Last year, an online store sold a large number of computers. 55% of the computers were made by company F, 30% were made by company G and 15% were made by company H. A random sample of 3 customers who each bought a computer from this store is chosen.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

1. [1] [0.55×0.3×0.15×3!=]0.1485, 297/2000
2. Accept 148500/1000000 condone 0.149.
3. [1] One term of the form 12Cx (p)^x (1-p)^(12-x), 0 < p < 1, x ≠ 0 or 12.
4. [1] Correct unsimplified expression, no terms omitted leading to final answer. Condone omission of last bracket '}' only.
5. [1] 0.9575 < p ≤ 0.958.
6. [Method 1 Answer] 1 - {12C10 0.55^10 0.45^2 + 12C11 0.55^11 0.45 + 0.55^12} = 1 - (0.0338529 + 0.0075229 + 0.0007662) = 0.957858 = 0.958
7. [Method 2 Answer] P(0,1,...,9) = 0.45^12 + 12C1 0.55^1 0.45^11 + ... + 12C9 0.55^9 0.45^3 = 0.958
8. [1] 21 and 17.85 (or 177/20) seen, allow unsimplified. May be in standardisation formula. (σ = √17.85, 4.224926...to at least 4SF implies correct variance). Withhold mark if variance clearly identified as standard deviation, condone N(21, √17.85 ) if standardisation formula correct or variance/standard deviation correctly stated as well.
9. [1] Substituting their μ and their σ into the ± standardisation formula (any number for 24.5), allow σ² or √σ.
10. [1] Use continuity correction 23.5 or 24.5 in their standardisation formula. Note: If no working ±3.5/√17.85 or ±3.5/4.225 seen gains M2 BOD.
11. [1] Appropriate area Φ, from final process, must be a probability. May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5.
12. [1] Final answer AWRT.
13. [Answer] Mean = 140 × 0.15 = 21. Variance = 140 × 0.15 × 0.85 = 17.85. P(X > 24) ≈ P(Z > (24.5 - 21)/√17.85) = P(Z > 3.5/4.2249) = P(Z > 0.8284) = 1 - Φ(0.8284) = 1 - 0.7961 = 0.2039. Rounded to 3 significant figures, P(X > 24) = 0.204.