paku sarawak 2013 m2(a)
TRANSCRIPT
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8/13/2019 PAKU SARAWAK 2013 M2(A)
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CONFIDENTIAL*/SULIT
STPM 950/2CONFIDENTIAL*
*This question paper is CONFIDENTIAL until the examination is over.
2
67
12
)24(2
1Median
th
th
5.57
2
5857
6
)24(4
1Q1
th
th
74
18
)24(4
3Q3
th
th
1M
c.
75.32
)5.5774(2
35.57
)(2
3QboundaryLower 131
QQ
75.106
)5.5774(2
374
)(2
3QboundaryUpper 131
QQ
Outlier = 1081 M
2M
4a)
Interquartile range = 13 QQ
= 74 -57.5 = 16.5
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5. 2M
6 a) GraphYear Quarter Unemployed 4-quarter Centred 4-quarter Deviation
3M
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school
leavers
moving
average
moving average
(t)
(y-t)
2010 1 22
2 12
43.75
3 110 43.625 66.375
43.504 31 45.25 -14.250
47.00
2011 1 21 52.00 -31.000
57.00
2 26 61.875 -35.875
66.75
3 150 70.375 79.625
74.00
4 70 75.250 -5.250
76.50
2012 1 50 76.000 -26.000
75.50
2 36 80.500 -44.50085.50
3 146
4 110
b) seasonal variation
Year Quarter 1 Quarter 2 Quarter 3 Quarter 4
2010 66.375 -14.250
2011 -31.000 -35.875 79.625 -5.250
2012 -26.000 -44.500
Unadjusted
seasonal
variation
-28.500 -40.188 73.000 -9.750
Correctionfactor
-1.3595 -1.3595 -1.3595 -1.3595
Seasonal
Variation
-27.14 -38.83 74.36 -8.39
C)The least square method
4822 x 66x 794182 y 784y 4627xy
12
)66(482
12)784)(66(4629
2
= 2.664 1M
[1M]
[2M]
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8/13/2019 PAKU SARAWAK 2013 M2(A)
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CONFIDENTIAL*/SULIT
STPM 950/2CONFIDENTIAL*
*This question paper is CONFIDENTIAL until the examination is over.
5
12
66664.2
12
784a =50.68
Regression line, y= 50.68+2.664x
For the first quarter of the year 2013,e
T = 50.68+2.664(13) = 85.312
Unemployed school leavers for the first quarter of the year 2013=85.312+ (-27.14) = 58.172 59
Or
Average quarterly increment =7
625.435.80 =5.2679
For the first quarter of the year 2013,e
T = 80.5 + 3 x 5.2679 = 96.304
Unemployed school leavers for the first quarter of the year 2013
= 96.304 + (-27.14)= 69.164 69
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7a Histogram
b) 5.62)125(21
21 N
minutes58.30)1(30
45)125(2
1
30
2
1
Median,
cf
FN
LMm
2M
x f fx 2fx Cumulative
frequency,F26.5 6 6
28.5 12 18
29.5 27 45
30.5 30 75
31.5 18 93
32.5 14 107
33.5 9 116
34.5 4 120
37.5 5 125
5.3861fx 25.1199052 fx
89.30125
5.3861
,Mean
f
fx
x
220.2)89.30(125
25.119905,DeviationStandard 2
2
f
fx
f
fx
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c. Pearson coefficient of skewness 1M
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=deviationstandard
median)3(mean= 4189.0
2.220
)58.303(30.89
The distribution is positively skewed or skewing to the right.
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7d)
.The number of times a bus arrives at town B between 0830 hours and 0836
hours = 122-45 = 77 2M
8a) Let X be the number of books read by an adult
Then, )5(0PX
a)
1404.0!3
)5()3(
35
e
XP
b) In 5 years, 2555 1M1M
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7
0.1
!2
)25(
!1
)25(
!0
)25(-1
)]2()1()0([1)2(
225125025
eee
XPXPXPXP
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b)(i)
0.0228
0.9772-1
2)P(z-1
2)P(z
10
5070)70(
zPXP
Hence, percentage of candidates = 0.0228 x 100% = 2.28%(ii)
76.44
524.010
50
0.524z
0.70z)P(Z
70.010
50
70.0)(
x
x
xZP
xXP
Hence, minimum marks required = 45
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8c) Expected number, 8)(E npx
Variance, Var (x) = npq = 1.6
8
6.1
np
npq
q = 0.2 p = 0.8 n = 10 )8.0,10(X (i) P( a teacher chosen at random owns a proton car) = 0.8(ii) P( exactly 4 teachers from the group own Proton cars)
= P(X = 4)
= 4104410 )2.0()8.0( C = 0.005505
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