2020 H2 Maths Paper 2 Q7 — Worked Solution

Part (i)

2 marks

r = - 0.848\ \text{(by GC)}

$\therefore$ the product moment correlation coefficient between $d$ and $t$ is $- 0.848$ , correct to 3 s.f.

Part (ii)

5 marks

Approach

the model $d = - a + bu$ is linear in $u$ , so we regress $d$ on $u = \frac{1}{t}$ (not on $t$ ) to find $a$ and $b$ .

Using $u = \frac{1}{t}$ , the data become

$u$	$\frac{1}{32}$	$\frac{1}{41}$	$\frac{1}{50}$	$\frac{1}{59}$	$\frac{1}{68}$
$d$	$172$	$57$	$27$	$19$	$14$

By GC, the regression line of $d$ on $u$ is

d = - 145 + 9460u\ \text{(by GC)}

so $- a = - 145$ and $b = 9460$ , giving

a = 145,\quad\quad b = 9460

both correct to 3 s.f.

r = 0.941\ \text{(by GC)}

$\therefore a = 145$ , $b = 9460$ , and the product moment correlation coefficient between $d$ and $u$ is $0.941$ , all correct to 3 s.f.

Part (iii)

1 mark

At $t = 86$ , using Lim’s model,

d = - 145 + \frac{9460}{86}

= - 35.0\ \text{(3 s.f.)}

This is negative, which is meaningless for a number of days, and very far from the actual value $32$ .

Explanation: $t = 86$ lies well outside the range of the given data ( $32 \leq t \leq 68$ ), so using the model here is an extrapolation; a negative predicted number of days shows the model does not fit this additional data, even though $|r| = 0.941$ is close to $1$ within the original data range.

$\therefore$ Lim’s model does not fit the additional data.

Part (iv)

Approach

we substitute $t$ in terms of $T$ into the equation from (ii). Rearranging $C = \frac{5}{9}(F - 32)$ for $F$ (with $F = t$ and $C = T$ here) gives $t$ in terms of $T$ , which we substitute into $d = - a + bu = - a + \frac{b}{t}$ .

Since $T = \frac{5}{9}(t - 32)$ ,

t = \frac{9}{5}T + 32

Substituting into $d = - 145 + \frac{9460}{t}$ from part (ii),

d = - 145 + \frac{9460}{\frac{9}{5}T + 32}

$\therefore d = - 145 + \frac{9460}{\frac{9}{5}T + 32}$ .

2020 H2 Maths Paper 2, Question 7

Part (i)

Part (ii)

Part (iii)

Part (iv)

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