2024 H2 Maths Paper 2 Q10 — Worked Solution

Part (a)

1 mark

Let $X$ be the mass, in grams, of a plate before drilling: $X \sim N\left( 200,\ {1.6}^{2} \right)$ .

Standardising,

Z = \frac{197.5 - 200}{1.6}

= - 1.5625

P(X > 197.5) = P(Z > - 1.5625)

= \Phi(1.5625)

= 0.941\ \left( 3\ \text{s.f.} \right)

$\therefore$ $P(X > 197.5) \approx 0.941$ .

Part (b)

4 marks

Approach

Drilling scales every plate’s mass by the same factor, so the drilled mass is a linear transformation of $X$ ; we find its distribution first, then use it to find the probability that one plate falls in $\lbrack 190,192\rbrack$ , and finally treat “at least 5 of 8 plates” as a binomial count.

Drilling reduces mass by 5%, so the drilled mass is $Y = 0.95X$ , where

Y \sim N\left( 0.95(200),\ {0.95}^{2}\left( {1.6}^{2} \right) \right) = N\left( 190,\ {1.52}^{2} \right)

Let $p = P(190 < Y < 192)$ , the probability that one drilled plate has mass between 190 g and 192 g.

Z_{1} = \frac{190 - 190}{1.52} = 0,\quad\quad Z_{2} = \frac{192 - 190}{1.52} = 1.3158

p = \Phi(1.3158) - \Phi(0)

= 0.9059 - 0.5

= 0.4059

Let $W$ be the number of the 8 plates, out of 8 chosen independently, with mass between 190 g and 192 g:

W \sim B(8,\ 0.4059)

Converting the tail to the cumulative form the GC uses,

P(W \geq 5) = 1 - P(W \leq 4)

= 1 - 0.8171

= 0.183\ \left( 3\ \text{s.f., by GC} \right)

To check your answer:

E(W) = 8(0.4059) = 3.25

, so

W \geq 5

is an upper-tail event; a probability of about

0.18

for it is a reasonable size. ✓

$\therefore$ $P(W \geq 5) \approx 0.183$ .

Part (c)

3 marks

Let $S$ be the total mass of 20 randomly chosen drilled plates. Since each drilled plate is $N\left( 190,\ {1.52}^{2} \right)$ (from (b)) and the plates are chosen independently,

S \sim N\left( 20(190),\ 20\left( {1.52}^{2} \right) \right) = N(3800,\ 46.208)

Z = \frac{3805 - 3800}{\sqrt{46.208}}

= 0.7358

P(S < 3805) = \Phi(0.7358)

= 0.769\ \left( 3\ \text{s.f.} \right)

To check your answer:

3805

is only

0.74

standard deviations above the mean

3800

, so a probability just above

0.5

(here

\text{≈}0.77

) is a reasonable size. ✓

$\therefore$ $P(S < 3805) \approx 0.769$ .

Part (d)

4 marks

Approach

A Value pack’s total mass is the sum of the pack of plates and all its screws and bolts, all independent; we find $E$ and $Var$ of this sum, then find the value $m$ that leaves $5\%$ of the normal distribution above it.

Each plate needs 4 screws and 2 bolts (given with the question), so a pack of 20 drilled plates needs $20 \times 4 = 80$ screws and $20 \times 2 = 40$ bolts.

Let $M$ be the total mass of a Value pack: $M = S + \left( \text{80 screws} \right) + \left( \text{40 bolts} \right)$ , where each screw is $N\left( 10,\ {0.3}^{2} \right)$ , each bolt is $N\left( 44,\ {1.1}^{2} \right)$ , and $S$ is the mass of the 20 plates from (c); all these masses are independent of one another.

E(M) = 3800 + 80(10) + 40(44)

= 3800 + 800 + 1760

= 6360

Var(M) = Var(S) + 80\left( {0.3}^{2} \right) + 40\left( {1.1}^{2} \right)

= 46.208 + 7.2 + 48.4

= 101.808

So $M \sim N(6360,\ 101.808)$ , with standard deviation $\sqrt{101.808} = 10.090$ .

We need $m$ such that $P(M > m) = 0.05$ , i.e. $m$ is $z$ standard deviations above the mean, where $P(Z > z) = 0.05 \Rightarrow z = 1.6449$ (by GC).

m = E(M) + z \cdot sd(M)

= 6360 + 1.6449(10.090)

= 6376.6

To check your answer: at

m = 6376.6

,

Z = \frac{6376.6 - 6360}{10.090} = 1.645

, and

P(Z > 1.645) = 0.05

. ✓

$\therefore$ the mass exceeded by just $5\%$ of Value packs is $m \approx 6377\ g$ (nearest gram).

2024 H2 Maths Paper 2, Question 10

Part (a)

Part (b)

Part (c)

Part (d)

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