A factory produces chocolate bars. The weights of the bars are normally distributed with mean 155 g and standard deviation 6 g. A random sample of 350 of these bars is chosen.

Mark scheme

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

1. [M1] Substituting 155, 6 and 148 or 160 into ± standardisation formula, not σ², not √σ, no continuity correction, Award -7/6 or 5/6 M1.
2. [M1] Calculating the appropriate probability area (leading to their final answer. Expect p > 0.5).
3. [A1] AWRT 0.676 SOI. If one or both M marks not awarded, SCB1 for AWRT 0.676 SOI.
4. [B1FT] Strict FT their at least 4 figure probability × 350 (Check with calculator). One integer answer only, either 236 or 237. No indication of 'approximation', e.g. ≈,=, about, 3SF. Condone use of 3SF probability if more accurate probability seen in workings.
5. [B1] 1.405 < z1 < 1.407, -1.407 < z1 < -1.405 seen.
6. [B1] 0.841 < z < 0.843, -0.843 < z < -0.841 seen.
7. [M1] Use of the standardisation formula once with μ, σ equating to a z-value (not 0.20, 0.80, 0.08, 0.92, 0.5793, 0.4207, 0.5319, 0.4681, 0.7881, 0.2119, 0.8212, 0.1788). Condone continuity correction ±0.05, not σ², not √σ.
8. [M1] Solve two equations in μ and σ using the elimination method, substitution method or other appropriate approach to obtain values for both μ and σ.
9. [A1] 104.5 < μ ≤ 105.5. 6.44 < σ < 6.475. If one or both the M marks have been withheld, SCB1 for both correct.