are less predictable [1] Examiner insight: Pair each centre with its spread and quote figures from both routes. A one-sided answer, or one with no numbers, leaves out half of the comparison.

10. (a) Mean of y = 60/40 = 1.5 [1]. Variance of y = 1 090/40 − 1.5² = 27.25 − 2.25 = 25 [1]. σ = 5 [1]. Mean of x = 50 + 1.5 = 51.5, and σ of x is the same as that of y, 5 [1] (b) Totals: 40 × 51.5 = 2 060 and 60 × 48.5 = 2 910 [1]. Combined mean = 4 970/100 = 49.7 [1] (c) Σx² for first set = 40 × (5² + 51.5²) = 107 090; for second set = 60 × (4² + 48.5²) = 142 095 [1]. Variance = 249 185/100 − 49.7² = 2 491.85 − 2 470.09 = 21.76 [1]. σ = √21.76 = 4.66 (3 s.f.) [1] Examiner insight: In extended questions, carry full-accuracy values from one part to the next. Rounding the mean to 49.7 is exact here, but rounding a standard deviation early can change the final digits.

11. (a) Mean fare = 0.8 × 20 + 5 = 21 dollars [1]. The added 5 does not change the spread, so the standard deviation = 0.8 × 4.5 = 3.60 dollars [2] (b) Median = 0.8 × 19 + 5 = 20.20 dollars [1]; IQR = 0.8 × 6 = 4.80 dollars [1] (c) The mean stays at 20 minutes because the new value equals the mean [1]. The sum of squared deviations is 30 × 4.5² = 607.5, unchanged by the new value (its deviation is 0) [1]. New variance = 607.5/31 = 19.6, so σ = 4.43 minutes, which is less than 4.5 [1] (d) A fare of 25 dollars comes from t = 25 minutes, since 0.8 × 25 + 5 = 25. For the fare: z = (25 − 21)/3.6 = 1.11 [1]. For the time: z = (25 − 20)/4.5 = 1.11, the same, because coding by a positive linear rule does not change how many standard deviations a value lies from the mean [1] Examiner insight: Part (c) asks you to show the decrease, so the working must justify it rather than just quote 4.43. Include units in your answers.

Where marks are usually lost

• Adding the constant to the standard deviation or the IQR.
• Giving a negative standard deviation or IQR after multiplying by a negative number.
• Averaging two means or two standard deviations without using group sizes.
• Finding a combined standard deviation without using Σx² for each group.
• Stopping at the variance and forgetting the square root.
• Comparing raw values from two different distributions instead of z-scores.
• Answering a comparison with only centre or only spread, or with no figures.
• Saying the median “must” change when a value is added, when it can stay the same.

Next steps

• Review the short summary in the revision notes.
• Re-read the explanations in the study guide.
• Go to the Cambridge IGCSE Statistics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class if you want help with any of these topics.

Official syllabus

Cambridge IGCSE Statistics 0479 syllabus for examination in 2027, version 1, Cambridge University Press & Assessment (Cambridge International Education), Topic 6, outcomes 6.1 to 6.6.