The time, T minutes, for a certain daily bus journey is normally distributed. The bus company claims that the mean of T is 45. A passenger believes that the mean of T is actually greater than 45. She notes the times taken for this journey on a random sample of 60 days. The results are summarised below. n = 60 ΣT = 2750 ΣT² = 127000

Mark scheme

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

1. [B1] Est(μ) = 2750/60 or 275/6 or 45.8 (3 sf)
2. [M1] Or (1/59) (127000 - 2750^2/60) Note: σ (4.03) can score M1 for correct expression Use of biased (15.97) M0.
3. [A1] =16.2 (3 sf) or 2875/177
4. [B1] H0: Population mean time = 45 H1: Population mean time > 45 Allow p = 0.25, p < 0.25. Allow 25%.
5. [M1] Standardise using their values from 3(a). Must have √60 (ignore cc).
6. [A1] = 1.602 to 1.595 Accept 3sf 1.6(0) if nothing better seen. (or area = 0.0546 to 0.0553). FT Biased in 3(a) scores A1 for 1.608 to 1.615.
7. [M1] '1.602' < 1.645 Or compare areas. i.e. 0.0546 to 0.0553 > 0.05.
8. [A1FT] [Accept H0] There is insufficient evidence [at 5% level] to reject the company's claim OR There is insufficient evidence to accept the passenger's belief OR There is insufficient evidence that the mean time is more than 45 minutes Note: accept cv method (45.856 > 45.83 or 44.98 < 45).