Calculate weighted normal probabilities and conditional areas
Apply different weights to independent normal inputs and calculate tails or restricted areas from the resulting distribution.
Section 6.2 · Linear combinations of random variables
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 squared independent variance coefficients.
- Intersect conditional events and divide by a positive condition.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What variance contribution does 2Y make when Var(Y) = 16?
What you will learn
- Form the distribution of a weighted normal combination.
- Calculate tails and conditional probabilities in the combined variable.
Weight each independent contribution
For independent X ~ N(30,9) and Y ~ N(12,16), W = 2X + 2Y + 4 has mean 60 + 24 + 4 = 88 and variance 4 × 9 + 4 × 16 = 100. Hence W ~ N(88,100), SD ten. P(W > 100) = 1 − Φ(1.2) = 0.1151 using Φ(1.2) = 0.8849.
For D = 2X − 2Y + 4, mean is forty and variance remains one hundred. P(D ≤ 30) = Φ(−1) = 0.1587 using Φ(1) = 0.8413. The same variance does not mean the same centre or event probability.
Condition on the combined observation
Given W > 88, finding P(W < 100 | W > 88) requires the intersection 88 < W < 100. Its area is Φ(1.2) − Φ(0), while the denominator P(W > 88) is one half.
Using the quoted table value, the conditional probability is (0.8849 − 0.5)/0.5 = 0.7698. Do not divide by P(W < 100) or assume these nested events are independent. The resulting normal variable is continuous, so no count-cell correction is needed.
Pause and explain
Which intersection is used for P(W < 100 | W > 88)?
Worked example
Find P(W > 100) for W = 2X + 2Y + 4 with independent X ~ N(30,9), Y ~ N(12,16).
Show the worked solution
- W has mean eighty-eight and variance one hundred.
- z = (100 − 88)/10 = 1.2.
- Using Φ(1.2) = 0.8849 gives the right tail 0.1151.
Answer 0.1151 using the quoted table value.
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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 N(30,9), N(12,16) inputs, find W = 2X + 2Y + 4’s model and P(W > 100), using Φ(1.2) = 0.8849.
Give me a hint
Calculate parameters before the tail.
Compare my reasoning
- W ~ N(88,100).
- The boundary standardises to 1.2.
- The upper area is 0.1151.
N(88,100); probability 0.1151.
Look for these in your work
- Square both weights.
- Use the combined SD ten.
For those inputs, find D = 2X − 2Y + 4’s model and P(D ≤ 30), using Φ(1) = 0.8413.
Give me a hint
Only the signed mean contribution changes.
Compare my reasoning
- D ~ N(40,100).
- z = (30 − 40)/10 = −1.
- The reflected left area is 0.1587.
N(40,100); probability 0.1587.
Look for these in your work
- Subtract mean contributions correctly.
- Keep positive variance.
Find P(W < 100 | W > 88), using the same W and Φ(1.2) = 0.8849.
Give me a hint
Use the interval intersection and positive conditioning area.
Compare my reasoning
- The numerator is 0.8849 − 0.5 = 0.3849.
- The condition has probability one half.
- Divide to obtain 0.7698.
0.7698 using the quoted table value.
Look for these in your work
- Intersect the events.
- Use the known condition as denominator.
Common mistakes
- Use the variances specified for this example.
- Square each weight.
- Condition on the stated event, not the requested one.
What can you explain now?
Why does replacing +2Y with −2Y keep variance the same?
Compare with the explanation
The squared coefficient is four in both cases.
Mean still changes because it uses the signed coefficient.
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Make it stick
Combine weighted parameters first; then translate the entire event, including any condition, into normal areas.
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.