Solution — 2016 Question 12 Normal Distribution

(a)

Let $B$ and $C$ be the masses (in grams) of a randomly chosen biscuit and empty box respectively.

\begin{matrix} B \sim N (20, {1.1}^{2}) \\ C \sim N (5, {0.8}^{2}) \end{matrix}

\begin{matrix} P (B < 19) & = 0.181 65 \\ = 0.182 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(B < 19\right)

(b)

Let $W = B_{1} + \dots + B_{12}$

\begin{matrix} W & \sim N (12 (20), 12 {(1.1)}^{2}) \\ W & \sim N (240, 14.52) \\ W + C & \sim N (240 + 5, 14.52 + {0.8}^{2}) \\ W + C & \sim N (245, 15.16) \end{matrix}

\begin{matrix} P (W + C > 248) & = 0.220 50 \\ = 0.221 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(W + C > 248\right)

(c)

\begin{matrix} 0.6 W & \sim N (0.6 (240), {0.6}^{2} (14.52)) \\ 0.6 W & \sim N (144, 5.2272) \\ 0.2 C & \sim N (0.2 (5), {0.2}^{2} {(0.8)}^{2}) \\ 0.2 C & \sim N (1, 0.0256) \\ 0.6 W + 0.2 C & \sim N (144 + 1, 5.2272 + 0.0256) \\ 0.6 W + 0.2 C & \sim N (145, 5.2528) \end{matrix}

\begin{matrix} P (142 < 0.6 W + 0.2 C < 149) & = 0.864 26 \\ = 0.864 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(142 < 0.6W + 0.2C < 149\right)