IB Maths AI HLStatistics ToolkitPaper 1 & 2~7 min read
Linear Transformations of Data
Sometimes you’ll want to change every value in a dataset — convert kilograms to grams, shift exam scores to a friendlier scale, or convert temperatures from Celsius to Fahrenheit. Any change of the form “multiply by a constant, then add a constant” is called a linear transformation: Y = aX + b. The key question is: how do the mean, variance, and standard deviation respond? Two simple rules cover everything. Adding a constant shifts the mean but leaves the spread alone. Multiplying by a constant scales both — except the variance gets squared. Memorise the two rules and you can solve every IB question on this topic in seconds.
📘 What you need to know
A linear transformation of data has the form Y = aX + b, where a and b are constants.
Mean rule: E(aX + b) = a E(X) + b. The mean is scaled by a and then shifted by b.
Variance rule: Var(aX + b) = a2 Var(X). The variance is multiplied by a2; the constant b has no effect.
Standard deviation rule: σnew = |a| × σ. SD is scaled by the absolute value of a (so the sign is dropped).
Adding a constant (just +b): the mean shifts, but the spread (variance, SD) is unchanged.
Multiplying by a constant (just ×a): the mean is multiplied by a, the SD by |a|, and the variance by a2.
Both formulas E(aX+b) and Var(aX+b) are in the HL formula booklet.
Also called: “effects of constant changes”, “linear coding”.
The two formulas — in the formula booklet
Linear transformation — mean
E(aX + b) = a E(X) + bin the formula booklet ✓
Linear transformation — variance
Var(aX + b) = a2 Var(X)
in the formula booklet ✓
From these, the standard deviation rule follows directly: σnew = √Var = √(a2 Var(X)) = |a| × σ.
how each rule splits the transformation
The “+ b” half shifts the whole distribution sideways — the mean moves, but the spread is unchanged. The “× a” half stretches the whole distribution — both the mean and the spread are scaled by |a|.
Why are the two rules different?
Adding +b
shifts only
Every value moves by the same amount. Distances between values stay the same — so spread is unchanged.
Multiplying ×a
scales everything
Every value is stretched. Distances between values grow by factor |a|, so SD grows by |a| and variance by a2.
🤔 Why does the variance multiply by a2, not by a?
Variance is the average of squared distances from the mean. When you multiply each value by a, every distance from the mean is also multiplied by a — but then those distances get squared in the variance formula. So variance gets multiplied by a2. Taking the square root brings the standard deviation back to a single factor of |a| (positive, because spread can’t be negative).
Quick summary table
Transformation
New mean
New variance
New SD
X + b
μ + b
σ2 (unchanged)
σ (unchanged)
X − b
μ − b
σ2 (unchanged)
σ (unchanged)
aX
aμ
a2σ2
|a|σ
aX + b
aμ + b
a2σ2
|a|σ
🧠 Memory aid — the “b ignores spread” rule
Adding a constant b is like sliding the entire dataset along a number line — every value moves the same distance, so gaps between values don’t change. That’s why + b changes the mean but never the variance or SD. Once you see that, the rest follows.
🧭 Recipe — applying a linear transformation
Identify a and b in the transformation Y = aX + b.
New mean: multiply the original mean by a, then add b.
New variance: multiply the original variance by a2. (b doesn’t appear.)
New SD: multiply the original SD by |a|.
Check the sign: if a is negative, the mean keeps the sign, but the SD becomes positive (|a|).
Worked examples
WE 1
Adding a constant
A dataset has mean 30 and standard deviation 6. Each value is increased by 4. Find the new mean and new standard deviation.
identify the transformationY = X + 4 → a = 1, b = 4new mean = aμ + b= 1 × 30 + 4 = 34new mean = 34new SD = |a| × σ= |1| × 6 = 6 (unchanged)new SD = 6adding a constant shifts the mean but leaves the spread completely unchanged.
WE 2
Multiplying by a constant
A dataset has mean 20 and standard deviation 5. Each value is multiplied by 3. Find the new mean, variance, and standard deviation.
identifyY = 3X → a = 3, b = 0new mean = aμ= 3 × 20 = 60new mean = 60new variance = a²σ²original variance = 5² = 25new = 3² × 25 = 9 × 25 = 225new variance = 225new SD = |a| × σ= |3| × 5 = 15new SD = 15multiplication scales both mean and spread — but variance scales by a², not by a.
WE 3
Combined transformation aX + b
A teacher marks his students’ tests. The raw mean is 31 marks and the standard deviation is 5 marks. The teacher standardises by doubling the raw score and adding 10.
(a) Calculate the mean of the standardised scores.
(b) Calculate the standard deviation of the standardised scores.
transformationY = 2X + 10 → a = 2, b = 10(a) new mean = aμ + b= 2 × 31 + 10 = 62 + 10(a) new mean = 72(b) new SD = |a| × σ= |2| × 5 = 10(b) new SD = 10use both rules: aμ + b for mean, |a|σ for SD. The b doesn’t appear in the SD calculation.
WE 4
Backwards — find a and b
A dataset has mean 50 and SD 8. After a transformation of the form Y = aX + b (with a > 0), the new mean is 110 and the new SD is 16. Find a and b.
Step 1: use SD to find a|a| × 8 = 16|a| = 2, and a > 0 ⇒ a = 2Step 2: use mean to find ba × 50 + b = 1102 × 50 + b = 110100 + b = 110b = 10a = 2, b = 10verifymean: 2(50)+10 = 110 ✓; SD: 2(8) = 16 ✓solve SD equation first to find |a|, then mean equation to find b.
WE 5
Temperature conversion — Celsius to Fahrenheit
The midday temperatures (in °C) over a week have mean 20°C and standard deviation 7°C. The conversion to Fahrenheit is given by F = 95C + 32.
Find the mean and standard deviation of the temperatures in °F.
identifyF = (9/5)C + 32 → a = 9/5, b = 32new mean = (9/5)μ + 32= (9/5)(20) + 32 = 36 + 32 = 68mean in °F = 68 °Fnew SD = (9/5) × σ= (9/5)(7) = 63/5 = 12.6SD in °F = 12.6 °Funit conversions are linear transformations — the rules apply directly.
WE 6
Negative multiplier
A dataset of profits (£) has mean 30 and variance 16. A new variable is defined by T = −2X + 50. Find the mean, variance, and standard deviation of T.
identifya = −2, b = 50new mean = aμ + b= (−2)(30) + 50 = −60 + 50new mean = −10new variance = a²σ²= (−2)² × 16 = 4 × 16 = 64new variance = 64new SD = |a| × σoriginal SD = √16 = 4new SD = |−2| × 4 = 8new SD = 8the negative sign survives in the mean but disappears in variance (squared) and SD (absolute value).
💡 Top tips
Memorise the two formulas: E(aX+b) = aE(X) + b and Var(aX+b) = a2Var(X). Both are in the HL formula booklet but quick recall saves time.
+ b never affects spread. If the question only adds or subtracts a constant, SD and variance don’t change at all.
SD uses |a|, not a. Always take the absolute value — SD is always positive.
Variance uses a2. Don’t forget the square.
For backwards problems: solve the SD equation first to find |a|, then the mean equation to find b.
The formula booklet uses E and Var notation. E(X) means the mean of X; Var(X) means the variance of X.
⚠ Common mistakes
Multiplying SD by a2. Variance uses a2; SD uses |a|.
Including b in the variance formula. Adding a constant doesn’t change spread, ever. Don’t write Var(aX+b) = a2Var(X) + b.
Forgetting absolute value when a < 0. A negative a still scales the SD positively; the sign only affects the mean.
Confusing variance and SD. If the problem gives you SD, square it to get variance first if needed (or work with SD directly).
Applying the SD/variance change to mode or median. These rules are for the mean and variance; mode, median, and IQR transform differently (though similarly for linear transformations).
Next up — Outliers. You’ve seen that outliers heavily affect the mean and SD but not the median and IQR. Now you’ll learn the formal IB rule for detecting outliers: a value is an outlier if it lies more than 1.5 × IQR below Q1 or above Q3. You’ll also see when outliers should — or shouldn’t — be removed from a dataset.
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