2024 H2 Mathematics Paper 2 Question 10

The Normal Distribution

Answers

$0.941.$

$0.183.$

$0.769.$

$6377 g .$

(a)

Let $X$ be the random variable of the masses, in grams, of a randomly chosen plate

X \sim N (200, {1.6}^{2})

\begin{aligned} P (X > 197.5) \\ = 0.94091 \\ = 0.941 (3 s.f.) ∎ \end{aligned}

(b)

Let $Y = 0.95 X$

\begin{aligned} Y & \sim N (0.95 (200), ({0.95}^{2}) (1 . 6^{2})) \\ Y & \sim N (190, 2.3104) \end{aligned}

\begin{aligned} P (190 < Y < 192) \\ = 0.40588 \end{aligned}

Let $N$ be the random variable of the number of plates with masses between $190 grams$ and $192 grams$ out of 8

N \sim B (8, 0.40588)

\begin{aligned} P (N \geq 5) \\ = 1 - P (N \leq 4) \\ = 0.18290 \\ = 0.183 (3 s.f.) ∎ \end{aligned}

(c)

Let $T_{p} = Y_{1} + Y_{2} + \dots + Y_{20}$

\begin{aligned} T_{p} & \sim N (20 (190), 20 (2.3104)) \\ T_{p} & \sim N (3800, 46.208) \end{aligned}

\begin{aligned} P (T_{p} < 3805) \\ = 0.76900 \\ = 0.769 (3 s.f.) ∎ \end{aligned}

(d)

Let $S$ and $B$ be the random variable of the mass of a randomly chosen screw and bolt, in grams, respectively

\begin{matrix} S \sim N (10, 0 . 3^{2}) \\ B \sim N (44, 1 . 1^{2}) \end{matrix}

Each pack consists of $20$ plates, $80$ screws, and $40$ bolts

Let $T_{s} = S_{1} + S_{2} + \dots + S_{80}$ and
$T_{b} = B_{1} + B_{2} + \dots + B_{40}$

\begin{aligned} T_{s} & \sim N (80 (10), 80 (0 . 3^{2})) \\ T_{s} & \sim N (800, 7.2) \end{aligned}

\begin{aligned} T_{b} & \sim N (40 (44), 40 (1 . 1^{2})) \\ T_{b} & \sim N (1760, 48.4) \end{aligned}

\begin{aligned} T_{p} + T_{s} + T_{b} & \sim N (3800 + 800 + 1760, 46.208 + 7.2 + 48.4) \\ T_{p} + T_{s} + T_{b} & \sim N (6360, 101.81) \end{aligned}

Let $m$ denote the mass exceeded by $5 %$ of the Value packs

\begin{aligned} P (T_{p} + T_{s} + T_{b} > m) & = 0.05 \\ m & = 6377 g (nearest gram) ∎ \end{aligned}