PSPM 2 Q&A

PSPM 2 SM025/1 Session 2018/2019

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.

SOLUTION

(5a)

= 39.3 = 0.9

~ (39.3, 0.92)

( > 40) = ( > 40 − 39.3
0.9 )

= ( > 0.78) CHOW

= 0.2177

(5b)

( < ) = 0.95
− 39.3

( < 0.9 ) = 0.95
− 39.3

( ≥ 0.9 ) = 0.05
:

( ≥ 1.65) = 0.05
− 39.3

0.9 = 1.65
= 40.785

(5c)
~ (5, 0.2177)
( ≥ 4) = ( = 4) + ( = 5)
= 5 4(0.2177)4(0.7826)1 + 5 5(0.2177)5(0.7826)0
= 0.00928

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PAPER 2 CHOW
SEMESTER 2
2018/2019

201

PSPM 2 SM025/2 Session 2018/2019

1. a) Evaluate ∫ (1 − )8 by using a suitable substitution.

b) Find the area of the region bounded by the curve = cos and − between

= 0 and = . Give your solution in term of .

2

2. Solve + 2( + 1) 2 = 0, given that = 1 when = 1. Express in terms of .



3. Use Newton-Raphson method to solve the equation − − 2 = 0 correct to four

decimal places by taking 1 = 1 as the first approximation.
4. An ellipse 2 + 2 + + + 1 = 0 passes through points (0,1), (1, −1) (2, 1).

a) Find the equation of the ellipse in the standard form. Hence, state the centre and

vertices of the ellipse.

b) Find the foci of th ellipse.

c) Sketch the graph of the ellipse.

5. The line 1 and 2 passes through the point (2, 4, −3) and (8, −5, 9)in the direction CHOW
of 2 − 3 + 4 − 2 + 3 , respectively.

a) State the equations for lines 1 and 2 in the vector form. Hence, calculate the
acute angle between the lines 1 and 2.

b) Find the equation of plane containing the line 1 and the point (7, −3, 5) in the
Cartesian form.

c) Determine whether the line 2 is parallel to the plane + 5 + 3 = 5.
6. TABLE 1 shows the frequency distribution of the diameter of 100 pebbles which are

measured to the nearest millimeter(mm).

TABLE 1

Diameter (mm) Frequency

10 – 14 22

15 – 19 20

20 – 24 25

25 – 29 15

30 - 34 18

Calculate the
a) mean.
b) mode.
c) median.

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PSPM 2 SM025/2 Session 2018/2019

7. A committee of 5 is to be selected from a group of 7 men and 6 women. How many

different committees could be formed if

a) there is no woman in the committee?

b) a particular man must be in the committee and the remaining has equal number

of men and women?

c) at least 3 men are in the committee?

8. The probabilities of events X and Y are given as ( ) = 3, ( ′| ) = 31 and

5 45

( ∩ ) = 2 .

25

a) Show that ( ) = 9

35

b) Find ( ∪ ′).

9. A discrete random variable X has the probability distribution function

+ 1 = 2, 3,4
16 ,
( ) = { , = 6, 8 CHOW
ℎ
0,

a) Show that = 1
56

b) Hence, calculate (3 ≤ < 8).

c) Determine the values of ( ) and ( ). Thus, evaluate (√3 − 1)

10. The number of batteries sold at a service center on any particular day follows a Poisson distribution

with mean .

a) If the probability of selling exactly 4 batteries divided by the probability of selling exactly 2
batteries is 21225, show that = 15.

b) On any particular day, calculate the probability that the service center sells between 5 and

14 batteries.

c) Given that the probability of selling less than batteries on any particular day is 0.917, find

the value of .

d) Find the probability that exactly 40 batteries are sold in 2 working days. Give your answer in

four decimal places.

END OF QUESTION PAPER

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PSPM 2 SM025/2 Session 2018/2019

1. a) Evaluate ∫ (1 − )8 by using a suitable substitution.
b) Find the area of the region bounded by the curve = cos and − between
= 0 and = . Give your solution in term of .

2

SOLUTION = 1 −
(a) ∫ (1 − )8

= 1 − 

= −1
= −

∫ (1 − )8 = ∫(1 − ) 8(− )

= − ∫(1 − ) 8

= − ∫ 8 − 9 CHOW

9 10
= − [ 9 − 10 ] +

(1 − )9 (1 − )10
= − [ 9 − 10 ] +

(1 − )10 (1 − )9
= 10 − 9 +

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

= cos CHOW
∫ = ∫ cos
= ∫2 cos
= sin
0

∫ cos

=

= 1
=

∫ = − ∫

= ( )(sin ) − ∫(sin ) ( )

= sin − (− cos )
= sin + cos



= ∫2 cos

0



= [ sin + cos ]02

= sin + cos − [0 sin 0 + cos(0)]
[2 (2) (2)]


= [2 + 0] − [0 + 1]

= − 1 2
2

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PSPM 2 SM025/2 Session 2018/2019

2. Solve + 2( + 1) 2 = 0, given that = 1 when = 1. Express in terms of .



SOLUTION

+ 2( + 1) 2 = 0


= −2( + 1) 2


−2 = −2( + 1)

∫ −2 = −2 ∫( + 1)

−1 2
−1 = −2 ( 2 + ) +

− 1 = − 2 − 2 +

CHOW
ℎ = 1, = 1

− 1 = −12 − 2(1) +
1

−1 = −1 − 2 +

= 2

− 1 = − 2 − 2 + 2


−1 = (− 2 − 2 + 2)
1

= 2 + 2 − 2

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PSPM 2 SM025/2 Session 2018/2019

3. Use Newton-Raphson method to solve the equation − − 2 = 0 correct to four
decimal places by taking 1 = 1 as the first approximation.

SOLUTION
− − 2 = 0
( ) = − − 2
′( ) = − 1

( )
+1 = − ′( )

− − 2
+1 = − − 1

1 = 1

1 − 1 − 2 CHOW
2 = 1 − 1 − 1 = 1.16395

3 = 1.16395 − 1.16395 − 1.16395 − 2 = 1.14642

1.163951 − 1

4 = 1.14642 − 1.14642 − 1.14642 − 2 = 1.14619

1.14642 − 1

4 = 1.14619 − 1.14619 − 1.14619 − 2 = 1.14619

1.14619 − 1

∴ = 1.1462

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4. An ellipse 2 + 2 + + + 1 = 0 passes through points (0,1), (1, −1) (2, 1).
a) Find the equation of the ellipse in the standard form. Hence, state the centre and
vertices of the ellipse.
b) Find the foci of th ellipse.
c) Sketch the graph of the ellipse.

SOLUTION
2 + 2 + + + 1 = 0
(0,1)

(02) + 12 + 0 + (1) + 1 = 0
= −2

(1, −1) = −2 CHOW
(12) + (−1)2 + (1) + (−2)(−1) + 1 = 0

+ 1 + + 2 + 1 = 0

+ = −4 ………………. (1)

(2,1) = −2
(2)2 + (1)2 + (2) + (−2)(1) + 1 = 0

4 + 1 + 2 − 2 + 1 = 0

4 + 2 = 0 ……………….. (2)

(1) 2

2 + 2 = −8 ……………….. (3)

(2) − (3)

2 = 8

= 4

= −8

ℎ :
4 2 + 2 − 8 − 2 + 1 = 0
4 2 − 8 + 2 − 2 = −1
4( 2 − 2 ) + ( 2 − 2 ) = −1
4( 2 − 2 + 12 − 12) + ( 2 − 2 + 12 − 12) = −1

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PSPM 2 SM025/2 Session 2018/2019

4[( − 1)2 − 1] + [( − 1)2 − 1] = −1
4( − 1)2 − 4 + ( − 1)2 − 1 = −1 ( − ℎ)2 ( − )2
4( − 1)2 + ( − 1)2 = 4 2 + 2 = 1
4( − 1)2 ( − 1)2 4

4 + 4 =4
( − 1)2 ( − 1)2

12 + 22 = 1

= 1, = 2, ℎ = 1, = 1 CHOW
: (ℎ, ) = (1,1)
: (ℎ, ± ) ⇒ 1(1,1 + 2), 2(1,1 − 2)

1(1,3), 2(1, −1)

(4b)
= √22 − 12 = √3
: (ℎ, ± ) = (1,1 ± √3)

(4c) 1(1,3)


1(1,1 + √3)

(1,1)



2(, 1,1 − √3)

2(1, −1)

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PSPM 2 SM025/2 Session 2018/2019

5. The line 1 and 2 passes through the point (2, 4, −3) and (8, −5, 9)in the direction
of 2 − 3 + 4 − 2 + 3 , respectively.
a) State the equations for lines 1 and 2 in the vector form. Hence, calculate the
acute angle between the lines 1 and 2.
b) Find the equation of plane containing the line 1 and the point (7, −3, 5) in the
Cartesian form.
c) Determine whether the line 2 is parallel to the plane + 5 + 3 = 5.

SOLUTION

(5a)

1: = 2 + 4 − 3 ; = 2 − 3 + 4
2: = 8 − 5 + 9 ; = − 2 + 3
= +

CHOW
1: = (2 + 4 − 3 ) + 1( 2 − 3 + 4 )
2: = (8 − 5 + 9 ) + 2( − 2 + 3 )

( )

= | .
|| |

(2 − 3 + 4 ). ( − 2 + 3 )
= |2 − 3 + 4 || − 2 + 3 |

(2)(1) + (−3)(−2) + (4)(3)
=

√(2)2 + (−3)2 + (4)2√(1)2 + (−2)2 + (3)2

20
=

√29√14
= 0.9926

= −1(0.9926)

= 6.98°

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PSPM 2 SM025/2 Session 2018/2019
(5b)

1
(7, −3,5)

(2, −3,4)
(2,4, −3)

CHOW
. = .

= 7 − 3 + 5
= ⃗ ⃗⃗ ⃗ ⃗⃗

= 2 − 3 + 4
⃗ ⃗⃗ ⃗⃗ ⃗ = ⃗⃗⃗ ⃗⃗ ⃗ − ⃗ ⃗⃗ ⃗⃗ ⃗

= (7 − 3 + 5 ) − (2 + 4 − 3 )
= ( − + 8 )

= | − |
−
= (−24 + 28) − (16 − 20) + (−14 + 15)
= 4 + 4 +


. = .
( + + ). (4 + 4 + ) = (7 − 3 + 5 ). (4 + 4 + )
4 + 4 + = 28 − 12 + 5
4 + 4 + = 21

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PSPM 2 SM025/2 Session 2018/2019
(5c)

3: + 5 + 3
=5

2


ℎ , , ℎ CHOW
= − 2 + 3
= + 5 + 3
. = ( − 2 + 3 ). ( + 5 + 3 )

= − +
=

. = , + + = .

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PSPM 2 SM025/2 Session 2018/2019

6. TABLE 1 shows the frequency distribution of the diameter of 100 pebbles which are

measured to the nearest milliType equation here.meter(mm).

TABLE 1

Diameter (mm) Frequency

10 – 14 22

15 – 19 20

20 – 24 25

25 – 29 15

30 - 34 18

Calculate the CHOW
a) mean.
b) mode.
c) median.

SOLUTION Class Boundary Mid Point Frequency Cummulative
Diameter (x) (f) Frequency
(mm) 9.5 – 14.5 (F)
14.5 – 19.5 12 22 22
10 – 14 19.5 – 24.5 17 20 42
15 – 19 24.5 – 29.5 22 25 67
20 – 24 30.5 – 34.5 27 15 82
25 – 29 32 18 100
30 - 34 100
∑

(6a)

∑
= ∑

= (12)(22) + (17)(20) + (22)(25) + (27)(15) + (32)(18)

100

2135
= 100

= 21.35

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PSPM 2 SM025/2 Session 2018/2019

(6b)

= + [ 1 1 2]
+

5
= 19.5 + [5 + 10] 5

= 21.17

(6c)

= + − −1
[2 ]

= 19.5 + [12002−5 42 5
]

= 21.1

CHOW

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PSPM 2 SM025/2 Session 2018/2019

7. A committee of 5 is to be selected from a group of 7 men and 6 women. How many
different committees could be formed if
a) there is no woman in the committee?
b) a particular man must be in the committee and the remaining has equal number
of men and women?
c) at least 3 men are in the committee?

SOLUTION
(7a)

ℎ ℎ .
7 5 6 0 =21

(7b) CHOW
1 1 6 2 6 2 = 225

(7c)

3 ℎ .

Men Women Total

3 2 7 3 6 2 = 525

41 7 4 6 1 =210

50 7 5 6 0 =21

Total 756

The number of different committee could be formed = 756

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PSPM 2 SM025/2 Session 2018/2019

8. The probabilities of events X and Y are given as ( ) = 3, ( ′| ) = 31 and

5 45

( ∩ ) = 2 .

25

a) Show that ( ) = 9

35

b) Find ( ∪ ′).

SOLUTION

(8a)

( ) = 3 ( ∩ ) = 2
5 25

( ′| ) = 31
45

( ′ ∩ ) 31 CHOW
( ) = 45

( ) − ( ∩ ) 31
= 45
( )

( ) − 2 31
25 45
=
( )

2 31
45
1 − 25 =
( )

2 31
45
25 = 1 −
( )

2 14
45
25 =
( )

( ) = 2 45
25 14

9
( ) = 35

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PSPM 2 SM025/2 Session 2018/2019

(8b)
( ∪ ′) = ( ) + ( ′) − ( ∩ ′)

( ′) = 1 − ( ) = 1 − 9 = 26
35 35

( ∩ ′) = ( ) − ( ∩ ) = 3 − 2 = 13
5 25 25

( ∪ ′) = 3 + 26 − 13
5 35 25

144
= 175

CHOW

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PSPM 2 SM025/2 Session 2018/2019

9. A discrete random variable X has the probability distribution function

+ 1
16 , = 2, 3,4
( ) = { ,
= 6, 8
0, ℎ

a) Show that = 1
56

b) Hence, calculate (3 ≤ < 8).

c) Determine the values of ( ) and ( ). Thus, evaluate (√3 − 1)

SOLUTION

+ 1 = 2, 3,4
16 ,
( ) = { , = 6, 8
ℎ
0, CHOW

2 3 4 6 8

( = ) 3 4 5 6 8

16 16 16

(9a)

∑ ( = ) = 1

345
16 + 16 + 16 + 6 + 8 = 1
12
16 + 14 = 1

1
14 = 4

1
= 56

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PSPM 2 SM025/2 Session 2018/2019

(9b)

2 3 4 6 8

( = ) 3 4 5 3 1
7
16 16 16 28

(3 ≤ < 8) = ( = 3) + ( = 4) + ( = 6)

453
= 16 + 16 + 28

75
= 112

(9c)

( ) = ∑ ( = )

( ) = (2) 3 + (3) 4 + (4) 5 + (6) 3 + (8) 1
(16) (16) (16) (28) (7)

233 CHOW
= 56

( 2) = ∑ 2 ( = )

( 2) = (2)2 3 + (3)2 4 + (4)2 5 + (6)2 3 + (8)2 1
(16) (16) (16) (28) (7)

= 21

( ) = ( 2) − [ ( )]2
233 2

= 21 − ( 56 )
= 3.688

( + ) = 2 ( )

2

(√3 − 1) = √3 3.688
= 11.065

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PSPM 2 SM025/2 Session 2018/2019

10. The number of batteries sold at a service center on any particular day follows a Poisson distribution
with mean .
a) If the probability of selling exactly 4 batteries divided by the probability of selling exactly 2
batteries is 21225, show that = 15.
b) On any particular day, calculate the probability that the service center sells between 5 and
14 batteries.
c) Given that the probability of selling less than batteries on any particular day is 0.917, find
the value of .
d) Find the probability that exactly 40 batteries are sold in 2 working days. Give your answer in
four decimal places.

SOLUTION CHOW
(10a)

For Poisson Distribution
−

( = ) = !

( = 4) 225
( = 2) = 12

− 4 225
12
4! =
− 2

2!

− 4 2! 225
( 4! ) ( − 2) = 12

2 225
12 = 12
2 = 225

= 15

(10b)
~ (15)
(5 < < 14) = ( ≥ 6) − ( ≥ 14)
= 0.9975 − 0.6368
= 0.3604

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PSPM 2 SM025/2 Session 2018/2019

(10c)

( < ) = 0.917
1 − ( ≥ ) = 0.917
( ≥ ) = 0.0830

From statistical table:
= 21

(10d)

~ (30)
−30(30)40

( = 40) = 40!
= 0.0139

CHOW

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PAPER 1 CHOW
SEMESTER 2
2019/2020

COVID 19

222

PAPER 2 CHOW
SEMESTER 2
2019/2020

COVID 19

223