Solution — 2021 Question 12 Normal Distribution

(a)

Let $A,$ $B$ and $I$ be the durations (in minutes) of randomly chosen songs by band A, songs by band B, and intervals respectively.

\begin{matrix} A \sim N (3.6, {0.5}^{2}) \\ B \sim N (4.2, {0.8}^{2}) \\ I \sim N (1.5, {0.4}^{2}) \end{matrix}

\begin{matrix} A - B & \sim N (3.6 - 4.2, {0.5}^{2} + {0.8}^{2}) \\ A - B & \sim N (- 0.6, 0.89) \end{matrix}

\begin{matrix} P (A > B) & = P (A - B > 0) \\ = 0.262 39 \\ = 0.262 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(A - B > 0\right)

(b)

Let $T = A_{1} + A_{2} + B_{1} + B_{2} + B_{3}$ be the total duration of $2$ songs by band A and $3$ songs by band B.

\begin{matrix} T & \sim N (2 (3.6) + 3 (4.2), 2 {(0.5)}^{2} + 3 {(0.8)}^{2}) \\ T & \sim N (19.8, 2.4200) \end{matrix}

\begin{matrix} P (18 < T < 22) & = 0.797 73 \\ = 0.798 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(18 < T < 22\right)

(c)

\begin{matrix} A_{1} + A_{2} + A_{3} + A_{4} & \sim N (4 (3.6), 4 {(0.5)}^{2}) \\ A_{1} + A_{2} + A_{3} + A_{4} & \sim N (14.4, 1) \\ B_{1} + B_{2} + B_{3} + B_{4} & \sim N (4 (4.2), 4 {(0.8)}^{2}) \\ B_{1} + B_{2} + B_{3} + B_{4} & \sim N (16.8, 2.5600) \\ I_{1} + I_{2} + \dots + I_{7} & \sim N (7 (1.5), 7 {(0.4)}^{2}) \\ I_{1} + I_{2} + \dots + I_{7} & \sim N (10.5, 1.12) \end{matrix}

Let $S$ be the total duration of the first half

\begin{matrix} S & \sim N (14.4 + 16.8 + 10.5, 1 + 2.5600 + 1.12) \\ S & \sim N (41.7, 4.6800) \end{matrix}

\begin{matrix} P (S > 40) & = 0.784 02 \\ = 0.784 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(S > 40\right)

(d)

As the two halves are independent,

\begin{matrix} P (both halves > 40) & = 0.784 02^{2} \\ = 0.615 ∎ (to 3 s.f.) \end{matrix}

(e)

The event in (d) is a proper subset of the event that the total of the two halves last for more than 80 minutes. For example, if the first half is 30 minutes and the second half is 55 minutes, the total will last more than 80 minutes but does not satisfy the event in (d) $∎$