TRIAL STPM MATHEMATICS M (MELAKA) –SMK Tinggi MELAKA QA

SECTION A : [ 45 MARKS] Answer all questions in this section

1. The table below shows the scores given by two judges in a talent competition for eight

contestants. The lower scores indicate poor performances and high scores indicate

good performances.

Contestants A B C D E F G H

Judge 1 14.50 17.00 12.40 9.50 15.80 10.60 11.00 18.80

Judge 2 10.00 9.60 13.80 14.50 16.40 15.00 12.20 17.00

Determine the Spearman rank correlation coefficient for Judge 1 and Judge 2. Comment

on your result. [4M]

2. A box contains green marbles and white marbles in the ratio 1 : 3. Three marbles are selected

at random from the box one at a time with replacement. Find the probability of getting

a. exactly two white marbles. [2M]

b. exactly two green marbles. [2M]

c. less than 2 white marbles. [2M]

3. The following table indicates the prices and values for five types of food for the year

2006 and 2008 where value is (price) x (quantity).

Food 2006 2008

Price (RM/kg) Value (RM) Price (RM/kg) Value (RM)

Mutton 12.50 8750 14.00 12600

Chicken 5.00 5500 6.50 8450

Fish 6.00 9000 7.50 15000

Prawn 25.00 6250 30.00 11400

Beef 8.00 7200 10.50 12600

Using 2006 as the base year,

a. find the Laspeyres and Paasche price indices for the year 2008. [4M]

b. which of the two price indices in (a) is more suitable to measure the changes in

prices of food ? Give a reason. [2M]

4. A hundred years ago, the occupational disease in one industry was such that the

workers had a 30% chance of suffering from it.

a. If six workers were selected at random, find the probability that less than three

of them contracted the disease . [3M]

b. Find the minimum number of workers could have been selected at random,

so that the probability that at least one of them contracted disease is greater than 0.9. [3M]

c. If there are 200 workers in the industry, calculate the probability that more than 50

workers were suffering from the disease. [3M]

1

5. X is a continuous random variable with probability density function

k ( 2

1

x - )x , 0 ≤ x ≤ 1

f(x) =

0 , otherwise

a. Find the value of k. [3M]

b. Calculate E(X) and Var(X). [6M]

6. The masses y (in kg) and ages t (in months) of a random sample of nine babies in a hospital

are recorded. The results obtained are summarized as follows :

∑t = 27 932 =∑ t ∑ = 6.59y 5.4082 =∑ y 3.191=∑ yt

a. Calculate the product-moment correlation coefficient between the mass and the age

of baby. Interpret your answer. [3M]

b. By using least square method, find the equation of regression line y on t in the

form of y = a + bt. Interpret the slope of the regression line. [6M]

c. Estimate the mass of a fifteen months old baby. [2M]

2

SECTION B [ 15 MARKS ] Answer any one question in this section

7. In an agricultural experiment, the gains in mass, in kilogram, of 100 cows during a

certain period were recorded as follows :

Gain in mass, kg 5-9 10-14 15-19 20-24 25-29 30-34 35-39

Frequency 4 12 29 32 13 7 3

a. Plot a histogram to illustrate the data above. Hence, find the median. [4M]

b. Calculate

i. mean, [2M]

ii. mode and [2M]

iii. standard deviation of the gains in mass of 100 cows. [2M]

c. Find the percentage of the gains in mass of cows in the range of one

standard deviation from the mean. [3M]

d. State whether the mean or median is more suitable as a representative value

of the above data. Justify your answer. [2M]

8. A and B are two events with P(A) = 0.60, P(B) = 0.7 and P(A U B) = 0.89.

a. Find

i. P(A ∩B) [2M]

i. P(A`∩B) [2M]

iii. P(A` U B`) [2M]

b. A group of students sit for both Mathematics and Economic papers in STPM. The results

are summarized as follows:

40% of the students pass Mathematics

70% of the students pass Economic if they pass Mathematics

68% of the students pass Economic if they fail Mathematics

A student is selected at random, find the probability that the student passes

i. Economic, [5M]

ii. Mathematics if he passes Economic. [2M]

Determine whether the event of ‘ a student passing Mathematics’ and the event

‘ he passes Economic ’ are independent. [2M]

3

Marking Scheme – Mathematics M Term 2 Trial Exam 2012/2013 SMK TINGGI MELAKA

No Answer Mark

1.

0714.0)18(8

)78(61

2=

−−=sr (3s.f.)

There is a weak positive Spearman rank correlation coefficient for the scores given by Judge 1 and Judge 2.

J1 5 7 4 1 6 2 3 8

J2 2 1 4 5 7 6 3 8

d2 9 36 0 16 1 16 0 0

B1

M1A1

B1 [4M]

2.

a.

b.

c.

P( White) = 4

3 and P( Green) =

4

1

P( W =2) = ()4

3( 2

4

1)

!2

!3= 64

27or 0.422 (3s.f.)

P( G = 2) = ()4

1( 2

4

3)

!2

!3= 64

9or 0.141 (3s.f.)

P( W < 2) = 1- P( W≥ 2) = 1- 64

27-

3)4

3( =

32

5or 0.156 (3s.f.)

M1A1

M1A1

M1A1 [6M]

3a.

b.

c.

Laspeyres Price Index

= 100)900(8)250(25)1500(6)1100(5)700(50.12

)900(50.10)250(30)1500(50.7)1100(50.6)700(14x

++++

++++

= 02.12310036700

45150=x

Paasche Price Index

= 100)1200(8)380(25)2000(6)1300(5)900(50.12

)1200(50.10)380(30)2000(50.7)1300(50.6)900(14x

++++

++++

= 93.12210048850

60050=x

Paasche Price Index is more suitable. It reflects the latest value since the

term of nqP0Σ changes as the year under study changes.

M1

A1

M1

A1

B1B1 [6M]

4a.

X ~ B ( 6 , 0.30 )

P ( X < 3 ) = P( X=0 ) + P( X =1 ) + P( X=2 )

= 426

2

516

1

606

0 )70.0()30.0()70.0()30.0()70.0()30.0( CCC ++

= 0.744 ( 3s.f. )

B1

M1 A1

4b.

4c.

P( X ≥ 1) > 0.9

1- P( X < 1 ) > 0.9

1-P( X = 0 ) > 0.9 P( X = 0 ) < 0.1

( 0.7 )n < 0.1

n lg (0.7) < lg ( 0.1 )

n > 6.4557

The minimum number of workers = 7

X ~ B ( 200 , 0.30 )

X ~ N ( 60 , 42 )

P( X > 50 ) = P( X > 50.5 ) = P ( Z > 42

605.50 − )

= P ( Z > -1.4659 )

= P ( Z < 1.4659 )

= 0.92866

= 0.929 (3 s.f.)

B1

M1

A1

B1

M1

A1 [9 M]

5a.

b.

k dxxx )(1

0

2

1

−∫ = 1

k 1]23

2[ 1

0

2

2

3

=−x

x

k[ 1]02

1

3

2=−−

k = 6

E(x) = 6 dxxxdxxxx )(6)( 21

0

2

31

0

2

1

−=− ∫∫ = 61

0

3

2

5

]35

2[

xx −

= 5

2

E(x2) = 6 dxxxdxxxx )(6)( 3

1

0

2

51

0

2

1

2 −=− ∫∫ = 61

0

4

2

7

]47

2[

xx −

= 14

3

Var (x) = 14

3 - (5

2 )2 =

350

19

M1

M1

A1

M1M1

A1

M1

M1A1 [9M]

6a.

b.

c.

r = .).3(971.0])6.59()5.408(9][)27()93(9[

)6.59(27)30.191(9

22fs=

−−

−

There is a strong linear correlation between the masses and ages of babies.

b = 04.1108

5.112=

a = .).3(50.34971.3)9

27(0417.1

9

6.59fs==−

Therefore, y = 3.50 + 1.04t. (3s.f.)

b = 1.04 indicates that the mass of a new born baby increases by 1.04 kg per

month.

y = 3.50 + 1.04 ( 15 ) = 19.1 (3s.f.) The mass of a 15 months baby is 19.1kg.

M1A1

B1

M1A1

M1A1

A1

B1

M1

A1 [11M]

7a.

b.

i.

ii.

iii.

c.

d.

0

5

10

15

20

25

30

35

4.5-9.5 9.5-

14.5

14.5-

19.5

19.5-

24.5

24.5-

29.5

29.5-

34.5

34.5-

39.5

Frequency

Mean = kg6.20100

2055=

Mode = 19.5 + (193

3

+)5 = 20.2kg ( 3s.f. )

Standard Deviation = 2)55.20(

100

46545−

= 6.57 kg ( 3.s.f.)

=±=± )5687.655.20(1σx ( 13.981 , 27.119 )

055.69)135

5.24119.27(3229)12

5

981.135.14( =

−+++

−xx

% of the gains in mass of cows = %1.69%100100

005.69=x

Median is more suitable because mean > mode,this dist. is tvely skewed.

D1 Label & Scale

D1 All correct

M1 Find Median

A1 Correct answer

M1A1

M1A1

M1

A1

M1

M1A1

B1B1 [15M]

8ai.

ii.

iii.

8b.

i.

ii.

P( A ∩ B ) = 0.60 + 0.70 – 0.89 = 0.410

P( A` ∩ B ) =P (B) –P(A ∩ B)

= 0.70 – 0.410 = 0.290

P( A` U B` ) = 1- P( A ∩ B )

= 1 – 0.410 = 0.590

P(M) = 0.40 P(E/M) = 0.70 P(E/M`) = 0.68

P( E/M) = 70.0)(

)(=

∩

MP

MEP

P (E∩M) = 0.280

P( E/M`) = 68.0`)(

`)(=

∩

MP

MEP

= 68.060.0

)()(=

∩− MEPEP

∴P( E ) = 0.408 + 0.280 = 0.688 (3s.f.)

P( M/E ) = 407.0688.0

280.0

)(

)(==

∩

EP

EMP

P(M) x P(E) ≠ P(M ∩ E )

0.40 x 0.688 ≠ 0.280 ∴Both events are not independent.

M1A1

M1A1

M1A1

B1

M1

A1

M1

A1

M1A1

M1 A1 [15M]

From the graph,

Median = 20.5 kg