Solution — 2022 Question 12 Normal Distribution

(a)

Let $S$ and $E$ be the masses (in kg) of randomly chosen Standard and Extra packs respectively.

\begin{matrix} S \sim N (1.21, {0.04}^{2}) \\ E \sim N (2.43, {0.06}^{2}) \end{matrix}

\begin{matrix} P (S > 1.19) & = 0.691 46 \\ = 0.691 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(S > 1.19\right)

(b)

\begin{matrix} P (S > 1.19) & = 1 - 0.691 46 \\ = 0.308 54 \end{matrix}

\begin{matrix} Required probability & = 0.308 54^{2} \\ = 0.0952 ∎ (to 3 s.f.) \end{matrix}

(c)

Let $S_{T} = S_{1} + \dots + S_{6}$ be the total mass of $6$ Standard packs and $E_{T} = E_{1} + E_{2} + E_{3}$ be the total mass of $3$ Extra packs

\begin{matrix} S_{T} & \sim N (6 (1.21), 6 {(0.04)}^{2}) \\ S_{T} & \sim N (7.26, 0.0096) \\ E_{T} & \sim N (3 (2.43), 3 {(0.06)}^{2}) \\ E_{T} & \sim N (7.29, 0.0108) \\ S_{T} - E_{T} & \sim N (7.26 - 7.29, 0.0096 + 0.0108) \\ S_{T} - E_{T} & \sim N (- 0.03, 0.0204) \end{matrix}

\begin{matrix} P (- 0.02 < S_{T} - E_{T} < 0.02) & = 0.108 95 \\ = 0.109 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(-0.02 < S{T} - E{T} < 0.02\right)

(d)

Let $S_{T_{8}} = S_{1} + \dots + S_{8}$ be the total mass of $8$ Standard packs and $E_{T_{4}} = E_{1} + E_{2} + E_{3} + E_{4}$ be the total mass of $4$ Extra packs

\begin{matrix} S_{T_{8}} & \sim N (8 (1.21), 8 {(0.04)}^{2}) \\ S_{T_{8}} & \sim N (9.68, 0.0128) \\ E_{T_{4}} & \sim N (4 (2.43), 4 {(0.06)}^{2}) \\ E_{T_{4}} & \sim N (9.72, 0.0144) \\ 4.8 S_{T_{8}} & \sim N (4.8 (9.68), {4.8}^{2} (0.0128)) \\ 4.8 S_{T_{8}} & \sim N (46.464, 0.294 91) \\ 4.5 E_{T_{4}} & \sim N (4.5 (9.72), {4.5}^{2} (0.0144)) \\ 4.5 E_{T_{4}} & \sim N (43.74, 0.291 60) \\ 4.8 S_{T_{8}} - 4.5 E_{T_{4}} & \sim N (46.464 - 43.74, 0.294 91 + 0.291 60) \\ 4.8 S_{T_{8}} - 4.5 E_{T_{4}} & \sim N (2.7240, 0.586 51) \end{matrix}

\begin{matrix} P (4.8 S_{T_{8}} - 4.5 E_{T_{4}} > 2) & = 0.827 76 \\ = 0.828 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(4.8S{T8} - 4.5E{T4} > 2\right)