TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

5
Answer Scheme for Mathematics (M) SMK SEAFIELD SUBANG Trial STPM 950/2 1 No. Answer scheme Marks Total 1(a) (b) Mean, 87 . 18 95 5 . 1792 = = x minutes Median, 78 . 15 10 32 29 5 . 47 10 = × + = m Standard deviation, 25 . 50831 2 = fx 38 . 13 95 5 . 1792 95 25 . 50831 2 = = s Pearson’s coefficient of skewness ( ) 6928 . 0 38 . 13 78 . 15 87 . 18 3 = = The distribution is positively skewed. M1A1 M1A1 B1 M1A1 M1 A1 B1 10 marks 2(a) (b) (c) P(the child is a girl who writes using left hand) = 0.052 × 0.58 = 0.03016 P(the child writes using left hand) = 0.03016 + 0.078 × 0.42 =0.06292 P(the child is a boy|the child writes using left hand) 5207 . 0 06292 . 0 42 . 0 078 . 0 = × = M1 A1 M1 A1 M1 A1 6 marks 3(a) (b) 0.15 + 0.40 + 2k + k = 1 k = 0.15 ( ) ( ) ( ) ( ) ( ) 15 . 0 2 30 . 0 1 40 . 0 0 15 . 0 1 + + + = X E = 0.45 ( ) ( ) ( ) ( ) ( ) ( ) 15 . 0 2 30 . 0 1 40 . 0 0 15 . 0 1 2 2 2 2 2 + + + = X E = 1.05 ( ) ( ) 2 45 . 0 05 . 1 Var = X = 0.8475 M1 A1 M1 A1 B1 M1 A1 7 marks

Transcript of TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

Page 1: TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

Answer Scheme for Mathematics (M) SMK SEAFIELD SUBANG Trial STPM 950/2

1

No. Answer scheme Marks Total

1(a)

(b)

Mean, 87.1895

5.1792==x minutes

Median, 78.151032

295.4710 =×

−+=m

Standard deviation,

25.508312 =∑ fx

38.1395

5.1792

95

25.508312

=

−=s

Pearson’s coefficient of skewness

( )

6928.0

38.13

78.1587.183

=

−=

The distribution is positively skewed.

M1A1

M1A1

B1

M1A1

M1

A1

B1

10 marks

2(a)

(b)

(c)

P(the child is a girl who writes using left hand)

= 0.052 × 0.58

= 0.03016

P(the child writes using left hand)

= 0.03016 + 0.078 × 0.42

=0.06292

P(the child is a boy|the child writes using left hand)

5207.0

06292.0

42.0078.0

=

×=

M1

A1

M1

A1

M1

A1

6 marks

3(a)

(b)

0.15 + 0.40 + 2k + k = 1

k = 0.15

( ) ( ) ( ) ( ) ( )15.0230.0140.0015.01 +++−=XE

= 0.45

( ) ( ) ( ) ( ) ( ) ( )15.0230.0140.0015.01 22222 +++−=XE

= 1.05

( ) ( )245.005.1Var −=X

= 0.8475

M1

A1

M1

A1

B1

M1

A1

7 marks

Page 2: TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

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4(a)

(b)

(c)

Negative correlation.

As the mortgage interest rate increases the housing sales index

decreases.

∑ ∑∑ ==== 985,1043,99,10 2 yxxn

∑ ∑ == 9355,1005252 xyy

Coefficient of determination,

( ) ( )( )[ ]( ) ( )[ ] ( ) ( )[ ]22

2

2

9851005251099104310

98599935510

−⋅−

−=r

= 0.7136

Hence, 71.36% of the variation in the number of houses sold is

accounted for by the variation in the mortgage interest rate.

D2

B1

B1

B1

M1

M1

A1

B1

9 marks

5(a)

(b)

Simple aggregate price index for February 2013

44.112

10080.3150.2468.5

40.3530.2899.5

=

×++++

=

The price of the seafood has increased by 12.44% from January

to February 2013.

Weighted average of price relatives for February 2013

27.111

10080.42400.28365.209

80.42480.31

40.35283

50.24

30.2865.209

68.5

99.5

=

×++

×+

×+

×=

The price of the seafood has increased by 11.27% from January

to February 2013.

M1

A1

B1

M1

A1

B1

6 marks

6(a)

(b)

(c)(i)

(ii)

The time series has an increasing trend.

The revenue is the highest in the fourth quarter each year but

rather low in the first three quarters.

An additive model is more suitable because the amplitude of

the seasonal variations is almost constant as the trend rises.

7439.0

00761.26427.06895.0

−=

=+−+−

k

k

2.0761 means that the revenue in the fourth quarter is

RM2.0761 million above the trend value.

B1

B1

B1

B1

M1

A1

B1

7 marks

Page 3: TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

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7(a)

(b)

(c)

(d)

( )

27

1

1033

=

=−

k

k

( ) ( )31327

11 −=>XP

27

8=

( ) ( )

−−=

,1

,327

11

,0

3xxF

3

30

0

>

≤<

x

x

x

Let X = the number of observations greater than 1.

Then,

27

8,3B~X .

( ) ( )011 =−=≥ XPXP

6515.0

27

191

3

=

−=

27

8,729B~X .

Hence, ( )152,216N~ɺX .

( ) ( )5.229229 >=> XPXP

( )1368.0

095.1

152

2165.229

=

>=

−>=

ZP

ZP

M1

M1

A1

B1

D2

B1

B1

M1

A1

B1

B1

B1

M1

A1

15 marks

2

1.5

1

0.5

-0.5

-1 1 2 3 4

0 3

1

( )xFy =

y

x

Page 4: TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

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8(a)

(b)

Year

Quarter

Sales

(×RM100,000),

Y

4-pt

moving

average

Centred

moving

average,

T

S=Y/T

1 5.3

2 4.1

3 6.8 5.725 5.6625 1.2009 2009

4 6.7 5.600 5.5625 1.2045

1 4.8 5.525 5.3750 0.8930

2 3.8 5.225 5.2375 0.7255

3 5.6 5.250 5.1875 1.0795 2010

4 6.8 5.125 5.1250 1.3268

1 4.3 5.125 5.1375 0.8370

2 3.8 5.150 5.0750 0.7488

3 5.7 5.000 5.1625 1.1041 2011

4 6.2 5.325 5.4250 1.1429

1 5.6 5.525 5.5750 1.0045

2 4.6 5.625 5.5125 0.8345

3 6.1 5.400 2012

4 5.3

Q1 Q2 Q3 Q4

2009 1.2009 1.2045

2010 0.8930 0.7255 1.0795 1.3268

2011 0.8370 0.7488 1.1041 1.1429

2012 1.0045 0.8345

Mean

Seasonal

Variation

0.9115 0.7696 1.1282 1.2247

Adjusting

factor 0.9916 0.9916 0.9916 0.9916

Adjusted

Seasonal

Variation

0.9038 0.7631 1.1187 1.2144

Adjusted seasonal variations are:

Q1 Q2 Q3 Q4

0.904 0.763 1.119 1.214

M1A1

(column

5)

B1

(column

6)

B1

M1

A1

Page 5: TRIAL STPM MATHEMATICS M 2 (SELANGOR) SMK Seafield Answer

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(c)

(d)

t Y S Deseasonalised

Series, y=Y/S

1 5.3 0.9038 5.8641

2 4.1 0.7631 5.3728

3 6.8 1.1187 6.0785

4 6.7 1.2144 5.5171

5 4.8 0.9038 5.3109

6 3.8 0.7631 4.9797

7 5.6 1.1187 5.0058

8 6.8 1.2144 5.5995

9 4.3 0.9038 4.7577

10 3.8 0.7631 4.9797

11 5.7 1.1187 5.0952

12 6.2 1.2144 5.1054

13 5.6 0.9038 6.1961

14 4.6 0.7631 6.0280

15 6.1 1.1187 5.4528

16 5.3 1.2144 4.3643

∑ ∑∑ ==== 7076.85,1496,136,16 2 yttn

∑ = 6733.719ty

( ) ( )( )( ) ( )2136149616

7076.851366733.71916

−=b

02600.0−=

( )

−−

=16

13602600.0

16

7076.85a

58.5=

Least squares regression equation is ty 0260.058.5 −= .

4th quarter of 2013, t = 20.

( )200260.058.5 −=y

= 5.06

2144.106.5ˆ ×=Y

= 6.145

Forecast sales = RM6.145 (×100,000)

M1A1

(column

4)

B1

M1

M1

A1

M1

M1

A1

15 marks