Solution — 2019 Question 11 Normal Distribution

(a)

Let $A$ and $B$ be the times (in minutes) taken by randomly chosen runners from the Arrows and Beavers respectively.

\begin{matrix} A \sim N (14.8, {0.55}^{2}) \\ B \sim N (15.2, {0.65}^{2}) \end{matrix}

\begin{matrix} P (A > 15) & = 0.358 06 \\ = 0.358 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(A > 15\right)

(b)

Let $T = A_{1} + A_{2} + B_{1} + B_{2} + B_{3}$ be the total time taken by $2$ randomly chosen runners from the Arrows and $3$ randomly chosen runners from the Beavers.

\begin{matrix} T & \sim N (2 (14.8) + 3 (15.2), 2 {(0.55)}^{2} + 3 {(0.65)}^{2}) \\ T & \sim N (75.2, 1.8725) \end{matrix}

\begin{matrix} P (T < 75) & = 0.441 90 \\ = 0.442 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(T < 75\right)

(c)

Let $S_{A} = A_{1} + \dots + A_{6}$ and $S_{B} = B_{1} + \dots + B_{6} .$

\begin{matrix} S_{A} & \sim N (6 (14.8), 6 {(0.55)}^{2}) \\ S_{A} & \sim N (88.8, 1.815) \\ S_{B} & \sim N (6 (15.2), 6 {(0.65)}^{2}) \\ S_{B} & \sim N (91.2, 2.535) \end{matrix}

\begin{matrix} S_{A} - S_{B} & \sim N (88.8 - 91.2, 1.815 + 2.535) \\ S_{A} - S_{B} & \sim N (- 2.4, 4.35) \end{matrix}

\begin{matrix} P (- 5 < S_{A} - S_{B} < 5) & = 0.893 53 \\ = 0.894 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(-5 < SA - SB < 5\right)

(d)

\begin{matrix} P (S_{A} \geq 90) & = 0.186 54 \end{matrix}

finding \mathrm{P}\left(S_A \geq 90\right)

\begin{matrix} P (S_{B} \geq 90) & = 0.774 48 \end{matrix}

finding \mathrm{P}\left(S_B \geq 90\right)

\begin{matrix} P (neither wins a medal) & = 0.186 54 (0.774 48) \\ = 0.144 ∎ (to 3 s.f.) \end{matrix}