2015 H2 Maths Paper 2 Q7 — Worked Solution

Part (i)

2 marks

Two assumptions needed, in context:

Errors occur independently of one another, i.e. an error occurring at one point on a page does not affect the probability of an error occurring at any other point on that page (or on another page).
Errors occur at a constant average rate of $1.3$ per page throughout the newspaper, i.e. the mean number of errors per page is uniform across all pages and does not vary, for example, from the front page to the later pages.

Part (ii)

3 marks

Let $T$ be the total number of errors on the first 6 pages of the newspaper on a given day. Since the number of errors per page is $Po(1.3)$ and errors on different pages are independent, the sum of 6 independent $Po(1.3)$ variables is itself Poisson, with mean $6 \times 1.3$ :

T \sim Po(7.8).

We want $P(T > 10)$ . Converting to cumulative form,

P(T > 10)

\Rightarrow 1 - P(T \leq 10)

= 1 - 0.835

= 0.165\ \text{(3 s.f., by GC).}

$\therefore$ $P(T > 10) = 0.165$ .

Part (iii)

2 marks

Let $T_{n}$ be the total number of errors on the first $n$ pages of the newspaper, so $T_{n} \sim Po(1.3n)$ .

We are given $P\left( T_{n} < 2 \right) < 0.05$ , i.e.

P\left( T_{n} = 0 \right) + P\left( T_{n} = 1 \right) < 0.05

e^{- 1.3n} + 1.3n\, e^{- 1.3n} < 0.05

e^{- 1.3n}(1 + 1.3n) < 0.05.

Approach

this inequality cannot be solved for $n$ algebraically, since $n$ appears both in the exponent and outside it. As $n$ must also be a positive integer (a number of pages), we instead evaluate the left side by GC for successive integer values of $n$ and find where it first drops below $0.05$ .

By GC:

n = 3:\ e^{- 3.9}(1 + 3.9) = 0.0992 > 0.05

n = 4:\ e^{- 5.2}(1 + 5.2) = 0.0342 < 0.05.

Since $e^{- 1.3n}(1 + 1.3n)$ is decreasing in $n$ , the least value of $n$ satisfying the inequality is $n = 4$ .

$\therefore$ the least possible value of $n$ is $4$ .

2015 H2 Maths Paper 2, Question 7

Part (i)

Part (ii)

Part (iii)

This is how we mark, and how we teach.