Solution — 2018 Question 12 Normal Distribution

(a)

Let $A$ and $B$ be the masses (in grams) of randomly chosen Type A and Type B components respectively.

\begin{matrix} A \sim N (250, 3^{2}) \\ B \sim N (240, 4^{2}) \end{matrix}

$\begin{matrix} P (A > 1.02 (250)) & = P (A > 255) \\ = 0.047 790 \\ = 0.0478 ∎ (to 3 s.f.) \end{matrix}$

[finding \mathrm{P}\left(A > 255\right)]

(b)

Let

\begin{matrix} A - B & \sim N (250 - 240, 3^{2} + 4^{2}) \\ A - B & \sim N (10, 25) \end{matrix}

$\begin{matrix} P (A > B) & = P (A - B > 0) \\ = 0.977 25 \\ = 0.977 ∎ (to 3 s.f.) \end{matrix}$

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

(c)

Let $T = A_{1} + \dots + A_{6} + B_{1} + \dots + B_{3}$ be the total mass of $6$ Type A and $3$ Type B components.

\begin{matrix} T & \sim N (6 (250) + 3 (240), 6 {(3)}^{2} + 3 {(4)}^{2}) \\ T & \sim N (2220, 102) \end{matrix}

$\begin{matrix} P (2190 < T < 2230) & = 0.837 46 \\ = 0.837 ∎ (to 3 s.f.) \end{matrix}$

[finding \mathrm{P}\left(2190 < T < 2230\right)]

(d)

Let $A_{T} = A_{1} + \dots + A_{10}$ and $B_{T} = B_{1} + \dots + B_{10}$

\begin{matrix} A_{T} & \sim N (10 (250), 10 {(3)}^{2}) \\ A_{T} & \sim N (2500, 90) \\ B_{T} & \sim N (10 (240), 10 {(4)}^{2}) \\ B_{T} & \sim N (2400, 160) \\ 0.02 A_{T} & \sim N (0.02 (2500), {0.02}^{2} (90)) \\ 0.02 A_{T} & \sim N (50, 0.036) \\ 0.03 B_{T} & \sim N (0.03 (2400), {0.03}^{2} (160)) \\ 0.03 B_{T} & \sim N (72, 0.144) \\ 0.03 B_{T} - 0.02 A_{T} & \sim N (72 - 50, 0.144 + 0.036) \\ 0.03 B_{T} - 0.02 A_{T} & \sim N (22, 0.18) \end{matrix}

$\begin{matrix} P (0.03 B_{T} - 0.02 A_{T} > 22.5) & = 0.119 30 \\ = 0.119 ∎ (to 3 s.f.) \end{matrix}$

[finding \mathrm{P}\left(0.03BT - 0.02AT > 22.5\right)]