Solution — 2020 Question 11 Normal Distribution

(a)

Let $C$ and $V$ be the service times (in minutes) for a randomly chosen car and van respectively.

\begin{matrix} C \sim N (100, 8^{2}) \\ V \sim N (150, 10^{2}) \end{matrix}

\begin{matrix} P (90 < C < 120) & = 0.888 14 \\ = 0.888 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(90 < C < 120\right)

(b)

\begin{matrix} P (V \leq t) & = 0.9 \\ t & = 162.82 \\ = 163 ∎ (to 3 s.f.) \end{matrix}

(c)

Let $C_{T} = C_{1} + C_{2} + C_{3}$ and $V_{T} = V_{1} + V_{2}$ be the total service time for $3$ cars and $2$ vans respectively

\begin{matrix} C_{T} & \sim N (3 (100), 3 {(8)}^{2}) \\ C_{T} & \sim N (300, 192) \\ V_{T} & \sim N (2 (150), 2 {(10)}^{2}) \\ V_{T} & \sim N (300, 200) \\ C_{T} + V_{T} & \sim N (300 + 300, 192 + 200) \\ C_{T} + V_{T} & \sim N (600, 392) \end{matrix}

\begin{matrix} P (C_{T} + V_{T} < 630) & = 0.935 14 \\ = 0.935 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(CT + VT < 630\right)

(d)

$$ 45 per hour = $ 0.75 per minute$
$$ 90 per hour = $ 1.5 per minute$

\begin{matrix} 0.75 C_{T} & \sim N (0.75 (300), {0.75}^{2} (192)) \\ 0.75 C_{T} & \sim N (225, 108) \\ 1.5 V_{T} & \sim N (1.5 (300), {1.5}^{2} (200)) \\ 1.5 V_{T} & \sim N (450, 450) \\ 0.75 C_{T} + 1.5 V_{T} & \sim N (225 + 450, 108 + 450) \\ 0.75 C_{T} + 1.5 V_{T} & \sim N (675, 558) \end{matrix}

\begin{matrix} P (0.75 C_{T} + 1.5 V_{T} > 700) & = 0.144 95 \\ = 0.145 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(0.75CT + 1.5VT > 700\right)