Combine independent normal sums and differences
Use normal closure under independence and retain additive variance even when means subtract.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.2. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Use E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X). Expectation is linear without independence, when the relevant moments exist. For independent X and Y, Var(aX + bY) = a²Var(X) + b²Var(Y); negative coefficients still give positive squared contributions. A constant changes the mean but not variance. SD scales by |a|, not a². Affine transforms of normal variables and linear combinations of independent normal variables are normal when the resulting variance is positive; zero variance is a constant, so handle it directly rather than standardising by zero. Independent Poisson counts have an exact Poisson sum with the parameters added; a scaled count or difference is not generally Poisson. Equal marginal distributions do not establish independence. Proofs are not required. Normal probabilities here use quoted cumulative values when supplied and otherwise unrounded standardisation with numerical areas. Exact distribution results are conditional on the stipulated models; real applications need evidence. Continuous density functions, sampling, estimation and hypothesis testing belong to later sections. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Use independent moment formulae.
- Standardise a normal boundary.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which variance belongs to the independent difference X − Y?
What you will learn
- State distributions for independent normal sums and differences.
- Use the combined SD to calculate probability.
Normality follows from the stipulated inputs
Let independent X ~ N(30,4) and Y ~ N(12,9). S = X + Y follows N(42,13). Both the normal shape and its parameters follow from the independent normal combination result. Means alone would not establish this shape.
P(S > 48) = 1 − Φ((48 − 42)/√13) ≈ 0.0480 using unrounded numerical normal areas. Do not add SDs two and three to get five; the correct combined SD is √13 ≈ 3.6056.
Subtract means but add independent variances
D = X − Y follows N(18,13). Its mean is eighteen, and its variance still thirteen because the negative coefficient squares to one. Symmetry gives P(D > 18) = 1/2.
If independence is absent, the theorem used here is unavailable. Marginal normal distributions alone do not generally determine the distribution of a combination; a supplied dependence relation may allow a separate calculation. Equal means or names do not settle that relationship.
Pause and explain
What extra fact justifies the normal combination result here?
Worked example
State the distribution of X − Y for independent N(30,4) and N(12,9).
Show the worked solution
- Independence and normal inputs give a normal difference.
- Mean = 30 − 12 = 18.
- Variance = 4 + 9 = 13.
Answer X − Y ~ N(18,13).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For independent X ~ N(30,4), Y ~ N(12,9), state S = X + Y’s distribution.
Give me a hint
Use normal closure and add both moment contributions.
Compare my reasoning
- Mean is forty-two.
- Variance is thirteen.
- Therefore S ~ N(42,13), SD √13.
N(42,13), SD √13.
Look for these in your work
- Justify normality and independence.
- Use variance as the second parameter.
For the same independent variables, find X − Y’s distribution and P(X − Y > 18).
Give me a hint
The mean subtracts but variance adds.
Compare my reasoning
- Difference mean is eighteen.
- Variance is thirteen, so D ~ N(18,13).
- Symmetry about eighteen gives probability one half.
N(18,13); probability 1/2.
Look for these in your work
- Keep signed mean contributions.
- Use normal symmetry.
Calculate P(X + Y > 48) and explain why SD five would be wrong.
Give me a hint
Standardise with the square root of the total variance.
Compare my reasoning
- S ~ N(42,13).
- z = 6/√13 ≈ 1.6641.
- Right area is approximately 0.0480; adding SDs ignores the squared-contribution rule.
Approximately 0.0480; SD √13, not 5.
Look for these in your work
- Show the full standardisation.
- Explain the spread calculation.
Common mistakes
- Normal inputs need the stated combination conditions.
- Subtracting means does not subtract independent variances.
- Add variances before taking a square root.
What can you explain now?
Is SD(X + Y) = 2 + 3 here?
Compare with the explanation
No; it is √13.
Independent variance contributions add, not their SDs.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Check independent normal inputs, combine signed means and squared variances, then use one combined SD.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.