1

MARKING SCHEME

PEPERIKSAAN PERCUBAAN PENGGAL 2 TAHUN 2014

SMK BUKIT MERTAJAM

MATHEMATICS M ( PAPER 2)

NO SCHEME MARKS

1.

[8]

a)

Time Frequency Width Midpoint Frequency

density

1200 x 1 120 60 0.5

180120 x 9 60 150 9

240180 x 15 60 210 15

300240 x 17 60 270 17

360300 x 13 60 330 13

600360 x 5 240 480 1.25

M1(width)

M1

(frequency

density)

D1(Label &

scale)

D1(All

correct)

2

b) Mean = 60

48053301327017210151509601

= 60

15840

= 264

c) Standard deviation = 226460

4674600

= 90.6

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2.

[6]

a) 5.0)(

1.0

BP

P(B) = 0.2

b) )()( BPABP A and B are not independent

c) 25.0)(

1.0

AP

P(A) = 0.4

)()()( ' BAPAPBAP

= 0.4 – 0.1

= 0.3

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3.

[6]

ai) r measure the strength of the relationship between x and y.

ii) 11 r

r = 1 mean a perfect positive relationship between x and y.

b)

8

)470(29450)(

8

)423(24479(

8

)470(42326520

22

r

= 0.847

There have a strong positive correlation between variable x and y.

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3

4.

[7]

a) p+0.25+2p+0.3+0.15=1

3p+0.7=1

P=0.1

b) E(x)=5.7

0(0.1)+2(0.25)+5(0.2)+n(0.3)+12(0.15) = 5.7

3.3+0.3n = 5.7

n = 8

c) E(X2) = 0

2(0.1)+2

2(0.25)+5

2(0.2)+8

2(0.3)+12

2(0.15)

= 46.8

Var (X) = 46.8-5.72

= 14.31

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5.

[10]

a)

Year Quarter

Time

column

, t

Number

of tourist

4-quarter

moving

average

Centred

4-quarter

moving

average

Deviation

2011

1 1 22.0

2 2 12.0

3 3 110.0 43.75 43.63 66.38

4 4 31.0 43.5 45.25 -14.25

2012

1 5 21.0 47 52.00 -31.00

2 6 26.0 57 61.88 -35.88

3 7 150.0 66.75 70.38 79.63

4 8 70.0 74 75.25 -5.25

2013

1 9 50.0 76.5 76.00 -26.00

2 10 36.0 75.5 80.50 -44.50

3 11 146.0 85.5

4 12 110.0

b)

Year Q1 Q2 Q3 Q4

2011 66.38 -14.25

2012 -31.00 -35.88 79.63 -5.25

2013 -26.00 -44.50

Unadjusted

seasonal

variation

-28.50 -40.19 73.00 -9.75

Correction

factor -1.359 -1.359 -1.359 -1.359

Seasonal

variation -27.14 -38.83 74.36 -8.39

M1

(4-quarter

moving

average)

A1 (Centred

4-quarter

moving

average)

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4

c)

Year Quarter Coded

quarter, t

Number

of

tourist, y

Seasonal

variation

, S

Deseaso

nalised,

T=y-S

t^2 tT

2011

1 1 22.0 -27.14 49.14 1 49.14

2 2 12.0 -38.83 50.83 4 101.66

3 3 110.0 74.36 35.64 9 106.92

4 4 31.0 -8.39 39.39 16 157.56

2012

1 5 21.0 -27.14 48.14 25 240.70

2 6 26.0 -38.83 64.83 36 388.97

3 7 150.0 74.36 75.64 49 529.48

4 8 70.0 -8.39 78.39 64 627.13

2013

1 9 50.0 -27.14 77.14 81 694.27

2 10 36.0 -38.83 74.83 100 748.28

3 11 146.0 74.36 71.64 121 788.05

4 12 110.0 -8.39 118.39 144 1420.69

Total 78 784.0 784.00 650 5852.84

12

78650

12

)784)(78(84.5852

2

b

= 5.293

)12

78(293.5

12

784a

= 30.93

tT 293.593.30

)13(293.593.301 T

= 99.74

1F = 99.74+(-27.14)

= 72.6

The predict number of tourist fot 1st quarter 2014=72.6X1000=72600

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6.

[8]

a) Simple aggregate price index

120= 1005126105.2

614123X

a

1005.35

35120 X

a

A=7.60

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5

b) 100)5(5)15(12)10(6)10(10)20(5.2

)5(6)15(14)10(6.7)10(12)20(3XLp

= 100415

496X

= 119.5

From January to July, the family’s expenditure on daily food has increased by

19.5%

c) 100)6(5)14(15)6.7(10)2(10)3(20

)4(6)10(14)9(6.7)8(12)10(3XPq

= 100496

4.358X

= 72.26

From January to July, there is a drop of 27.74% in the quantity of daily food

bought by the family.

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7.

[15]

a) p=0.6 , q = 0.4

i) )3( XP

53

3

8 )4.0()6.0(C

=0.124

ii) )1()0(1)2( XPXPXP

71

1

880

0

8 )4.0()6.0()4.0()6.0(1 CC

= 1- 0.000655-0.00786

= 0.9914

iii) E(X) = 8X0.4

= 3.2

= 3

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6

iv) qx

pxn

xXP

xXP

)1(

)(

)(

)1(

If P(X = x+1) > P(X = x)

(n-x)p > (x+1)q

(8-x)(0.6) > (x+1)(0.4)

x < 4.4

Hence, P(X=5) > P(X=4) > …. >P(X=0)

If P(X = x) > P(X = x+1)

Then x > 4.4

Hence, P(X=5) > P(X=6) > …. >P(X=8)

P(X=5) has the highest probability, Hence, number of students that most

likely to pass is 5.

b) 99.0)1( XP

1 – P(X=0) > 0.99

P(X=0) < 0.1

01.0)4.0()6.0( 0

0 nnC

01.04.0 n

01.0log4.0log n

4.0log

01.0logn

02.5n

n = 6

c) 6or 36)4.0)(6.0(150 and 90)6.0(150 npqnpqnp

36) ,90(~ NX

P( X> 85) = )5.85( XP

= )6

905.85(

ZP

= )75.0( ZP

= )75.0(1 ZP

= 1-0.2266

= 0.7734

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7

8.

[15]

. a)

Boys Stem Girls

0 8

1 6

9 6 2 4 5

9 8 6 6 5 5 5 5 4 4 1 0 3 1 2 4 4

7 5 3 0 0 0 4 0 0 0 2 4 5 7 8 8

5 0

Key: 6 2 means hours

Key: 0│8 means 8 hours

b) The distribution for the boys is almost symmetrical while the girls is

negatively skewed.

c) Boys: IQR Girls: IQR

d)

Boys:

mean hours

number of hours

Girls:

mean hours

number of hours

e) The dispersion of distribution is larger for the girls.

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A1A1

M1M1

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A1A1

A1