Solution — 2017 Question 12 Normal Distribution

(a)

Let $X$ and $Y$ be the journey times (in minutes) by bus and train respectively.

\begin{matrix} X \sim N (45, 4^{2}) \\ Y \sim N (42, 3^{2}) \end{matrix}

\begin{matrix} P (X < 48) & = 0.773 37 \\ = 0.773 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(X < 48\right)

(b)

\begin{matrix} P (X > 48) & = 1 - 0.773 37 \\ = 0.226 63 \end{matrix}

\begin{matrix} Required probability & = 0.226 63^{2} \\ = 0.0514 ∎ (to 3 s.f.) \end{matrix}

(c)

The event in part (b) is a proper subset of the event in part (c). In particular, for example, if two bus journeys take 50 and 47 minutes respectively, then this will satisfy the condition for part (c) but not for part (b) $∎$

(d)

\begin{matrix} X_{1} + X_{2} + X_{3} & \sim N (3 (45), 3 {(4)}^{2}) \\ X_{1} + X_{2} + X_{3} & \sim N (135, 48) \\ Y_{1} + Y_{2} & \sim N (2 (42), 2 {(3)}^{2}) \\ Y_{1} + Y_{2} & \sim N (84, 18) \end{matrix}

Let $S = X_{1} + X_{2} + X_{3} + Y_{1} + Y_{2}$

\begin{matrix} S & \sim N (135 + 84, 48 + 18) \\ S & \sim N (219, 66) \end{matrix}

\begin{matrix} P (S > 210) & = 0.866 03 \\ = 0.866 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(S > 210\right)

(e)

$B = 0.12 X$ and $T = 0.15 Y$

\begin{matrix} B & \sim N (0.12 (45), {0.12}^{2} {(4)}^{2}) \\ B & \sim N (5.4, 0.2304) \\ T & \sim N (0.15 (42), {0.15}^{2} {(3)}^{2}) \\ T & \sim N (6.3, 0.2025) \\ 3 B & \sim N (3 (5.4), 3^{2} (0.2304)) \\ 3 B & \sim N (16.2, 2.0736) \\ 2 T & \sim N (2 (6.3), 2^{2} (0.2025)) \\ 2 T & \sim N (12.6, 0.81) \\ 3 B - 2 T & \sim N (16.2 - 12.6, 2.0736 + 0.81) \\ 3 B - 2 T & \sim N (3.6000, 2.8836) \end{matrix}

\begin{matrix} P (3 B - 2 T < 3) & = 0.361 92 \\ = 0.362 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(3B - 2T < 3\right)

The answer means that the probability that three times the cost of a bus journey exceed 2 times the cost of a train journey by less than $$ 3$ is $0.362.$