2019 singapore- cambridge a level h2 math p2 suggested

2019 Singapore-

CAMBRIDGE

A Level

H2 Math P2

Suggested Answer

Key (9758)

Written and Prepared by Mr Mitch Peh

Preface

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2019 A Level H2 Math P2 Answers Done by Mr Mitch Peh

3

Overall Remarks

This is a manageable paper consisting of a good balance of standard questions and thinking questions. For the

standard questions, you would have to be careful to avoid careless mistakes and losing unnecessary marks.

For the section on Pure Math, the mathematical manipulations are generally tedious just like those of previous

years. My personal advice for students who are easily intimidated by tedious manipulations is to start working

on the section on Statistics first where manipulations are generally more manageable.

Also, you have to be very familiar with the O Level of skill of factorising expressions with exponents in order

to do Q1(iii) and to find the expression for volume in Q3.

The sketching of graph in Q2 can be a little tricky as the portion of the graph beyond x=2 including the minimum

point is difficult to observe on the GC. It would be an advantage if you had come across questions where you

have to sketch graphs for similar algebraic expression before during your practices.

In addition, most of the Pure Math topics which have not been tested extensively in Paper 1 were tested here:

Integration techniques, maxima differentiation and Maclaurin series.

However, there are some exceptions. Even though the topic of solving inequalities by graphical method has

already been tested in Paper 1 Question 4, it is again tested in Paper 2 question 2.

Question 4 which is on the topic of vectors is also testing concepts similar to what has been tested in Paper 1

question 12. For paper 1 question 12, you are required to find the intersection point between line and plane

while for paper 2 question 4, you are required to find the intersection point between 2 lines for 2 parts of the

question. There are many areas in Vectors that A Level examiners could have tested instead such as the dot

product formula, distributive properties, ratio theorem etc. but were left out of this year’s A Level paper.

For the section on Statistics, the parts which involve manipulations are largely standard where you should strive

to secure most of the marks while the parts which require explanations are a little more unconventional and

need you to think quite abit.

o The explanation questions in 6(i) and (ii) are interesting while 6(iii) is a standard combination question.

o Q7 is a pretty standard binomial distribution question where the distribution has even been stated explicitly

for you in the question. You should strive to obtain full credit here.

o Q8 is a probability question which requires you to interpret the information provided in the table well. You

need to be familiar with techniques such as separating into the various cases for (iii) and using algebraic

manipulations to form equation in (iv).

o Q9 is a hypothesis testing question where you must notice that the population variance value has been

provided so there is no need to compute the unbiased estimate of the population variance from the value of

sample variance. The explanation question in (iii) will require you to think abit.

o Q10 a correlation and regression question where you should be familiar in computing the product moment

correlation coefficient value and obtaining the equation of the least square regression line with the use of the

GC. The explanation question in (d) will need you to know how a regression line is being constructed and

the relationship between the least square regression line and the sum of squared residuals.

o Similar to 2018 paper 2, the last question in this paper is based on the topic of normal distribution. The

computations required here are pretty standard so you should strive to obtain full credit here.

Similar to 2018 paper 2, the marks weightage for the topic of permutation and combination is low in this paper,

tested in Q6(iii). In the Syllabus document, Permutation and Combination (P&C) is considered as part of the

topic Probability. Since the topic of Probability has been tested extensively in Q8, this could be the reason why

the mark weightage for P&C is low here. In fact, concepts for P&C are required to compute some of the

probability values in Q8.

For the A Level Paper 2 in subsequent years, you can continue to expect some focus to be on explanation

questions in the Statistics Section.


4

2019 A Level 9758 H2 Math P2 Suggested Solution

Section A: Pure Mathematics [40 marks]

1 (i) Topic: Integration by parts

1

21I x x dx

We will differentiate x and integrate 1

21 x

1 3

2 22

, 1 1, 13

dv duu x x v x

dx dx

1 3 3

2 2 22 2

1 1 13 3

I x x dx x x x dx

3 3

2 22 2

1 13 3

x x x dx

3 5

2 2

3 5

2 2

2 2 21 1

3 3 5

2 41 1

3 15

x x x C

x x x C

Comments

Over here, we have 2 algebraic functions. We should differentiate x because after differentiating,

we will only be left with 1 algebraic function within the next integral function.

We have to be careful while performing the integration to avoid careless mistakes.

[2]

(ii) Topic: Integration with the use of substitution

2 1u x 21x u and

1

21u x

Differentiating with respect to x,

2 1 2du

u dx ududx

1

221 1 2I x x dx u u u du

2 42 2u u du

3 5

1

2 2

3 5u u C

3 5

2 21

2 21 1

3 5x x C

Comments

After differentiating2 1u x , we can leave dx in terms of u and du and substitute into the

equation for I.

[2]


5

(iii) Topic: Algebraic manipulation

Difference between answers to part (i) and (ii)

3 5 3 5

2 2 2 21

2 2 2 41 1 1 1

3 5 3 15x x C x x x C

3 3 5

2 2 21

2 2 21 1 1

3 3 3x x x x C C

3

21

21 1 1

3x x x C C

3

21

21 0

3x C C

1C C which is a constant (shown)

Comments

You need to be strong with algebraic manipulations involving exponents for this question.

Be careful as the terms 3

22

13

x and 3

22

13

x x are different by a factor of x.

The technique here is to factorise 3

22

13

x out from the expressions which are in terms of x and

simplify. We should notice that the terms will then cancel out to be 0.

[2]

2 (i) Topic: Graph Sketching

2

2 2

3 5 8 3 8 1

x xy

x x x x

Hence, the equations of the asymptotes are 8

3x , 1x , 0y

To find the intersection point with the x axis, set y=0

2

0 23 8 1

xx

x x

Hence, the curve intersects the x axis at (2, 0)

To find the intersection point with the y axis, set x=0

2 1

8 0 1 4y

Hence, the curve intersects the y axis at (0, -1/4)

[4]


6

Comments

To factorise 23 5 8x x , we can either use the cross method or use our calculator to find the

roots of the equation and then deduce the factorised expression.

For the sketching of the curve, the portion of the graph beyond x=2 including the minimum point

is difficult to observe on the GC.

Even though this question does not require us to find the stationary points, we should know that

we find them by using the maximum and minimum function in GC.

Maximum point: (-0.160,-0.248), Minimum point: (4.16, -0.0334)

Alternatively, we can know that the curve has 2 stationary points with the use of differentiation.

2

22 2

1 3 5 8 2 6 52

3 5 8 3 5 8

x x x xx dyy

x x dx x x

2 2

22

3 5 8 10 7 6

3 5 8

dy x x x x

dx x x

2

22

3 12 2

3 5 8

x x

x x

Setting 0dy

dx , using GC, we will similarly obtain x = -0.160 and x = 4.16.

(ii) Topic: Solving inequalities using graphical method

Based on the graph above,

the inequality 2

20

3 5 8

x

x x

holds when

8

3x or 1 2x .

Comments

Since this question is only worth 1 mark, we should not attempt to solve the inequality manually.

[1]


7

To obtain the correct answer here, we would have to sketch the graph in (i) correctly.

(iii) Topic: Solving Inequalities

Given 2

20

3 5 8

x

x x

,

Multiplying by -1 throughout,

2

20

3 5 8

x

x x

Hence, observing from the graph above, this inequality is satisfied when 8

13

x or 2x .

Comments

Again, we should not attempt to solve the inequality manually but make use of the graph obtained

in (i).

[1]

3 Topic: Maxima differentiation

Surface area of cylinder = 22 2 900r rh

22 900 2rh r

450h r

r …..(1)

Volume of cylinder, 2V r h …..(2)

Subst (1) in (2),

2 3450450V r r r r

r

2450 3dV

rdr

For 0,dV

dh

150r

since r>0.

2

26

d Vr

dr . When

150r

,

2

20

d V

dr

Hence, when 150

r

, we have the maximum possible volume of the cylinder.

Maximum volume of the cylinder is given by 1 3

2 23 150 150

450 450V r r

1

2

3

150 150450

150300

V

cm

[7]


8

361500 cm

2

2450 450

r r r

h rr

r

Subst value of 150

r

in the above expression,

150

150 1

150 300 2450

r

h

Hence, the ratio r:h which gives the maximum volume is 1:2.

Comments

This is an unguided but fairly routine maxima differentiation problem. We have to recall the formulae of

the total surface area of a cylinder and total volume of a cylinder correctly.

Then we replace h in the volume equation to be in terms of r before performing differentiation with

respect to r.

We need to perform second derivative test to verify that when 150

r

, we have the maximum volume

of the cylinder.

After we have obtained 150

r

for maximum volume to occur, we should simplify the expression for

the maximum volume of the cylinder through factorisation in our answers and not simply leave it as 1 3

2 2150 150450V

. This is a similar technique employed in question 1(iii).

For the ratio of r : h, we will have to substitute in the value of r to obtain the exact value.


9

4 (i) Topic: Maclaurin Series

Given sec2f x x ,

'( ) 2sec2 tan2f x x x

2 3

2 2

2 2

2

''( ) 2 2sec 2 tan 2 2sec 2

4sec 2 tan 2 sec 2

4sec 2 tan 2 tan 2 1

4sec 2 2 tan 2 1

f x x x x

x x x

x x x

x x

When x=0, 1

0 1cos0

f

'(0) 2sec0tan0 0f

24''(0) 2 tan 0 1 4

cos0f

Hence, 21 2 ...f x x

Comments

It is important to get the answer for (i) correct as it will affect our answers in (ii) and (iv).

To differentiate sec2f x x , we can refer to MF26. Then we should remember to apply chain

rule here.

To differentiate '( ) 2sec2 tan2f x x x , we need to remember to apply product rule.

Then, we need to know that cos0 =1, tan0 = 0 and then apply the general formula for Maclaurin

series expansion given in MF26 to find the Maclaurin Series.

[5]

(ii) Topic: Maclaurin series and integration 0.02 0.02

2

0 0

0.02

3

0

sec 2 1 2

2

3

xdx x dx

x x

32

0.02 0.02 03

0.02001 (5 decimal places)

Comments

A standard question where you should know that you should replace the function sec2x with the

Maclaurin series found in (i) before performing the integration.

Do remember to leave your answer to 5 decimal places as required by the question.

[2]


10

(iii) Topic: Integration

0.02 0.02

0 0

1sec2

cos2xdx dx

x 0.02001 (5 decimal places)

Comments

As the GC is unable to integrate the function of sec2x directly, we need to know1

sec2cos 2

xx

,

then we can perform the integration with our GC.

The integration function in the GC is under Math → 9:fnInt

We need to ensure that our GC is set to radian mode here.

Again, we need to remember to leave our answer to 5 decimal places as required by the question.

[1]

(iv) Topic: Maclaurin Series involving polynomial approximation

Since the two values are the same up to 5 decimal places, the approximation is very accurate. This

is because the values of x are between 0 and 0.02 which are small in magnitude such that x3 and

higher powers of x can be neglected.

Comments

You need to know the condition for Maclaurin series approximation to be accurate which is that

the range of x values should be small here.

[2]

(v) Topic: Maclaurin Series

1( ) cos 2

sin 2g x ec x

x

When x=0, sin0=0 which will cause g(0) to be undefined.

Comments

You need to know that the condition for Maclaurin series to be applicable is the function must be

defined when x=0.

[1]

5 (i) Topic: Vectors on finding point of intersection between 2 lines.

Point X is the point of intersection between line AC and BD

2 4

4

AC OC OA

a b a

a b

Hence, : 4ACl r a a b where

5

5

BD OD OB

b a b

a

Hence, : 5BDl r b a where

At the point of intersection between AC and BD,

4 5 a a b b a

1 4 5 a b a b

Comparing the coefficients of a and b, we have:

5 1 ……(1)

[4]


11

1

4 14

Subst 1

4 in (1):

1 15 1

4 4

Subst µ value in the equation of line BD,

1 5

54 4

OX b a a b

Comments

The approach to this question may not be obvious to some students but if you have been practising

some of the past year prelim papers, you would have come across questions which require similar

approach.

Alternatively, we can substitute λ value in the equation of line AC, we will obtain the same answer

for OX .

1 5

44 4

OX a a b a b

(ii) Topic: Vectors on finding point of intersection between 2 lines

Point Y is the point of intersection between the extended line of OX and line CD

If points O, X and Y are collinear, 5

4OY kOX k

a b where k

5 2 4

3 3

3

CD OD OC

b a a b

a b

a b

Hence, : 2 4CDl r a b a b where

At the point of intersection between OY and CD,

5

2 44

k

a b a b a b

5

2 44

k k a b a b

Comparing the coefficients of a and b,

5 52 2

4 4k k …….(1)

4 4k k ……(2)

Solving with our GC,

8

3k and

4

3

Substituting 8

3k in the equation of line OY,

8 5 10 8

3 4 3 3OY

a b a b

[6]


12

1 3

8

OX OX

OY kOX k

Hence, OX:OY = 3:8

Comments

This question requires the same approach as (i). Hence, if you do not know how to solve (i), it is

unlikely that you will know how to solve this part of the question as well, causing you to be heavily

penalised.

Also, we need to know the implication of having points O, X and Y being collinear to one another.

To obtain vector OY , we can also substitute value of α in the equation of line CD. We will obtain

the same answer.

4 10 8

2 43 3 3

OY a b a b a b

Section B: Probability and Statistics (60 marks)

6 (i) Topic: Sampling

Since Alice is only interested in training approaches in Division One and there are altogether 22

clubs in Division One, so the 22 clubs form a population.

Comments

This is an interesting question.

A population set is a set of similar items which are of interest for an experiment. Since Alice is is

only interested in clubs in Division One, the 22 clubs in Division One form a population.

Many students may think that 22 clubs in Division One is a sample as it is a subset of 100 clubs.

This is only true if Alice is interested in all the clubs in the 4 divisions.

Hence, the key to identifying whether the clubs form a sample or a population is to identify the

purpose of the experiment and the list of items of interest.

[1]

(ii) Topic: Sampling

Dilip should send out questionnaires regarding the facilities for supporters to 11 clubs in division

One, 12 clubs in Division Two, 13 clubs in Division Three and 14 clubs in division Four.

The clubs from each division are chosen randomly using a computer.

Comments

A rather ambiguous question where you need to write down the steps to conduct the investigation.

Some knowledge of stratified random sampling is required here to ensure that the sample is

representative of the population.

To ensure that the numbers selected from each Division are whole numbers, our sample size will

need to be 50 as shown above.

If the sample size is smaller e.g. 20, Dilip will need to send out questionnaire to22

20 4.4100

clubs in Division One and 28

20 5.6100

clubs in Division Four which is not feasible.

[2]


13

(iii) Topic: Permutation and Combination

No. of different possible samples of 20 football clubs, with 5 clubs chosen from each division 22 24 26 28

5 5 5 5C C C C

187.24 10 samples

Comments

Simple but high weightage question where the order of arrangement does not matter here.

[3]

7 (i) Topic: Binomial Distribution

The probability of mugs being faulty is constant at 8%.

The event of any chosen mug being faulty is independent of the event that any other mug is also

faulty.

Comments

We should not mix the 2 assumptions together and say that the probability of a mug being faulty is

independent of the probability that another mug is faulty.

[2]

(ii) Topic: Binomial Distribution

F~B(50, 0.08)

( 7) 1 ( 6)P F P F

0.10187

0.102(3 . .)s f

Comments

Take note that when at least 7 faulty mugs are found, the case of having 7 faulty mugs is included.

We should know how to adjust ( 7)P F so that we can key in the expression into our GC to find

the probability.

We key into our GC: 1-binomcdf(50, 0.08, 6) to find the probability value.

We need the correct probability results for (iii) of this question.

[2]

(iii) Topic: Binomial Distribution

Let X be the number of days out of 5 days where at least 7 faulty mugs are found.

X~B(0.10187)

( 2) 0.99097P X

0.991(3 . .)s f

Comments

No more than 2 days means that the scenarios of having 0, 1 and 2 days where at least 7 faulty mugs

are found are included here.

We key into our GC: binomcdf(5, 0.10187, 2).

Note that the more exact value of probability of having a randomly chosen day where at least 7

faulty mugs are found should be used here, which is 0.10187.

[2]


14

(iv) Topic: Binomial Distribution

Let X be the number of faulty saucers in a random sample of 10 saucers.

X~B(10, p)

P(X=2)

810 2

2

82

1

45 1

C p p

p p

Comments

We have to use the binomial distribution formula here which can also be found in MF26.

[1]

(v) Topic: Binomial Distribution and probability

Case 1: No faulty items out of the 2 randomly chosen mugs and 2 randomly chosen saucers

P(No faulty items)

=(0.92)2 (1-p)2

=0.8464(1-p)2

Case 2: One faulty mug out of the 2 randomly chosen mugs and 2 randomly chosen saucers

P(One faulty mug)

2

2

2 0.08 (0.92) 1

0.1472 1

p

p

Case 3: One faulty saucer out of the 2 randomly chosen mugs and 2 randomly chosen saucers

P(One faulty saucer)

2(0.92) 2 1

1.6928 1

p p

p p

Hence, the equation that is satisfied by p is:

2 2

0.14720. 1 1.6928 1 0.9846 74 1 p p p p

Using GC, p=0.0689(3s.f.)

Comments

Since the probability of the mug and the saucer being faulty are different, we cannot simply lump

them together as a single binomial distribution where the number of trials is 4.

Instead, we should first find the probabilities of all the possible cases which will result in at most

1 faulty item carefully.

Then form an equation to sum the probabilities of the various cases together which will give us

probability value of 0.97.

Note that we can simply solve the equation formed with the use of our GC and there is no need to

manipulate the equation around manually.

[4]


15

8 (i)

(a)

Topic: Probability

From the table, there is 9 horses and 14 riders.

Hence, probability that this item is either a Horse or a Rider

23

56

Comments

Easy question where we just have to extract the relevant information from the table.

[1]

(b) Topic: Probability

From the table, there is 11 non-white dogs and 15 non-white birds.

Hence, probability that this item is either a Dog or a Bird but the item is not White

26 13

56 28

Comments

Easy question where we just have to extract the relevant information from the table

[1]

(ii)

(a)

Topic: Probability

From the table, there are 8 non-orange horses.

Hence, probability that both of the items are Horses, but neither of the items is Orange

8 7 1

56 55 55

Comments

Note that the 2 items are chosen without replacement here.

Another easy question where we just have to extract the relevant information from the table

[1]

(b) Topic: Probability

For Gerri’s two items to include exactly one Dog and exactly one item that is Yellow

o Case 1: One yellow dog, one non-yellow character that is not dog

o Case 2: One yellow character that is not dog, one non-yellow dog

There are 7 yellow dogs and 32 non-yellow character that is not dog

P(One yellow dog, one non-yellow character that is not dog)

7 32 82!

56 55 55

There are 7 yellow characters which are not dogs, and 10 non-yellow dogs

P(One yellow character that is not dog, one non-yellow dog)

7 10 12!

56 55 22

Hence, the probability that Gerri’s two items include exactly one Dog and exactly one item that is

Yellow

8 1 21

55 22 110

[3]


16

Comments

The technique here is to consider the 2 different cases which will cause us to have two items

which include exactly one Dog and exactly one item that is Yellow

(iii) Topic: Probability

Let the number of the first favourite and second favourite colour characters among the 56 items be x

and y respectively.

P(Choosing the two favourite colour characters at random)

22!

56 55 3080

x y xy

2 1 40

3080 77 3080

xy

Hence, 20xy

From the table, x and y must take on the values of 4 and 5.

Thus the possibilities for Gerri’s two favourite colour/character combinations are:

1. White horse (4) and white rider (5)

2. White horse (4) and yellow bird (5)

3. Orange bird (4) and white rider (5)

4. Orange bird (4) and yellow bird (5)

Comments

This is a thinking question where we should try to choose algebraic manipulations to resolve and

then deduce the possible combinations.

Do not forget about multiplying by 2! when finding the probability of choosing the two favourite

colour characters at random.

[3]

9 (i) Topic: Hypothesis Testing

The manager should carry out a two tailed test since he wishes to test whether the mean resistance

of these resistors is in fact 750 ohms. Hence, the alternative hypothesis includes the cases of mean

resistance of the resistors being both higher and lower than 750 ohms since he did not indicate a

preference to test if the population mean is higher or lower.

Let X be the resistance of resistors rated at 750 ohms.

Let µ be the population mean of the resistance of resistors rated at 750 ohms

X~N(µ, 100)

H0:µ =750 and H1:µ ≠750

Comments

Note that we should define the population mean, µ in this question. The question has also explicitly

requested us to define any symbols that we use.

[2]


17

(ii) Topic: Sampling

Method 1 : Use GC to find sample mean Method 2: Find sample mean manually

Using GC, 756X

742 771 768 738 769 752 742 766

8

756

X

H0:µ =750 and H1:µ ≠750

Under H0, 100

~ N 750,8

X

and 756

~ (0,1)10

8

XZ N

Reject H0 if p value < 0.05

value =2P 756 0.089685p X

0.0897(3 . .)s f

Since p value= 0.0897 is greater than 0.05, there is insufficient evidence to reject H0 at the 5% level of

significance. Thus, the production manager is unable to conclude that the mean resistance of the

resistors is not 750 ohms.

Comments

For the GC method, when we perform 1-Var Stats, our frequency list should be empty.

Note that we have to be careful when computing the value of the variance of the sample mean

We should use the population variance of 100 that has been provided divided by 8, and not the

unbiased estimate of population variance:

213.990

8 that can be obtained through the sample.

Another common mistake when computing the value of the variance of the sample mean is to

forget about dividing by 8.

In the conclusion, we should state explicitly the level of significance used in this question again.

Take note that it is incorrect to conclude by saying that there is sufficient evidence to suggest the

mean resistance is 750 ohms.

[5]

(iii) Topic: Sampling and hypothesis testing

Firstly, since the distribution of the 1250 ohms resistors is unknown, the sample size cannot be just

8. Instead, it will have to be at least 30 so that the sample size is large enough such that central limit

theorem can be applied to approximate the sample mean distribution as a normal distribution. Then

the z test can be carried out.

Secondly the null hypothesis and alternative hypothesis will be different.

Ho: µ = 1250 and H1: 1250

Thirdly, since the distribution of the resistances of these resistors is unknown, the population

variance is unknown as well. Hence, we have to find the unbiased estimate of the population

variance from the sample variance value.

Comments

The question is a little vague on what it wants you to discuss about exactly. One way is to re-look

at the steps of the hypothesis testing which you have conducted in (ii) and analyse the changes

which have to be made.

[2]


18

10 (i)

(a),

(b)

Topic: Correlation and Regression on scatter diagram and graph plotting

Let ei be the residual for each point, for i=1, 2, 3, 4, 5

[2]

(c) Topic: Correlation and Regression on finding sum of the squares of residuals

Sum of the squares of the residuals for Abi’s line 2 2 2 20.33 0.33 0.33 1 1.3267

Comments

We should leave the sum of the squares of the residuals as the exact value here, and not round off to

3 significant figures.

[1]

(d) Topic: Correlation and Regression

Residual is the difference between f(a) and b, which can be 0, positive or negative.

With the points (a, f(a)) plotted, we would then want to construct a best fit line such that we have

minimal vertical deviations between the points and the best fit line itself.

When we find the sum of residuals, the residuals may end up cancelling each other to become 0

even though the vertical distance between residuals and the line may be large.

However, the square of the residuals will always be positive so we can then draw the best fit line

where the sum of the square of the residuals is the minimum.

[1]


19

Comments

We need to understand how a regression line is being constructed to answer this question.

(ii) Topic: Correlation and Regression

Bhani’s model gives a better fit because the squares of the residuals for Bhani’s line is smaller

compared to that of Abi’s line.

[1]

(iii) Topic: Correlation and Regression

40 45 50 55 6050

5x

and

22 20 18 17 1618.6

5y

The least squares regression line must pass through the point 50,18.6 .

Comments

We need to know that the least squares regression line must pass through the point with

coordinates being the mean value of x and y.

To find the values of x and y , we can also key in the values of x and y into the Stat table of our

GC and use 2-Var Stats to find.

[1]

(iv) Topic: Correlation and Regression

Using our GC, product moment correlation coefficient, r = -0.985(3s.f.)

Equation of least squares regression line of y on x is:

0.3 33.6y x

Comments

We need to recall that for a least square regression line of y on x, x is the independent variable

and y is the dependent variable.

We key into the GC the following:

[3]

(v) Topic: Correlation and Regression

Given 0.3 33.6y x , when 30x ,

0.3 30 33.6 24.6y kilometres per litre

The estimate is not reliable as speed of 30 kilometres per hour is outside the data range of

40 60x . Hence, the linear relationship found in (iv) may not be valid outside the data range.

Comments

Simple substitution of x value into the regression line

Do not miss out that there is a second part of the question. This is a standard explanation of

whether the estimate obtained from the regression line is reliable.

[2]

(vi) Topic: Correlation and Regression

Since the sum of the squares of the residuals for her line is 0, it means that each individual square

of the residual is 0.

Hence, all the data points lie on the regression line itself.

[1]


20

Comments

It is also good to know that in such a situation, there is a perfect linear correlation between the

variables where the magnitude of the product moment correlation coefficient is 1.

11 (i) Topic: Normal Distribution

Let X be the mass of a randomly chosen white ball.

X~N(110, 42)

E(X1+X2+X3+X4) = 4E(X) = 440

Var(X1+X2+X3+X4) = 4Var(X) =4(42)=82

X1+X2+X3+X4~N(440,82)

1 2 3 4( 425) 0.96960P X X X X

0.970 (3s.f.)

Comments

To calculate the probability, we key into our GC: normalcdf(425, 1E99, 440, 8)

Be careful when finding the variance value of X1+X2+X3+X4 in this question.

[2]

(ii) Topic: Normal Distribution

Let Y be the mass of randomly chosen black ball.

Y~N(55, 22)

E(X+Y) = E(X) + E(Y) = 110 +55 = 165

Var(X+Y) = Var(X) + Var(Y)

=42+22=20

Hence, X+Y~N(165, 20)

(161 175) 0.80178P X Y

0.802 (3 s.f.)

Comments

We key into our GC normalcdf(161,175, 165, sqrt(20))

[2]

(iii) Topic: Normal Distribution

Given X~N(110, 42) and Y~N(55, 22),

E(X1+X2+Y1+Y2+Y3) = 2E(X) + 3E(Y)

2 110 3 55 385

Var(X1+X2+Y1+Y2+Y3)=2Var(X) + 3Var(Y)

2 22 4 3 2 44

Hence, X1+X2+Y1+Y2+Y3~N(385, 44)

1 2 1 2 3( ) 0.271X X Y Y YP M

From GC, 380.96 381M (3s.f.)

Comments

A standard normal distribution question where we first have to find the mean and variance of

X1+X2+Y1+Y2+Y3 so that we can define its distribution.

To obtain M value, we key into our GC: invNorm(0.271, 385, sqrt(44))

[4]


21

(iv) Topic: Normal Distribution

Let W be the mass of a connecting rod.

W~N(20, 0.92)

Let A be the mass of a methane molecule in grams.

We need to find the distribution of

A = 0.7X1 + 0.9Y1 + 0.9Y2 + 0.9Y3 + 0.9Y4 + W1 + W2 + W3 +W4

( ) 0.7 110 4 0.9 55 4 20 355E A

2 2 2 2 2( ) 0.7 4 4 0.9 2 4 0.9 24.04Var A

A~N(355, 24.04)

( 350) 0.84608P A

0.846 (3s.f.)

Comments

As there are many items in this question, make sure that we account for the mass of 1 white ball, 4

black balls and 4 connecting rods after drilling has been done.

It is good to define the mass of a methane molecule with a variable such as A to make our

presentation simpler as we do not have to keep writing 0.7X1 + 0.9Y1 + 0.9Y2 + 0.9Y3 + 0.9Y4 + W1

+ W2 + W3 +W4 in our working.

To find the probability, we key into our GC: normalcdf(350, 1E99, 355, sqrt(24.04))

[4]

End of Solutions

2019 singapore- cambridge a level h2 math p2 suggested

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