2019 H2 Maths Paper 2 Q11 — Worked Solution

Let $W$ grams be the mass of a randomly chosen white ball, so $W \sim N\left( 110,\ 4^{2} \right)$ .

Let $B$ grams be the mass of a randomly chosen black ball, so $B \sim N\left( 55,\ 2^{2} \right)$ .

Part (i)

2 marks

Let $S = W_{1} + W_{2} + W_{3} + W_{4}$ be the total mass of $4$ randomly and independently chosen white balls.

E(S) = 4(110) = 440

Var(S) = 4\left( 4^{2} \right) = 64

S \sim N(440,\ 64)

By GC,

P(S > 425) = 0.970\ (3\ s.f.)

Part (ii)

2 marks

Let $T = W + B$ be the total mass of a randomly chosen white ball and a randomly chosen black ball, chosen independently.

E(T) = 110 + 55 = 165

Var(T) = 4^{2} + 2^{2} = 20

T \sim N(165,\ 20)

By GC,

P(161 < T < 175) = 0.802\ (3\ s.f.)

Part (iii)

4 marks

Let $U = W_{1} + W_{2} + B_{1} + B_{2} + B_{3}$ be the total mass of $2$ randomly and independently chosen white balls and $3$ randomly and independently chosen black balls.

E(U) = 2(110) + 3(55) = 385

Var(U) = 2\left( 4^{2} \right) + 3\left( 2^{2} \right) = 44

U \sim N(385,\ 44)

We are given $P(U < M) = 0.271$ . Standardising,

P\left( Z < \frac{M - 385}{\sqrt{44}} \right) = 0.271,\ \text{where}\ Z \sim N(0,1).

By GC (inverse normal),

\frac{M - 385}{\sqrt{44}} = - 0.610\ (3\ s.f.)

M = 385 - 0.6098\sqrt{44}

M = 381\ (3\ s.f.)

To check your answer: since

0.271 < 0.5

, we expect

M < 385 = E(U)

, and indeed

M = 381 < 385

. ✓

$\therefore M = 381$ (to 3 s.f.).

Part (iv)

4 marks

Approach

Drilling the holes scales each ball’s mass by a constant factor (not a fixed reduction), so the drilled ball masses are found by multiplying the original normal random variables $W$ and $B$ by constants; this keeps them normal, but scales both the mean and the variance.

Drilling reduces the mass of each black ball by $10\%$ , so a drilled black ball has mass $0.9B$ , where

E(0.9B) = 0.9(55) = 49.5

Var(0.9B) = {0.9}^{2}\left( 2^{2} \right) = 3.24

0.9B \sim N(49.5,\ 3.24)

Drilling reduces the mass of each white ball by $30\%$ , so a drilled white ball has mass $0.7W$ , where

E(0.7W) = 0.7(110) = 77

Var(0.7W) = {0.7}^{2}\left( 4^{2} \right) = 7.84

0.7W \sim N(77,\ 7.84)

Let $R$ grams be the mass of a randomly chosen connecting rod, so $R \sim N\left( 20,\ {0.9}^{2} \right)$ .

The model is made from $4$ drilled black balls, $1$ drilled white ball and $4$ connecting rods, all chosen independently. Let $X$ be the total mass, in grams, of the model.

X = \left( 0.9B_{1} + 0.9B_{2} + 0.9B_{3} + 0.9B_{4} \right) + 0.7W + \left( R_{1} + R_{2} + R_{3} + R_{4} \right)

E(X) = 4(49.5) + 77 + 4(20)

= 355

Var(X) = 4(3.24) + 7.84 + 4(0.81)

= 24.04

X \sim N(355,\ 24.04)

By GC,

P(X > 350) = 0.846\ (3\ s.f.)

2019 H2 Maths Paper 2, Question 11

Part (i)

Part (ii)

Part (iii)

Part (iv)

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