# A Level Maths Resource Site

## Elegance articulated. A WHITE GROUP MATHEMATICS SUPPLEMENTARY WEBSITE

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#### Need clarifications on University Statistics course notes

 Posted by Whitecorp on October 7, 2013 at 7:20 AM comments (0)

(This comes from a former student of mine now in her second year at Singapore Management University (SMU)-I received this e-mail earlier on 9 September 2013):

Hi Mr Koh!

XXX  here. How are you? School has started and stats has come back to haunt. Have a couple of questions, hope you won't mind answering them!

For starters, what is this?

Attached are my lesson slides for context if needed.

(Stats 101 Powerpoint slides)

Thanks so much and hope to hear from you soon!

Warm regards,

Former Student X

Hi ,

Ah so the new semester has started. Nice ppt slides.

(i) E(Y) when Y is a discrete variable

Let's say we have Y=X^2 and the probability distribution function (pdf) of X is as follows:

X             0               1             2           3          4

P(X=x)         0             1/6         1/3       1/12       5/12

Then the probability distributionfunction (pdf) of Y would be:

Y              0                 1           4          9          16

P(Y=y)          0                1/6       1/3        1/12     5/12

In that regard, E(Y) = 0(0) +(1)(1/6) + 4(1/3) + 9 (1/12) + 16 (5/12)

= 8.9167 (shown)

(ii) E(Y) when Y is a continuous random variable

Let's say X is defined by the following probability distribution function:

f(x) =  2x + 1        0 ≦ x ≦ 1,         0    otherwise

and Y is related to X such that   Y=X^2,

then E(Y) = E(X^2)  is obtained by evaluating the integral   ∫ (x^2) * (2x+1) dx  from x=0 to x=1.

Hope this clarifies. Welcome back  to Statistics.

Best Regards,

Mr Koh

#### How to obtain the taylor series for this function?

 Posted by Whitecorp on September 20, 2013 at 8:40 PM comments (0)

Hi,

We're currently learning about Taylor series and I'm having a lot of trouble understanding the concept and the math involved in it. I typed into Wolfram Alpha the function e^-(x^2) at point 0 and order 4 and have attached the output:

My understanding is you start by differentiating the function 4 times but I don't understand how to get from there to the equation Wolfram Alpha has given me. I would really appreciate some help with this as the web-sites and books I have looked through so far have only managed to confused me further.

Student X

The Taylor series is explained by

f(x) = f(a) + f'(a) * (x-a) +  f" (a) *( x-a)^2 /2 ! +  f''' (a) *( x-a)^3 / 3! +........

When the function to be represented is about x=0 as mentioned in your problem, the taylor series is reduced to that of a Maclaurin's Series,

ie f(x)= f(0) + f'(0) * x +  f" (0) *x^2/ 2! + f'''(0) * x^3/3! +........

Let's work things out to obtain a clearer picture.

Assuming y =f(x)= e^-(x^2) , and therefore implying  f'(x)=dy/dx, f"(x)= d^2 y/ dx^2  so on and so forth,

differentiating this once wrt x on both sides gives

dy/dx = -2x * e^-(x^2) = -2xy         (Note: The e^-(x^2) component is replaced simply by y)

Differentiating a second time wrt x on both sides

d^2 y/ dx^2 = -2 [ x *dy/dx +y] = -2x*dy/dx -2y

Differentiating a third time wrt x on both sides,

d^3 y/ dx^3 = -2[ x* (d^2 y/ dx^2) +dy/dx] -2 dy/dx

= -2x * (d^2 y/ dx^2) -4 dy/dx

Differentiating one final time wrt x on both sides,

d^4 y/ dx^4  =  -2 [ x* (d^3 y/ dx^3) + (d^2 y/ dx^2)] -4 (d^2 y/ dx^2)

Now, we seek to find the values of the various orders of derivatives when x=0, ie

f(0)= value of y when x=0,

f'(0) = value of dy/dx when x=0,

f"(0)= value of (d^2 y/ dx^2) when x=0, so on and so forth

By making a series of substitutions into the above block of differential equations,

When x=0,

y=1, dy/dx= 0, d^2 y/ dx^2 =-2 ,  d^3 y/ dx^3 = 0  and  d^4 y/ dx^4 = 12

Hence,

the expansion of the series is given by

f(x)= 1 +  0 * x + (-2) *x^2 /2!  + 0 *x^3/ 3! + (12) *x^4 /4! +.......

= 1 - x^2 + 0.5 * x^4 -.........

This may take a while to internalize, so be patient.

Hope it helps. Peace.

Best Regards,

Mr Koh

#### Help needed with word-number problem

 Posted by Whitecorp on August 27, 2013 at 11:30 PM comments (0)

Hello,

My math teacher gave us a worksheet with six 4-letter words.

Book = 13

Full= 19

Sail= 20

Boil = 11

Silk = 18

Foil = 10

Each letter in the word has a value and they all should equal the number after the word.

also, the B for instance in the word Book has to be the same value as the B in the word Boil, and etc.

I cannot figure this out after hours of trying. How do I do this? Thank you for your time!

Student X

You can formulate a system of linear equations to solve this.

If we let the values of the individual letters be represented by the letter variable themselves,

then

B+ 2O + K = 13

F+ U+ 2L = 19

S + A + I + L = 20

B + O + I + L = 11

S + I + L + K= 18

F + O + I + L =10

The resulting augmented matrix can be written as such:

A   B   F   I   K   L   O  S  U

0   1   0   0   1   0   2  0   0       13

0   0   1   0   0   2   0  0   1       19

1   0   0   1   0   1   0  1   0       20

0   1   0   1   0   1   1  0   0       11

0   0   0   1   1   1   0  1   0       18

0   0   1   1   0   1   1  0   0       10

A first tier reduction of the matrix gives

A= 12 - 1/2 O - 1/2 S

B= 3 - 3/2 O + 1/2 S

F= 2 - 3/2 O + 1/2 S

I= -1/2 - 1/4 O - 1/4 S + 1/2 U

K= 10 - 1/2 O - 1/2 S

L= 17/2 + 3/4 O - 1/4 S - 1/2 U

O=  O

S = S

U=  U

Assuming all letters are associated with integer values, then a possible choice would be to let O=2, S=4 and U=6.

Then  A=9, B= 2,  F=1, I=1, K=7 and L=6.

Naturally there will be many other possible solution sets since this system involves 9 variables yet is only described by 6 equations.

Hope this helps. Peace.

Best Regards,

Mr Koh

 Posted by Whitecorp on August 14, 2013 at 7:00 AM comments (0)

Hi,

I'm taking Mathematical Modelling this semester and am having trouble making sense of a question given to us on a worksheet to complete. I've attached the question and the solution given to us in the hope that you may be able to help me make sense of the solution as I'm struggling to understand how the equilibrium was found and the conclusion was reached that the equilibrium is unstable. Thanks a million in advance.

Student X

Attachment:

An equilibrium value exists when the bird population neither grows nor shrinks, ie the net rate of change of population is simply zero (dP/dt =0). This value as computed happens to be b/a.

Perhaps solving the differential equation might yield greater insight into what's exactly happening.

Since dP/dt=aP-b,

∫1/(aP-b) dP = ∫ dt

1/a* ∫ a/(aP-b) dP = ∫ dt

1/a* ln |aP-b| = t + C   where C is a constant of integration

ln |aP-b|= at+B (where B is another constant= aC)

aP-b = e^(at+B)=e^B * e^(at) = Ae^(at)   (where A is another constant =e^B)

P = 1/a *[Ae^(at)+ b ]

= De^(at) +(b/a)    (where D is another constant =A/a)

Hence, at the onset when t=0,  initial population Po = D + b/a

CASE 1

If Po is greater than b/a, then D + b/a > b/a ===>  D > 0

Since e^(at) is an exponential growth factor, as t approaches infinity in the long run,

the component of P which is  De^(at) will be unbounded and grows infinitely huge as well,

which therefore implies the same thing about P.

CASE 2

If Po is lesser than b/a, then D + b/a < b/a ===>  D < 0

Since e^(at) is an exponential growth factor, as t approaches infinity in the long run,

the component of P which is  De^(at) will ultimately decay towards zero (due to the negative

nature of D), which therefore implies P will head downwards as well.

Hope this clarifies. Peace.

Best Regards,

Mr Koh

#### Interesting American Math Problem Queries

 Posted by Whitecorp on August 5, 2013 at 8:40 PM comments (0)

Hi ! So I have this summer math packet to complete before school starts, which is on August 7th. I reached a couple of problems which left me stumped because I didn't know what to do. Your help would be so appreciated. Oh, when you help me with these, would you mind telling me the process and everything? Thank you.

78. Working together, Natalie and Paul can pick forty bushels of apples in 4.8 hours. Had he done it alone, it would have taken Paul 12 hours. Find how long it would have taken Natalie.

81. An aircraft carrier left the pier traveling east eight hours before a fishing boat. The fishing boat traveled in the opposite direction for four hours, going 5 mph slower than the aircraft carrier. The ships were then 380 miles apart. What was the aircraft carrier's speed?

Student  X

Here are the detailed solutions:

78. Paul's rate of picking bushels (alone) = 12 /40 = 0.3 hours per bushel

Let Natalie's rate of picking bushels (alone) = x hours per bushel

If their efforts were combined, then Natalie would have picked 4.8/x bushels while Paul

would have picked 4.8/0.3 = 16 bushels in a span of 4.8 hours.

Since 40 bushels would have been picked in a combined effort scenario,

16 + 4.8/x = 40

4.8/x  = 24

x = 4.8/24 = 0.2

Hence, if Natalie was left to pick all 40 bushels by herself,

she would have taken 40*0.2  = 8 hours (shown)

81. Let the aircraft carrier's speed be x mph, then the fishing boat's speed would be (x-5)  mph.

After the first eight hours, the aircraft carrier has traveled 8x miles and the distance between the two

ships at this instant would simply be  8x  miles.

In the next four hours,  the aircraft carrier would have traveled another 4x  miles east while the

fishing boat  4(x-5) = 4x-20  miles west in the opposite direction.

Hence, the total distance separating both ships after 12 hours would be

8x + 4x + 4x - 20 = 16x -20

Since this is equivalent to 380 miles,

16x-20 = 380  =====>  x =400/16 = 25 mph (shown)

Hope this helps. Peace.

Best Regards,

Mr Koh

#### Change in hypothesis testing circumstances and associated p-value variation

 Posted by Whitecorp on July 25, 2013 at 11:30 PM comments (0)

For hypothesis Z-testing where a certain p-value is arrived at based on Ho: μ = μo against

H1: μ > μo;    if we recalibrate the test to that of  Ho: μ = μo against  H1: μ ≠ μo  but utilise the same

data set ( x̄ , σ, n,  μo) as previously, how would the p-value change?

Student X

You should expect the new p-value to be twice that of the original.

For the original one-tail test, the p-value is simply the result of computing P( Z  >  (x̄ -μo) / (σ/ √n )  ) ; adjusting this to a two-tail test using the same set of data would imply the value of the test-statistic Z= (x̄ -μo) / (σ/ √n ) remains unchanged, and in similar regard  P( Z  >  (x̄ -μo)/ (σ/ √n )  ) .

Since a two-tail test includes consideration of the left tail, by symmetry  it is therefore also known that    P( Z  <  (x̄ -μo) / (σ/ √n )  ) = original p-value. Adding p-values attributable to both sides together, recognize that the overall new p-value is twice that of the original.

Hope this clarifies. Peace.

Best Regards,

Mr Koh

#### Do you teach primary school maths?

 Posted by Whitecorp on July 18, 2013 at 11:00 PM comments (0)

Dear Mr. Koh,

I’d like to inquire whether you offer tuition for grade 6 (Primary 6) math.

My daughter currently attends an international school and is having great difficulty in math, especially with word problems.

If you do teach this level, how much is your rate and will you be willing to travel to Upper Bukit Timah area?

Best regards,

Parent X

Hi ,

Thanks for your inquiry. Unfortunately I only teach junior college maths, so I am not able to help your child in that department. However, you may wish to check out another website owned by me which hosts the details of high calibre tutors whom you can contact directly at http://www.domainofexperts.com .  All the best, and have a good weekend. God bless.

Best Regards,

Mr Koh

#### Help: Quitting Polytechnic to join JC

 Posted by Whitecorp on July 11, 2013 at 3:10 AM comments (0)

Dear Mr Koh,

I am writing to you today to seek more information in regards to your tuition services.First of all,let me first state down my personal information. I am 17 this year, and also was a student from Ngee Ann Polytechnic- Banking and Financial Services.I left the school after just 2 months there, mainly because i have lost interest in my course and wouldn't want to continue with that career path for the rest of my life. Due to my lack of research before the selection of my choices after my 'O' levels, I made a choice without giving it much needed thoughts. Additionally, I found it harder for polytechnic student to enter local universities,once again due to my lack of research,I made the wrong choice of picking a polytechnic over a JC as i have wished to enter NUS long before I have taken my 'O' level papers.Furthermore, i yet to have found my preferred occupation in the future,entering a JC would allow me more time to find one.I have scored a raw L1R5 score of 11 for my 'O' levels last year and have planned to enter TPJC next year.

With already 1 year wasted, I wouldn't want to spend more time retaking JC1 or the 'A'levels so I am prepared to work hard in order to excel. With the vast amount of time i have,i want to start preparing for the JC syllabus before enrolling in it. So I wish you could provide me your services. I live in Location X. Here are my o lvl scores for my respective subjects .English - B3 , A maths - A1 , E maths - A1 , pure physics - B3 , chemistry - A1 , Combined humanities(History + SS) - A1 , Chinese B - Passed. I plan to take H2 maths,chem,physics and also h1 econs for my 'A' levels. I would also like to ask about your tuition fee rates.

Thank you for taking time to read my email.

Sincerely,

Student X

Hi,

Thanks for your mail. Indeed it would be rather tough on you to make such a major transition, however if you have truly made up your mind about where to go from here, proceed ahead and don't look back. Chance favors the bold and brave. Given your reasonably good O level grades, I reckon the barriers of entry into a junior college wouldn't be much of an issue, though it is definitely recommended you consult the relevant admission departments on the necessary switching procedures involved.

I noticed you made mention of having previously taken Additional Maths in secondary school (and scoring a distinction for it), so pursuing H2 Maths wouldn't be problematic administratively speaking. However, be warned life isn't going to be a bed of roses as there is an enormous gap between the content/style of the H2 Maths and A Maths syllabi. You have to remain constantly motivated and cannot afford to feel discouraged for prolonged periods despite all the potential setbacks you shall possibly encounter during your learning journey. Hang in there regardless, and you will eventually see light at the end of the tunnel.

Bear in mind these days competition for places in the local universities is extremely fierce, so scoring well in Mathematics alone will not count for much. Inject sufficient time and effort into ensuring you become equally proficient in all other subjects. To my existing knowledge, H2 Chemistry in particular is becoming a great pain for a large number of students, so do keep a watchful eye on it and seek assistance the moment you feel you are slipping.

Should you wish to engage my tutoring services, please be advised my current rate is \$----/hr. Feel free to discuss with me any other concerns you have, and I will try to address them to the best of my abilities. Good luck, and god bless.

Best Regards,

Mr Koh

#### Unable to visualize vectors problem in space

 Posted by Whitecorp on June 27, 2013 at 8:05 AM comments (0)

Hi Mr Koh,

Here is a prelim question which I attempted recently:

While I was able to obtain all the correct answers, I am unable to visualize the above context in   3-dimensional space. Could you please help me out?

Student X

I have provided the diagram below (took me quite a while to draw this), hope it helps. Peace.

Best Regards,

Mr Koh

#### A small Normal Distribution problem

 Posted by Whitecorp on June 19, 2013 at 10:05 AM comments (0)

Hi Mr Koh,

Here is a problem from my tutorial which I am not able to solve:

Given that X~N( -12, 6) ,   find  k such that P( |X+12| < k) =0.85

Student X

I will offer two slightly different methods of solving your question.

1. The distribution of X+12 will have a different mean value; however its variance remains unchanged.

ie  X+12~N( 0, 6)

P( |X+12| < k) =0.85 ====> P (-k < X+12 < k ) =0.85

If you draw the normal distribution curve for X+12, and pencil in the limits -k / +k , you will notice

that the area under the curve between -k and +k is equal to 0.85. Since this curve is symmetrical

about the y-axis, you can further reinterpret the probability structure to arrive at P( X+12 < -k) =

(1-0.85)/2 = 0.075.

Thereafter, using the Inverse Normal computation, you should have k=3.5261  (shown)

2.  P( |x+12| < k) =0.85 ====> P (-k < X+12 < k ) =0.85  or  P (-k-12 < X < k -12 ) =0.85  ----------(1)

Consider standidization to the standard normal Z variable where Z~ N(0, 1),

(1) becomes  ( -k/ sqrt(6) < Z <  k/sqrt(6) ) =0.85

Since the standard normal distribution curve is symmetrical about the y-axis, then it is also

alternatively true to state that P( Z< -k/ sqrt(6) )  = (1-0.85)/2 =0.075.

Thereafter, using the Inverse  Normal computation, you should have -k/sqrt(6) = -1.4395

===> k= 3.5261  (shown)

Hope this helps. Peace.

Best Regards,

Mr Koh