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YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 1 of 8
Qn Solution
1 235
( 2) 2 5 0k x kx has no real solution i.e. discriminant 2 3
5(2 ) 4( 2)( 5) 0D k k
2 35
5( 2) 0k k 2 3 10 0 ( 5)( 2) 0k k k k
Therefore 5 2k .
2 3 62
6
4 24 2
64 2
x xy y
i.e. 6 6 2 3 3x y y x
lg( ) 1 lg( 1)y x x
lg(2 3) 1 lg( 1)x x
lg(2 3) lg( 1) 1x x
lg (2 3)( 1) lg10x x (2 3)( 1) 10x x
22 5 7 0x x (2 7)( 1) 0x x
1x or 3.5
Reject 1x since lg( 1)x is undefined.
When 3.5x , from 3 3 10.5 3 7.5y x
Therefore 3.5x and 7.5y
3i
3ii 1.000x and 1.501x (3 d.p.)
k 5 2
(0, 1)
y
x
( 1,0)
(0,ln2)
(1,0)
2 1y x
ln( 2)y x
2x
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 2 of 8
3iii Area bounded by the curves
1.5010175392
1ln( 2) ( 1) dx x x
2.926338462
2.926 (3 d.p.)
4 212 1 2
x xy y x x
x
2
2
d 11 1
d
yx
x x
When 3
2x ,
1
6y and
d 4 51
d 9 9
y
x
Equation of tangent: 1 5 3
6 9 2y x
5 2
9 3y x or 9 5 6y x
Equation of normal: 1 9 3
6 5 2y x
9 43
5 15y x or 15 27 43y x
Point T : When
20
3x y
Point N : When 43
015
x y
Area of 1 43 2 3
2 15 3 2PNT
53
20 or 2.65
5a 2
2 2 4 43 d 3 2 dx x x x x xe e e x e e e x
5 33 2 dx x xe e e x 5 3
3 25 3
x xxe ee C
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 3 of 8
5b
1 12 22 2d 13 2 7 3 2 7 6 2
d 2x x x x x
x
1223 2 7 3 1x x x
or
122
3 1
3 2 7
x
x x
3
32
2 00
3 1d 3 2 7
3 2 7
xx x x
x x
28 7
7 3 3
2 20 0
9 3 3 1d 3 d
3 2 7 3 2 7
x xx x
x x x x
3 7
5c At
11,
2
, 2d
3 0 3d
yx ax b a b
x
At 2
,3
k
, 2 4
3 3a b
Solving, 5 5
13 3
a a and 2b
2d 3 2d
yx x
x
23 2 y x x dx 3 21
22y x x x C
At 12
(1, ) , 1 12 2
1 2 2C C
Therefore, equation of curve is 3 21
2 22
y x x x
At 2
,3
k
, 8 2 4 76
227 9 3 27
k
6
A B
0.13 0.28
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 4 of 8
6i ( ) ( ) ( ')P A P A B P A B
0.35 ( ) 0.28P A B
Therefore ( ) 0.07P A B
6ii ( ) ( ) ( ' )P A B P A P A B
0.35 0.13 0.48
Therefore ( ' ') 1 ( ) 0.52P A B P A B
6iii ( ) ( ) ( ')P B P B A P B A
0.07 0.13
0.2
6iv ( )( | )
( )
P A BP A B
P B
0.07
0.2
0.35
Since ( | ) ( ) 0.35P A B P A , events A and B are independent.
7i
7ii P(correctly diagnosed)
0.08(0.95) 0.92(0.98)
0.9776 or 611
625
7iii P(has diabetes given incorrectly diagnosed)
( ') 0.08(0.05)
1 0.9776 0.0224
P A B
5
28
A
A
B
B
B
B
0.08
0.92
0.95
0.05
0.02
0.98
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 5 of 8
8a
Age Group
(Yrs)
Uses
Public
Transport
Uses
Private
Transport
Below 26 87 11
26 to 54 124 32
Above 54 43 3
The required number of people are to be selected randomly from each stratum.
8b Advantage: Sample obtained is a good representation of the population.
Disadvantage: Requires precise and accurate information about the population.
9i Let X be the number of ripe watermelons out of 30 shipped out. ~ (30,0.85)X B
P(all 30 are ripe) ( 30)P X
0.0076307596
0.00763
9ii P(fewer than 25 are ripe) ( 24)P X
0.289424
0.289
Let Y be the number of crates, each with fewer than 25 ripe watermelons, out of 50
crates.
~ (50,0.28942)Y B
Since 50n is large, 50(0.28942) 5np and (1 ) 50(1 0.28942) 5n p
~ (14.471,10.28280318)Y N approx.
Required probability
(15 45) (15.5 44.5)ccP Y P Y
0.374146
0.374
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 6 of 8
10i Let X be the mass (kg) of an adult trout in the lake.
The unbiased estimate for the population mean is ( 3.2)
3.2100
xx
31.23.2
100
3.512 kg (4 sfs)
The unbiased estimate for the population variance is
2
2 2( 3.2)1
( 3.2)100 1 100
xs x
21 31.215.551
99 100
0.05875354
0.05875 (4 sfs)
10ii 0 :H 3.47 kg
1 :H 3.47 kg
Suppose 0H is true, test statistic ~ (0,1)X
Z Ns
n
approx. by CLT where
3.512x kg, 0.05875s kg and 100n .
Level of significance: 5% From GC, p-value 0.04157
Since p-value100
0.05 , reject 0H .
There is significant evidence at 5% level that the marine biologist has understated the
mean mass of the adult trouts in the lake.
0 :H 3.47 kg
1 :H 3.47 kg
Suppose 0H is true, test statistic ~ (0,1)X
Z Ns
n
approx. by CLT where
0.0625s kg and 100n .
Level of significance: 1% From GC, critical or rejection region is where 2.32635z
Since 0H is not rejected, the test-value is such that 3.47
2.326350.25
10
x
0 3.52815875x
Maximum 3.528x kg (3 d.p.)
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 7 of 8
11i Let X and Y be the mass (g) of a cheese biscuit and a ginger biscuit respectively.
By Central Limit Theorem, since sample size is large, 2
1 2 44... ~ (44 10,44 0.3 )Y Y Y N approx.
i.e. 1 2 44... ~ (440,3.96)Y Y Y N approx. 2
1 2 74... ~ (74 6,74 0.2 )X X X N approx.
i.e. 1 2 74... ~ (444,2.96)X X X N approx.
1 2 44 1 2 74... ...Y Y Y X X X ~ ( 4,6.92)N approx.
1 2 44 1 2 74( ... ... )P Y Y Y X X X
1 2 44 1 2 74( ... ... 0)P Y Y Y X X X
0.9358163198
0.936
11ii 1 2 7... 4X X X Y ~ (7 6 4 10,7 0.04 16 0.09)N
i.e. 1 2 7... 4X X X Y ~ (2,1.72)N
1 2 7( ... 4 )P X X X Y
1 2 7( ... 4 0)P X X X Y
0.9363686479
0.936
12i Let X be the mass (g) of a bag of cashew nuts and 2~ (128.5,1.5 )X N .
( 130) 0.1586552596 0.159P X
12ii Let W be the mass (g) of a carton of 50 bags of cashew nuts. 2~ (50 128.5 550,50 1.5 )W N
i.e. ~ (6975,112.5)W N
( 7000)P W
0.0092110461 0.00921
12iii ~ (6975,112.5)W N
112.5~ 6975,
30W N
i.e. ~ 6975,3.75W N
(6970 6980) 0.9901766644P W
0.990
YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM
Page 8 of 8
12iv Let Y be the mass (g) of a bag of cashew nuts and 2~ ( , )Y N .
Given ( 130) 0.18P Y
130 0.18P Z or 130 0.82P Z
Using GC, 130
0.915365082
0.915365082 130 ------------ (1)
Given ( 126) 0.03P Y
126 0.03P Z
Using GC, 126
1.88079361
1.88079361 126 ---------- (2)
Solving, 128.691 128.7 g (4 sfs)
and 1.43053 1.431 g (4 sfs)
13i
The scatter diagram reveals a negative linear correlation between the temperature and the
pH of skimmed milk where the pH of skimmed milk decreases with an increase in
temperature.
13ii Linear correlation coefficient, 0.9647177563 0.965r
Least square regression line of y on x is
6.84992649 0.0071895151y x
i.e. 6.85 0.00719y x
13iii When 20 Cx , 6.71y
This estimate is reliable as 20 Cx is within the range of data given and the correlation
coefficient is close to 1 .
13iv The equation obtained in (ii) can be used to estimate the temperature when given the pH
of skimmed milk because:
a) temperature, x , is the only independent variable,
b) the linear correlation coefficient, r , is close to 1 .
Temp. ( C)x
(4,6.85)
(63,6.36)
pH ( )y
Temperature (x)
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