2022 H2 Maths Paper 2 Q11 — Worked Solution

Part (a)

3 marks

Let $Z \sim N\left( 80,2^{2} \right)$ be Zhou’s time and $T \sim N\left( 79,3^{2} \right)$ be Tan’s time, independent. Zhou wins the race if his time is shorter, i.e. if $Z < T$ , equivalently if $D = Z - T < 0$ .

Since $Z$ and $T$ are independent normal random variables,

D = Z - T \sim N\left( 80 - 79,\ 2^{2} + 3^{2} \right) = N(1,13)

Standardising,

P(D < 0)

= P\left( Y < \frac{0 - 1}{\sqrt{13}} \right),\quad Y \sim N(0,1)

= \Phi\left( - \frac{1}{\sqrt{13}} \right)

= 0.391\ \left( \text{3 s.f., by GC} \right)

To check your answer:

- \frac{1}{\sqrt{13}} \approx - 0.2774

, and

\Phi( - 0.2774) \approx 0.391

, a value just under

0.5

— this makes sense, since Zhou’s mean time (

80

) is only slightly slower than Tan’s (

79

), so his probability of winning should be just under one-half. ✓

$\therefore$ $P\left( \text{Zhou wins} \right) = \Phi\left( - \frac{1}{\sqrt{13}} \right) = 0.391$ (3 s.f.).

Part (b)

2 marks

\overline{x} = 79.21,\quad s^{2} = 14.7\ \left( \text{3 s.f.} \right)

Explanation: with $n = 30$ , $\sum x = 2376.3$ , $\sum x^{2} = 188653.7$ , the unbiased estimate of the population mean is

\overline{x} = \frac{\sum x}{n} = \frac{2376.3}{30} = 79.21

and the unbiased estimate of the population variance is

s^{2} = \frac{1}{n - 1}\left( \sum x^{2} - \frac{(\sum x)^{2}}{n} \right) = \frac{1}{29}\left( 188653.7 - \frac{{2376.3}^{2}}{30} \right) = 14.7\ \left( \text{3 s.f.} \right)

To check your answer:

{2376.3}^{2} = 5646787.69

, so

\frac{{2376.3}^{2}}{30} = 188226.2563

, giving

188653.7 - 188226.2563 = 427.4437

, and

\frac{427.4437}{29} \approx 14.7233

. ✓

Part (c)

5 marks

Let $\mu$ be the population mean time, in seconds, for Zhou’s $100$ -metre freestyle swims after his exercise regime. We test:

H_{0}:\mu = 80\quad\quad H_{1}:\mu < 80

at the $5\%$ level of significance (a one-tail test, since we are testing specifically whether his times have reduced).

Since the population variance is unknown and $n = 30$ is reasonably large, we use the unbiased estimate $s^{2} = 14.7233$ from (b) and, by the Central Limit Theorem, the test statistic

Z = \frac{\overline{x} - \mu_{0}}{s/\sqrt{n}} = \frac{79.21 - 80}{\sqrt{14.7233/30}}

\approx - 1.128

To check your answer:

\sqrt{14.7233/30} = \sqrt{0.49078} \approx 0.7006

, so

Z \approx \frac{- 0.79}{0.7006} \approx - 1.128

. ✓

The critical value for a one-tail test at the $5\%$ level is $z = - 1.6449$ , so we reject $H_{0}$ if $Z < - 1.6449$ .

Since $- 1.128 > - 1.6449$ , $Z$ does not fall in the critical region, so we do not reject $H_{0}$ .

To check your answer: the

p

-value is

\Phi( - 1.128) \approx 0.130

, which is greater than

0.05

— confirming, via the equivalent

p

-value approach, that we do not reject

H_{0}

. ✓

$\therefore$ there is insufficient evidence, at the $5\%$ level of significance, that Zhou’s mean swim time has reduced.

Part (d)

1 mark

A $2$ -tail test.

Explanation: Tan is testing whether his mean time has changed, which could mean either an increase or a decrease with no direction stated in advance, so the alternative hypothesis is non-directional.

Part (e)

2 marks

Since Tan’s sample size is small ( $n = 6$ ), the Central Limit Theorem cannot be relied on to justify treating the sample mean as approximately normal, so Tan must assume:

Assumption 1: Tan’s individual swim times are normally distributed, so that the sample mean itself is exactly normal even for this small sample size.

Assumption 2: the population variance of Tan’s times is unchanged, still $3^{2} = 9$ (its previously known value), so that a $Z$ -test using this known variance can be carried out, since only the sample mean — not a sample variance — was recorded.

$\therefore$ (1) Tan’s times are normally distributed; (2) the population variance is still $3^{2} = 9$ .

2022 H2 Maths Paper 2, Question 11

Part (a)

Part (b)

Part (c)

Part (d)

Part (e)

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