CONFIDENTIAL 950/2 Mathematics M Mei 2013 1½ Hours
SEK. MEN. KEB. TINGGI MELAKA (Malacca High School. Estd.1826) Ke Arah Kecemerlangan Kecemerlangan Pendidikan Pendidikan
STPM TRIAL EXAMINATION UPPER SIX 2013
MATHEMATICS M (1½
HOURS)
Instruction Instruction to candidates:
Answer all questions in section A and any one questions in section B. The marks for questions or or parts of questions questions are given in brackets ( ) . All necessary working must be shown clearly to obtain marks. Non-exact numerical answers may be given correct to three significant figures, or one decimal place in the case of angles in degrees ,unless a different level l evel of accuracy is specified in the question.
This question paper consists of 4 printed pages. pages.
Year 2013 This question paper is CONFIDENTIAL CONFIDENTIAL until the t he examination is over.
Turn over CONFIDENTIAL
SECTION A : [ 45 MARKS] Answer all questions in this section
1. The table below shows the scores given by two judges in a talent c ompetition for eight contestants. The lower scores indicate poor performances and high scores indicate good performances. Contestants Judge 1 Judge 2
A 14.50 10.00
B 17.00 9.60
C 12.40 13.80
D 9.50 14.50
E 15.80 16.40
F 10.60 15.00
G 11.00 11.00 12.20 12.20
H 18.80 17.00
Determine the Spearman rank correlation coefficie nt for Judge 1 and Judge 2. Comment on your result. [4M]
2. A box contains green green marbles and white marbles 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. b. exactly two green marbles. c. less than 2 white marbles.
[2M] [2M] [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 Mutton Chicken Fish Prawn Beef
2006 Price (RM/kg) 12.50 5.00 6.00 25.00 8.00
Value (RM) 8750 5500 9000 6250 7200
2008 Price (RM/kg) 14.00 6.50 7.50 30.00 10.50
Value (RM) 12600 12600 8450 15000 11400 12600
Using 2006 as the base year, a. find the Laspeyres and Paasche price indices for the year 2008. b. which of the two price indices in (a) is more suitable to measure the changes in prices of food ? Give a reason.
[4M] [2M]
4. A hundred years 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 . b. Find the minimum number of workers could have been selected at random, so that the probability probability that at least one of them contracted disease is greater greater than 0.9. c. If there are 200 workers in the industry, industry, calculate the probability that more than 50 workers were suffering from the disease. 1
[3M] [3M] [3M]
5. X is a continuous random variable with probability density function 1
k ( x 2 -
x )
,
0 x 1
,
otherwise
f(x) = 0
a. Find the value of k. b. Calculate E(X) and Var(X).
[3M] [6M]
6. The masses y (in kg) and ages t (in months) of a random random sample of nine babies in a hospital are recorded. The results obtained are summarized as follows :
t = 27
t
2
93
y 59.6 y
2
408.5
yt 191.3
a. Calculate the product-moment correlation coefficient between the mass and the age of baby. Interpret your answer. 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. c. Estimate the mass of a fifteen months old baby.
2
[3M] [6M] [2M]
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 Frequency
5-9 4
10-14 12
15-19 29
20-24 32
25-29 13
30-34 7
a. Plot a histogram to illustrate the data above. Hence, find the median. b. Calculate i. mean, ii. mode and iii. standard deviation of the gains in mass of 100 cows. c. Find the percentage of of the gains in mass of cows in the range of one standard deviation from the mean. d. State whether the mean or median is more suitable as a representative value of the above data. Justify your answer.
35-39 3 [4M] [2M] [2M] [2M] [3M] [2M]
8. A and B are two events with P(A) = 0.60, 0.60, P(B) = 0.7 and P(A U B) B) = 0.89. a. Find i. P(A B) i. P(A` B) iii. P(A` U B`)
[2M] [2M] [2M]
b. A group of students sit for both Mathematics 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 Determine whether the event of ‘ a student passing Mathematics’ and the event ‘ he passes Economic ’ are independent. independent.
3
[2M]
Marking Scheme – Scheme – Mathematics Mathematics M Term 2 Trial Exam 2012/2013 No
Answer
1.
J1 J2 d
2
r s 1
5 2 9
7 1 36
6(78) 8(82 1)
4 4 0
1 5 16
Mark
6 7 1
2 6 16
3 3 0
8 8 0
B1
0.0714 (3s.f.)
M1A1
There is a weak positive Spearman rank correlation correlation coefficient coefficient for the scores given by Judge 1 and Judge 2. 2.
3
P( White) =
a.
and P( Green) =
4 3
2
P( W =2) = ( ) (
4
b.
1
2
P( G = 2) = ( ) (
4
1 4 3 4
)
)
3! 2! 3! 2!
=
=
27 64 9 64
c.
P( W < 2) = 1- P( W 2) = 1-
3a.
Laspeyres Price Index =
27 64
[4M]
1 4 M1A1
or 0.422 (3s.f.)
M1A1
or 0.141 (3s.f.)
3
3
- ( ) =
4
5 32
or 0.156 (3s.f.)
14(700) 6.50(1100) 7.50(1500) 30(250) 10.50(900)
12.50(700) 5(1100) 6(1500) 25(250) 8(900) 45150 = x100 123.02 36700 b.
B1
x100
M1A1
[6M]
M1
A1
Paasche Price Index =
14(900) 6.50(1300) 7.50(2000) 30(380) 10.50(1200)
12.50(900) 5(1300) 6(2000) 25(380) 8(1200) 60050 x100 122.93 = 48850
x100
M1
A1
c.
Paasche Price Index is more suitable. It reflects reflects the latest value since the term of P 0 qn changes as the year under study changes.
B1B1
4a.
X ~ B ( 6 , 0.30 )
B1
P ( X < 3 ) = P( X=0 ) + P( X =1 ) + P( P( X=2 ) 6
0
= 0 C (0.30) (0.70) = 0.744 ( 3s.f. )
6
16C (0.30)1 (0.70)5 26C (0.30) 2 (0.70) 4
M1 A1
[6M]
4b.
P( X ≥ 1) > 0.9 1- P( X < 1 ) > 0.9 1-P( X = 0 ) > 0.9 P( X = 0 ) < 0.1
B1
n
4c.
( 0.7 ) < 0.1 n lg (0.7) < lg ( 0.1 ) n > 6.4557 The minimum number of workers = 7
A1
X ~ B ( 200 , 0.30 ) X ~ N ( 60 , 42 )
B1
P( X > 50 ) = P( X > 50.5 ) = P ( Z >
M1
50.5 60 42
)
M1
= P ( Z > -1.4659 ) = P ( Z < 1.4659 ) = 0.92866 = 0.929 (3 s.f.)
k ( x 2 x)dx = 1 0
3
2
k [ x k[
3 2 3
2
x
2 1 ]0 1
2
M1 M1
1
0] 1 2
A1
k=6 1
1
b.
E(x) = 6 x( x
2
0
1
2
E(x ) = 6 x ( x
2
0
3
5
3
M1M1
1
]0
A1 5
2
2
3
7
7 2
x4 4
1
]0
14
19
M1A1
350
2
27(59.6)
(27) ][9(408.5)
2
0.971(3 s. f .)
(59.6) ]
There is a strong li near correlation between the masses and ages of babies. b.
108
a=
1.0417(
27
) 3.4971 3.50(3 s. f .)
9 9 Therefore, y = 3.50 + 1.04t. (3s.f.)
b = 1.04 indicates that the mass mass of a new born born baby increases increases by 1.04 kg per month. c.
[9M]
M1A1 B1 M1A1
b = 112.5 1.04 59.6
M1
3
3
5 0
9(191.30) [9(93)
x
x)dx 6 ( x x )dx = 6 [ x
- ( 2 ) =
r=
5
1
2
14
2
2
= Var (x) =
5
2
2
2
0
1
1
2
3
x)dx 6 ( x x )dx = 6 [ x =
6a.
[9 M]
1
1
5a.
A1
y = 3.50 + 1.04 ( 15 ) = 19.1 (3s.f.) The mass of a 15 months baby is 19.1kg.
M1A1 A1 B1
M1 A1
[11M]
7a.
Frequency
D1 Label & Scale
35 30 25
D1 All correct
20 15
From the graph, Median = 20.5 kg
10 5
M1 Find Median A1 Correct answer
0
4.5-9.5 4.5-9.5 9.5-
14.5-
19.5-
24.5-
29.5-
34.5-
14.5
19.5
24.5
29.5
34.5
39.5
b. i.
Mean =
ii.
Mode = 19.5 + (
iii.
2055 100
M1A1
20.6kg 3 3 19
46545
Standard Deviation =
M1A1
)5 = 20.2kg ( 3s.f. )
100
(20.55) 2
M1
= 6.57 kg ( 3.s.f.)
A1
x 1 (20.55 6.5687) ( 13.981 , 27.119 ) 14.5 13.981 27.119 24.5 ( x12) 29 32 ( x13) 69.055 5 5 69.005 % of the gains in mass of cows = x100% 69.1% 100
c.
d.
Median is more suitable because mean > mode,this dist. is tvely skewed.
8ai.
P( A
ii.
P( A`
B ) = 0.60 + 0.70 – 0.70 – 0.89 0.89 = 0.410
B1B1
[15M]
B ) =P (B) – (B) – P(A P(A B)
iii.
P( A` U B` ) = 1- P( A B ) = 1 – 1 – 0.410 0.410 = 0.590
8b.
P(M) = 0.40 P( E/M) =
P(E/M) = 0.70 P ( E M ) P ( M )
P( E/M`) = =
M1A1 M1A1
P(E/M`) = 0.68
B1 M1
0.70
P (E M) = 0.280
A1
P ( E M `)
0.68 P ( M `) P ( E ) P ( E M )
0.68
0.60
P( E ) = 0.408 + 0.280 = 0.688 (3s.f.) ii.
M1A1
M1A1
= 0.70 – 0.70 – 0.410 0.410 = 0.290
i.
M1
M1 A1
P( M/E ) = P ( M E ) 0.280 0.407
M1A1
P(M) x P(E) P(M E ) 0.40 x 0.688 0.280 Both events are not i ndependent. ndependent.
M1 A1
P ( E )
0.688
[15M]
