2020 H2 Maths Paper 2 Q6 — Worked Solution

Part (i)

2 marks

Since $T \sim N\left( 5,\ {1.2}^{2} \right)$ , the density is a symmetric bell curve centred at $t = 5$ , with the period from 8 am to 8.10 am corresponding to $t \in \lbrack 0,10\rbrack$ .

Part (ii)

1 mark

James leaves later than 8.06 am means $T > 6$ .

P(T > 6) = 0.202\ \text{(by GC)}

$\therefore P(T > 6) = 0.202$ , correct to 3 s.f.

Part (iii)

2 marks

Approach

James leaves at $T$ minutes past 8 am and then walks for $W$ minutes, so his arrival time is $T + W$ minutes past 8 am. He is late if this exceeds $30$ (i.e. past 8.30 am). Since $T$ and $W$ are independent normal random variables, $T + W$ is also normal, with mean and variance found by adding.

Let $T + W$ be James’s arrival time, in minutes past 8 am, on a day he walks. Since $T \sim N\left( 5,\ {1.2}^{2} \right)$ and $W \sim N\left( 21,\ 3^{2} \right)$ are independent,

T + W \sim N\left( 5 + 21,\ {1.2}^{2} + 3^{2} \right)

= N(26,\ 10.44)

James is late if $T + W > 30$ , so

P(T + W > 30) = 0.108\ \text{(by GC)}

$\therefore$ the probability that James is late for work on a day he walks is $0.108$ , correct to 3 s.f.

Part (iv)

5 marks

Approach

we are asked for $P\left( \text{fine} \mid \text{late} \right)$ , a reversed conditional probability, so we use Bayes’ theorem. This needs $P\left( \text{late} \right)$ overall, which splits into the walking case (already found in (iii)) and a driving case that we find the same way.

Let $F$ be the event that the weather is fine on a given day, and $L$ be the event that James is late for work that day. We are given $P(F) = 0.7$ , so $P(F') = 0.3$ .

On a day the weather is not fine, James drives, arriving $T + D$ minutes past 8 am, where $T \sim N\left( 5,\ {1.2}^{2} \right)$ and $D \sim N\left( 19,\ 6^{2} \right)$ are independent, so

T + D \sim N\left( 5 + 19,\ {1.2}^{2} + 6^{2} \right)

= N(24,\ 37.44)

giving

P(L \mid F') = P(T + D > 30) = 0.163\ \text{(by GC)}

From part (iii), $P(L \mid F) = 0.10786\ldots$ (unrounded). By the law of total probability,

P(L)

= P(L \mid F)P(F) + P(L \mid F')P(F')

= (0.10786\ldots)(0.7) + (0.16340\ldots)(0.3)

= 0.12452\ldots

By Bayes’ theorem,

P(F \mid L)

= \frac{P(L \mid F)P(F)}{P(L)}

= \frac{(0.10786\ldots)(0.7)}{0.12452\ldots}

= 0.606\ldots

$\therefore P(F \mid L) = 0.606$ , correct to 3 s.f.

2020 H2 Maths Paper 2, Question 6

Part (i)

Part (ii)

Part (iii)

Part (iv)

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