A company sells bags of pasta. The masses of large bags of pasta are normally distributed with mean 2.50kg and standard deviation 0.12kg. (a) Find the probability that the mass of pasta in a randomly chosen large bag is less than 2.65 kg. [2] A restaurant manager buys 160 of these large bags of pasta. (b) Find the number of bags for which you would expect the mass of pasta to be more than 1.65 standard deviations above the mean. [3] The masses of small bags of pasta sold by the company are normally distributed with mean μ kg and standard deviation σ kg. Tests show that 77% of these bags have masses greater than 1.26 kg, and 44% have masses less than 1.35 kg. (c) Find, in either order, the value of μ and the value of σ. [5]

Mark scheme

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

1. [M1] [P(X<2.65) =] P(Z< (2.65-2.50)/0.12) [ = P(Z< 1.25)]. Substituting 2.65, 2.50 and 0.12 into the standardisation formula allow σ² or √σ. Condone continuity correction ±0.005.
2. [A1] = 0.894 (0.894 < p < 0.8944).
3. [M1] P(Z> (2.50+1.65×0.12 -2.50)/0.12) = P(Z> 1.65). Seeing 2.50 + 1.65 × 0.12 or 2.698 and 2.50 and 0.12 substituted into the standardisation formula with a direct attempt to find the probability area. Accept numerator of 1.65 × 0.12. Accept Z linked to 1.65 with a direct attempt to find the probability area. Do not allow σ² or √σ. No continuity correction.
4. [A1] = 1-0.9505 = 0.0495 AWRT.
5. [B1 FT] Expected number = 0.0495 × 160 = 7.92, so 7 or 8. Strict FT their at least 4-DP calculated probability, seen anywhere (not a z-value). Final answer must be a single positive integer value, no approximation or rounding stated.
6. [B1] 1.26 - μ / σ = -0.739. (0.738 < z1 < 0.740 or -0.740 < z1 < -0.738 seen.)
7. [B1] 1.35 - μ / σ = -0.151. (0.150 < z2 < 0.152 or -0.152 < z2 < -0.150 seen.)
8. [M1] Use of the standardisation formula once with μ, σ equating to a z-value (not 0.77, 0.23, 0.44, 0.56, 0.7794, 0.5910, 0.6700, 0.7123, 0.2206, 0.4090, 0.3300, 0.2877). No continuity correction, not σ², not √σ.
9. [M1] Solve 2 equations in μ and σ using the elimination method, substitution method or other appropriate approach to obtain values for both μ and σ.
10. [A1] μ = 1.37, σ = 0.153 AWRT μ=1.37, σ = 0.153. (If one or both the M marks have been withheld, SCB1 for both correct.)