Amir believes that 20% of the students at his college are left-handed. His friend believes that the true proportion, p, is less than 20%. Amir plans to use the binomial distribution to test the null hypothesis, H₀: p = 0.2, against the alternative hypothesis, H₁: p < 0.2. He decides to choose 35 students at random. If 3 or fewer of these students are left-handed, Amir will reject his belief.

Mark scheme

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

1. [B1] B(35, 0.2) used May be implied.
2. [M1] 0.8^35 + 35×0.8^34×0.2 + 35C2×0.8^33×0.2² + 35C3×0.8^32×0.2³ OR 0.0004056 + 0.0035494 + 0.015085 + 0.0414838 No end errors. Accept fully correct sigma notation.
3. [A1] = 0.0605 [so significance level is] 6.05% or 6.1% or 6% (or accept anything in range 6.05% to 14.3%) As final answer. Must see 0.0605. SC Unsupported correct answer 6.05% scores B1 B1.
4. [B1] 0.0605 Correct or FT their <=3 (from Bin) in 5(a) must be 3sf.
5. [B1] B(35, 0.05) used May be implied.
6. [M1] 1 - (0.95^35 + 35×0.95^34×0.05 + 35C2×0.95^33×0.05² + 35C3×0.95^32×0.05³) 1 - (0.1661+0.3059 +0.2737 + 0.1585) No end errors. Accept fully correct sigma notation.
7. [A1] = 0.0958 (3 sf) accept 0.0957 SC Unsupported correct answer scores B1 B1.