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Combinations of Random Variables

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Edexcel Further Stats 2 · Chapter 4

Combinations of Random Variables

Big idea: if you add or subtract normal random variables (each multiplied by a number), the result is STILL normal — you just add up the new means and, separately, add up the new variances (never standard deviations directly).

Summary — What This Chapter Is Really About

You already know E(X+Y) = E(X)+E(Y) and similar rules for expectation — this chapter reuses them.
For independent variables, variances always add — even when you're subtracting the variables themselves.
If X and Y are each normally distributed and independent, then aX + bY (or aX − bY) is also normally distributed.
The new normal distribution's mean and variance are found using simple formulas — then you standardise with Z exactly like any other normal distribution question.
When you add several copies of the same distribution (like 7 identical bags of sugar), there's a shortcut: just multiply the mean and variance by n.
The single biggest exam trap: mixing up variance and standard deviation when combining. This chapter is built almost entirely around avoiding that one mistake.

1. Combining Random Variables — The Expectation Rules

Let's start with something you've seen before, just to warm up. If X and Y are any two random variables (they don't even need to be independent, and they don't even need to be normal), then expectation behaves exactly like you'd hope — it's "linear," meaning it distributes nicely over addition, subtraction, and multiplication by constants.

Expectation rules

E(X + Y) = E(X) + E(Y)

E(X − Y) = E(X) − E(Y)

E(aX + bY) = aE(X) + bE(Y)

E(aX − bY) = aE(X) − bE(Y)

In plain words: to find the average of a combination of variables, just combine the averages the same way. Multiplying a variable by a constant "a" scales its mean by a too.

Why this feels obvious If the average apple weighs 125g and the average orange weighs 250g, then obviously the average weight of "3 apples + 3 oranges" is 3(125) + 3(250). You're not doing anything clever here — you're just scaling and adding averages the same way you'd scale and add the actual quantities.

2. Combining Random Variables — The Variance Rules (the important one)

This is where people trip up, so slow down here. Variance rules only work this simply when X and Y are independent — meaning the outcome of one doesn't affect or relate to the outcome of the other (e.g. the mass of an apple and the mass of an orange in the same bag are independent; your maths score and your physics score on the same day probably aren't).

Variance rules — for independent X and Y only

Var(X + Y) = Var(X) + Var(Y)

Var(X − Y) = Var(X) + Var(Y)

Var(aX + bY) = a²Var(X) + b²Var(Y)

Var(aX − bY) = a²Var(X) + b²Var(Y)

⚠️ The #1 exam mistake Notice that Var(X − Y) = Var(X) + Var(Y) — the minus sign disappears and becomes a plus! This feels wrong the first time you see it, but think about it intuitively: subtracting two independent random things doesn't cancel out uncertainty, it adds MORE uncertainty (now you have two separate sources of randomness contributing to the spread of the result). Variance never subtracts when the variables are independent.

Why does "a" become "a²"? Because variance measures spread using squared deviations from the mean. If you scale every possible outcome of X by a, you scale the spread by a too — but since variance is already a squared quantity, that scaling factor gets squared as well. This is exactly the same idea as Var(aX) = a²Var(X), which you'll have met before this chapter.

3. Combining Two Normal Distributions

Here's the big result of this chapter. Suppose X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²), and X and Y are independent. Then any linear combination aX + bY (or aX − bY) is also normally distributed. You find its mean using the expectation rules from Section 1, and its variance using the variance rules from Section 2 — then you just plug both into a new normal distribution.

Core formula — memorise this exactly

aX + bY ~ N(aμ₁ + bμ₂, a²σ₁² + b²σ₂²)

aX − bY ~ N(aμ₁ − bμ₂, a²σ₁² + b²σ₂²)

Notice again: the mean has a "−" for subtraction, but the variance is ALWAYS "+". This is the same trap as Section 2, just dressed up with normal distribution notation.

How to read N(μ, σ²) Remember the second number inside N(...) is always the variance (σ²), not the standard deviation (σ). This trips people up constantly in this exact chapter because the PMT source material sometimes writes X ~ N(68, 5) meaning σ² = 5 (variance), not σ = 5. Always double check which one you're being given in a question, and always work in variances when combining — only take a square root at the very last moment when you standardise to find a probability.

Worked Example — Full Method

Worked Example

X ~ N(68, 5) and Y ~ N(82, 10) are independent. Let C = 2X + Y. Find P(C < 176).

Step 1 — Find the mean of C	E(C) = 2(68) + 1(82) = 136 + 82 = 218
Step 2 — Find the variance of C	Var(C) = 2²(5) + 1²(10) = 20 + 10 = 30
Step 3 — Write the new distribution	C ~ N(218, 30)
Step 4 — Standardise using Z	Z = (176 − 218) / √30 = −7.667
Step 5 — Use the normal tables / calculator	P(C < 176) = P(Z < −7.667) ≈ 0.0000 (essentially 0, since 176 is over 7 standard deviations below the mean)

Note: this uses the same numbers as the classic "Example 1" from the PMT notes, but done fully correctly end-to-end — always divide by the square root of the variance to get the standard deviation before you standardise.

1 Practice Question

The independent random variables X and Y have distributions X ~ N(40, 6) and Y ~ N(55, 9). Let W = 3X − 2Y. Find the distribution of W, and hence find P(W > 10).

2 Practice Question

The mean mass of a small cake is 60g (σ = 3g) and the mean mass of a large cake is 150g (σ = 6g), independently. A gift box contains 2 small cakes and 1 large cake. Find the probability that the box (cakes only) weighs less than 260g.

4. The "Difference" Trick — Turning P(3X > Y) into a Single Distribution

A very common exam phrasing is something like "find P(3X > Y)" instead of directly asking about a combined variable. This looks unfamiliar, but it's just Section 3 in disguise. The trick: rearrange the inequality so everything is on one side, turning it into a statement about a single new variable.

The rearrangement trick P(3X > Y) is exactly the same event as P(3X − Y > 0). So define a new variable D = 3X − Y, find its distribution using the Section 3 formulas, and then just find P(D > 0). Any inequality between two variables can be rearranged this way — always aim to get "0" (or whatever target value) alone on one side.

Worked Example

X ~ N(16, 3) and Y ~ N(41, 4) are independent. Find P(3X > Y).

Step 1 — Rearrange	P(3X > Y) = P(3X − Y > 0)
Step 2 — Define D = 3X − Y	μ_D = 3(16) − 41 = 48 − 41 = 7
Step 3 — Find variance	σ²_D = 3²(3) + 1²(4) = 27 + 4 = 31
Step 4 — Write distribution	D ~ N(7, 31)
Step 5 — Standardise	Z = (0 − 7) / √31 = −1.257
Step 6 — Read from tables	P(D > 0) = P(Z > −1.257) = 0.8962

3 Practice Question

Independent variables X ~ N(20, 5) and Y ~ N(35, 8). Find P(2X < Y − 3).

5. Combining Several Copies of the SAME Distribution

Sometimes every variable you're adding comes from the identical distribution — like 7 randomly chosen bags of flour all drawn from the same population, X ~ N(58, 4). Instead of writing out the full sum formula seven times, there's a shortcut: multiply the mean by n, and multiply the variance by n (not by n²! — this is different from multiplying a single variable by a constant).

Sum of n identical independent variables

If X₁, X₂, ..., Xₙ ~ X ~ N(μ, σ²) independently, then:

∑Xᵢ ~ N(nμ, nσ²)

This is really just Section 3's formula applied n times with every coefficient equal to 1 — the algebra simplifies to "multiply by n" because you're adding n independent copies of the same variance.

Don't confuse this with "nX" This is the single most important distinction in this whole chapter. X₁ + X₂ + ... + Xₙ (n separate, independent variables) is NOT the same random variable as nX (n copies of the same single outcome, scaled up).

• Sum of n independent copies: ∑Xᵢ ~ N(nμ, nσ²) — variance multiplies by n.
• One single X scaled by n: nX ~ N(nμ, n²σ²) — variance multiplies by n².

Ask yourself: "am I picking n different random items, or taking one random item and multiplying its value by n?" That answer tells you which formula to use.

Worked Example

X₁, X₂, ..., X₇ all have distribution X ~ N(58, 4), independently. Find the distribution of R = ∑Xᵢ (i = 1 to 7).

Step 1 — Multiply mean by n = 7	Mean = 7 × 58 = 406
Step 2 — Multiply variance by n = 7	Variance = 7 × 4 = 28
Step 3 — Write distribution	R ~ N(406, 28)

4 Practice Question

The volume of soft drink in a can is X ~ N(330, 6), independently between cans. A pack contains 12 cans. Find the probability the total volume in the pack exceeds 4000ml.

What to Memorise

Rule	Formula
Expectation, addition/subtraction	E(aX ± bY) = aE(X) ± bE(Y)
Variance, independent variables	Var(aX ± bY) = a²Var(X) + b²Var(Y) (always +)
Combining two independent normals	aX ± bY ~ N(aμ₁ ± bμ₂, a²σ₁² + b²σ₂²)
Sum of n independent identical normals	∑Xᵢ ~ N(nμ, nσ²)
One variable scaled by n (different!)	nX ~ N(nμ, n²σ²)
Rearranging inequalities	P(X > Y) = P(X − Y > 0)
N(μ, σ²) notation	Second value is always variance, not SD — square root it before standardising

Concepts Checklist

I can state and apply E(aX ± bY) = aE(X) ± bE(Y)
I know variances ADD even when the variables are being subtracted (for independent variables)
I can find the distribution of aX ± bY when X and Y are independent normal variables
I can standardise a combined normal distribution using Z = (value − mean) / √variance
I can rearrange P(X > Y)-style questions into P(D > 0) form before solving
I can find the distribution of a sum of n identical independent variables using N(nμ, nσ²)
I can tell the difference between "sum of n independent copies" (nσ²) and "one variable scaled by n" (n²σ²)
I always double-check whether N(μ, σ²) is giving me variance or standard deviation before combining

Exam Tips — Where Marks Are Lost

Trap 1: Subtracting variances By far the most common error. Var(X − Y) = Var(X) + Var(Y), never a minus. If you ever end up subtracting variances, stop and check your working.

Trap 2: Forgetting to square the coefficients In Var(aX − bY) = a²Var(X) + b²Var(Y), students often forget to square "a" and "b" — a very easy mark to drop under time pressure.

Trap 3: Confusing σ and σ² Always check whether the question (or the N(μ, σ²) notation given) is stating a variance or a standard deviation. Combine using variances; only square-root at the final standardisation step.

Trap 4: nσ² vs n²σ² "Total of 7 separately chosen bags" uses nσ². "One bag's contents multiplied by 7" uses n²σ². Read the question carefully — are you summing independent items, or scaling a single one?

What examiners reward Full method marks usually require you to clearly state the new mean, clearly state the new variance, write the resulting distribution in N(μ, σ²) form, THEN standardise. Skipping straight to a Z-value without showing the combined distribution often loses marks even if your final answer is correct.

Independence must be stated All these variance/normal-combination rules only work because the variables are independent. If a question doesn't explicitly say "independent," check carefully — it's usually implied by context (e.g. randomly selected items) but examiners expect you to recognise why the rule applies.

Also in the full note

4. The "Difference" Trick — Turning P(3X > Y) into a Single Distribution

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