2025 TMJC P2 Q11

2025 TMJC P2 Q11

15 marks

Anastasia leaves her house to travel to school every school day. On each school day, she will first take \(C\) minutes to cycle from her house to the train station, followed by taking a train ride of \(T\) minutes, and finally taking a bus ride of \(B\) minutes, before she reaches her school. You should assume that all waiting and walking times are negligible. It is given that \(C\), \(T\) and \(B\) follow independent normal distributions, with means and standard deviations as shown in the table.

Mean (minutes)Standard Deviation (minutes)
\(C\)\(8\)\(0.2\)
\(T\)\(17\)\(\sigma\)
\(B\)\(15\)\(2.1\)

Anastasia is supposed to reach school by \(7.30\) am so that she is not late for school.

  1. Given that \(10\)% of the train rides will take at most \(15\) minutes, find the value of \(\sigma\).[2]

For the rest of the question, use \(\sigma = 1.4\).

  1. Find the probability that Anastasia spends between \(35\) and \(50\) minutes travelling from her house to school on a randomly chosen school day.[3]
  2. Show that the probability that Anastasia is late for school if she leaves house at \(6:48\) am is \(0.215\), correct to \(3\) significant figures.[1]
  3. Find the latest time that Anastasia needs to leave her house if she wants to be at least \(90\)% confident that she will not be late for school.[2]
  4. Find the probability that the total time Anastasia spends cycling from her house to the train station on five randomly chosen school days exceeds twice the time she spends on the train ride on a randomly chosen school day.[3]
  5. Anastasia attends school five days a week and she leaves house at \(6:48\) am every day for school. Find the probability that she is late for school twice in the first two weeks and her \(3^{rd}\) late coming occurs in the \(6^{th}\) week.[4]

Solution:

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Answer:(a) \(1.56\). (b) \(0.976\). (c) \(0.215\) (shown). (d) Approximately \(6{:}46{:}45\) am; \(6{:}46\) am if restricted to whole-minute departures. (e) \(0.983\). (f) \(0.00560\), or \(0.00558\) using the rounded \(p=0.215\).

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