Wendy's journey to work consists of three parts: walking to the train station, riding on the train and then walking to the office. The times, in minutes, for the three parts of her journey are independent and have the distributions N(15.0, 1.1²), N(32.0, 3.5²) and N(8.6, 1.2²) respectively. If Wendy's journey takes more than 60 minutes, she is late for work.
- (a)Find the mean and variance of the total time for Wendy's journey.[2]
- (b)Find the probability that, on a randomly chosen day, Wendy will be late for work.[3]
- (c)Find the probability that the mean of Wendy's journey times over 15 randomly chosen days will be less than 54.5 minutes.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [B1] (a) Mean = 15.0+32.0+8.6 [= 55.6]; Allow unsimplified
- [B1] (a) Var = 1.1² + 3.5² + 1.2² [= 14.9]; Allow unsimplified
- [M1] (b) (60-"55.6") / √"14.9" [= 1.140]; FT their 55.6 and 14.9; Ignore continuity correction
- [M1] (b) 1 – Φ("1.140"); For correct probability area consistent with their working
- [A1] (b) 0.127 (3 sf); CWO
- [M1] (c) (54.5-"55.6") / √( "14.9"/15 ) or (817.5-834) / √223.5 [= -1.104]; FT their 55.6 and 14.9; No mixed methods
- [M1] (c) 1 – Ф("1.104"); For correct probability area consistent with their working
- [A1] (c) 0.135 (3 sf); As final answer
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