### 2014 A-level H1 Mathematics (8864) Question 12 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X, Y be the output daily from mines A and B, respectively.
$X ~\sim~ \mathrm{N}(50, \sigma^{2})$

$\mathrm{P}(X>75) = 0.0189$

$latex \mathrm{P}(X<75) = 0.9811$ $latex \Rightarrow \mathrm{P}(Z< \frac{25}{\sigma}) = 0.9811$ $latex \frac{25}{\sigma} = 2.07701689$ $latex \sigma^{2} = 144.88 \approx 145$ (ii) $latex Y ~\sim~ \mathrm{N}(75, 64)$ $latex Y_1 + \ldots +Y_7 ~\sim~ \mathrm{N}(525,448)$ $latex \mathrm{P}(Y_1 + \ldots +Y_7 < 500) = 0.119$ (iii) $latex Y_1 + \ldots + Y_7 - 2(X_1 + \ldots +X_5) ~\sim~ \mathrm{N} (25, 3348)$ $latex \mathrm{P}[Y_1 + \ldots + Y_7 - 2(X_1 + \ldots +X_5) > 0] = 0.667$

Some students forgot to change the (i)’s probability before applying inverse normal. This proved to be costly as the whole question was ruined. Some students applied the expectations and variance wrongly too.

### 2014 A-level H1 Mathematics (8864) Question 11 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

$\mathrm{P} (L) = \frac{90}{290+x}$

$\mathrm{P} (G) = \frac{100}{290+x}$

Since L and G are independent events,

$\mathrm{P} (L \cap G) = \mathrm{L} \times \mathrm{G}$

$\frac{30}{290+x} = (\frac{90}{290+x})(\frac{10}{290+x})$

$x = 10$

(ii)
$\mathrm{P} (L \cup T) = \frac{235}{300} = \frac{47}{60}$

(iii)

$\mathrm{P} (T \cap G) = \frac{142}{300} = \frac{71}{150}$

(iv)

$\mathrm{P} (L | G) = \mathrm{P}(L)$ since L and G are independent events.

$\therefore, \mathrm{P} (L | G) = \frac{90}{300} = \frac{3}{10}$

(v)

Required Probability $= \frac{37}{300} \times \frac{36}{299} = 0.0148$

Some students commented that (v) should be done via binomial distribution, but this is a misguided approach as they should realise that the probability is not constant. Thus, a binomial model will not be appropriate.

### 2014 A-level H1 Mathematics (8864) Question 8 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

(ii)
Using Graphing Calculator, $r = -0.926$.

Since $|r|$ is close to 1, there is a strong negative correlation between the number of hours spent travelling to and fro work and the number of hours spent watch television.

(iii)
Using Graphing Calculator, equation of regression line of $y \mathrm{~on~} x$ is $y = -0.9021x+16.15$

(iv)
When $x = 13.2, y = 4.24$ hours

The estimate is not reliable since $x =13.2$ is outside the range of sample data.

Students must check what they key in very carefully. And for the drawing of regression line, students must learn how to store them properly so as to draw as precisely as possible.

### 2014 A-level H1 Mathematics (8864) Question 7 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Order a list of all 5000 households from 1 to 5000.

$\frac{5000}{100} = 50$

Randomly select a number from 1 to 50 and taking that as a starting point, sample at continuous interval of 50. Eg, 10, 60, 110, …

(ii)

Use simple random sampling within each stratum to obtain the sample.

(iii)
Stratified sampling is more appropriate since it is more representative of the users of different age and shopping methods.

Some students overlook that for systematic sampling, we need to RANDOMLY select. For (ii), they can draw a nice table to present their answers.

### 2014 A-level H1 Mathematics (8864) Question 10 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X denotes the length of a leaf in cm.

$H_0: \mu = 7$

$latex H_0: \mu < 7$ Under $latex H_0, \bar{X} ~\sim~ N(7, \frac{4.4}{50})$ approximately by Central Limit Theorem since n is large. From Graphing Calculator, p-value $latex = 0.0459 < 0.05$. Thus we reject $latex H_0$, and conclude with sufficient evidence at 5% level of significant that the mean length of the leaf is less than 7 cm. (ii) Unbiased estimate of $latex \mu = \frac{310.4}{50} = 6.208$ Unbiased estimate of $latex {\sigma}^{2} = \frac{1}{49}[2209.2 - \frac{310.4^{2}}{50}] = 5.76$ (iii) $latex H_0: \mu = 7$ $latex H_0: \mu \ne 7$ Under $latex H_0, \bar{X} ~\sim~ N(7, \frac{5.79935}{50})$ approximately by Central Limit Theorem since n is large. From Graphing Calculator, p-value $latex = 0.019624$. Since $latex H_0$ is rejected, p-values $latex \le \alpha \%$ Required set of values $latex = \{ \alpha \in \mathbb{R}: 1.97 \le \alpha \le 100 \} ### KS Comments Nothing special about this question. Students still forget to introduce Central Limit Theorem when doing the hypothesis testing though. Lastly, the question asks for the set of values, so students should use precise set notation. ### 2014 A-level H1 Mathematics (8864) Question 9 Suggested Solutions All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) (a) Let X be the number of cakes containing fruit, out of 6. $X ~\sim~ \mathrm{B} (6, 0.4)$ $\mathrm{P} (X = 0) = 0.0467$ (b) $\mathrm{P} (X \le 2) = 0.544$ (ii) Let Y be the number of packets with at most 2 cakes containing fruit, out of 8. $Y ~\sim~ \mathrm{B} (8, 0.54432)$ $\mathrm{P} (Y \ge 4) = 1 - \mathrm{P}(Y \le 3) = 0.729$ (iii) Let W be the number of packets with at most 2 cakes containing fruit, out of 150. $W ~\sim~ \mathrm{B}(150, 0.54432)$ Since $n = 150$ is large, $np = 81.648 > 5$ and $nq = 68.352 > 5$, $W ~\sim~ \mathrm{N}(81.648, 37.20536)$ approximately. $\mathrm{P}(W > 75) = \mathrm{P}(W > 75.5)$ by continuity correction $= 0.843$ ### KS Comments Students must call all variables out carefully as they denote different things. And many still forget to perform continuity correction. ### 2014 A-level H1 Mathematics (8864) Question 6 Suggested Solutions All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) $X~\sim~N(142.2, 36)$$latex \mathrm{P}(X<146) = 0.737$(ii)$latex \mathrm{P}(137.2KS Comments

A very straightforward question. 🙂

### 2014 A-level H1 Mathematics (8864) Question 5 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

5.
(i)
$\frac{dy}{dx} = 3x^{2} + 2kx + 7$

When $x = 1, ~ \frac{dy}{dx} = 0$

$\Rightarrow 3 + 2k + 7 = 0$

$k = -5$

$\Rightarrow 1^{3} - 5 (1)^{2} + 7 + c = 2$

$\therefore, c = -1$

(ii)
$\frac{dy}{dx} = 3x^{2} - 10 x + 7 = 0$

$x= \frac{7}{3} \mathrm{~or~} x = 1$

When $x = \frac{7}{3}, y = \frac{22}{27}$

$\therefore, (\frac{7}{3}, \frac{22}{27})$

(iii)

(iv)
Area $= \int_1^2 x^{3} - 5 x^{2} + 7x - 1 dx$

$= \frac{x^{4}}{4} - \frac{5x^{3}}{3} + \frac{7x^{2}}{2} - x \biggl|_1^2$

$= \frac{19}{12}$

Students must be careful to leave answers in fractions and not just decimals. They should also check their answers with the Graphing Calculator.

### 2014 A-level H1 Mathematics (8864) Question 4 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

4.
(i)
By Pythagoras’ Theorem, $(2x)^{2} + (y-2)^{2} = (2\sqrt{65})^{2}$

$\therefore, 5x^{2} + y^{2}-2xy = 260$

(ii)

Perimeter $= 6x+2y = 60$

$\Rightarrow 5x^{2} + (30-3x)^{2}-2x(30-3x) = 260$

$x^{2} -12x +32 = 0$

$x = 4 ~\mathrm{or}~ x= 8$

When $x = 4, y = 18$

When $latex x = 8, y = 6 < x$, which is not possible given the question. Thus, rejected. $latex \therefore, x=4, y=18$

Students need to be alert to reject the appropriate x-value. Alternatively, they can also check the second order derivative to determine which is the suitable x value.

### 2014 A-level H1 Mathematics (8864) Question 3 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

(ii)
$\frac{dy}{dx}= -e^{1-2x}(-2) = 2 e^{1-2x}$

Sub $(1, 1- \frac{1}{e}), ~ \frac{dy}{dx} = \frac{2}{e}$

Tangent: $y - (1-\frac{1}{e}) = \frac{2}{e}(x-1)$

$\therefore, y = \frac{2}{e}x - \frac{3}{e} +1$