2022 H2 Maths Paper 2 Q7 — Worked Solution

Part (a)

1 mark

The 53 employees who replied form a sample, not the population.

Explanation: the population is the entire group under study — here, all $75$ employees in the Staffing department, since the questionnaire was sent to every one of them. The $53$ who replied are only part of that group (a self-selecting subset, since not everyone asked chose to respond), so they are a sample of the department, not the department itself.

Part (b)

2 marks

She should take a random sample of employees — for example a stratified sample, grouping employees by department and then taking a simple random sample within each department, in proportion to that department’s size.

Explanation: she wants to know the views of employees across the whole company but cannot survey everyone, so a random sample is needed to give every employee a known, non-zero chance of selection and so avoid bias. Stratifying by department further ensures each department is fairly represented, so the results can be reliably generalised to the whole company’s workforce.

Part (c)

5 marks

Approach

Let $a,p,m,s$ be the numbers chosen from Administration, Production, Marketing and Staffing respectively, so $a + p + m + s = 8$ with $1 \leq a \leq 7,\ 1 \leq p \leq 6,\ 1 \leq m \leq 4,\ 1 \leq s \leq 3$ (at least one from each department) and $a > p,\ a > m,\ a > s$ (Administration strictly the largest). Since $p + m + s \geq 3$ , we have $a \leq 5$ ; since $a$ must exceed each of $p,m,s \geq 1$ , $a \geq 2$ . We work through $a = 5,4,3,2$ in turn.

Case $a = 5$ : then $p + m + s = 3$ with each $\geq 1$ and $\text{<}5$ , forcing $p = m = s = 1$ :

\binom{7}{5}\binom{6}{1}\binom{4}{1}\binom{3}{1} = 21 \times 6 \times 4 \times 3

= 1512

Case $a = 4$ : then $p + m + s = 4$ with each $\geq 1$ and $\text{<}4$ , so $\{ p,m,s\}$ is $2,1,1$ in some order:

\binom{6}{2}\binom{4}{1}\binom{3}{1} + \binom{6}{1}\binom{4}{2}\binom{3}{1} + \binom{6}{1}\binom{4}{1}\binom{3}{2}

= 180 + 108 + 72

= 360

Multiplying by $\binom{7}{4} = 35$ for the choice of Administration members,

35 \times 360 = 12600

Case $a = 3$ : then $p + m + s = 5$ with each $\geq 1$ and $\leq 2$ , so $\{ p,m,s\}$ is $2,2,1$ in some order:

\binom{6}{2}\binom{4}{2}\binom{3}{1} + \binom{6}{2}\binom{4}{1}\binom{3}{2} + \binom{6}{1}\binom{4}{2}\binom{3}{2}

= 270 + 180 + 108

= 558

Multiplying by $\binom{7}{3} = 35$ ,

35 \times 558 = 19530

Case $a = 2$ : then $p + m + s = 6$ with each $\geq 1$ and $\text{=}1$ only (since each $\text{<}2$ ), but $1 + 1 + 1 = 3 \neq 6$ , so this case is impossible and contributes $0$ .

Adding all cases,

1512 + 12600 + 19530 + 0 = 33642

To check your answer: a direct enumeration over every valid

(a,p,m,s)

with

a + p + m + s = 8

in the stated ranges and

a > p,\ a > m,\ a > s

, summing

\binom{7}{a}\binom{6}{p}\binom{4}{m}\binom{3}{s}

over each, gives

33642

, matching this calculation ✓

$\therefore$ the number of ways is $33642$ .

2022 H2 Maths Paper 2, Question 7

Part (a)

Part (b)

Part (c)

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