The lengths of Western bluebirds are normally distributed with mean 16.5cm and standard deviation 0.6 cm. A random sample of 150 of these birds is selected. The lengths of Eastern bluebirds are normally distributed with mean 18.4cm and standard deviation σcm. It is known that 72% of Eastern bluebirds have length greater than 17.1 cm. A random sample of 120 Eastern bluebirds is chosen.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [M1] Use of ± standardisation formula once with 16.5, 0.6 and either 15.4 or 16.8 substituted.
2. [M1] Calculating the appropriate probability area (leading to their final answer, expect > 0.5). 0.6915-(1-0.9666) or (0.6915-0.5) + (0.9666 – 0.5) OE are alternatives.
3. [A1] 0.658 ≤ p < 0.6585 . If A0 scored, SC B1 for 0.658 < p < 0.6585 .
4. [B1FT] [Expected number =] 0.6581 × 150 = 98, 99. FT their 4SF (or better) probability from a normal calculation. Must be a positive single integer answer. No approximation notation.
5. [B1] 0.5825 < z < 0.583 or -0.583 < z < -0.5825 seen.
6. [M1] Use of the ± standardisation formula with 17.1, 18.4, σ and a z-value (not 0.28, 0.72, 0.4175, 0.2358, 0.7642, 0.6103, 0.3897, ...). Condone continuity correct ± 0.05, not σ²,√σ.
7. [A1] σ = 2.23. AWRT
8. [B1] [Mean = 120 × 0.72 =] 86.4 AND [Var = 120 × 0.72 × 0.28 =] 24.192. 86.4, 842/5 and 24 24/125, 24.192 to at least 3SF seen, allow unsimplified. May be seen in standardisation formula. (4.918 ≤ σ ≤ 4.919 implies correct variance) Incorrect notation is penalised.
9. [M1] Substituting their mean (not 18.4) and their positive 4.9185 into ± standardisation formula (any number for 79.5), condone their 4.918² and √their 4.918 .
10. [M1] Using continuity correction 79.5 or 80.5 in their standardisation formula.
11. [M1] Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5 . Note: correct final answer implies this M1.
12. [A1] 0.0804. 0.0803 < p ≤ 0.0804