### 2011 A-level H2 Mathematics (9740) Paper 2 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)
P(faulty)
= P(made by A and faulty) + P(made by B and faulty)
= 0.6(0.05) + 0.4(0.07)
= 0.058

(b)
= $\frac{\mathrm{P(made~by~A~and~faulty)}}{\mathrm{P(faulty)}}$
= $\frac{0.6(0.05)}{0.058}$
= $\frac{15}{29}$

(ii)
(a)
P(exactly one of them is faulty)
= $0.058 \times (1 - 0.058) \times 2!$
= 0.109272 (exact)

(b)
P(both were made by A | exactly one is faulty)
= $\frac{\mathrm{P(both~were~made~by~A~and~exactly~one~is~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}$
= $\frac{\mathrm{P(one~is~made~by~A~and~faulty,~the~other~is~made~by~A~and~not~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}$
= $\frac{0.6(0.05) \times 0.6(0.95) \times 2!}{0.109272}$
= $\frac{1425}{4553}$

Question can be easily solved by drawing a tree diagram. Do take note that we only wrong off NON exact answers to 3sf, so for (iia), we keep the full exact answer.

### 2011 A-level H2 Mathematics (9740) Paper 2 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 $\mu$ be the population mean time taken to install an electronic component.

$H_0 : \mu = 38$

$H_1 : \mu \textless 38$

(ii)
Under $H_0, \bar{T} \sim \mathrm{N}(38, \frac{25}{n})$

Test Statistic, $Z = \frac{\bar{T}-38}{\frac{5}{\sqrt{n}}} \sim N(0, 1)$

At 5% level of significance, we reject $H_0$ if $z \le -1.6449$

$\Rightarrow \frac{\bar{t}-38}{\frac{5}{\sqrt{n}}} \le -1.6449$

$\therefore, \{ \bar{t} \in \mathbb{R} | 0 \textless \bar{t} \le 36.8 \}$

(iii)
At 5% level of significance, we do not reject $H_0$ if $z > -1.6449$

$\Rightarrow \frac{37.1-38}{\frac{5}{\sqrt{n}}} \textless -1.6449$

$n \textless 83.5$

$\therefore, \{n \in \mathbb{R}| 0 \textless n \le 83 \}$

Some students forget to define $\mu$ appropriately. (ii), a few students actually do not realised that the rejection rule is when $p \le \alpha$ instead of $latex p < \alpha$, and did not consider the correct inequality. (iii), Students overlook that n refers to sample size and sample size must be an integer. The answers are required to be given in sets here. Some students still have trouble writing sets.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

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

(i)
$\mathrm{P}(R = 4) = \frac{{^{18}\!C_{4}}{^{12}\!C_{6}}}{{^{30}\!C_{10}}} \approx 0.0941$

(ii)
$\mathrm{P}(R = r) > \mathrm{P}(R = r+1)$

$\Rightarrow \frac{{^{18}\!C_r}{^{12}\!C_{10-r}}}{{^{30}\!C_{10}}} > \frac{{^{18}\!C_{r+1}}{^{12}\!C_{9-r}}}{{^{30}\!C_{10}}}$

$(\frac{18!}{r!(18-r)!})(\frac{12!}{(10-r)!(2+r)!}) > (\frac{18!}{(r+1)!(17-r)!})(\frac{12!}{(9-r)!(3+r)!})$

$(r+1)! (17-r)! (9-r)! (r+3)! > r! (18-r)! (10-r)! (r+2)!$

$\frac{(r+1)!(r+3)!}{r!(r+2)!} > \frac{(18-r)!(10-r)!}{(17-r)!(9-r)!}$

$(r+1)(r+3) > (18-r)(10-r)$

$r^2 + 4r +3 > 180 -28 +r^2$

$r > 5.53$

Since r is an integer, $r = 6$

(ii) is honestly not easy. Some students didn’t know the formula which is puzzling as the formal is given in MF15. I think it is important that students knows what formulas they are given in exams so they can utilise they properly.

### 2011 A-level H2 Mathematics (9740) Paper 2 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 denote the number of people joining the queue in 4 minutes.

$X \sim \mathrm{Po}(4.8)$

$\mathrm{P}(X \ge 8) = 1 - \mathrm{P}(X \le 7) \approx 0.113$

(ii)
Let Y denotes number of people joining the queue in t seconds.

$Y \sim \mathrm{Po}(0,02t)$

$\mathrm{P}(Y \le 1) = 0.7$

$\mathrm{P}(Y = 0) + \mathrm{P}(Y = 1) = 0.7$

$\frac{e^{-0.02t}(0.02t)^0}{0!} + \frac{e^{-0.02t}(0.02t)^1}{1!}=0.7$

$e^{-0.02t}(1+0.02t) = 0.7$

Using the Graphing Calculator, t = 55

(iii)
Let W denote the number of people joining the queue in 15 minutes.

$W \sim \mathrm{Po}(18)$

Since $\lambda = 18 > 10, ~W \sim \mathrm{N}(18, 18)$ approximately.

Let V denote the number of people leaving the queue in 15 minutes.

$V \sim \mathrm{Po}(27)$

Since $\lambda = 27 > 10, ~V \sim \mathrm{N}(27, 27)$ approximately

$\Rightarrow W - V \sim \mathrm{N}(-9,45)$ approximately

$\mathrm{P}(W -V \ge -11)$

$= \mathrm{P}(W-V > -11.5)$ by continuity correction

$\approx 0.645$

(iv)
The average number of people joining the queue per minute may not remain constant throughout a period of long hours as they have different traveling time. There might be no flight departing too. Thus, it would not appropriate.

The question is pretty standard Poisson. So long as students are careful with continuity correction, I don’t see much problems with this questions. They should answer (iv) contextually.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 7 Suggested Solutions

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

(i)
Firstly, the number of different friends is a fixed constant.
Secondly, the probability of successfully contacting a friend is constant.

(ii)
The probability of contacting a friend will vary from friend to friend, as some friend might not be free or contactable during that evening.

(iii)
$\mathrm{R} \sim \mathrm{B}(8,0.7)$

$\mathrm{P}(R \ge 6) = 1 - \mathrm{P}(R \le 5) \approx 0.552$

(iv)
$\mathrm{R} \sim \mathrm{B}(40, 0.7)$

Since $n = 40 \mathrm{~is~large,~} np = 28 > 5 \mathrm{~and~} nq = 12 > 5 ~ \mathrm{R} \sim \mathrm{N}(28, 8.4) \mathrm{approximately}$

$\mathrm{P}(R \textless 25)$

$= \mathrm{P}(R \textless 24.5)$ by continuity correction

$\approx 0.114$

Students should be careful to write all assumptions in the context of question. I cannot stress the importance of this. Conditions for Binomial Approximation have to be well stated and explained. Lastly, remember to do continuity correction.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 6 Suggested Solutions

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

(i)
Stratify the residents of the city suburb into different age groups.
Stand on a street corner and ask people who walk past until the required number of people in each age group have been interviewed.

(ii)
The sample obtained is non-random as not every resident in the city suburb has an equal chance of being selected.
Not everyone that walks there is a resident.

(iii)
Stratified Sampling

It would not be realistic as we will not be able to obtain the sampling frome.

Most questions today require students to be able to answer in the required context. So students need to understand the sampling methods so as to phrase them appropriately.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 5 Suggested Solutions

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

$X \sim \mathrm{N}(\mu, \sigma^2)$

$\mathrm{P}(X \textless 40.0) = 0.05$

$\Rightarrow \frac{40 - \mu}{\sigma} = -1.6449 \rightarrow (1)$

$\mathrm{P}(X \textless 70.0) = 0.05$

$\Rightarrow \frac{70 - \mu}{\sigma} = 1.9600 \rightarrow (1)$

Solving, $\mu = 53.7$ and $\sigma = 8.32$

A very straightforward question here, as it didn’t even try to trick unsuspecting students who often forget to change the signs before applying inverse norm function. Please use a Graphing Calculator to solve if you aren’t good with simultaneous equations.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 3 Suggested Solutions

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

(i)
Let $y = \mathrm{ln} (2x+1) + 3$

$x = \frac{e^{y-3}-1}{2}$

$\therefore, f^{-1}(x) = \frac{e^{x-3}-1}{2}, x \in \mathbb{R}$

$D_f = \mathbb{R}$

$R_{f^{-1}} = D_f = (-\frac{1}{2}, \infty)$

(ii)

(iii)
$f(x) = f^{-1}(x) = x$

$\Rightarrow f(x) = x$

$\mathrm{ln}(2x+1) + 3 = x$

$\mathrm{ln}(2x+1) = x - 3$

Using Graphing calculator, $x = -0.4847 \mathrm{~or~} 5.482$

Students should be careful to draw the curves according to the given domain. And since the question requests for equations of asymptotes, we should have label them too.

The last part, students can also use the inverse function to solve.

This question, in general did not pose much of a problem for students.

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

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

(i)
$V = PQ \times QR \times \mathrm{height}$

$= (2n-2x)(n-2x)x$

$= 2n^2x-6nx^2+4x^3$

(ii)
$\frac{dV}{dx}=2n^2 - 12nx +12x^2$

Let $\frac{dV}{dx}=0$

$\Rightarrow 6x^2 -6nx + n^2 = 0$

$x = (\frac{3 \pm \sqrt{3}}{6})n$

Since $2x \textless n, ~ x \textless \frac{n}{2} \Rightarrow x = (\frac{3- \sqrt{3}}{6})n$

We reject $x=\frac{3+\sqrt{3}}{6}n$ since $\frac{3+\sqrt{3}}{6}n > \frac{n}{2}$

$\therefore$ Required stationary value of $x = \frac{3-\sqrt{3}}{6}n$