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CATHOLIC JUNIOR COLLEGE

H2 MATHEMATICS

JC2 PRELIMINARY EXAMINATION PAPER 2 SOLUTIONS 2015

Section A: Pure Mathematics

1 System of Linear Equations

No.

Assessment Objectives Solution Feedback

Formulating a system of linear

equations from a problem situation.

[A02]

Finding table values in SGD,

Cheese/kg Chocolate/kg Candy/kg

Price/SGD 4 6 6

Price/SGD 8 10 4

Price/SGD 8 5 7

Every 3 in 4 students were able to

form the 3 equations correctly.

However, some students did not

answer the question fully, leaving out

the last part about the total number of

packs.

A number of students gave incorrect

definitions, resulting in an error in the

total number of packs.

Common mistakes

Giving total as 3 (5 3 5) 39

or 3 or 5

Misinterpreting the question and

defining the variables as mass of

each commodity

Incorrect use of matrices and

failure to explicitly write out the 3

equations

Finding the numerical solution of

equations using a GC

Answering the question

Let x, y, z be the number of three kg packs bought from Denmark,

England and Russia respectively.

4 8 8 84

6 10 5 85

6 4 7 77

x y z

x y z

x y z

5,x 3,y 5z

She should buy 13 packs in total.

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2 Transformation + Sketch y = f ‘ (x) graph

No.


Able to identify the sequence of

transformations and sketch the graph

of 2 fy x when given the graph of

fy x . [A02]

(i)(a) 2f ( ) f ( ) f ( )y x y x y x

Common mistakes:

Some of them were confused with the

sequence of transformations, where

they performed reflection last.

Some of them lost the marks for the

shape where the curve cuts the x axis.

The vertical asymptote x = -1 were

shown but marks were not deducted.

Able to sketch the graph of f 'y x

when given the graph of fy x .

[A01]

(b) f '( )y x

Generally well done.

Common mistake:

Coordinates of the turning point

becomes (1, -0.5).

x

y

O

(1, 0)

1x

x

y

O

(2, 0) (–2, 0)

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Able to sketch the graph of fy x

when given the graph of fy x

and use of graphical method to find

number of roots. [A02]

(ii) Ans: 5

Badly done.

x

y

O

3x 1x

1

2y

Alternative

x

y

O

1

2y

1

2y

fy x

fy x

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Able to identify the sequence of

transformations and describe each

transformation. [A01]

(iii) Method :

f ( )

I : translation of 1 units in the negative -direction

f ( 1)

1II : scaling parallel to -axis by a scale factor of

2

f 2( ) 1

III : translation of 3 units in the negative -direction

f (2 1) 3

y x

x

y x

x

y x

y

y x

Method :

f ( )

1I : scaling parallel to -axis by a scale factor of

2

f (2 )

1II : translation of units in the negative -direction

2

1f 2

2

III : translation of 3 units in the negative -direction

f (2 1) 3

y x

x

y x

x

y x

y

y x

Generally well done.

Common mistake:

Some students used the terms like

shift, transform, move, transfer, etc to

describe translation.

Scaling of 0.5 units.

Scale with a factor of 0.5 then

translation of -1 unit instead of -0.5

unit.

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3 Complex Numbers

No.


Calculation of modulus and argument

of a complex number. [A01]

(i) 1 i1 i 2

11 i 1 i 2

a

2 2 21

1 i 1 i 2b

1 iarg arg

1 ia

arg 1 i arg 1 i

4 4

π π

2

π

2arg arg

1 ib

arg 2 arg 1 i

π

04

4

π

Most students were able to handle this part of

the Complex Number question.

Other students who failed to do so mainly

committed manipulation error and careless

mistakes.

Students must understand that although the

question states explicitly that a calculator is

NOT to be used in answering this question,

they SHOULD actually use the calculator to

check that their moduli and argument of a and

b is correct before continuing to solve the rest

of the question.

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Carry out division of complex

numbers [A02] (ii)

π2 4

2 8

ππ

BOC

π π 3πarg

8 4 8a b

2

2

1 i 1 i 1 2i i 2ii

1 i 1 i 1 i 2a

2

2 1 i2 1 i 21 i

1 i 1 i 1 i 2b

2 2 2i 1 i 1 i

2 2 2a b

21

3 22tan 1 1 22 2

2

π8

(shown)

Since the question states explicitly that a

calculator is NOT to be used in answering

this question, all workings must be clearly

shown.

Students must know that

arg arg arga b a b .

Many students simply conclude that

3arg arg t

πan

8a b c

without explain

how they obtained 3π8

.

Re

Im

O

A

B

C

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Sketch the loci of simple equations in

an Argand diagram.

Conics - Equation of circle [A02]

(iii) (a) 0 2i i2z z

Locus is a circle, centre (0, 1) and radius 2

(b) πarg

2z b

2 2arg i

2 2 2

πz

Locus is a half-line from the point 2 2

,2 2

(excluding the angle itself) making an angle of

π2

to the positive Re-axis direction.

Equation of circle: 22 21 2x y

A handful of students drew the locus on

SEPARATE Argand diagrams which clearly

does not satisfy the question requirement.

A number of students showed poor

understanding when they shaded the locus of

z .

Many students could identify the first

locus to be a circle but gave the wrong

centre. In addition, they could identify the

second locus to be a half-line but failed to

put an open circle at 2 2

,2 2

.

Students who attempted to find the exact

complex number z using Cartesian

equations identified the equation of circle

to be 22 1 2yx . They failed to

realise that they can actually check the

equation of circle using GC: Apps –

Conics.

Re

Im

O

1

z

y

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When 2

2x ,

2

2 221 2

2y

2

2

7

2

7 71 or 1 (rej. is positive)

2

11 4

2

71

2

2

1

y

y

y

y y

2 71 i

2 2z

Alternative for finding y

From diagram, using Pythagoras’ theorem, 2

2 2 71 2 1

2 2y

A number of students performed

manipulation error at this juncture leading

to inaccurate answer.

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4 Differential Equations

No.


Solve first order differential equations

by separation of variables

Apply integration techniques: partial

fractions

Use an initial condition to find a

particular solution

[A01]

(i) d0.64 1

d 10

P PP

t

d0.064 10

d

PP P

t

1 d

0.06410 d

P

P P t

1

d 0.064d10

P tP P

1 1 1+ d 0.064d

10 10P t

P P

1 1+ d 0.64d

10P t

P P

ln ln 10 0.64P P t C

ln 0.6410

Pt C

P

0.64e10

t CP

P

0.64e , where 10

et cPA

PA

0.6410 1e , where tP

B BP A

0.64101 e t

PB

0.64

10

1 e tP

B

When 0t , 1P ,

0

101

1 eB

9B

This question was poorly done in general.

It is urgent for many students to upgrade

their skills in applying techniques of

integration in the context of differential

equations.

When there is a mixture of decimal numbers

and fractions, some students have failed in

algebraic manipulations before separating

the variables successfully for integration.

The integral on the variable P can be

resolved by either partial fractions or

completing the square. However, many

forgot to keep P-10, which is negative, in

modulus.

Once the integration is successful, most

students could obtain the particular solution

without much difficulty.

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0.649

10

1 e tP

Solution curve

[A02] Interpretation of the solution in terms

of the problem situation

[A03]

The population will approach 10 asymptotically in the

long run.

Most students who obtained the particular

solution could present the graph and the

comment correctly.

A few students substituted in a few values

for t because they did not understand what a

“solution curve” means.

Solve first order differential equations

by separation of variables

Apply integration techniques:

logarithms

Use an initial condition to find a

particular solution [A01]

(ii) d 100.4 ln

d

PP

t P

1 d0.4

(ln10 ln ) d

P

P P t

10.4d

(ln10 ln )dP t

P P

ln ln10 ln 0.4P t C

0.4ln10 ln e t CP 0.4ln10 ln , wheree e t CP AA 0.4ln ln10 e tAP

0.4e10etAP

The integration involved in this part was not

frequently encountered for most students.

However, there are a dozen of students who

could identify the correction method to

integrate successfully.

10

1

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Solution curves [A02]

Further interpretation of the solution

in terms of the problem situation

[A03]

When 0t , 1P , 0e1 10e A

ln10A 0.4(ln10)e10e

t

P

(Similarity)

The population approaches 10 asymptotically in the long

run in both models.

(Difference)

The population in the second model would increase

faster first then slower compared to the first model.

OR

The Gompertz function will take a longer time to

approach 10 as compared to the Logistic function.

Most students skipped to this part to secure

this easy score for the particular solution.

But many failed to get the mark because they

did not present the solution of P in terms of t

properly. Some others missed out the

negative sign when they tried to plug in the

value of A back to the expression of P.

Among the students who managed to find

both solution curves, a few did not read the

requirement of the question carefully so they

did not put the two curves in the same

diagram for comparison. Nonetheless, with

the two curves in place, the students could

give reasonable comments on similarity and

difference.

1

10

Gompertz

Logistic

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Section B: Statistics

5 Premutations and Combinations + Probability + Sampling Methods

No.


Identify Simple random

sampling and its

disadvantages

(i) Simple random sampling method.

Possible answers for disadvantages:

not representative of the gender make-up of the class.

Many students managed to get the full credit

for this part.

Common mistakes:

1) Quoting “textbook” answers like “not

representative of the population” were not given credit as it did not answer in

context.

Selecting 3 females and 2

males (ii) Method :

P 3 females and 2 males are selected

P female,female,female,male5!

3!2!

10 9 5!

2

,ma

2 21

le

15 14 13

2 3!2!

195

506

0.385 3 s.f

5 24 23

This part was not well attempted. Common

mistakes:

1) Just putting 15 14 13

25 24 23

10 9

22 21 , this

is only one possible case. There are a

total of 5!

2!3!

Method :

25

5n CS

To select 3 females and 2 males

Stage 1: Select 3 females out of 15, no, of ways 15

3C

Stage 2: Select 2 males out of 10, no. of ways 10

2C

Total number of ways 15 10

3 2C C

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15 10

3 2

25

5

P 3 females and 2 males are selecte

195

506

0.3

d

C C

C

85 3 s.f

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2 particular students are

selected (iii) Method :

P Amy and Bertrand are selected

P Amy,Bertrand,any student,any stud5!

3!

1 1 23 22 21 5!

25 24 23 22 21 3!

1

30

0.0

ent,any

333 3 s.

s t

f

tuden


mistakes:

1) Just putting 1 1

25 24 , this is only one

possible case. There are a total of 5!

3!

cases.

2) 10 15 14 9

1 1 2 1

25

5

C C C C

C where

15

1C is for

selecting Amy and 10

1C is for selecting

Bertrand. This is wrong since Amy and

Bertrand are already selected.

Method :

To select Amy, Bertrand and 3 other students

Stage 1: Select Amy 1

Stage 2: Select Bertrand 1

Stage 3: Select any 3 students 23

3C

Total number of ways 23

3C

23

3

25

5

P Amy and Bertrand are select

1

30

0.033

ed

3 3 s

C

f

C

.

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Conditional probability (iv) Method :

P Amy and Bertrand are selected | 3 females and 2 males are selected

P Amy and Bertrand are selected 3 females and 2 males are selected


P Amy,Bertrand,female,female,male

5!

2!195

506

1 14 13 9 5!

24 23 22 21 2!195

1

25

39

2530

506

195

50

1 or 0 04

6

.25


mistakes:

1) Many students did not realized that this is

a conditional probability question.

2) Many students just assumed that (iii)/(ii)

will give correct answers.

3) In the numerator

P P PA B A B were seen.

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Method :

14 9

2 1

25

5




1 C C

C

1

195

506

195

5

39

2530

1 or 0.04

25

06

Note: 14 9

2 11 C C1 is select Amy, Select Bertrand, Select 2 females

out of 14 and select 1 male out of 9

Method : Reduced sample space

15 10

3 2n ' C C 20475S

Number of ways Amy, Bertrand, 2 females and 1 male are selected 14 9

2 11 C C1 819




819

20475

1 or 0.04

25

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(v) The teacher can write each student’s name (all the students’ names are distinct) on a small piece paper of the same size, folded it into half. He

can then place all the females’ students name into a large bowl. He then shook the bowl and took out 3 pieces of paper one by one without

replacement. Next, he placed all the male students’ name into a large bowl. He then shook the bowl and took out 2 pieces of paper one by one

without replacement.

This part of the question was well attempted.

Identifying that the

condition probability is

the answer for stratified

sampling

Method :

1 or 0.04

25 same answers in (iv)

Method :

Many students did not realized that the

answer is the same as (iv).

For students who got (iv) wrong and they

had shown that they realized that the answer

is the same as part (iv), credit was given.

Quite a number of students put

2 1

1 2

3

5 0 51

It can be proven that a particular person is

chosen in a simple random sample of r

persons out of n is r

n. Since this part of the

question is a “state” questions. Credit was

given.

14 9

2 1

15 10

3 2

P Amy and Bertrand are selected

1 C C

C C

1 or 0.04

25

1

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6 Binomial Distribution

Thi

sN

o.


Recognise assumptions under which

the binomial distribution is a suitable

model

(i)

Assume that the event of obtaining a rotten cherry is

independent of another.

Assume that the probability of obtaining a rotten cherry

is constant.

Many students were unable to recognize the

difference between conditions vs assumptions

under which the binomial distribution is a

suitable model.

These were the conditions given in the

question – fixed number of cherries and the

cherries were either rotten or not rotten.

There were also some students who did not

contextualize the assumptions.

There were many students who mistakenly

stated “the probability of rotten cherries must be independent of each other”. This is a misconception, thinking that “probabilities” are independent when we only look at events

that are independent.

To find the mode of the binomial

distribution (ii)

Let X be the random variable denoting the number of

rotten cherries out of 26.

~ B 26,0.08X

X P X x

1 0.25868

2 0.28117

3 0.19560

The most likely number of rotten cherries is 2.

This was mostly well done, except for some

who were confused and wrote:

Mode/mostly like number = E(X) =26x0.08

2 . This was an unacceptable method even

though the mode = 2.

Apply

P C where 0, 1, 2, ...,n x n x

xX x p q x n

Use of graph to solve for unknown

number of trials.

(iii)

Let Y be the random variable denoting the number of

rotten cherries out of n.

~ B ,0.08Y n

P 0 P 1 0.1Y Y

0 1 1

0 1C 0.08 1 0.08 C 0.08 1 0.08 0.1n nn n

10.92 0.08 0.92 0.1

n nn

Method :

Most students were able to form the inequality

P( 1) 0.1Y . However, there were many

who did not form the inequality in terms of n.

The majority of the students used the GC

successfully to obtain the probabilities of

P( 1)Y . However, there were some students

who did not realise that n has to be the

least/min/smallest value.

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The least number of n is 48.

Method :

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n P 1Y

47 0.10104 (> 0.1)

48 0.09454 (< 0.1)

49 0.08844 (< 0.1)

The least number of n is 48.

Check criteria to identify Poisson

Distribution as an approximation to

Binomial Distribution

(iv)

Let W be the random variable denoting the number of

rotten cherries out of 60.

~ B 60,0.08W

Since 60n , n is large.

60 0.08 4.8 5np

~ Po 4.8 approximatelyW

P 1 0.0477325329

0.0477 to 3 s.f.

W

This part was generally well done. Most of the

students were able to accurately state the

conditions for the use of Poisson as an

approximation.

There were some students who checked the

value of nq and used Normal as an

approximation. This is incorrect and

unnecessary because np<5.

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Establish a Poisson Distribution.

Check criteria to identify Normal

Distribution as an approximation to

Poisson Distribution

Apply continuity correction

(v)

Let C be the random variable denoting the number of

boxes of cherries thrown away in a year.

~ Po 12C

Since 12 10 , is large.

~ N 12,12 approximatelyC

. .P 2 5 P 1.5 5.5c cC C

P 1.5 5.5 0.0291 to 3 s.f.C

The question was generally well done.

However, there were some students who did

not use any suitable approximation. There

were a number of students who performed

continuity correction incorrectly.

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7 Normal Distribution

No.


Distinguish the difference between

1 3

2

2 ... ~ N ,nX X X X n n

and 2 2~ N ,nX n n

Use of invNorm to find the unknown

range of mean value

(i) Let C and O be the random variable denoting the masses

of a randomly selected carrot and onion, in grams,

respectively.

22 ~ N ,5C c and 2O ~ N 75,3

1 2 3 4 5

2 2

2

~ N 2 5 75 ,4 5 5 3

C O O O O O

c

1 2 3 4 52 ~ N 2 375,145C O O O O O c

1 2 3 4 5

1 2 3 4 5

P 2 0.9

P 2 0 0.9

C O O O O O

C O O O O O

0 2 375P 0.9

145

cZ

2 375P 0.1

145

cZ

2 3751.281551567

145

c

195 to 3 s.f.c

The distributions of the masses of all carrots and onions

are independent of one another.

This part is not well done.

Many students encounter problems performing

standardization and use the wrong inequality.

Some students use GC Table to simulate the

value of c, not realizing that c is not an integer

value.

Quite a number of students gave the correct

assumption.

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Sampling distribution (ii)

1 2 1 2 3 ~ N 125, 3.085

C C O O O

1 2 1 2 3P 1305

0.00219 3 s.f.

C C O O O

Common mistake:

1 2 31 2P 1302 3

O O OC C

Some students calculated the variance wrongly

despite stating the correct random variable.

Associate 2 2~ N ,nX n n to

converting mass to price.

Use of consistent unit in gram and

price per gram

Application and evaluation of

modulus

(ii)

1 2 3 1 2 3 4

2 2 2 2

0.0018 0.0015

0.0018 3 200 0.0015 4 75~ N

, 0.0018 3 5 0.0015 4 3

C C C O O O O

1 2 3 1 2 3 40.0018 0.0015

~ N 0.63,0.000324

C C C O O O O

1 2 3

1 2 3 4

0.0018P 0.6

0.0015

C C C

O O O O

1 2 3

1 2 3 4

0.6 0.0018P

0.0015 0.6

C C C

O O O O

0.0478 to 3 s.f.

Some students use cents instead of dollar.

This resulted in the inconsistent and wrong

calculation in the subsequent part of the

solution.

Many students quoted the wrong random

variable such as 0.0018 3 0.0015 4C O , but

arrive at the correct mean, variance and final

answer. These are signs of misconception and

students were penalized.

Some students did not realize the need to apply

modulus.

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8 Hypothesis Testing + Estimates

No.


Able to use t-test in hypothesis

testing. (i) Let X be the random variable denoting the mass of peanut

butter in a random jar.

Sample mean = x

n =

2695

10 = 269.5 (exact)

s2 =

1

n - 1 [ x

2 -

( x )2

n ] =

1

9 [ 726313 -

(2695)2

10 ]

= 7

6 = 1.166666667

H0 : = 270

H1 : < 270

At 10% level of significance, reject H0 if p < 0.10

Hypothesis Testing was not well done

as many students did not seemed to

have grasp this topic well.

(i)

- Quite a number of students used z-test

instead of t-test for this part.

- There were many cases of wrong

notation used to express H0 and H1 .

- Quite a few did not understand the

question and used H1 : > 270 and did

not realise that the resultant p-value is

larger than 0.5 which is unusual.

Since n = 10 is small and population variance is unknown,

assume X is normally distributed and use t-test.

Under H0, test-statistic,

T ~ t(9)

From GC, t = - 1.463850109

p = 0.0886338513 < 0.10

Hence reject H0 and conclude at 10% level of significance

there is sufficient evidence that the manufacturer has

overstated the mean mass of peanut butter in each jar.

- Many students could not get p-value

correct because they keyed in the s

value incorrectly e.g either without

square rooting 7

6 or using

7

60 .

- Quite a number of students could not

write the conclusion correctly i.e.” p-

value < 0.10 and hence do not reject

H0” .or vice versa.

- Some are confused as to what is the

claim i.e. is it the manufacturer claim or

the shopkeeper’s suspection, showing that they did not read carefully and fully

understand the situation.

Since n is too small for CLT to be used, we need to assume

that X is normally distributed and then t-test can be used. - Many could not state the assumption

with the explanation correctly. A few

wrote that it was because the population

size is small. Some mentioned assume

the use of CLT.

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Able to adjust p-value to a two-tailed

test. In this case,

H0 : = 270

H1 : 270

So p-value = 2 0.0886338513

= 0.1772677026 > 0.10

So the conclusion would change to do not reject H0 and that

there is insufficient evidence at 10% level that the mean mass

differs from 270 g.

- Many did not use the original p-value

to explain but instead obtained another

p-value using the GC to perform

another t-test.

- Some knew that the p-value would be

larger but vaguely say that thus there

will be a change in the conclusion

without considering whether the

magnitude of their vague “change in the

p-value” would be large enough to

result in a change in the conclusion.

Find o in a z-test. (ii) In this case,

H0 : = o

H1 : o

Under H0, since X is normally distributed and population

variance is given, ~ N(o , 1.1

2

10 )

Since Ho is not rejected at 5% level of significance,

- 1.959963986 < 269.5 - o

1.1

10

< 1.959963986

-1.959963986 1.1

10 <269.5 - o <1.959963986

1.1

10

- 269.5- 1.959963986 1.1

10 < - o <- 269.5+1.959963986

1.1

10

- 270.1817745 < - o < - 268.8182255

268.8182255 < o < 270.1817745

268.8 < o < 270.2 (to 1 d.p.)

or 268.9 ≤ o ≤ 270.1 (to 1 d.p.)

- Many did not read carefully and still

used the above scenario left-tail test

i.e. H1 : < 270

- Many continued to follow part (i) and

used t-test, although the question states

that the situation has changed.

- Many also continued to use s2 =

7

6

instead of using the given population

variance 1.12.

- many used the wrong test-statistics

formula i.e. o - 269.5

1.1

10

or thought that

the unknown is the sample mean instead

of o.

- Some broke the inequality into two

parts but could not put it back correctly

in the final answer.

- Some handled the negatives in the

inequality sign incorrectly.

Note: As the standard deviation is very

small, many students could still get the

final answer using the wrong critical

values or variance when rounded up to

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1 d.p. but their method is incorrect.

Their first line of inequality will

indicate if they are on the right track.

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9 Correlation and Linear Regression

No.


Able to draw scatter diagram (i)

Generally well done. Some students forgot

to label axes or labelled wrongly as y on x

or t on w (but drew w on t).

Some students also forgot to label the max

and min values.

Care has to be taken when plotting points

to show (25, 45) to be above (20, 42). (30,

21) and (35, 21) should be of equal height.

Able to find r value and comment. (ii)

For r - 0.964 (to 3 s.f.)

From the product moment correlation coefficient value, r is

close to – 1 indicating that there is a strong negative linear

relationship between w and t, i.e. as the number of days

increases, the amount of fertilizer retained decreases.

Most students are able to obtain this value.

Minority gave 2r or r as the answer, and

many did not adhere to 3 s.f.

Only a handful got the full answer. Most

mentioned strong negative linear

relationship, but did not go on to

contextualize.

Amount of

fertilizer, w mg

Period of days, t days 0 5 40

89

18

of 29

Able to find line of regression. (iii) Regression line w on in the form w = a + bt:

w = 96.32142857 – 2.180952381t

w = 96.3 – 2.18t (to 3 s.f.)

Most students able to answer this, but some

did not give answers to 3 s.f.

Regression line has to be accurately drawn,

taking into account positions of the points

in the scatter plot. i.e. Points above the line

should not be seen below the line and vice

versa.

Understand the relevance of the least

square regression coefficients. (iv) b: This value gives the estimated rate at which the fertilizer

is lost from the soil, i.e. for every additional day, the

amount of fertiliser lost from the soil is approximately

2.18 mg.

About half of the students were able to

answer this. Some who didn’t manage to did not contextualize answer, simply

giving gradient of the regression line as the

answer.

Use of regression line to estimate

values and comment on its validity.

(v) When w = 30.7 mg,

Using w = 96.32142857 – 2.180952381t

30.7 = 96.32142857 – 2.180952381t

t = 30.08842795

30.1 days

Many students rounded their answer to the

nearest day when this was not asked for.

Non exact answers should be given to 3 s.f.

Since w = 30.7 mg is within the data range 18 ≤ w ≤ 89 and r = - 0.964 is close to – 1 showing a strong negative

correlation between t and w, the estimate is reliable.

Many students managed to state that w is

within the data range but did not write that

r is close to -1.

0 40

89

18 Period of days, t

Amount of fertilizer, w mg

of 29

Understand that ,t w lies on the

regression line.

(vi)

Some students used the product moment

correlation formula in the MF15, resulting

in very long workings and frequently

wrong answers.

Answer 47.25 is an exact value and should

not be rounded to 3 s.f.

Among students who did not obtain the

correct values, many did manage to plot

the point to lie on the line of regression,

attaining the credit for that.

Since ,t w is on the line of regression, so the additional

point is (22.5, 47.25).

Amount of

fertilizer, w mg

Period of days, t days 0 5 40

89

18

(22.5, 47.25)

section a: pure mathematicssupergrades.com.sg/wp-content/uploads/2016/07/cjc_prelim2015p2sol.pdfjc2...

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