IB Maths AA HL Topic 4 — Statistics & Probability Paper 1 & 2 ~6 min read

Linear Transformations of Data

When every value in a data set is rescaled by the same rule (y = ax + b), the mean and standard deviation transform predictably — but not the same way. The mean follows the rule literally; the SD ignores the constant shift but multiplies by |a|.

📘 What you need to know

Linear transformation: every value x is replaced by y = ax + b (a single rule applied to all data).
Mean: ⎯y = a ⎯x + b — both the multiplier and the shift apply.
Variance: σ²_y = a²σ²_x — the multiplier squared; the shift has no effect.
Standard deviation: σ_y = |a| σ_x — absolute value because SD is non-negative.
Why does the shift not affect spread? Adding a constant moves every value by the same amount → no change in distance between values.
HL formula booklet states this as E(aX + b) = aE(X) + b and Var(aX + b) = a² Var(X).
Common contexts: standardising scores, unit conversions, applying scaling factors.

The transformation rules

Mean E(aX + b) = a E(X) + b

Variance Var(aX + b) = a² Var(X)

Standard deviation σ_y = |a| σ_x

Effect of each operation

Operation	Effect on mean	Effect on SD / variance
Multiply by a	multiply by a	SD: × \|a\|; variance: × a²
Add b	add b	no change
Both: y = ax + b	a ⎯x + b	SD: × \|a\|; variance: × a²

Intuition: shifting moves the whole distribution; rescaling stretches or shrinks it. Spread cares about distances between values, which are unaffected by shifts but scaled by stretches.

🧭 Recipe — apply a linear transformation

Identify a and b from the rule y = ax + b.
New mean = a × (old mean) + b.
New SD = |a| × (old SD).
New variance = a² × (old variance).
If working backwards (given new, find old): solve the same equations for the unknowns.

Worked examples

WE 1

Apply a basic linear transformation

A data set has mean 24 and standard deviation 6. Each value is transformed using y = 3x + 5. Find the new mean and standard deviation.

Step 1: Identify a = 3 and b = 5 Step 2: New mean = a × old mean + b ⎯y = 3(24) + 5 = 72 + 5 = 77 Step 3: New SD = |a| × old SD σ_y = 3 × 6 = 18 New mean = 77; new SD = 18 the +5 doesn’t enter the SD calculation — only the multiplier matters for spread

WE 2

Transformation with a negative coefficient

A test produced raw scores with mean 25 and SD 4. The teacher rescales scores using y = −2x + 80 (so high raw scores become low scaled scores). Find the new mean and SD.

Step 1: a = −2, b = 80 Step 2: New mean ⎯y = (−2)(25) + 80 = −50 + 80 = 30 Step 3: New SD = |a| × old SD σ_y = |−2| × 4 = 2 × 4 = 8 New mean = 30; new SD = 8 absolute value matters: SD never goes negative even when a is negative

WE 3

Find the original mean and SD given the transformed values

After applying the transformation y = 4x − 7, a data set has mean 65 and standard deviation 12. Find the original mean and standard deviation.

Step 1: Apply mean rule, work backwards 65 = 4 ⎯x − 7 → 4 ⎯x = 72 → ⎯x = 18 Step 2: Apply SD rule, work backwards 12 = |4| × σ → σ = 12/4 = 3 Original mean = 18; original SD = 3 verify: 4(18) − 7 = 65 ✓; |4| × 3 = 12 ✓

WE 4

Adding a constant to every value

The daily temperatures recorded over a week have mean 17 °C and standard deviation 2.5 °C. Each temperature is increased by 8 (e.g., to model a heatwave shift). State the new mean and SD.

Step 1: Transformation is y = x + 8 (so a = 1, b = 8) Step 2: New mean ⎯y = 17 + 8 = 25 °C Step 3: New SD σ_y = |1| × 2.5 = 2.5 °C (unchanged) New mean = 25 °C; new SD = 2.5 °C (unchanged) adding a constant shifts the centre but doesn’t change how spread out the values are

WE 5

Unit conversion — multiply only

Heights of plants measured in centimetres have mean 36 cm and standard deviation 4 cm. The data is converted to millimetres (multiply each value by 10). Find the new mean and SD.

Step 1: y = 10x (so a = 10, b = 0) Step 2: New mean ⎯y = 10 × 36 = 360 mm Step 3: New SD σ_y = 10 × 4 = 40 mm New mean = 360 mm; new SD = 40 mm scaling preserves the relative spread — both mean and SD scale by the same factor

WE 6

Multi-part: mean, variance, and SD after transformation

A teacher’s marks have mean 62 and variance 81. To rescale, the teacher applies y = 0.8x + 5. Find (a) the new mean, (b) the new variance, and (c) the new standard deviation.

Original SD = √81 = 9 a = 0.8; b = 5 (a) New mean ⎯y = 0.8(62) + 5 = 49.6 + 5 = 54.6 (b) New variance = a² × old variance σ²_y = (0.8)² × 81 = 0.64 × 81 = 51.84 (c) New SD σ_y = |0.8| × 9 = 7.2 Cross-check: √51.84 = 7.2 ✓ (a) 54.6; (b) 51.84; (c) 7.2 multiplier < 1 reduces spread; variance scales by a² so the effect is more pronounced

💡 Top tips

Spread ignores shifts — only the multiplier a affects SD and variance.
Variance gets a²; SD gets |a| — match the right rule to the right measure.
Always use absolute value for SD — works seamlessly with negative a.
For unit conversions: just multiply (b = 0); both mean and SD scale by the same factor.
The HL formula booklet states E(aX+b) and Var(aX+b) — you can look these up if you forget.

⚠ Common mistakes

Adding b to the SD — only multiplication affects spread; addition does not.
Multiplying SD by a² instead of |a| — that’s the variance rule, not the SD rule.
Forgetting the absolute value on negative coefficients — leads to a “negative SD”.
Confusing variance and SD — variance has the squared scaling.
Applying the rule once instead of to each value — every data point is transformed, not just one.

Next: Outliers. The 1.5 × IQR rule formally defines what counts as an “extreme” value, but the question of whether to remove outliers is more interesting — sometimes they’re errors, sometimes they’re genuine and important. Context decides.

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