The masses, in grams, of small and large bags of flour have the distributions N(510, 100) and N(1015, 324) respectively. André selects 4 small bags of flour and 2 large bags of flour at random.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [B1] 6(a) E(T) = 4 × 510 + 2 × 1015 [= 4070]
2. [B1] 6(a) Var(T) = 4 × 100 + 2 × 324 [= 1048] or √(4×100+2×324) [=32.4(3sf)] .
3. [M1] 6(a) 4130-4070/√'1048' [= 1.853] Standardising with their values. Variance must be from a combination attempt. M1 can be implied by correct final answer.
4. [M1] 6(a) Φ('1.853') For area consistent with their values. M1 can be implied by correct final answer.
5. [A1] 6(a) = 0.968 (3 sf) As final answer.
6. [B1] 6(b) E(D) = 4 × 510 – 2 × 1015 [= 10]
7. [B1] 6(b) Var(D) = 4 × 100 + 2 × 324 [= 1048] Or √(4×100+2×324) [=32.4(3sf)].
8. [M1] 6(b) 0-'10'/√'1048' [= -0.309] Standardising with their values. Variance must be from a combination attempt. M1 can be implied by correct final answer.
9. [M1] 6(b) 1 – Ф(‘−0.309’) = Φ(‘0.309’) For area consistent with their values. M1 can be implied by correct final answer.
10. [A1] 6(b) = 0.621 As final answer.