2015 H2 Maths Paper 2 Q10 — Worked Solution

Part (i)

1 mark

Part (ii)

3 marks

By GC, entering the ten pairs of values and computing the product moment correlation coefficient for each transformation of $h$ :

Part (a)

$r = - 0.9807$

Part (b)

$r = - 0.9748$

Part (c)

3 marks

$r = - 0.9986$

Part (iii)

2 marks

Approach

the “most appropriate” model is the one whose PMCC is closest to $- 1$ , since a value of $r$ nearer $\text{±}1$ indicates a stronger linear relationship between the transformed variable and $P$ — and only a variable that is (nearly) linearly related to $P$ should be used for linear regression.

Since $| - 0.9986|$ is closest to $1$ among the three values in part (ii), the model using $\sqrt{h}$ and $P$ is the most appropriate.

By GC, the regression line of $P$ on $\sqrt{h}$ is

P = 34.789 - 0.146865\sqrt{h}

$\therefore$ the equation which best models air pressure at different heights is $P = 34.8 - 0.147\sqrt{h}\ \text{(3 s.f.)}$ , where $h$ is in feet and $P$ is in inches of mercury.

Part (iv)

Approach

the equation in part (iii) takes $h$ in feet. Since $1$ metre $\text{=}3.28$ feet, a height of $h_{m}$ metres corresponds to $h = 3.28h_{m}$ feet, and we substitute this into the equation from part (iii).

Let $h_{m}$ be the height in metres, so $h = 3.28h_{m}$ .

P = 34.789 - 0.146865\sqrt{3.28h_{m}}

= 34.789 - 0.146865\sqrt{3.28}\sqrt{h_{m}}

= 34.789 - 0.265984\sqrt{h_{m}}

$\therefore$ $P = 34.8 - 0.266\sqrt{h_{m}}\ \text{(3 s.f.)}$ , where $h_{m}$ is the height in metres.

2015 H2 Maths Paper 2, Question 10