Two random variables X and Y have the independent distributions N(7, 3) and N(6, 2) respectively. A random value of each variable is taken. Find the probability that the two values differ by more than 2.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [1] E(X-Y) = 1 Var(X – Y) = 5 Seen or implied, OE e.g. X - Y−2.
2. [1] (2-1)/√5 [= 0.447] (Or (-2-1)/√5 [= -1.342]) Standardising with their values must come from a combination.
3. [1] 1 – Φ('0.447') (Or Φ('-1.342') = 1 – Ф('1.342)) Correct probability area consistent with their values.
4. [1] = 0.327 or 0.328 (Or = 0.0898 or 0.0899) Seen or implied.
5. [1] Probability that difference is more than 2 = 0.417 (3 sf) or 0.418
6. [1] E(X) = 62 + 1.5(42) [= 125] OE.
7. [1] Var(X) = 158 + 1.5² × 108 [= 401] Correct expression OE.
8. [1] (90-"125") / √"401" [=-1.748] Correct standardisation using their E(X) and Var(X). Must both be from a combination attempt. Ignore any attempted continuity correction.
9. [1] Φ('1.748') Correct probability area consistent with their stated values.
10. [1] = 0.960 or 96.0% (3 sf) Allow 0.96 or 96%.
11. [1] Unlikely. A candidate who does well in Theory is likely to do well in Practical. Need both. Accept 'unlikely', 'not independent', 'dependent', 'not realistic', or similar; and accept 'both testing knowledge from the same syllabus', 'theory and practical share same content' or similar statement.