### 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.

### Differential Equations (Applications)

When Mr. Teng retired on 1 January 2018, he put a sum of $10,000 into a senior citizen fund that has a constant rate of return of 5% at the end of every month. Starting in February 2018, he withdraws$500 at the start of each month for groceries. Denote the amount of money that Mr. Teng has at the time $n$ years by $\x$.

(i) The differential equation relating $x$ and $n$ can be written in the form of $\frac{dx}{dn}= kx+c$. State the values of $k$ and $c$.

(ii) Solve the differential equation and find the amount of money that Mr. Teng has after 15 months.

(iii) In which month will Mr. Teng no longer be able to withdraw the full \$400?

### 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

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

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

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

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

### All the best for 2017 Release of A’levels Results!

Will like to take this opportunity to wish all students receiving your results tomorrow, all the best! And do not worry too much and feel free to ask for any advice!
Students will find the following useful 🙂

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

NTU IGP

NUS IGP

Take note, the courses treat GP & PW as a “C”.

### Release of Results for 2017 A’levels

Students should note that the release of A’levels 2017 is on 23rd Feb 2018. More details can be found here.

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

### Solutions to December Holiday Revision

Hopefully you find your holidays meaningful. Take the time to unwind and relax. Also, if you have completed the Set A and Set B. The solutions can be found here.

If you do have any questions, please WhatsApp me. 🙂

Solutions to Set A

Solutions to Set B

Relevant Materials: MF26

### Solutions to Set B

Hopefully, you guys have started on the Set B. You will find the following solutions useful. Click on the question. Please do attempt them during this December Holidays. 🙂

If you do have any questions, please WhatsApp me. 🙂

Relevant Materials: MF26