The weights of male leopards in a particular region are normally distributed with mean 55kg and standard deviation 6kg.

Mark scheme

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

1. M1 46 or 62, 55 and 6 substituted into ±standardisation formula once. Condone 6² and continuity correction ±0.5
2. B1 Both standardisation values correct, accept unsimplified
3. M1 Calculating the appropriate area from stated Φs of z-values, must be probabilities.
4. A1 0.8115 < p < 0.812
5. P(46 < X < 62) = P((46-55)/6 < Z < (62-55)/6)
6. = P(-1.5 < Z < 7/6)
7. = P(-1.5 < Z < 1.1666...)
8. = Φ(1.1666...) - Φ(-1.5)
9. = Φ(1.1666...) - (1 - Φ(1.5))
10. = 0.8784 + 0.9332 - 1 = 0.8116
11. B1 CAO, critical z-value
12. M1 36 and 42 substituted in ±standardisation formula, no continuity correction, not σ², √σ, equated to a z-value
13. A1 WWW. Only dependent on M.
14. P(F < 36) = 0.25
15. P(Z < (36-42)/σ) = 0.25
16. P(Z < -6/σ) = 0.25
17. From tables, z = -0.674
18. -6/σ = -0.674
19. σ = 8.899 ≈ 8.90
20. M1 FT value from part (a) or Correct: 1-Φ((46-55)/6), condone continuity correction, σ², √σ, and probability found. Condone unsupported correct value stated.
21. M1 46, 42 and their 4(b) σ (or correct σ) substituted in ±standardisation formula, condone continuity correction, σ², √σ, and probability found Condone 4/their 8.90
22. M1 Product of their 2 probabilities (0 < both < 1) Not 0.25 or their final answer to 4(a) used.
23. A1 0.0449 < p < 0.0450
24. P(male < 46) = P(Z < (46-55)/6) = P(Z < -1.5) = 1 - Φ(1.5) = 1 - 0.9332 = 0.0668
25. P(female < 46) = P(Z < (46-42)/σ) = P(Z < 4/8.90) = P(Z < 0.4495) = Φ(0.4495) = 0.6736
26. P(both < 46) = 0.0668 × 0.6736 = 0.045009