2023 H2 Maths Paper 2 Q9 — Worked Solution

Part (a)

1 mark

Let $X$ be the length, in metres, of a randomly chosen Long plank: $X \sim N\left( 1.82,\ {0.2}^{2} \right)$ .

P(X < 1.79) \approx 0.440\ \left( 3\text{ s.f.} \right)

Explanation: standardising,

P(X < 1.79) = P\left( Z < \frac{1.79 - 1.82}{0.2} \right)

= P(Z < - 0.15)

\approx 0.440\ \text{(by GC)}

Part (b)

3 marks

Let $S = X_{1} + X_{2} + \cdots + X_{8}$ be the total length of $8$ randomly chosen Long planks. Since each $X_{i} \sim N\left( 1.82,\ {0.2}^{2} \right)$ independently,

S \sim N\left( 8(1.82),\ 8\left( {0.2}^{2} \right) \right)

= N(14.56,\ 0.32)

P(S > 14.5) = P\left( Z > \frac{14.5 - 14.56}{\sqrt{0.32}} \right)

= P(Z > - 0.106)

\approx 0.542\ \left( 3\text{ s.f.} \right)

To check your answer: the mean total

14.56

is only just above

14.5

, so the probability of exceeding

14.5

should be just over

0.5

✓

$\therefore\ P(S > 14.5) \approx 0.542$ .

Part (c)

2 marks

Let $Y$ be the length, in metres, of a randomly chosen Regular plank: $Y \sim N\left( 1.22,\ {0.3}^{2} \right)$ .

P(Y > 1.25) = P\left( Z > \frac{1.25 - 1.22}{0.3} \right)

= P\left( Z > 0.1\overline{6} \right)

\approx 0.460\ \left( 3\text{ s.f.} \right)

Let $W$ be the number of Sylvio’s $120$ planks that are longer than $1.25$ m. Since each plank independently has probability $\text{≈}0.4602$ of being longer than $1.25$ m, $W \sim B(120,\ 0.4602)$ approximately, so

E(W) = 120 \times 0.4602

\approx 55.2

$\therefore\ E(W) \approx 55.2$ , i.e. about $55$ planks.

Part (d)

3 marks

Approach

We combine the two independent total-length variables as a single normal variable $D = L - R$ , then use the symmetric interval $P( - 0.65 < D < 0.65)$ — this interval is not centred at $E(D)$ , so we standardise both endpoints separately.

Let $L$ = total length of $10$ randomly chosen Long planks and $R$ = total length of $16$ randomly chosen Regular planks.

L \sim N\left( 10(1.82),\ 10\left( {0.2}^{2} \right) \right)

= N(18.2,\ 0.4)

R \sim N\left( 16(1.22),\ 16\left( {0.3}^{2} \right) \right)

= N(19.52,\ 1.44)

Let $D = L - R$ , the difference between the two total lengths. Since $L$ and $R$ are independent,

D \sim N(18.2 - 19.52,\ 0.4 + 1.44)

= N( - 1.32,\ 1.84)

We require $P\left( |D| < 0.65 \right) = P( - 0.65 < D < 0.65)$ :

P( - 0.65 < D < 0.65) = P\left( \frac{- 0.65 - ( - 1.32)}{\sqrt{1.84}} < Z < \frac{0.65 - ( - 1.32)}{\sqrt{1.84}} \right)

\Rightarrow P(0.494 < Z < 1.45)

\approx 0.237\ \left( 3\text{ s.f.} \right)

To check your answer:

E(D) = - 1.32

lies outside

( - 0.65,\ 0.65)

, so the interval is not centred on the mean, and a probability well below

0.5

is expected ✓

$\therefore\ P\left( |D| < 0.65 \right) \approx 0.237$ .

Part (e)

5 marks

Approach

A Short plank cut from a Long plank has length $\frac{X}{3}$ ; one cut from a Regular plank has length $\frac{Y}{2}$ . We find the distribution of each, then of their difference $D_{2} = \frac{X}{3} - \frac{Y}{2}$ , and evaluate $P\left( D_{2} > 0 \right)$ .

A Short plank cut from a Long plank has length $\frac{X}{3}$ , where $X \sim N\left( 1.82,\ {0.2}^{2} \right)$ :

\frac{X}{3} \sim N\left( \frac{1.82}{3},\ \frac{{0.2}^{2}}{9} \right)

= N(0.607,\ 0.00444)\ \left( 3\text{ s.f.} \right)

A Short plank cut from a Regular plank has length $\frac{Y}{2}$ , where $Y \sim N\left( 1.22,\ {0.3}^{2} \right)$ :

\frac{Y}{2} \sim N\left( \frac{1.22}{2},\ \frac{{0.3}^{2}}{4} \right)

= N(0.61,\ 0.0225)

Let $D_{2} = \frac{X}{3} - \frac{Y}{2}$ . Since $\frac{X}{3}$ and $\frac{Y}{2}$ are independent,

D_{2} \sim N(0.6067 - 0.61,\ 0.004444 + 0.0225)

= N( - 0.00333,\ 0.0269)

We require $P\left( D_{2} > 0 \right)$ :

P\left( D_{2} > 0 \right) = P\left( Z > \frac{0 - ( - 0.00333)}{\sqrt{0.0269}} \right)

= P(Z > 0.0203)

\approx 0.492\ \left( 3\text{ s.f.} \right)

To check your answer:

E\left( D_{2} \right) \approx - 0.0033

is very close to

0

, so the probability should be very close to, but just under,

0.5

✓

$\therefore\ P\left( \frac{X}{3} > \frac{Y}{2} \right) \approx 0.492$ .

Part (f)

The Long plank needs two cuts to produce three equal Short pieces, while the Regular plank needs only one cut to produce two equal pieces. If each cut wastes a small amount of length, the Short plank from a Long plank loses length from two cuts, while the Short plank from a Regular plank loses length from only one — reducing $E\left( \frac{X}{3} \right)$ by more than $E\left( \frac{Y}{2} \right)$ .

$\therefore$ the mean of $\frac{X}{3}$ decreases by more than the mean of $\frac{Y}{2}$ , so the probability in part (e) would decrease.

2023 H2 Maths Paper 2, Question 9