Solution — 2023 Question 12 Normal Distribution

(a)

Let $S,$ $M$ and $L$ be the masses (in grams) of randomly chosen small, medium and large candles respectively.

\begin{matrix} S \sim N (0.24, {0.04}^{2}) \\ M \sim N (0.35, {0.05}^{2}) \\ L \sim N (0.6, {0.08}^{2}) \end{matrix}

\begin{matrix} P (0.21 < S < 0.25) & = 0.372 08 \\ = 0.372 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(0.21 < S < 0.25\right)

(b)

\begin{matrix} S + M + L & \sim N (0.24 + 0.35 + 0.6, {0.04}^{2} + {0.05}^{2} + {0.08}^{2}) \\ S + M + L & \sim N (1.19, 0.0105) \end{matrix}

\begin{matrix} P (S + M + L > 1.25) & = 0.279 09 \\ = 0.279 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(S + M + L > 1.25\right)

(c)

\begin{matrix} S_{1} + S_{2} + \dots + S_{6} & \sim N (6 (0.24), 6 {(0.04)}^{2}) \\ S_{1} + S_{2} + \dots + S_{6} & \sim N (1.44, 0.0096) \end{matrix}

\begin{matrix} M_{1} + M_{2} + M_{3} + M_{4} & \sim N (4 (0.35), 4 {(0.05)}^{2}) \\ M_{1} + M_{2} + M_{3} + M_{4} & \sim N (1.4, 0.01) \end{matrix}

Let $D = (M_{1} + \dots + M_{4}) - (S_{1} + \dots + S_{6}) .$

\begin{matrix} D & \sim N (1.4 - 1.44, 0.01 + 0.0096) \\ D & \sim N (- 0.04, 0.0196) \end{matrix}

\begin{matrix} P (M_{1} + \dots + M_{4} < S_{1} + \dots + S_{6}) & = P (D < 0) \\ = 0.612 45 \\ = 0.612 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(D < 0\right)

(d)

\begin{matrix} S_{1} + S_{2} + \dots + S_{400} & \sim N (400 (0.24), 400 {(0.04)}^{2}) \\ S_{1} + S_{2} + \dots + S_{400} & \sim N (96, 0.64) \end{matrix}

\begin{matrix} M_{1} + M_{2} + \dots + M_{200} & \sim N (200 (0.35), 200 {(0.05)}^{2}) \\ M_{1} + M_{2} + \dots + M_{200} & \sim N (70, 0.5) \end{matrix}

\begin{matrix} L_{1} + L_{2} + \dots + L_{50} & \sim N (50 (0.6), 50 {(0.08)}^{2}) \\ L_{1} + L_{2} + \dots + L_{50} & \sim N (30, 0.32) \end{matrix}

$Let T = S_{1} + \dots + S_{400} + M_{1} + \dots + M_{200} + L_{1} + \dots + L_{50}$

\begin{matrix} T & \sim N (96 + 70 + 30, 0.64 + 0.5 + 0.32) \\ T & \sim N (196, 1.46) \end{matrix}

$$ 0.02$ per gram is equal to $$ 20$ per kg

\begin{matrix} 20 T & \sim N (20 (196), 20^{2} (1.46)) \\ 20 T & \sim N (3920, 584) \end{matrix}

\begin{matrix} P (20 T < 3950) & = 0.892 77 \\ = 0.893 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(20T < 3950\right)