2016 H2 Maths Paper 2 Q10 — Worked Solution

Part (i)

2 marks

Two conditions needed, in context:

Dandelion plants occur independently of one another, i.e. a dandelion plant growing at one point in the field does not affect the probability of a dandelion plant growing at any other point in the field.
Dandelion plants occur singly, at random, and at a constant average rate throughout the field, i.e. the mean number of dandelion plants per unit area does not vary from one part of the field to another.

Part (ii)

2 marks

Let $X$ be the number of dandelion plants in a randomly chosen $1\ \text{m}^{2}$ of Mia’s field. Since the average number of dandelions per $\text{m}^{2}$ is $1.5$ ,

X \sim Po(1.5).

We want $P(X \geq 2)$ . Converting to cumulative form,

P(X \geq 2)

\Rightarrow 1 - P(X \leq 1)

= 1 - 0.558

= 0.442\ \text{(3 s.f., by GC).}

$\therefore\ P(X \geq 2) = 0.442$ .

Part (iii)

2 marks

Let $Y$ be the number of dandelion plants in a randomly chosen $4\ \text{m}^{2}$ of Mia’s field. Since the numbers of dandelions in disjoint areas are independent, the sum of $4$ independent $Po(1.5)$ variables is itself Poisson, with mean $4 \times 1.5$ :

Y \sim Po(6).

P(Y \leq 3) = 0.151\ \text{(3 s.f., by GC).}

$\therefore\ P(Y \leq 3) = 0.151$ .

Part (iv)

4 marks

Approach

the number of dandelions in $80\ \text{m}^{2}$ is Poisson with a large mean ( $\lambda = 120 > 10$ ), so we approximate it by a normal distribution, using a continuity correction since we are approximating a discrete distribution by a continuous one.

Let $W$ be the number of dandelion plants in an $80\ \text{m}^{2}$ area of the field. Then

W \sim Po(120),

since the mean number of dandelions is $1.5 \times 80 = 120$ .

Since $\lambda = 120 > 10$ is large, we use the approximation

W \sim N(120,\ 120)\ \text{approximately,}

as $E(W) = Var(W) = \lambda = 120$ for a Poisson variable.

Applying a continuity correction (since $W$ is discrete),

P(110 \leq W \leq 140)

\approx P(109.5 < W < 140.5)

= 0.800\ \text{(3 s.f., by GC).}

To check your answer: the exact Poisson value is

P(110 \leq W \leq 140) = 0.798

(3 s.f., by GC), close to the normal approximation

0.800

. ✓

$\therefore\ P(110 \leq W \leq 140) = 0.800$ .

Part (v)

4 marks

Approach

let $N_{1}$ be the number of daisies in $1\ \text{m}^{2}$ and $N_{2}$ the number in $2\ \text{m}^{2}$ , so $N_{2} \sim Po(2\lambda)$ . The given condition equates $P\left( N_{1} \leq 2 \right)$ with $P\left( N_{2} > 2 \right)$ ; writing both sides out in terms of $\lambda$ gives a transcendental equation in $\lambda$ (it cannot be rearranged algebraically, since $\lambda$ appears both inside and outside an exponential), so we solve it numerically by GC.

Let $N_{1} \sim Po(\lambda)$ be the number of daisies in $1\ \text{m}^{2}$ of the field, and $N_{2} \sim Po(2\lambda)$ be the number of daisies in $2\ \text{m}^{2}$ of the field. We are given $P\left( N_{1} \leq 2 \right) = P\left( N_{2} > 2 \right)$ , i.e.

e^{- \lambda}\left( 1 + \lambda + \frac{\lambda^{2}}{2} \right) = 1 - e^{- 2\lambda}\left( 1 + 2\lambda + 2\lambda^{2} \right).

Solving this equation for $\lambda > 0$ by GC,

\lambda = 1.85\ \text{(3 s.f.).}

$\therefore\ \lambda = 1.85$ .

2016 H2 Maths Paper 2, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

This is how we mark, and how we teach.