2018 H2 Maths Paper 2 Q10 — Worked Solution

Part (i)

2 marks

Let $X$ be the mass, in grams, of a randomly chosen light bulb. Then $X \sim N\left( 50,\ {1.5}^{2} \right)$ , so the density is a symmetric bell curve centred at $x = 50$ .

Part (ii)

1 mark

Standardising, with $Z = \frac{X - 50}{1.5} \sim N(0,1)$ ,

P(X < 50.4)

= P\left( Z < \frac{50.4 - 50}{1.5} \right)

= 0.605\ \text{(3 s.f., by GC).}

$\therefore P(X < 50.4) = 0.605$ , correct to 3 s.f.

Part (iii)

2 marks

Approach

“total mass of 4 randomly chosen empty boxes” is a sum of 4 independent normal random variables, not 4 times a single box’s mass — the mean scales by $4$ but the variance also scales by $4$ (not $16$ ), since the boxes are independent.

Let $Y$ be the mass, in grams, of a randomly chosen empty box, so $Y \sim N\left( 75,\ 2^{2} \right)$ . Let $T = Y_{1} + Y_{2} + Y_{3} + Y_{4}$ be the total mass of 4 independently, randomly chosen empty boxes. Since the $Y_{i}$ are independent,

T \sim N\left( 4(75),\ 4\left( 2^{2} \right) \right) = N(300,\ 16).

P(T > 297) = 0.773\ \text{(3 s.f., by GC).}

$\therefore P(T > 297) = 0.773$ , correct to 3 s.f.

Part (iv)

3 marks

Let $W = X + Y$ be the total mass, in grams, of a randomly chosen light bulb together with a randomly chosen box. Assuming the bulb’s mass and the box’s mass are independent,

W \sim N\left( 50 + 75,\ {1.5}^{2} + 2^{2} \right) = N(125,\ 6.25).

P(124.9 < W < 125.7) = 0.126\ \text{(3 s.f., by GC).}

$\therefore P(124.9 < W < 125.7) = 0.126$ , correct to 3 s.f.

Part (v)

4 marks

Approach

the padding’s mass is $30\%$ of the bulb’s mass, i.e. $0.3X$ — a fixed multiple of $X$ , not a separate independent random variable — so the total mass of box, bulb and padding is $Y + X + 0.3X = Y + 1.3X$ . We find the distribution of $Y + 1.3X$ , then use the given tail probability to solve for $k$ .

Let $U = Y + X + 0.3X = Y + 1.3X$ be the total mass, in grams, of a box containing a bulb and its padding. Since $X$ and $Y$ are independent,

1.3X \sim N\left( 1.3(50),\ {1.3}^{2}\left( {1.5}^{2} \right) \right) = N(65,\ 3.8025)

U = Y + 1.3X \sim N\left( 75 + 65,\ 2^{2} + 3.8025 \right) = N(140,\ 7.8025).

We are given $P(U > k) = 0.9$ . Standardising,

P\left( Z > \frac{k - 140}{\sqrt{7.8025}} \right) = 0.9,\ \text{where}\ Z \sim N(0,1).

By GC (inverse normal),

\frac{k - 140}{\sqrt{7.8025}} = - 1.2816\ldots

\Rightarrow k = 140 - 1.2816\ldots \times \sqrt{7.8025}

= 136\ \text{(3 s.f.)}

To check your answer: since

P(U > k) = 0.9 > 0.5

,

k

should be below the mean

140

;

k = 136

is indeed below

140

. ✓

$\therefore k = 136$ , correct to 3 s.f.

Part (vi)

3 marks

Approach

as in (iii), “total mass of 4 boxes, each containing a bulb and padding” sums 4 independent copies of $U$ from (v).

Let $S = U_{1} + U_{2} + U_{3} + U_{4}$ be the total mass of 4 independently, randomly chosen boxes, each containing a bulb and padding. Since the $U_{i}$ are independent,

S \sim N\left( 4(140),\ 4(7.8025) \right) = N(560,\ 31.21).

P(S > 565) = 0.185\ \text{(3 s.f., by GC).}

$\therefore P(S > 565) = 0.185$ , correct to 3 s.f.

2018 H2 Maths Paper 2, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

Part (vi)

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