Model totals with fixed fees and random components
Translate a cost or duration into a linear variable and distinguish a fixed charge from an independent random contribution.
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 linear expectation.
- Add independent squared variance contributions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In C = 3X + 2Y + 10, which term is a fixed fee?
What you will learn
- Build a linear model from quantities and units.
- Calculate moments while distinguishing fixed and random fees.
Write the quantity before its moments
Suppose independent X,Y have means four and two and variances one and four. A stipulated cost C = 3X + 2Y + 10 has mean 12 + 4 + 10 = 26 and variance 9 + 16 = 25, SD five. Coefficients convert each quantity to cost units, while ten is a fixed fee.
A duration model T = X + 2Y instead has mean eight and variance 1 + 4 × 4 = 17. Explain why Y is multiplied twice in the quantity but four times in variance. The same input moments do not imply the same total spread for every weighting.
A random fee is a new variable
If the fee is instead an independent variable F with mean ten and variance nine, C = 3X + 2Y + F still has mean twenty-six but variance 9 + 16 + 9 = 34. The mean alone hides the added uncertainty.
Independence of all three inputs is stipulated for this example. If the fee depends on X or Y, the additive variance calculation is not justified by these marginal summaries alone. No distribution shape or real-world independence is established solely by the cost notation.
Pause and explain
An independent random fee has mean 10 and variance 9. What changes from the fixed-fee cost?
Worked example
Independent X,Y have means 4,2 and variances 1,4. Find moments of C = 3X + 2Y + 10.
Show the worked solution
- Mean = 12 + 4 + 10 = 26.
- Variance = 9 × 1 + 4 × 4 = 25.
- SD = 5; the fixed fee contributes no variance.
Answer Mean 26; variance 25; SD 5.
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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 X,Y with means 4,2 and variances 1,4, find moments of 3X + 2Y + 10.
Give me a hint
Scale each input and keep the fixed fee out of variance.
Compare my reasoning
- Mean is twenty-six.
- Variance is 9 × 1 + 4 × 4 = 25.
- SD is five.
Mean 26; variance 25; SD 5.
Look for these in your work
- Translate coefficients correctly.
- Exclude the fixed fee from variance.
For those inputs, find moments of T = X + 2Y.
Give me a hint
Use signed scales for mean and squared scales for variance.
Compare my reasoning
- Mean = 4 + 2 × 2 = 8.
- Variance = 1 + 4 × 4 = 17.
- SD = √17.
Mean 8; variance 17; SD √17.
Look for these in your work
- Use the correct weighting.
- State transformed units appropriately.
Replace the fixed fee ten in the cost by independent F with mean 10 and variance 9. Compare the moments.
Give me a hint
Treat F as a random term, not a constant.
Compare my reasoning
- The expected fee remains ten, so total mean stays twenty-six.
- Variance becomes 25 + 9 = 34.
- Without independence, the stated marginal moments alone would not justify this sum.
Mean 26; variance 34 under the stipulated independence.
Look for these in your work
- Distinguish fixed and random amounts.
- Check independence of every contribution.
Common mistakes
- Write the model before calculating.
- Fixed and random fees differ.
- Independence needs evidence in applications.
What can you explain now?
Does replacing the fixed fee with a random one of the same mean preserve variance?
Compare with the explanation
Not generally; its randomness can add spread.
Under the stipulated independence, add its variance.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Form the quantity in consistent units, separate fixed from random terms, then calculate mean and variance with their own rules.
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.