### Modal value & Expected value

Let us look at the difference between modal value and expected value. We shall start by saying they are different, albeit close.

Modal value refers to the mode, that is, the value that has the highest probability (chance) of occurring.

Expected value refers to the value, we expect to have, on average.

Before we start, I’ll do a fast recap on Binomial Distribution, $X \sim \text{B}(n, p)$ by flashing the formulae that we can find on MF26.

$\text{P}(X = x) = ^n C_x (p)^x (1-p)^{n-x}$

$\mathbb{E}(X) = np$

$\text{Var}(X) = np(1-p)$

The expected value is simply given by $\mathbb{E}(X)$.

Now to find the modal value, we have to go through a slightly nasty and long working. You may click and find out.

We have that $\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{(n-r)}{(r+1)} \frac{p}{1-p}$. This is what we call the recurrence formula. We consider this to give us the ratio between successive probabilities. And to illustrate how this works, nothing beats an example question.

Consider candies are packed in packets of 20. On average the proportion of candies that are blue-colored is $p$. It is know that the most common number of blue-colored candies in a packet is 6. Use this information to find exactly the range of values that $p$ can take.

First, most common number is the same as saying the modal/ highest frequency.

This means that $\text{P}(X=6)$ is the highest/ largest probability… Let us turn our attention to the recurrence formula now. If $\text{P}(X=6)$ is the largest, then it means that $\text{P}(X=6) \textgreater \text{P}(X=7)$ and also $\text{P}(X=6) \textgreater \text{P}(X=5)$.

Lets start by looking at the first one… $\text{P}(X=6) \textgreater \text{P}(X=7)$

$\text{P}(X=6) \textgreater \text{P}(X=7)$

$1 > \frac{\text{P}(X=7)}{\text{P}(X=6)}$

$\frac{\text{P}(X=7)}{\text{P}(X=6)} \textless 1$

But hold on! This looks like the recurrence formula. (ok, in exams, its either you use the recurrence formula or derive on the spot. Both works!)

Now I’ll advice you try the second one (before clicking on answer) on your own, that is, $\text{P}(X=6) > \text{P}(X=5)$.

Now, if the question simply says that the expected number of blue-colored candies in a packet of 20 is 6. Then

$\mathbb{E}(X) = 6$

$(20)p = 6$

$p = \frac{3}{10}$

We observe that this value actually falls in the range of $p$ we found.

### DRV questions with a twist

I want to share a question that is really old (older than me, actually). It is from the June 1972 paper. As most students know, Maclaurin’s series was tested in last year A’levels Paper 1 as a sum to infinity. And this DRV did the same thing. Here is the question.

A bag contains 6 black balls and 2 white balls. Balls are drawn at random one at a time from the bag without replacement, and a white ball is drawn for the first time at R th draw.

(i) Tabulate the probability distribution for R.

(ii) Show that E( R ) = 3, and find Var( R ).

If each ball is replaced before another is drawn, show that in this case E( R ) = 4.

### Solutions to Review 1

Question 1
(i)
$y = f(x) = \frac{x^2 + 14x + 50}{3(x+7)}$

$3y(x+7) = x^2 + 14x + 50$

$x^2 + (14-3y)x + 50 - 21 y = 0$

$\text{discriminant} \ge 0$

$(14-3y)^2 - 4(1)(50-21y) \ge 0$

$196 - 84y + 9y^2 - 200 + 84y \ge 0$

$9y^2 - 4 \ge 0$

$(3y - 2)(3y + 2) \ge 0$

$y \le - \frac{2}{3} \text{~or~} y \ge \frac{2}{3}$

(ii)
Using long division, we find that

$y = \frac{x^2 + 14x + 50}{3(x+7)} = \frac{x}{3} + \frac{7}{3} + \frac{1}{3(x+7)}$

So the asymptotes are $y = \frac{x}{3} + \frac{7}{3}$ and $x = -7$

Question 2
(i)
$x^2 - 9y^2 + 18y = 18$

$x^2 - 9(y^2 - 2y) = 18$

$x^2 - 9[(y-1)^2 - 1^2] = 18$

$x^2 - 9(y-1)^2 + 9 = 18$

$x^2 - 9(y-1)^2 = 9$

$\frac{x^2}{9} - (y-1)^2 = 1$

This is a hyperbola with centre $(0, 1)$, asymptotes are $y = \pm \frac{x}{3} + 1$, and vertices $(3, 1)$ and $(-3, 1)$.

$y = \frac{1}{x^2} + 1$ is a graph with asymptotes $x = 0$ and $y=1$.

Use GC to plot.

(ii)
$\frac{x^2}{9} - (y-1)^2 = 1$—(1)

$y = \frac{1}{x^2} + 1$ —(2)

Subst (2) to (1),

$\frac{x^2}{9} - (\frac{1}{x^2} + 1 - 1)^2 = 1$

$\frac{x^2}{9} - (\frac{1}{x^2})^2 = 1$

$x^2 - \frac{9}{x^4} = 9$

$x^6 - 9 = 9x^4$

$x^6 - 9x^4 - 9 = 0$

(iii)
From graph, we observe two intersections. Thus, two roots.

Question 3
(ai)
$\sum_{r=1}^n (r+1)(3r-1)$

$= \sum_{r=1}^n (3r^2 + 2r -1)$

$= \sum_{r=1}^n 3r^2 + \sum_{r=1}^n 2r - \sum_{r=1}^n 1$

$= 3 \sum_{r=1}^n r^2 + 2 \sum_{r=1}^n r - \sum_{r=1}^n 1$

$= 3 \frac{n}{6}(n+1)(2n+1) + 2 \frac{n}{2}(1 + n) - n$

$= \frac{n}{2}(n+1)(2n+1) + n(1+n) - n$

$= \frac{n}{2}(n+1)(2n+1) + n^2$

(aii)
$2 \times 4 + 3 \times 10 + 4 \times 16 + ... + 21 \times 118$

$= 2 [2 \times 2 + 3 \times 5 + 4 \times 8 + ... + 21 \times 59]$

$= 2 [(1+1) \times (3 \cdot 1 - 1) + (2+1) \times (3 \cdot 2 -1) + (3+1) \times (3 \cdot 3 -1) + ... + (20+1) \times (3 \cdot 20 -1) ]$

$= 2 \sum_{r=1}^{20} (r+1)(3r-1)$

$= 2 [\frac{n}{2}(n+1)(2n+1) + n^2 ]$

$= n(n+1)(2n+1) + n^2$

$= n(2n^2 + 3n + 1) + n^2$

$= 2n^3 + 4n^2 + n$

(bi)
$\frac{2}{(r-1)(r+1)} = \frac{A}{r-1} - \frac{B}{r+1}$

$2 = A(r+1) - B(r-1)$

Let $r = -1$

$2 = - B(-2) \Rightarrow B = 1$

Let $r = 1$

$2 = A(2) \Rightarrow A = 1$

$\therefore \frac{2}{(r-1)(r+1)} = \frac{1}{r-1} - \frac{1}{r+1}$

(bii)
$\sum_{r=2}^n \frac{1}{(r-1)(r+1)}$

$= \frac{1}{2} \sum_{r=2}^n \frac{2}{(r-1)(r+1)}$

$= \frac{1}{2} \sum_{r=2}^n (\frac{1}{r-1} - \frac{1}{r+1})$

$= \frac{1}{2} [ 1 - \frac{1}{3}$

$+ \frac{1}{2} - \frac{1}{4}$

$+ \frac{1}{3} - \frac{1}{5}$

$...$

$+ \frac{1}{n-3} - \frac{1}{n-1}$

$+ \frac{1}{n-2} - \frac{1}{n}$

$+ \frac{1}{n-1} - \frac{1}{n+1}]$

$= \frac{1}{2} [1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}]$

$= \frac{1}{2} (\frac{3}{2} - \frac{n+1+n}{n(n+1)})$

$= \frac{3}{4} - \frac{2n+1}{2n(n+1)}$

(biii)
As $n \to \infty$, $\frac{1}{n} \to 0$ and $\frac{1}{n+1} \to 0$, the sum of series tends to $\frac{3}{4}$, a constant. Thus, series is convergent.

(biv)

$\sum_{r=5}^{n+3} \frac{1}{(r-3)(r-1)}$

Replace $r$ by $r + 2$. Then we have

$\sum_{r=3}^{n+1} \frac{1}{(r-1)(r+1)}$

$= \sum_{r=2}^{n+1} \frac{1}{(r-1)(r+1)} - \frac{1}{(2-1)(2+1)}$

$= \frac{3}{4} - \frac{2(n+1)+1}{2(n+1)[(n+1)+1]} - \frac{1}{3}$

$= \frac{5}{12} - \frac{2n+3}{2(n+1)(n+2)}$

Its been awhile since we last posted. And it is good to know that JC1s have all been well inducted or settled into their schools. Of course, I do hear that many schools are severely overcrowded recently. Anyway, I thought of sharing how students can get ready for JC. I contemplated sharing how to cope with JC Mathematics, but decided to be more general this time round.

Firstly, JC life can be quite rigorous. With CCA and different subject commitments piling, students must try their best to stay healthy (get enough sleep) and juggle time (skip some dramas) efficiently. For science subjects (not just H2 Mathematics), students should avoid procrastinating. The schools do not go back to teaching the subjects again, maybe just refresh using questions or tests. Thus, seek help if you need and do not just sweep it off. For J1, your A’levels is pretty much in 22 months while for J2, it is 10 months. So the clock started ticking.

Secondly, last year’s papers were intuitive and some questions were driven to see if students do understand their content and can think on their feet. And we have a name for such questions, it is application questions. For H2 Mathematics, they have made an effort to allocate about 25% of the total marks to application questions. Thus, students need to shift their focus from doing to learning. It is important for them to appreciate the concepts in each topic.

Thirdly, I understand some students enter JC and realise that there are really some (or a lot of) smarter peers around. Do not feel pressured and just stay focus. Some of them might have found help, or developed better intuition for certain things. Comparing with your neighbour will only make yourself more stressed. This is unnecessary stress.

Lastly, JC is the last “school” you have. So do enjoy yourself. Pick a CCA that you really want to try. 🙂

Since Mr Teng teaches H2 Mathematics, here are some little tips for H2 Mathematics as I told my J1s this year.

1. Some topics from High School are still very relevant, which is why I gave a proper review test. These topics are considered under assumed knowledge for H2 Mathematics, and you can find them here. A good understanding of these topics will allow you to follow classes better. You will learn that schools are constantly rushing to clear topics.

2. Learn the topics. You do not need to master them, but learn and find out what is going on. Because you can memorise the entire Ten Years Series and realise that it will not save you.

3. You will learn that time is very precious during exams. In general, 1 mark is 1.5 mins. And you should not go beyond it for questions. Rather learn the hard way to time manage well during exams, start with your normal practices at home. Thats why I encourage my students in class to do fast. Your papers will be two 3-hour papers, so during that 3-hour, you must exhibit sufficient tenacity.

P.S. I’ve spent the last few months getting a lot of application questions up. Aside from sharing them with students in my classes, I’ll also put them here. So do check in. 🙂

Happy CNY!

### JC Talk 2018

Over the weekends, we had the privilege of conducting a mini JC talk which saw Mr. Teng and Ms. Christine share their knowledge with parents and students of O’levels 2017.

The lessons for J1 2018 started on the first week of January and the schedules can be found here.

The following are the grade profiles of local universities, NUS and NTU.

NTU IGP

NUS IGP

We are very thankful for your attendance and do hope that the information was beneficial. If you do have more questions, you can contact Mr. Teng at +65 9815 6827

### Call for Registration 2018!

Registration for classes in 2018 has been opened. You can find out more about the class schedules here. Do note that JC1 class will commence in first week of January as there are registrations from IP students. For O’levels students, you can treat it like a head-start! We will also be holding a workshop for Post-O’levels Release of Results. So do stay tuned!

Students/ Parents can call Mr. Teng at +65 9815 6827, or the centre at +65 6567 3606  for further enquiries!

Lessons will be held at:
Newton Apple Learning Hub
Blk 131, Jurong Gateway Road #03-263/265/267 Singapore 600131
Tel: +65 6567 3606

### Thoughts on the H2 Mathematics (9758) Papers 2017

Solutions can be found here.

Personal Thoughts: The paper isn’t tedious. Students can do them so long as they know their stuffs. There are several generalising of questions, like question 6 of paper 1. We also saw how conditional probability was actually tested subtly, this tests students’ abilities to reason with guidance (not sure if after this first trial year, will they still guide the students.) Application questions were not tough and well guided. Students can solve it easily if they read it well. Statistics was well crafted and neat.

To be blunt, I’ll give credit to the 9740 H2 Mathematics paper that run concurrently, since it is too tough to set two sets of papers. Its easy to acknowledge that the 9740 (2016) paper was way harder than 9740 (2017). Next year won’t be the same.

Advice: Students should be careful when you revise, make sure you learn, and not do. Understand what you’re doing. The 2017 paper was an inquisitive paper, examiners were watching closely if you pay attention to details, and know your definitions well.

I’ll do an analysis for the paper, you can click on the individual question and read. For students that took the paper, I hope it doesn’t demoralise you.

Paper 1

Paper 2

### 2017 A-level H1 Mathematics (8865) Paper 1 Suggested Solutions

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

Numerical Answers (click the questions for workings/explanation)

Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6: $\mu = 1.69, \sigma^2 = 0.0121$
Question 7: $0.254; 0.194; 0.908$
Question 8: $40320; 0.0142; \frac{1}{4}$
Question 9: $\text{r}=0.978; a=0.182, b=2.56$; \$293
Question 10: $0.0336; \bar{y}=0.64, s^2 = 0.0400$; Sufficient evidence.
Question 11: $\frac{48+x}{80+x}, \frac{32+x}{80+x}; x= 16; \frac{25}{32}; \frac{7}{16}; \frac{341}{8930}$
Question 12: $0.773; 0.0514; 0.866; 0.362$

MF26

### Solutions to the modified A’levels Questions

Students of mine who have been diligently doing the modified TYS I sent them, and have difficulties with the questions that were added in to make the paper a full 3 hour paper, will find the following solutions helpful. Please try to do them in a single 3 hour seating, these are modified to cater to the 9758 syllabus…

The rest of the solutions (that are questions from the original TYS) can be found here.

2012/P1/Q10

2012/P2/Q2

20112/P2/Q7

2012/P2/Q7

### 2017 A-level H2 Mathematics (9758) Paper 2 Suggested Solutions

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

Numerical Answers (click the questions for workings/explanation)

Question 1: $2 \sqrt{15}; xy=6$
Question 2: $d = 1.5;~ r \approx 1.21 \text{~or~} r \approx -1.45;~n=42$
Question 3: $(\frac{1}{2}a, 0), (0,b);~ (a+1, 0,);~ (\frac{a+1}{2}, 0);~ (0, a), (b, 0);~ a = 1;~ gg(x) = x, x \in \mathbb{R}, x \neq 1 , ~ g^{-1}(x) = 1 - \frac{1}{1-x}, x \in \mathbb{R}, x \neq 1;~b= 2 \text{~or~}0$
Question 4: $15.1875;~ \frac{\pi}{2a(a-1)};~ b = \frac{1}{2} + \frac{1}{2}\sqrt{1-a+a^2}$
Question 5: $\frac{5}{12}, \frac{5}{14}, \frac{5}{28}, \frac{1}{21};~ \mathbb{E}(T) = \frac{20}{7}, \text{Var}(T) = \frac{75}{98};~ 0.238$
Question 6: $955514880;~ 1567641600;~ \frac{1001}{3876}$
Question 7: $31.8075, 0.245;~ p = 0.0139$; Do not reject $h_0$, Not necessary.
Question 8: Model (D); $a \approx 4.18, b \approx 74.0;~ r \approx 0.981$
Question 9: $0.632;~ 1.04 \times 10^{-4};~ 0.458;~ 0.421;~ 0.9408$
Question 10: $0.345;~ 0.612;~ \mu = 12.3, \sigma = 0.475;~ k \approx 55.7$

MF26