PSPM 2 Q&A

PSPM 2 QS 025/1 Session 2017/2018

7. Given the line : −1 = −3 = −2 and the planes 1: 2 − − 2 = 17 and 2: −4 −
2 −1 −3

3 + 5 = 10.

Find

a) The intersection point between and 1.
b) The acute angle between 1 and 2.
c) The parametric equations of the line that passes through the point (2, −1, 3) and

perpendicular to the plance 2.

SOLUTION

a) : −1 = −3 = −2 ………… (1)

2 −1 −3 CHOW

1: 2 − − 2 = 17 ………… (2)

From (1), the parametric equation of L:

= 1 + 2 ; = 3 − = 2 − 3 ………… (3)

(3) (2)

2(1 + 2 ) − (3 − ) − 2(2 − 3 ) = 17

2 + 4 − 3 + − 4 + 6 = 17

11 = 22

= 2 ………… (4)

(4) (3)

= 1 + 2(2); = 3 − 2 = 2 − 3(2)

= 5 = 1 = −4

Page 15

150

PSPM 2 QS 025/1 Session 2017/2018

ℎ 1 (5, 1, −4)

b) 1: 2 − − 2 = 17  = 2 − − 2

2: −4 − 3 + 5 = 10  = −4 − 3 + 5

cos = .
| || |

. = (2 − − 2 ). (−4 − 3 + 5 )

= ( )(− ) + (− )(− ) + (− )( )

= − + −

= − CHOW

| | = √(2)2 + (−1)2 + (−2)2
=3

| | = √(−4)2 + (−3)2 + (5)2

= √50

cos = .
| || |

−15
=

3√50

= −0.7071

= −1 (−0.7071)

= 135°

∴ = 180° − 135° = 45°

Page 16

151

PSPM 2 QS 025/1 Session 2017/2018
c) = −4 − 3 + 5
2: −4 − 3 + 5 = 10

:

= + = + = +

= = − − +  a=-4; b= -3; c=5 CHOW

= − +  = ; = − ; =
= +
∴ = − = − −

(2, −1,3))

Page 17

152

PSPM 2 QS 025/1 Session 2017/2018

8. Express 6 2− +7 in partial fractions. Hence, show that ∫01 6 2− +7 = 1 + ln 2.
(4−3 )(1+ )2 (4−3 )(1+ )2

SOLUTION

6 2 − + 7
(4 − 3 )(1 + )2 = (4 − 3 ) + (1 + ) + (1 + )2

6 2 − + 7 (1 + )2 + (4 − 3 )(1 + ) + (4 − 3 )
(4 − 3 )(1 + )2 =
(4 − 3 )(1 + )2

6 2 − + 7 = (1 + )2 + (4 − 3 )(1 + ) + (4 − 3 )

ℎ = −1 CHOW
6 2 − + 7 = (1 + )2 + (4 − 3 )(1 + ) + (4 − 3 )
6(−1)2 − (−1) + 7 = (1 + (−1))2 + (4 − 3(−1))(1 + (−1)) + (4 − 3(−1))

6 + 1 + 7 = (0)2 + (4 − 3(−1))(0) + (4 + 3)

14 = 7

= 2

4
ℎ = 3

6 2 − + 7 = (1 + )2 + (4 − 3 )(1 + ) + (4 − 3 )

42 4 42 44 4
6 (3) − (3) + 7 = [1 + (3)] + [4 − 3 (3)] [1 + (3)] + [4 − 3 (3)]

16 4 72 4
6 ( 9 ) − (3) + 7 = [3] + [0] [1 + (3)] + [0]

32 4 21 49
3 − (3) + 3 = 9

Page 18

153

PSPM 2 QS 025/1 Session 2017/2018
49 49
3 = 9
49 9
= ( 3 ) (49)
=3

= 3, = 2, = 0

6 2 − + 7 = (1 + )2 + (4 − 3 )(1 + ) + (4 − 3 )

7 = 3(1)2 + (4)(1) + 2(4)

7 = 3 + 4 + 8

4 = −4

= −1 CHOW

6 2 − + 7 3 1 2
(4 − 3 )(1 + )2 = (4 − 3 ) − (1 + ) + (1 + )2

6 2 − + 7 312
∫ (4 − 3 )(1 + )2 = ∫ (4 − 3 ) − (1 + ) + (1 + )2

312
= ∫ (4 − 3 ) − ∫ (1 + ) + ∫ (1 + )2

= − ∫ (4 −3 − ∫ (1 1 ) + ∫ 2(1 + )−2
− 3 ) +

= − ln(4 − 3 ) − ln(1 + ) − 1 2
+

Page 19

154

PSPM 2 QS 025/1 Session 2017/2018

1 6 2 − + 7 21
∫ (4 − 3 )(1 + )2 = [− ln(4 − 3 ) − ln(1 + ) − 1 + ]0

0

= [− ln(4 − 3(1)) − ln(1 + (1)) − 1 2 1] − [− ln(4 − 3(0)) − ln(1 + (0)) − 1 2
+ + (0)]

22
= [− ln(1) − ln(2) − 2] − [− ln(4) − ln(1) − 1]

= [0 − ln 2 − 1] − [− ln(4) − 0 − 2]

= 0 − ln 2 − 1 + ln 4 + 2

= − ln 2 − 1 + ln 22 + 2

= − ln 2 + 2ln 2 + 1

= 1 + ln 2

CHOW

Page 20

155

PSPM 2 QS 025/1 Session 2017/2018

9. a) Find the general solution of the differential equation = 2 −2 . Give your



answer in the form = ( ).

b) Find the particular solution of the differential equation + = √1 + 2, given
1+ 2

that y = 1 when x = 0.

solution

a) = 2 −2



= −2
2

∫ 1 = ∫ −2 CHOW
2

Integration by part

∫ −2 = ∫ −2

1 = = ∫ −2
− = − ∫ =1 −2

= −2

=

− 1 = ( ) −2 − ∫ −2

() −2
−2

1 −2 1
− = −2 + 2 ∫ −2

1 −2 1 −2
− = −2 + 2 [−2] +

1 −2 −2
− = − +
−2 4

1 −2 −2
= + −
2
4

Page 21

156

PSPM 2 QS 025/1 Session 2017/2018

1 2 −2 −2 4
= +−
4
44

1 2 −2 + −2 − 4
= 4

4
=

2 −2 + −2 − 4

b) + = √1 + 2
1+ 2

+ ( ). = ( )
+ 2)
+ (1 = √1 + 2

( ) = (1+ 2) ( ) = √1 + 2 CHOW

( ) = ∫ ( )

= ∫1+ 2

= 1 ∫1+2 2

2

= 1 ln(1+ 2)

2

1 =

= ln(1+ 2)2

1

= (1 + 2)2

( ). = ∫ ( ) ( )

11

(1 + 2)2. = ∫(1 + 2)2. √1 + 2

1

(1 + 2)2. = ∫ 1 + 2

Page 22

157

PSPM 2 QS 025/1 Session 2017/2018

(1 + 1 = + 3 +
3
2)2.

ℎ = 0, = 1

(1 + 1 (1) = 0 + 03 +
3
02)2.

= 1

∴ ℎ : √1 + 2 = + 3 + 1
3

CHOW

Page 23

158

PSPM 2 QS 025/1 Session 2017/2018

10. Given the curve 2 = and the line = −2 + 1.
a) Determine the points of intersection between the curve and the line.
b) Sketch the curve and the line on the same axes. Shade the region bounded by
the curve and the line. Label the points of intersection.
c) Find the area of the region .
d) Calculate the volume of the solid generated when the region is rotated 2
radian about the y-axis.

SOLUTION

a) 2 = and the line = −2 + 1.

2 = …………. (1)

= −2 + 1 …………. (2) CHOW

(1) (2)

= −2 2 + 1

2 2 + − 1 = 0

(2 − 1)( + 1) = 0

= 1 or = −1

2

= 2

= 1 = 1

4

∴ ℎ : (1, −1), 1 , 1
(4 2)

Page 24

159

PSPM 2 QS 025/1 Session 2017/2018
(b)

11 2 =
(4 , 2) = −2 + 1

(1, −1)

1 − 1 CHOW
( −2 )
( ) = 2 − ( 2)

∫

−1

1 − 1 + 2 2 )
2
=∫ (
−2
−1

1 1
−2
= 2 2 2 + − 1

∫

−1

1 2 3 2 1

2

= −2 [ 3 + 2 − ] −1

= 1 2 (21)3 + (21)2 − 1 − 2(−1)3 + (−1)2 − (−1)]}
−2 {[ 3 2 (2)] [3 2

1 1 1 1 −2 1
= −2 {[12 + 8 − 2] − [ 3 + 2 + 1]}

Page 25

160

PSPM 2 QS 025/1 Session 2017/2018

1 2 + 3 − 12 −4 3 6
= −2 {[ 24 ] − [ 6 + 6 + 6]}

1 −7 5
= −2 (24 − 6)

1 −7 20
= −2 (24 − 24)

1 −27
= −2 ( 24 )

1 −9
= −2 ( 8 )

= 9 2
16

(d) = 1 ( −1)2 − ( 2)2 CHOW

∫−21 −2

= 1 2 − 2 + 1 − 4
2 4

∫

−1

1 2 − 2 + 1 − 4 4
2
= ∫
4
−1

1
4
= 2 −4 4 + 2 − 2 + 1

∫

−1

1

−4 5 3 2 2 2
= 4 [ 5 + 3 − 2 + ]

−1

−4 5 3 1
[5 3
= + − 2 + 2
4
]

−1

−4 (12)5 + (21)3 − 12 + 1 − −4(−1)5 + (−1)3 − (−1)2 + (−1))
=4 5 3 (2) (2) (5 3

{( ) }

Page 26

161

PSPM 2 QS 025/1 Session 2017/2018

−1 1 1 1 4 1
= 4 {(40 + 24 − 4 + 2) − (5 − 3 − 1 − 1)}

−24 + 40 − 240 + 480 12 − 5 − 15 − 15
= 4 {( 960 ) − ( 15 )}

256 23
= 4 {(960) + (15)}

256 + 1472
= 4 ( 960 )

1782
= 4 ( 960 )

= 9 3
20

CHOW

Page 27

162

PAPER 2 CHOW
SEMESTER 2
2017/2018

163

PSPM 2 QS025/2 Session 2017/2018

1. A sample of positive integers is arranged in ascending order as follow:

3 + 2, 40, 4 , 2 , 59, 3 − 9

If the mean and median of the sample are 49 and 47 respectively, determine the values

of x and y. Hence, rewrite the sample in ascending order.

2. Let A and B be two events where ( ′) = 0.7, ( ) = 0.4 ( ∪ ) = 0.6.

Determine ( ∩ ) and then evaluate ( | ).

Hence, state with reason whether A and B are independent events.

3. The number of text messages received by Rosnaida during a fixed time interval is

distributed with a mean of 6 messages per hour.

a) Find the probability that Rosnaida will receive exactly 8 messages between

16:00 and 18:00 on a particular day.

b) It is known that Rosnaida has received at least 10 messages between 16:00 and

18:00 on a particular day, find the probability that she received 13 text

messages during that time interval. CHOW

4. A probability distribution for discrete random variable X is as shown in the table

below.

X -2 -1 0 1 2 3

P(X=x) p 0.1 0.3 q 0.2 p+q

Where p and q are constants. If E(X)=0.65, determine the values of p and q. Hence,

calculate the standard deviation of X.

5. The probability distribution function of a discrete random variable X is given as:

( ) =
17 , = 1,2,3

34 , = 4,5,6,7

{ 0 , ℎ

a) Calculate (2 ≤ ≤ 5)

b) Determine the value of Var(X).

Hence, calculate the standard deviation of = (√5 − 1).

6. The frequency distribution for 80 employees at a supermarket according to their daily

wage class is as shown below.

Page 2

164

PSPM 2 QS025/2 Session 2017/2018

Class Boundary of Daily Wage (RM) Frequency
15 – 20 4
20 – 25 12
25 – 30 20
30 – 35 32
35 – 40 8
40 – 45 4

a) Find the median and standard deviation of the sample.

b) Calculate and interpret the Pearson’s coefficient of skewness for the data.

c) Determine the daily wage k where 80% of the workers earn at most k ringgit

per day.

7. A box contains 12 cups of similar size and shape but in different colours. There are 5

blue cups, 4 red cups and 3 yellow cups. A rack can only take up 6 cups. In how many CHOW

ways can:

a) Any 6 cups be placed on the rack?

b) An equal number of coloured cups that could be placed on the rack?

c) An equal number of coloured cups that could be placed on the rack with cups of

the same colour being side by side?

d) An equal number of yellow and red cups be placed on the rack?

8. A survey was implemented on 400 students at a private university in order to collect

information on the popular choice of minor subjects (Language, Statistics and

Information Technology) by students of various major of study (Medicine, Engineering

and Economics). The following table describes the data collected from the survey.

Minor Major Total
Medicine Engineering Economic

Language 30 80 30 140

Statistics 10 30 10 50

Information Technology 60 120 30 210

Total 100 230 70 400

If a student from this group is selected at random, what is the probability that he:
a) Is either majoring in Medicine or doing a minor in Information Technology?
Page 3

165

PSPM 2 QS025/2 Session 2017/2018

b) Is a non-Medicine student who does a minor in Language? CHOW
c) Chooses a minor in Statistics knowing that he is an Economics student?
d) Is neither an Engineering student who does a minor Statistics nor is he an

Economics student who does a minor in Language?
9. In every delivery of cupcakes to a particular restaurant, 30% will be returned due to

not favoured by cupcakes lovers.
a) Suppose 20 of the cupcakes are randomly selected from a delivery. What is the
probability that at most 5 will be returned?
b) Suppose the restaurant will be holding an event which requires an order of 200
cupcakes from the same supplier.
i. Approximate the probability that between 56 and 62 of the cupcake will
be returned.
ii. If the probability of observing less that n number of cupcakes among
thos delivered which are returned is 0.992, use the normal
approximation to determine the value of n.

10. Continuous random variable X has a density probability function given by
, 0 ≤ ≤ 1


( ) = {3 (4 − ) , 1 ≤ ≤ 4

0 , ℎ
where a is a constant.

a) Find the value of a.
b) Find the ( ) and (2 − 3 ).
c) Evaluate ( − ( ) < ).
d) Estimate the median.

END OF QUESTION PAPER

Page 4

166

PSPM 2 QS025/2 Session 2017/2018

1. A sample of positive integers is arranged in ascending order as follow:
3 + 2, 40, 4 , 2 , 59, 3 − 9

If the mean and median of the sample are 49 and 47 respectively, determine the values
of x and y. Hence, rewrite the sample in ascending order.

SOLUTION

3 + 2, 40, 4 , 2 , 59, 3 − 9

Mean = 49

= 3 + 2 + 40 + 4 + 2 + 59 + 3 − 9

6

49 = 7 + 5 + 92

6

7 + 5 + 92 = 294

7 + 5 = 202 …………… (1)

Median = 47 CHOW

3 + 2, 40, 4 , 2 , 59, 3 − 9

4 + 2
2 = 47

4 + 2 = 94

2 + = 47

= 47 − 2 …………… (2)

(2) (1)

7 + 5(47 − 2 ) = 202

7 + 235 − 10 = 202

33 = 3

= 11

= 47 − 2(11) = 25

∴ = 11; = 25

3 + 2, 40, 4 , 2 , 59, 3 − 9

3(11) + 2, 40, 4(11), 2(25), 59, 3(25) − 9

35, 40, 44, 50, 59, 66

Page 5

167

PSPM 2 QS025/2 Session 2017/2018

2. Let A and B be two events where ( ′) = 0.7, ( ) = 0.4 ( ∪ ) = 0.6.
Determine ( ∩ ) and then evaluate ( | ).
Hence, state with reason whether A and B are independent events.

SOLUTION ( ∪ ) = 0.6
( ′) = 0.7, ( ) = 0.4

( ∪ ) = ( ) + ( ) − ( ∩ )

( ∩ ) = ( ∪ ) − ( ) − ( )

= 0.6 − (1 − 0.7) − 0.4

= 0.1

( | ) = ( ∩ )
( )

0.1 CHOW
= 0.4
= 0.25

( | ) ≠ ( ), ℎ .

Independent Events

( ∩ ) = ( ). ( )

( | ) = ( ∩ )
( )

( ∩ ) = ( | ). ( )

:

, ( | ) = ( )

Page 6

168

PSPM 2 QS025/2 Session 2017/2018

3. The number of text messages received by Rosnaida during a fixed time interval is
distributed with a mean of 6 messages per hour.
a) Find the probability that Rosnaida will receive exactly 8 messages between
16:00 and 18:00 on a particular day.
b) It is known that Rosnaida has received at least 10 messages between 16:00 and
18:00 on a particular day, find the probability that she received 13 text
messages during that time interval.

SOLUTION CHOW
= 6 ℎ
a) 16: 00 18: 00  2 hours  = 12 ℎ
~ (12)
( = 8) = ( ≥ 8) − ( ≥ 9)
= 0.9105 − 0.8450
= 0.0655

b) ( = 13| ≥ 10) = ( =13 ∩ ≥10)

( ≥10)

( = 13)
= ( ≥ 10)

( ≥ 13) − ( ≥ 14)
= ( ≥ 10)

0.4240 − 0.3185
= 0.7576
= 0.1393

( = 13 ∩ ≥ 10) = ( = 13)
= ሼ13ሽ, = ሼ10, 11, 12, 13, … ሽ

( ∩ ) = ሼ13ሽ

Page 7

169

PSPM 2 QS025/2 Session 2017/2018

4. A probability distribution for discrete random variable X is as shown in the table

below.

X -2 -1 0 1 2 3

P(X=x) p 0.1 0.3 q 0.2 p+q

Where p and q are constants. If E(X)=0.65, determine the values of p and q. Hence,
calculate the standard deviation of X.

SOLUTION
: ∑ ( = ) = 1

+ 0.1 + 0.3 + + 0.2 + + = 1

2 + 2 = 0.4 CHOW

+ = 0.2 …………….. (1)

Given that E(x)=0.65

( ) = ∑ ( = )

(−2)( ) + (−1)(0.1) + (0)(0.3) + (1)( ) + (2)(0.2) + (3)( + ) = 0.65

−2 − 0.1 + 0 + + 0.4 + 3 + 3 = 0.65

+ 4 = 0.35 …………….. (2)

(2) – (1)
3 = 0.15
= 0.05

(1):
= 0.2 −
= 0.2 − 0.05
= 0.15

Page 8

170

PSPM 2 QS025/2 Session 2017/2018

X -2 -1 012 3
P(X=x) 0.15 0.1 0.3 0.05 0.2 0.2

( ) = ( 2) − [ ( )]2

( 2) = ∑ 2 ( = )

( 2) = (−2)2(0.15) + (−1)2(0.1) + (0)2(0.3) + (1)2(0.05) + (2)2(0.2) + (3)2(0.2)
= 0.6 + 0.1 + 0 + 0.05 + 0.8 + 1.8
= 3.35

( ) = 3.35 − (0.65)2
= 2.9275

, = √ ( ) CHOW
= √2.9275
= 1.711

Page 9

171

PSPM 2 QS025/2 Session 2017/2018

5. The probability distribution function of a discrete random variable X is given as:

( ) =
17 , = 1,2,3

34 , = 4,5,6,7

{ 0 , ℎ

a) Calculate (2 ≤ < 5)

b) Determine the value of Var(X).

Hence, calculate the standard deviation of = (√5 − 1).

SOLUTION

( ) =
17 , = 1,2,3

34 , = 4,5,6,7

{ 0 , ℎ

a) (2 ≤ < 5) = ( = 2) + ( = 3) + ( = 4) CHOW

234
= 17 + 17 + 34

14
= 34

7
= 17
) ( ) = ( 2) − [ ( )]2

1234567
( ) = (1) (17) + (2) (17) + (3) (17) + (4) (34) + (5) (34) + (6) (34) + (7) (34)

1 4 9 16 25 36 49
= 17 + 17 + 17 + 34 + 34 + 34 + 34

154
= 34

77
= 17

( 2) = (1)2 1 + (2)2 2 + (3)2 3 + (4)2 4 + (5)2 5 + (6)2 6 + (7)2 7
(17) (17) (17) (34) (34) (34) (34)

1 8 27 64 125 216 343
= 17 + 17 + 17 + 34 + 34 + 34 + 34

820 410
= 34 = 17

Page 10

172

PSPM 2 QS025/2 Session 2017/2018

( ) = ( 2) − [ ( )]2 1. ( ) = 0
2. ( + ) = 2 ( )
410 77 2
= 17 − (17)

1041
=

289

When = (√5 − 1)
( ) = (√5 − 1)

2

= (√5) ( )
1041

= 5 ( 289 )
5205

=
289

, = √5205 CHOW
289

= 4.244

Page 11

173

PSPM 2 QS025/2 Session 2017/2018

6. The frequency distribution for 80 employees at a supermarket according to their daily
wage class is as shown below.

Class Boundary of Daily Wage (RM) Frequency
15 – 20 4
20 – 25 12
25 – 30 20
30 – 35 32
35 – 40 8
40 – 45 4

a) Find the median and standard deviation of the sample. CHOW
b) Calculate and interpret the Pearson’s coefficient of skewness for the data.
c) Determine the daily wage k where 80% of the workers earn at most k ringgit

per day.

SOLUTION Mid Point Frequency Cummulative
Class X (f) Frequency
(F) 1225
Boundary of 17.5 4 6075
Daily Wage 22.5 12 4 70 15125
27.5 20 16 270 33800
(RM) 32.5 32 36 550 11250
15 – 20 37.5 8 68 1040 7225
20 – 25 42.5 4 76 300 74700
25 – 30 80 170
30 – 35 2400
35 – 40 Total
40 – 45

(a) = + ( 2 − −1)



= 30; = 80; −1 = 36; = 32; = 5

Page 12

174

PSPM 2 QS025/2 Session 2017/2018
= 30 + (8203−236) 5
= 30.625

= √∑ 2 − (∑ )2
− 1

= √74700 − (2400)2
80

80 − 1

= 5.846

(b) Pearson’s coefficient of skewness, = 3( − ) CHOW


∑
, ̅ = ∑

2400
= 80

= 30

3( − )
=

3(30 − 30.625)
= 5.846

= −0.321

ℎ , ℎ ℎ ( )

(c) = 80

= + − −1
(100 )

Page 13

175

PSPM 2 QS025/2 Session 2017/2018

Class Mid Point Frequency Cummulative
Boundary of X (f) Frequency
Daily Wage (F)
17.5 4
(RM) 22.5 12 4
15 – 20 27.5 20 16
20 – 25 32.5 32 36
25 – 30 37.5 8 68
30 – 35 42.5 4 76
35 – 40 80
40 – 45

= + − −1 CHOW
(100 )

= 30; = 80; = 80; −1 = 36; = 32; = 5

(80)(80) − 36
)
80 = 30 + ( 100 5
32

= 34.375

∴ = 34.375

Page 14

176

PSPM 2 QS025/2 Session 2017/2018

7. A box contains 12 cups of similar size and shape but in different colours. There are 5
blue cups, 4 red cups and 3 yellow cups. A rack can only take up 6 cups. In how many
ways can:
a) Any 6 cups be placed on the rack?
b) An equal number of colored cups that could be placed on the rack?
c) An equal number of colored cups that could be placed on the rack with cups of
the same colour being side by side?
d) An equal number of yellow and red cups be placed on the rack?

SOLUTION
5B, 4R, 3Y: B, B, B, B, B, R, R, R, R, Y, Y, Y

(a)

Blue Red Yellow Permutation CHOW
510
501 6! =6
420 5! 1! 0! =6
411 = 15
402 6! = 30
330 5! 0! 1! = 15
321 = 20
312 6! = 60
303 4! 2! 0! = 60
240 = 20
231 6! = 15
222 4! 1! 1! = 60
213 = 90
141 6! = 60
4! 0! 2! = 30

6!
3! 3! 0!

6!
3! 2! 1!

6!
3! 1! 2!

6!
3! 0! 3!

6!
2! 4! 0!

6!
2! 3! 1!

6!
2! 2! 2!

6!
2! 1! 3!

6!
1! 4! 1!

Page 15

177

PSPM 2 QS025/2 Session 2017/2018

1 3 2 6! = 60
1! 3! 2!

1 2 3 6! = 60
1! 2! 3!

0 4 2 6! = 15
0! 4! 2!

0 3 3 6! = 20
0! 3! 3!

Total 642

ℎ = 642

(b) An equal number of colored cups that could be placed on the rack?

Blue Red Yellow Permutation
330
303 6! = 20 CHOW
222 3! 3! 0! = 20
033 = 90
6! = 20
3! 0! 3!

6!
2! 2! 2!

6!
0! 3! 3!

Total 150

ℎ = 150

Page 16

178

PSPM 2 QS025/2 Session 2017/2018

(c) An equal number of colored cups that could be placed on the rack with cups of the
same colour being side by side?

Blue Red Yellow Permutation

3 3 0 3! 3! 3! 3! =2
3! 3! 3! 3! 2!

2!

3 0 3 3! 3! 3! 3! =2
3! 3! 3! 3! 2!

2!

2 2 2 2! 2! 2! 2! 2! 2! =6 CHOW
2! 2! 2! 2! 2! 2! 3!

3!

0 3 3 3! 3! 3! 3! =2
3! 3! 3! 3! 2!

2!

Total 12

ℎ = 12

Page 17

179

PSPM 2 QS025/2 Session 2017/2018

(d) An equal number of yellow and red cups be placed on the rack?

Blue Red Yellow Permutation
411
222 6! = 30
033 4! 1! 1! = 90
= 20
6!
2! 2! 2!

6!
0! 3! 3!

Total 140

ℎ = 140

CHOW

Page 18

180

PSPM 2 QS025/2 Session 2017/2018

8. A survey was implemented on 400 students at a private university in order to collect

information on the popular choice of minor subjects (Language, Statistics and

Information Technology) by students of various major of study (Medicine, Engineering

and Economics). The following table describes the data collected from the survey.

Major Medicine Engineering Economic Total
(Me) (En) (Ec)
Minor

Language (L) 30 80 30 140

Statistics (S) 10 30 10 50

Information Technology (I) 60 120 30 210

Total 100 230 70 400

If a student from this group is selected at random, what is the probability that he: CHOW
a) Is either majoring in Medicine or doing a minor in Information Technology?
b) Is a non-Medicine student who does a minor in Language?
c) Chooses a minor in Statistics knowing that he is an Economics student?
d) Is neither an Engineering student who does a minor Statistics nor is an
Economics student who does a minor in Language?

SOLUTION

−
−
−
−
−
−

(a) ( ∪ ) = ( ) + ( ) − ( ∩ )

100 210 60
= 400 + 400 − 400

250
= 400

5
=8

Page 19

181

PSPM 2 QS025/2 Session 2017/2018

(b) ( ′ ∩ ) = ( ) − ( ∩ )
140 30

= 400 − 400
110

= 400
11

= 40

(c) ( | ) = ( ∩ )

( )

10

= 400
70

400

10 400
= 400 70

1 CHOW
=7

(d) [( ∩ )′ ∩ ( ∩ )′)] = [( ∩ ) ∪ ( ∩ )]′

= 1 − [( ∩ ) ∪ ( ∩ )]

= 1 − 30 + 30
(400 400)

17
= 20

Page 20

182

PSPM 2 QS025/2 Session 2017/2018

9. In every delivery of cupcakes to a particular restaurant, 30% will be returned due to
not favoured by cupcakes lovers.
a) Suppose 20 of the cupcakes are randomly selected from a delivery. What is the
probability that at most 5 will be returned?
b) Suppose the restaurant will be holding an event which requires an order of 200
cupcakes from the same supplier.
i. Approximate the probability that between 56 and 62 of the cupcake will
be returned.
ii. If the probability of observing less that n number of cupcakes among
those delivered which are returned is 0.992, use the normal
approximation to determine the value of n.

SOLUTION

a) = 0.3, = 20 CHOW
~ (20, 0.3)
( ≤ 5) = 1 − ( ≥ 6)
= 1 − 05836
= 0.4164

b) = 200; = 0.3

~ (200, 0.3)

= 60 > 5, = 140 > 5, ℎ

~ ( , ) ⟹ ~ [ , ]

~ (200, 0.3) ⟹ ~ [(200)(0.3), (200)(0.3)(0.7)]

~ (200, 0.3) ⟹ ~ (60,42)

i.

~ (200, 0.3) ⟹ ~ (60,42) ⟹ ~ (0,1)

(56 < < 62) = (56.5 ≤ ≤ 61.5) = ( 56.5 − 60 ≤ ≤ 61.5 − 60 )

√42 √42

= (−0.54 ≤ ≤ 0.23)

= 1 − ( ≥ 0.54) − ( ≥ 0.23)

= 1 − 0.2946 − 0.4090

= 0.2964

Page 21

183

PSPM 2 QS025/2 Session 2017/2018

ii. ( < ) = 0.992

( < − 0.5) = 0.992

( < ( − 0.5) − 60 ) = 0.992

√42

( < − 60.5 ) = 0.992

√42

− 60.5
=

√42

( < ) = 0.992

( ≥ ) = 1 − 0.992

( ≥ ) = 0.008

= 2.41

− 60.5
2.41 =

√42

= 76.1 CHOW

= 76

Page 22

184

PSPM 2 QS025/2 Session 2017/2018

10. Continous random variable X has a probability density function given by

, 0 ≤ ≤ 1
{3
( ) = (4 − ) , 1 ≤ ≤ 4

0 , ℎ

where a is a constant.

a) Find the value of a.

b) Find the ( ) and (2 − 3 ).

c) Evaluate ( − ( ) < ).

d) Estimate the median.

SOLUTION

, 0 ≤ ≤ 1
{3
( ) = (4 − ) , 1 ≤ ≤ 4

0 , ℎ CHOW

a) For probability density function  ∫−∞∞ ( ) = 1

∫−∞∞ ( ) = 1

∫−0∞ 0 + ∫01 + ∫14 (4 − ) + ∫4∞ 0 = 1
3

[ 2]1 + [4 − 2]4 = 1
20 3 21

[(12) − (02)] + [(4(4) − 42) − (4(1) − 12)] = 1
2 23 2 2

1 7
2 + 3 [8 − 2] = 1

1 9
2 + 3 [2] = 1

13
2 + 2 = 1

2 = 1

1
= 2

Page 23

185

PSPM 2 QS025/2 Session 2017/2018
b) ( ) = ∫−∞∞ . ( )

( ) = 1 1
6 2 , 0 ≤ ≤ 1

(4 − ) , 1 ≤ ≤ 4

{ 0 , ℎ

0 11 41 ∞

( ) = ∫ (0) + ∫ (2 ) + ∫ [6 (4 − )] + ∫ (0)

−∞ 0 1 4

= 1 1 + 1 4 − 2
2 6
∫ 2 ∫ 4

0 1

1 3 1 1 4 2 3 4
= [ ] + [ − ]
2 3 6 2 3
0 1

1 13 03 1 4(4)2 (4)3 4(1)2 (1)3 CHOW
= 2 [( 3 ) − ( 3 )] + 6 [( 2 − 3 ) − ( 2 − 3 )]

11 1 64 1
= 2 (3) + 6 [(32 − 3 ) − (2 − 3)]

1 1 32 5
= 6 + 6 [ 3 − 3]

1 1 27
= 6+6[ 3 ]

10
=6

5
=3

( 2) = 0 + 1 1 + 4 1 (4 − )] + ∞
(2 ) [6
∫ 2(0) ∫ 2 ∫ 2 ∫ 2(0)

−∞ 0 1 4

= 1 1 + 1 4 − 3
2 6
∫ 3 ∫ 4 2

0 1

1 4 1 1 4 3 4 4
= 2[4] +6[ 3 − 4]

01

1 14 04 1 4(4)3 44 4(1)3 14
= 2 [( 4 ) − ( 4 )] + 6 [( 3 − 4 ) − ( 3 − 4 )]

1 1 256 256 4 1
= 8 + 6 [( 3 − 4 ) − (3 − 4)]

Page 24

186

PSPM 2 QS025/2 Session 2017/2018

1 1 1024 − 768 16 − 3
= 8 + 6 [( 12 ) − ( 12 )]

1 1 256 13
= 8 + 6 [( 12 ) − (12)]

1 1 243
= 8 + 6 ( 12 )

1 81
= 8 + 24

84
= 24

7
=2

( ) = ( 2) − [ ( )]2 ( + ) = 2 ( ) CHOW
7 52

= 2 − (3)
7 25

=2− 9
63 − 50

= 18
13

= 18

(2 − 3 ) = (−3)2 ( )
13

= 9 (18)
13

=2

c) ( − ( ) < ) = ( − 5 < 1)

32

15
= ( < 2 + 3)

13
= ( < 6 )

13
6

= ∫ ( )

−∞

Page 25

187

PSPM 2 QS025/2 Session 2017/2018

0 11 13

∫ 0 + + 16 4 −
= ∫ ∫
−∞ 2 6
0 1

1 2 1 1 13

2 6
= [ ] + [4 − ]
2 2 6 2
0 1

= 1 12 − 02 + 1 [(4 13 − (1623)2) − (4(1) − 12
2 [( 2 ) ( 2 )] 6 (6) 2 )]

1 1 52 169 1
= 4 + 6 [( 6 − 72 ) − (4 − 2)]

1 1 624 − 169 7
= 4 + 6 [( 72 ) − (2)]

1 1 455 7
= 4 + 6 [ 72 − 2]

1 1 455 − 252 CHOW
= 4 + 6 [ 72 ]

1 1 203
= 4 + 6 [ 72 ]

1 203
= 4 + 432

108 + 203
= 432

311
= 432

d) Median, m  ( ) = 0.5
( ≤ ) = 0.5



∫ ( ) = 0.5

−∞

0 + 1 1 + 1 − ) = 0.5
2 6
∫0 ∫ ∫ (4

−∞ 0 1

1 2 1 1 2
[ ] + [4 − ] = 0.5
2 2 6 2
0 1

11 2 12
4 + 6 [(4 − 2 ) − (4(1) − 2 )] = 0.5

Page 26

188

PSPM 2 QS025/2 Session 2017/2018

11 2 1
4 + 6 [(4 − 2 ) − (4 − 2)] = 0.5

1 1 8 − 2 7
4 + 6 [( 2 ) − (2)] = 0.5

1 8 − 2 − 7
4 + 12 = 0.5

8 − 2 − 7 1
12 = 0.5 − 4

8 − 2 − 7 1
12 = 4

8 − 2 − 7 = 3

2 − 8 + 10 = 0

− ± √ 2 − 4 CHOW
= 2

−(−8) ± √(−8)2 − 4(1)(10)
= 2(1)

= 8 ± √64 − 40
2

= 8 ± √24
2

= 6.45 = 1.55

1 ≤ ≤ 4, ℎ = 1.55

Page 27

189

PAPER 1 CHOW
SEMESTER 2
2018/2019

190

PSPM 2 SM025/1 Session 2018/2019

1. The size of a population of insects is increasing at a rate proportional to the number of insects, N, in

time t days which satisfies the equation = , where > 0. Given that the number of insects at


the beginning of an observation is 0 and is double in 2 days, find the number of insects after 5 days.

2. Sketch and shade the region bounded by the curve = 4 − , the straight line = 4 − , −

= 3. Hence, find the area of the shaded region by using trapezoidal rule with five

ordinates. Give your answer correct to four decimal places.

3. Given a circle 2 + 2 + + 6 + 8 = 0, where is a positive constant.

a) Determine the value of and the centre of the circle if the radius is √13 unit.
2

b) Find the points of intersection of the circle with straight line − + 2 = 0. Hence, obtain

one of the tangent equation at the point of intersection.

4. The continuous random variable X has the cumulative distribution function

0, ≤ 0
0 ≤ ≤ 3
( ) = 1 (2 2 − 3
9 3) ≥ 3
CHOW
{ 1,

a) Find the median

b) Determine the probability density function of X.

c) Hence, find the mode and the mean.

d) State the skewness of the distribution with a reason.

5. The amount of cement packed by a machine is normally distributed with mean 39.3kg and standard

deviation 0.9kg. A bag of cement is randomly selected.

a) Find the probability that the bag weighs more than 40kg.

b) If the probability of the bag weighs not more than m kg is 0.95, determine the value of m.

c) A total of 5 bags of cement are chosen at random. Find the probability that at least 4 bags

weigh more than 40kg.

END OF QUESTION PAPER

Page 2

191

PSPM 2 SM025/1 Session 2018/2019

1. The size of a population of insects is increasing at a rate proportional to the number of insects, N, in

time t days which satisfies the equation = , where > 0. Given that the number of insects at


the beginning of an observation is 0 and is double in 2 days, find the number of insects after 5 days.

SOLUTION CHOW

=

=

∫ = ∫
ln = +
= +
=

Given that when

= 0; =
= (0)
=

= 2; = 2 , =

2 = (2)

2 = 2


2 = 2

2 = ln 2

ln 2
= 2 = 0.3466
= 0.3466

ℎ = 5:
= 0.3466(5)
= 5.66

Page 3

192

PSPM 2 SM025/1 Session 2018/2019

2. Sketch and shade the region bounded by the curve = 4 − , the straight line = 4 − , −
= 3. Hence, find the area of the shaded region by using trapezoidal rule with five
ordinates. Give your answer correct to four decimal places.

SOLUTION
= 3

4

= 4 −


= 4 −

3 CHOW

= ∫ (4 − ) − (4 − )

0

= 4

3−0
ℎ = 4 = 0.75

= 4 − − 4 −

0 = 0 0
1 = 0.75
2 = 1.5 1.36053
3 = 2.25
4 = 3.0 1.60748

1.32840

0.80085

Total 0.80085 4.29641

= ℎ [( 0 + 4) + 2( 1 + 2 + 3)]
2

= 0.75 [0.80085 + 2(4.29641)]
2

= 3.5226 2

Page 4

193

PSPM 2 SM025/1 Session 2018/2019

3. Given a circle 2 + 2 + + 6 + 8 = 0, where is a positive constant.

a) Determine the value of and the centre of the circle if the radius is √13 unit.
2

b) Find the points of intersection of the circle with straight line − + 2 = 0. Hence, obtain

one of the tangent equations at the point of intersection.

SOLUTION

(3a)
2 + 2 + 2 + 2 + = 0
2 + 2 + + 6 + 8 = 0 ℎ

2 = 2 = 6 = 8 = √ 2 + 2 −
, = (− , − )
= = 3

2

= √ 2 + 2 −

√13 = √32 + 2 − 8 CHOW
2 (2)

√13 = √1 + 2
2 4

13 2
4 =1+ 4
2 13
4 = 4 −1
2 9
4 =4
2 = 9

= 3 ( > 0)

= (− , − ) = (− 3 , −3)
2

Page 5

194

PSPM 2 SM025/1 Session 2018/2019

(3b)

2 + 2 + 3 + 6 + 8 = 0 …………………… (1)



− + 2 = 0 ………………… (2)
= − 2

Substitute (2) into (1)

2 + ( − 2)2 + 3 + 6( − 2) + 8 = 0

2 + 2 − 4 + 4 + 3 + 6 − 12 + 8 = 0

2 2 + 5 = 0

(2 + 5) = 0 CHOW

= 0 5
= −2 = − 2

9
= − 2

ℎ ℎ (0, −2) 59
(− 2 , − 2).

( , − ) + + + + =
1 + 1 + ( + 1) + ( + 1) + = 0

1 = 0; 1 = −2; = 3; = 3, = 8

2

3
(0) + (−2) + 2 ( + 0) + 3( − 2) + 8 = 0

3
−2 + 2 + 3 − 6 + 8 = 0

3
+ 2 + 2 = 0

2 + 3 + 4 = 0



Page 6

195

PSPM 2 SM025/1 Session 2018/2019

(− , − + + + + =
)

1 = − ; 1 = − ; = 3 ; = 3, = 8
2

5 93 5 9
(− 2) + (− 2) + 2 ( − 2) + 3 ( − 2) + 8 = 0

5 9 3 15 27
− 2 − 2 + 2 − 4 + 3 − 2 + 8 = 0

−10 − 18 + 6 − 15 + 12 − 54 + 32 = 0

−4 − 6 − 37 = 0

4 + 6 + 37 = 0

CHOW

Page 7

196

PSPM 2 SM025/1 Session 2018/2019

4. The continuous random variable X has the cumulative distribution function

0, ≤ 0
0 ≤ ≤ 3
( ) = 1 (2 2 − 3
9 3) ≥ 3

{ 1,

a) Find the median

b) Determine the probability density function of X.

c) Hence, find the mode and the mean.

d) State the skewness of the distribution with a reason.

SOLUTION
(4a)

0, ≤ 0
0 ≤ ≤ 3
( ) = 1 (2 2 − 3
9 3) ≥ 3

{ 1, CHOW

:

1
( ) = 2

1 (2 2 − 3 = 1
9 3) 2

2 2 − 3 = 9
3 2

12 2 − 2 3 = 27

2 3 − 12 2 + 27 = 0



= 5.564 = 1.7907 = −1.3548

∴ = 1.7907

Page 8

197

PSPM 2 SM025/1 Session 2018/2019
(4b)

≤ 0
0 ≤ ≤ 3 ( ) = (0) = 0

≥ 3 ( ) = 1 (2 2 − 3
9 3)

= 1 (4 − 2)
9

= 4 − 1 2
9 9

( ) = (1) = 0


( ) = { 4 − 1 2 , 0 ≤ ≤ 3
9 9

0 , ℎ CHOW

(4c)

( ) = 4 − 1 2
9 9

14
= − 9 , = 9 ; = 0

:


= − 2

4

= − 9 19)
(−
2

=2

∴ : = 2

∞

= ( ) = ∫ ( )

−∞

= ( ) = 0 (0) + 3 4 − 1 2) + ∞ (0)
(9 9
∫ ∫ ∫

−∞ 0 3

Page 9

198

PSPM 2 SM025/1 Session 2018/2019

= 3 4 − 1 2)
(9 9
∫

0

= 3 4 2 − 1 3
9 9
∫

0

= 4 3 − 1 4 3
[27 36
]

0

= 4 33 − 1 34) − (0)
(27 36

9
=4−4

7
=4

(4d) CHOW
< , ℎ ℎ ℎ .

< , ℎ ℎ ℎ .

Note:

Skewed to Skewed to
the right the left

Page 10

199