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

This guide covers Topic 6, Transformations of data sets, in the Cambridge IGCSE Statistics (0479) syllabus for examination in 2027. It teaches outcomes 6.1 to 6.6: the effect of adding or multiplying by a constant, the effect of adding or removing values, combining two data sets, scaling and standardising, and comparing distributions. It builds on the averages and measures of spread from Topics 4 and 5.

Keep these pages open together: the revision notes, the practice set, the Statistics hub and the printable checklist. To find your weak spots first, try the free 10-minute diagnostics. For calculating means and standard deviations from tables, see Frequency Distributions.

What this topic covers

Outcome	What you must be able to do
6.1	Say what happens to the mean and standard deviation when a constant is added to every value or every value is multiplied by a constant
6.2	Say the same for the median and the interquartile range (IQR)
6.3	Say whether a measure increases, decreases or stays the same when values are added or removed
6.4	Find the mean and standard deviation of two combined data sets, including finding the second mean from the combined mean
6.5	Scale data to a given mean and standard deviation, and standardise to mean 0 and standard deviation 1
6.6	Compare distributions using a measure of centre and a measure of spread

The standard deviation in this syllabus is the version that divides by n:

standard deviation = √( Σ(x − x̄)² / n ) = √( Σx²/n − (Σx/n)² )

I write x̄ for the mean and σ for the standard deviation.

6.1 Adding a constant and multiplying by a constant

Suppose every value x is changed to y = ax + b. This is called coding.

• Adding b shifts every value along by b. The data moves, but the gaps between values do not change.
• Multiplying by a stretches the data from zero, so the gaps are multiplied by a.

Change	New mean	New standard deviation
Add b	x̄ + b	σ (unchanged)
Multiply by a	a × x̄
Both: y = ax + b	a × x̄ + b

The standard deviation never takes b into account, and it is never negative. If a = −2, the spread is multiplied by 2, not −2.

Worked example 1. The data 8, 11, 14, 9, 13 is coded using y = 3x − 5. Find the mean and standard deviation of y.

Mean of x = (8 + 11 + 14 + 9 + 13)/5 = 55/5 = 11
Deviations: −3, 0, 3, −2, 2
Squares:     9, 0, 9, 4, 4   → sum 26
σx = √(26/5) = √5.2 = 2.28 (3 s.f.)
Mean of y = 3 × 11 − 5 = 28
σy = 3 × 2.28 = 6.84 (3 s.f.)

Check by coding each value: 19, 28, 37, 22, 34. Their mean is 140/5 = 28, which agrees.

Why this matters. Coding lets you work with small numbers. If the raw values are 1 203, 1 207 and 1 212, subtract 1 200 first, work with 3, 7, 12, then add 1 200 back to the mean only.

6.2 Median and interquartile range

The median is one value in the ordered list, so it is changed in exactly the same way as a single data value. The IQR is a difference between two values, so the added constant cancels out.

Change	New median	New IQR
Add b	median + b	IQR (unchanged)
Multiply by a (a > 0)	a × median	a × IQR
y = ax + b	a × median + b

Negative multipliers need care. If a is negative, the order of the data reverses. The old upper quartile becomes the new lower quartile. The median still follows y = ax + b, but the IQR is multiplied by |a|.

Worked example 2. The ordered data is 2, 5, 6, 8, 11, 12, 15, 19. Find the median and IQR, then the median and IQR after (a) y = 0.5x + 10 and (b) y = 20 − 2x.

Median = (8 + 11)/2 = 9.5
Lower half: 2, 5, 6, 8      → Q1 = (5 + 6)/2 = 5.5
Upper half: 11, 12, 15, 19  → Q3 = (12 + 15)/2 = 13.5
IQR = 13.5 − 5.5 = 8

(a) median = 0.5 × 9.5 + 10 = 14.75,  IQR = 0.5 × 8 = 4
(b) median = 20 − 2 × 9.5 = 1,         IQR = 2 × 8 = 16

For (b), the new quartiles are 20 − 2(13.5) = −7 and 20 − 2(5.5) = 9. The new lower quartile is −7 and the new upper quartile is 9, so IQR = 9 − (−7) = 16.

6.3 Adding or removing values

Here you decide whether a measure goes up, down or stays the same. Compare the new value with the current measure.

• Mean: adding a value above the mean raises it. Adding a value below the mean lowers it. Adding a value equal to the mean leaves it unchanged.
• Median: adding a value above the median raises it or leaves it the same. Adding a value below the median lowers it or leaves it the same. It cannot move the wrong way.
• Range: adding a value can only keep the range the same or increase it. Removing a value can only keep it the same or decrease it.
• Standard deviation: adding a value far from the mean increases the spread. Adding a value equal to the mean decreases the standard deviation, because the sum of squares is the same but is shared over more values.

Worked example 3. The data 10, 12, 14, 16 has mean 13 and σ = √5 = 2.24. A fifth value, 13, is added.

New mean = (10 + 12 + 14 + 16 + 13)/5 = 65/5 = 13   → stays the same
Sum of squared deviations is still 9 + 1 + 1 + 9 = 20 (the new value adds 0)
New σ = √(20/5) = 2                                    → decreases

For the median, take the data 3, 5, 8, 12, 20 with median 8. Adding 2 gives a median of 6.5 (decreases). Adding 8 leaves it at 8 (same). Adding 30 gives a median of 10 (increases).

When a question asks for a reason, give the comparison in words: “The new value is greater than the mean, so the mean increases.”

6.4 Combining two data sets

Two groups have sizes n₁ and n₂, means x̄₁ and x̄₂, and standard deviations σ₁ and σ₂.

Combined mean. Add the totals and divide by the total number:

combined mean = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)

You cannot just average the two means unless n₁ = n₂.

Combined standard deviation. Use Σx² for each group. From the formula σ² = Σx²/n − x̄², you get Σx² = n(σ² + x̄²). Then:

combined variance = (Σx² for both groups)/(n₁ + n₂) − (combined mean)²
combined σ = √(combined variance)

Do not average the two standard deviations.

Worked example 4. Group A has n = 20, mean 62, σ = 8. Group B has n = 30, mean 70, σ = 6. Find the mean and standard deviation of all 50 values.

Total A = 20 × 62 = 1 240        Total B = 30 × 70 = 2 100
Combined mean = 3 340/50 = 66.8

Σx² for A = 20 × (8² + 62²) = 20 × 3 908 = 78 160
Σx² for B = 30 × (6² + 70²) = 30 × 4 936 = 148 080
Combined variance = (78 160 + 148 080)/50 − 66.8²
                  = 4 524.8 − 4 462.24 = 62.56
Combined σ = √62.56 = 7.91 (3 s.f.)

The combined variance (62.56) is more than the weighted average of the two variances, (20 × 64 + 30 × 36)/50 = 47.2. The extra comes from the gap between the two group means.

Finding the second mean. If you know the combined mean, work with totals backwards.

Worked example 5. Group P has 12 values with mean 45. Group Q has 18 values. The 30 values together have mean 51. Find the mean of Q.

Total of all 30 = 30 × 51 = 1 530
Total of P = 12 × 45 = 540
Total of Q = 1 530 − 540 = 990
Mean of Q = 990/18 = 55

6.5 Scaling and standardising

Standardising converts a value x from a distribution with mean μ and standard deviation σ into a z-score:

z = (x − μ)/σ

The z-score says how many standard deviations a value is above (positive) or below (negative) the mean. Standardised data has mean 0 and standard deviation 1. It is just a coded set with a = 1/σ and b = −μ/σ.

Scaling to a given mean and standard deviation. To move data to a new mean M and new standard deviation S, standardise first and then rebuild:

new value = M + S × (x − μ)/σ

Worked example 6. A test has mean 54 and σ = 12. The marks are rescaled to mean 60 and σ = 10. Find the new value of a mark of 66.

z = (66 − 54)/12 = 1
new mark = 60 + 10 × 1 = 70

To go backwards, undo each step: a scaled mark of 55 has z = (55 − 60)/10 = −0.5, so the original was 54 + 12 × (−0.5) = 48.

Using z-scores to compare. Raw marks from different tests cannot be compared directly.

Worked example 7. Amir scores 72 in maths (mean 60, σ = 8) and 68 in English (mean 55, σ = 6.5). In which subject did he do better relative to the class?

Maths:   z = (72 − 60)/8   = 1.5
English: z = (68 − 55)/6.5 = 2.0

His English result is 2.0 standard deviations above the mean, compared with 1.5 for maths, so he did relatively better in English even though his raw mark is lower.

6.6 Comparing distributions

Always make two comparisons, each in context: one for centre and one for spread.

• Larger values generally: the distribution with the larger mean (or median).
• More consistent: the distribution with the smaller standard deviation (or smaller IQR).

Worked example 8. Two machines fill bags of rice. Machine A: mean 502 g, σ = 1.8 g. Machine B: mean 498 g, σ = 4.6 g. Compare them.

Machine A fills bags with more rice on average (502 g compared with 498 g). Machine A is also more consistent, because its standard deviation (1.8 g) is smaller than machine B’s (4.6 g).

Pair the measures properly: the mean goes with the standard deviation, and the median goes with the IQR. If the data has outliers, say that the median and IQR are less affected by them.

Common errors

• Adding b to the standard deviation or the IQR. Neither changes.
• Writing a negative standard deviation after multiplying by a negative number.
• Averaging two means or two standard deviations without using group sizes.
• Taking the “combined standard deviation” as the mean of the two standard deviations.
• Comparing the raw marks from different tests instead of z-scores.
• Writing “A is better” without saying whether you mean larger values or more consistent values.
• Forgetting that the median can stay the same when a value is added.

Where to go next

Work through the practice set, then use the revision notes for a short summary.

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.