Kestrels are birds whose adult wingspans are normally distributed with mean 74.8cm and standard deviation 3.2 cm. A random sample of 120 adult kestrels is selected.

Mark scheme

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

1. [M1] Use of standardisation formula once with 74.8, 3.2 and either 72.4 or 76.3. No continuity correction, not σ², not √σ.
2. [M1] Calculating the appropriate probability area, must be a probability (assume a 4SF value in the calculation it is a probability, assume a 3SF value in the calculation it is not a probability). (leading to their final answer, expect < 0.5). Unless there is a misread, the probability area calculation must be structured as shown.
3. [A1] AWRT 0.454 WWW.
4. [B1] FT Strict FT their 4-figure probability × 120 (check with calculator). One integer answer. No indication of approximation, e.g. ≈, , about, 2SF.
5. [B1] 0.841 < z1 < 0.843 or -0.843 < z1 < -0.841 seen.
6. [B1] 0.582 < z2 < 0.584 or -0.584 < z2 < -0.582 seen.
7. [M1] Use of the standardisation formula once with μ, σ equating to a z-value (not 0.20, 0.80, 0.28, 0.72, 0.5793, 0.4207, 0.6103, 0.3897, 0.7881, 0.2119, 0.7642, 0.2358, -0.417, 0.417, -0.158 0.158, etc). No continuity correction, not σ², not √σ.
8. [M1] Solve 2 equations in μ and σ with an attempt at the elimination method, substitution method or other appropriate approach to obtain values for both μ and σ.
9. [A1] AWRT μ=0.192, σ = 0.0119. There must be consistency with signs in the solution, e.g. σ = 0.017 / -1.425 = 0.0119 is not acceptable, A0. If one or both the M marks have been withheld, SCB1 for both correct WWW.
10. Method 1: [M1] One term of the form 10Cx (p)x(1-p)10-x, 0<p<1,x≠0 or 10. [A1] Correct un-simplified expression. [B1] 0.677 < p < 0.678.
11. Method 2: [M1] One term of the form 10Cx (p)x(1-p)10-x, 0<p<1,x≠0 or 10. [A1] Correct un-simplified expression. [B1] 0.677 < p ≤ 0.678.