2015 H2 Maths Paper 2 Q12 — Worked Solution

Part (i)

2 marks

Let $X$ be the mass, in grams, of a randomly chosen apple, so $X \sim N\left( 300,20^{2} \right)$ .

Let $T = X_{1} + X_{2} + X_{3} + X_{4} + X_{5}$ be the total mass of $5$ randomly chosen apples, where $X_{1},\ldots,X_{5}$ are independent and each distributed as $X$ .

E(T) = 5(300) = 1500

Var(T) = 5\left( 20^{2} \right) = 2000

so $T \sim N(1500,2000)$ .

Standardising,

P(T > 1600) = P\left( Z > \frac{1600 - 1500}{\sqrt{2000}} \right)

= 0.0127\ \text{(3 s.f., by GC).}

$\therefore$ $P(T > 1600) = 0.0127$ .

Part (ii)

3 marks

Let $Y$ be the mass, in grams, of a randomly chosen pear, so $Y \sim N\left( 200,15^{2} \right)$ .

Let $S = Y_{1} + \cdots + Y_{8}$ be the total mass of $8$ randomly chosen pears, where $Y_{1},\ldots,Y_{8}$ are independent and each distributed as $Y$ .

E(S) = 8(200) = 1600

Var(S) = 8\left( 15^{2} \right) = 1800

so $S \sim N(1600,1800)$ .

Approach

the question asks for $P(T > S)$ , a comparison between two sums. We handle this by working with the single random variable $D = T - S$ , whose distribution follows from $T$ and $S$ both being normal.

Let $D = T - S$ , the total mass of the $5$ apples minus the total mass of the $8$ pears.

E(D) = E(T) - E(S) = 1500 - 1600 = - 100

Var(D) = Var(T) + Var(S) = 2000 + 1800 = 3800

Note: $Var(D) = Var(T) + Var(S)$ , not $Var(T) - Var(S)$ , since $T$ and $S$ are independent.

so $D \sim N( - 100,3800)$ .

P(T > S) = P(D > 0)

= 0.0524\ \text{(3 s.f., by GC).}

$\therefore$ $P(T > S) = 0.0524$ .

Part (iii)

4 marks

Approach

peeling and coring an apple scales its mass by a constant factor $0.85$ (a reduction of $15\%$ ), and a pear’s mass by $0.9$ (a reduction of $10\%$ ). Since a linear transformation of a normal random variable is normal, the prepared masses are still normally distributed, with mean and variance scaled accordingly — the variance scales by the square of the factor.

Let $X'= 0.85X$ be the mass, in grams, of a prepared apple, and $Y'= 0.9Y$ be the mass, in grams, of a prepared pear.

E(X') = 0.85(300) = 255,\quad\quad Var(X') = {0.85}^{2}\left( 20^{2} \right) = 289

so $X'\sim N(255,289)$ .

E(Y') = 0.9(200) = 180,\quad\quad Var(Y') = {0.9}^{2}\left( 15^{2} \right) = 182.25

so $Y'\sim N(180,182.25)$ .

Let $U = X'_{1} + \cdots + X'_{5} + Y'_{1} + \cdots + Y'_{8}$ be the total prepared mass of $5$ apples and $8$ pears, where all $13$ masses are independent.

E(U) = 5(255) + 8(180) = 1275 + 1440 = 2715

Var(U) = 5(289) + 8(182.25) = 1445 + 1458 = 2903

so $U \sim N(2715,2903)$ .

P(U < 2750) = P\left( Z < \frac{2750 - 2715}{\sqrt{2903}} \right)

= 0.742\ \text{(3 s.f., by GC).}

To check your answer:

E(U) = 2715

is already close to and below

2750

, and

2750

is under one standard deviation (

\sqrt{2903} \approx 53.9

) above the mean, so a probability just above

0.5

(here

0.742

) is a sensible size. ✓

$\therefore$ $P(U < 2750) = 0.742$ .

2015 H2 Maths Paper 2, Question 12

Part (i)

Part (ii)

Part (iii)

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