PSPM 2 Q&A

PSPM I QS 025/1 Session 2015/2016

6. (a) Given 1( ) = 2x and 2( ) = − ln .
i. Without using curve sketching, show that = 1( ) and = 2( ) intersect on
the interval of [0.1, 1].
ii. Use Newton-Raphson’s method to estimate the intersection point of = 1( )
and = 2( ), with the initial value 1 = 1. Iterate until | ( )| < 0.005. Give
your answer correct to three decimal places.

b) By using the tropezoidal rule, find the approximate valur for ∫01 √ + 1 dx when n = 4,
correct to four decimal places.

SOLUTION

(ai)

1( ) = 2x , 2( ) = − ln CHOW

( ) = 1( ) − 2( ) Intermediate Value Theorem
( ) = 2x − (− ln )
http://mathinsight.org/intermediate_value_theorem_location_roots_refresher

( ) = 2x + ln

(0.1) = 2(0.1) + ln(0.1) = −2.013 < 0

(1) = 2(1) + ln 1 = 2 > 0

Since f(0.1)<0 and f(1)>0, by using intermediate value theorem, there are at least one root
in the interval [0.1, 1].

(aii) Newton-Raphson’s method

( ) = 2x + ln

′( ) = 1
2 +

+1 = − ( )
′( )

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+1 = − 2 + ln
2 +
1


1 = 1

2 = 1 − 2(1) + ln(1) = 0.3333

2 + 1
(1)

3 = 0.3333 − 2(0.3333) + ln(0.3333) = 0.4197

2 + 1
(0.3333)

4 = 0.4197 − 2(0.4197) + ln(0.4197) = 0.4263

2 + 1
(0.4197)

5 = 0.4263 − 2(0.4263) + ln(0.4263) = 0.4263

2 + 1 CHOW
(0.4263)

∴ = 0.426

ℎ = 0.426

1( ) = 2x
1(0.426) = 2(0.426) = 0.852
∴ (0.426, 0.852) ℎ 1( ) 2( )

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PSPM I QS 025/1 Session 2015/2016
(b) Trapezoidal Rule
Trapezoidal Rule
∫01 √ + 1 dx when n = 4
1−0 ℎ [ 0 2( 1 −1)]
∫ ( ) ≈ + + + 2 + ⋯ +
ℎ = 4 = 0.25 2


−
ℎ =

( ) = √ +

0.00 0.00000

0.25 0.27951

0.50 0.61237 CHOW
0.75 0.99216
1.00
1.41421

Total ( + ) = . ( + + ⋯ + − ) = .

By trapezoidal rule:

1ℎ [( 0 ) 2( 1 + −1)]
∫ √ + 1 dx = + + + 2 ⋯ +
2
0

= . [( . ) + 2( . )]
2

= 0.64779

= 0.6478

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PSPM I QS 025/1 Session 2015/2016

7. (a) If = 3 − + 2 and = 2 + 2 − , show that

| x |2 = | |2| |2 − ( . )2

(b) Given a triangle ABC with ⃗ ⃗ ⃗⃗ ⃗ = 2 and ⃗ ⃗ ⃗⃗ ⃗ = 3 . Use the result in part (a), show that the
area of the triangle is 3√| |2| |2 − ( . )2. Hence, deduce the area of the triangle if =
and =

SOLUTION

(a) = 3 − + 2 = 2 + 2 −

CHOW
x = |3 −1 2 |

2 2 −1

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

= −3 − (−7) + (8)

= −3 + 7 + 8

| x | = √(−3)2 + (7)2 + (8)2

= √122
| x | = 122

| | = √(3)2 + (−1)2 + (2)2

= √14
| | = 14

| | = √(2)2 + (2)2 + (−1)2

= √9
| | = 9

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PSPM I QS 025/1 Session 2015/2016
C
. = (3 − + 2 ). (2 + 2 − )
= (3)(2) + (−1)(2) + (2)(−1)

. = 2

| |2| |2 − ( . )2 = (14)(9) − (2)2
= 122

∴ | x |2 = | |2| |2 − ( . )2

(b) CHOW
3b

A B
2a
( ) = ( )
, = | ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗ | 2 3 = (2)( )
= (2 3)( )
= (6)( )
1
= 2 |2 3 |

= 3| |

= 3| |

= 3√| |2| |2 − ( . )2

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PSPM I QS 025/1 Session 2015/2016

ℎ = =
Area = 3√| |2| |2 − ( . )2
= 3√| |2| |2 − ( . )2
= 3√122

CHOW

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PSPM I QS 025/1 Session 2015/2016

8. Given a line : = 2 − , = −3 + 4 , = −5 − 3 , and two planes 1: 2x − y + 7z = 53
and 2: 3x + y + z = 1. Find
a) The point of intersection between the line and the plane 1.
b) The acute angle between the line and the plane 1.
c) The acute angle between planes 1 and 2.

SOLUTION

: = 2 − , = −3 + 4 , = −5 − 3 ,

1: 2x − y + 7z = 53 CHOW
2: 3x + y + z = 1
(a)

= 2 − , = −3 + 4 , = −5 − 3 …………………. (1)

2x − y + 7z = 53 …………………. (2)

Substitute (1) into (2)

2(2 − ) − (−3 + 4 ) + 7(−5 − 3 ) = 53

4 − 2t + 3 − 4t − 35 − 21t = 53

−28 − 27t = 53

27t = −81

t = −3

ℎ = −3

= 2 − = 2 + 3 = 5

= −3 + 4 = −3 − 12 = −15

= −5 − 3 = −5 + 9 = 4

∴ ℎ (5, −15, 4)

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( )

= 2 − , = −3 + 4 , = −5 − 3 …………………. (1)

2x − y + 7z = 53 …………………. (2)

ℎ :

= − + 4 − 3

ℎ :

= 2 − + 7

ℎ ℎ 1 CHOW
.

cos = | |. | |

. = (− + 4 − 3 ). (2 − + 7 )
= (−1)(2) + (4)(−1) + (−3)(7)
= −27

| | = √(−1)2 + (4)2 + (−3)2
= √26

| | = √(2)2 + (−1)2 + (7)2
= √54

.
cos = | |. | |

−27
=

√26. √54
= −0.7206

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PSPM I QS 025/1 Session 2015/2016

= −1 .
| |. | |

= −1(−0.7206)

= 136.10°

ℎ ℎ 1
= 136.10° − 90°

= 46.10°

( )

1: 2x − y + 7z = 53 2: 3x + y + z = 1 CHOW

ℎ 1 ℎ 2

= −1 .
| |. | |

= 2 − + 7

= 3 + +

. = (2 − + 7 )(3 + + )

= (2)(3) + (−1)(1) + (7)(1)

= 12

| | = √(2)2 + (−1)2 + (7)2

= √54

| | = √(3)2 + (1)2 + (1)2

= √11

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PSPM I QS 025/1 Session 2015/2016

ℎ 1 ℎ 2

= −1 .
| |. | |

= −1
√54. √11

= −1(0.4924)

= 60.50°

CHOW

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PSPM I QS 025/1 Session 2015/2016

9. (a) Find the equation in standard form of an ellipse which passes through the point (-1, 6)
and having foci at (-5, 2) and (3, 2).

(b) From the result obtained in part (a), sketch the graph of the ellipse.

SOLUTION
(a)

(-1, 6)

x 1(−5, 2) x 2(3, 2)

ℎ : CHOW
( − ℎ)2 ( − )2
2 + 2 = 1
ℎ 2 − 2 = 2

−5 + 3 2 + 2
, (ℎ, ) = ( 2 , 2 )

−2 4
=( 2 ,2)
= (−1,2 )
:
2 = 3 − (−5)
=8
= 4

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2 − 2 = 2 …………………… (1)
2 − 2 = 42
2 − 2 = 16

ℎ :

( + 1)2 ( − 2)2
2 + 2 = 1

(−1, 6)

(−1 + 1)2 (6 − 2)2
2 + 2 = 1

16 CHOW
0 + 2 = 1

2 = 16 …………………… (2)

(2) (1)

2 − 16 = 16

2 = 32

( + 1)2 ( − 2)2
∴ 32 + 16 = 1

y

(b)
(-1, 6)
x

1(−5, 2) x (-1, 2) x 2(3, 2)
x

x

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PSPM I QS 025/1 Session 2015/2016

10. (a) Sketch and shade the region R bounded by the curve = √ , line = 2 − and y –
axis. Hence, find the area of the region R.

(b) If 1 is a region bounded by the curve = √ , line = 2 − and -axis, deduce the
ratio of : 1.

(c) Find the volume of the solid generated when the region R is rotated through 3600
about the -axis.

SOLUTION = 2 −
(a) y

= √ ,

2 CHOW
R

1

2 x

= √ …………………………… (1)

= 2 − …………………………… (2)

(1) (2)

√ = 2 −
= (2 − )2
= 4 + 2 − 4

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2 − 5 + 4 = 0
( − 1)( − 4) = 0
= 1 = 4
∴ = 1

1

, = ∫ (2 − ) − √

0

11

= ∫ 2 − − 2

0

31

2 2 2
= [2 − 2 − 3 ]

0

33
12 2(1)2 02 2(0)2
= [2(1) − 2 − 3 ] − [2(0) − 2 − 3 ]

12 CHOW
= [2 − 2 − 3] − 0

5
=6

(b)

15
1 = 2 (2)(2) − 6

7
=6

:

5

= 6
7

6

56
= 6 7

5
=7

∴ : = :

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PSPM I QS 025/1 Session 2015/2016

(c)
Volume of the solid generated when the region R is rotated through 3600 about the -axis.

, = 1 − (√ )2

∫ (2 − )2

0

1

= ∫ (4 + 2 − 4 ) − ( )

0

1

= ∫ ( 2 − 5 + 4)

0

3 5 2 1

= [ 3 − 2 + 4 ]

0

(1)3 5(1)2 (0)3 5(0)2
= [( 3 − 2 + 4(1)) − ( 3 − 2 + 4(0))]
CHOW
15
= [(3 − 2 + 4) − 0]

11
= 6

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PAPER 2 CHOW
SEMESTER 2
2015/2016

65

PSPM I QS 025/2 Session 2015/2016

1. Weights (kg) of a random sample of 90 female students and 105 male student is summarised as
∑( − 50) = −234 and ∑( − 63) = 367.5 respectively. Calculate the mean weight of all the
students.

2. The length of newborn babies at a hospital for a particular year is normally distributed with mean
of 52 cm and standard deviation of 2.5 cm. A baby’s length is considered normal if it is between
46cm and 56 cm. From a list of 100 birth records selected randomly for that particular year at the
hospital, how many babies are expected to have normal lengths?

3. A car rental company has 7 cars available for rental each day. Assuming that each rental is for the CHOW
whole day and that the number of demands has a mean of 3 cars per day. Find the probability that
a) The company cannot meet the demand in any one day.
b) Less than 5 cars are rented in a period of 3 days.

4. Given ( ) = 0.37, ( | ) = 0.13 ( ′ ∩ ) = 0.47. Find
a) ( ∩ ).
b) ( ) and hence calculate ( ∪ )′

5. Given a set of digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
a) Find the number of different ways to choose two prime digits from the set.
b) Four-digit numbers are to be formed from the set and the numbers do not start with
digit 0. Find the possible number of ways of getting
i. Even numbers between 6000 and 7000 if every digit can be repeated.

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ii. Numbers greater than 6000 that end with digit 5 and the digits can only be used
once.

iii. Numbers that contain exactly two odd digits and they must be next to each other
with no repetitions of digits allowed.

6. The time taken for 70 students to walk from the hostel to class in a certain college are shown in the
following table.

Time (minute) Number of student CHOW
2–4 5
5–7 9
8 – 10 19
21
11 – 13 12
14 – 16 4
17 - 19

a) Find the mean and mode.
b) Determine the 40th percentile.
c) Find the standard deviation.
d) Calculate the Pearson’s coefficient of skewness. Interpret your answer.

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7. A car insurance company offers two types of insurance plan for privately-owned cars, namely Plan
X and Plan Y. For a random sample of 60 clients for each insurance plan, the number of claims is
given in the following table.

Claim
Plan

Yes No

X 38 22

Y 45 15

Let CHOW

A = the event that no claim is made by the client

B = the event that the customer takes Plan X.
a) Find ( ∩ ).
b) Find ( ′ ∩ )
c) Given that the chosen client did not make any claim, find the probability that the insurance

plan taken was Plan X.
d) Determine whether the events “make a claim” and “the type of each insurance plan taken” are

independent. Give reason for your answer.

8. It is known that 37% of the students at a college do not take breakfast regularly. A random sample
of 20 students is chosen.
a) Find the probability that there are at least two students who do not take breakfast regularly.
b) Use normal approximation to calculate the probability that there are more than 10 students
who do not take breakfast regularly. Verify that the distribution can be approximated by a
normal distribution.

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9. The continuous random variable X has cumulative distribution function F(x) given by

0 , ≤ 0
2 , 0 ≤ ≤ 2

( ) = 6 , 2 ≤ ≤ 3
2 , ≥ 3
− 3 + 2 − 2
{1



a) (1 < < 2.2) CHOW
b) ℎ
c) The probability density function of X.
d) The expected value of X.
e) The variance of X, given that ( 2) = 169.

10. Two dice are thrown and the numbers x and y obtained from each dice are noted. The discrete
random variable W is defined as

= ( ) = {| , =
− |, ≠

a) Write all the outcomes for W=4 and hence show that

5
( = 4) = 36
b) Construct a table of the probability distribution of the random variable W. Hence, show that W
is a discrete random variable.
c) Find ( > 9).
d) Find the mode of W.
e) Find ( ) and hence, calculate (3 – 4 ).

END OF QUESTION PAPER

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PSPM I QS 025/2 Session 2015/2016

1. Weights (kg) of a random sample of 90 female students and 105 male student is summarised as
∑( − 50) = −234 and ∑( − 63) = 367.5 respectively. Calculate the mean weight of all the
students.

SOLUTION

∑( − 50) = −234

90

∑( − 50) = −234

=1

( 1 − 50) + ( 2 − 50) + ( 3 − 50) + ⋯ + ( 89 − 50)+( 90 − 50) = -234

( 1 + 2 + 3 + ⋯ + 89 + 90) + (−50) (90) = -234

∑ − 4500 = −234 CHOW

∑ = −234 + 4500

∑ = 4266 ……………………………………………………… (1)

105

∑( − 63) = 367.5

=1

( 1 − 63) + ( − 63) + ( 3 − 63) + ⋯ + ( 104 − 63)+( 105 − 63) = 367.5
( 1 + 2 + 3 + ⋯ + 104 + 105) + (−63) (105) = 367.5

∑ − 6615 = 367.5

∑ = 6982.5 ……………………………………………………… (2)
∑ + ∑

ℎ =
4266 + 6982.5

= 90 + 105
= 57.68

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2. The length of newborn babies at a hospital for a particular year is normally distributed with mean
of 52 cm and standard deviation of 2.5 cm. A baby’s length is considered normal if it is between
46cm and 56 cm. From a list of 100 birth records selected randomly for that particular year at the
hospital, how many babies are expected to have normal lengths?

SOLUTION

− ℎ

= 52, = 2.5 ~ ( , 2 )  ~ (0, 1 ) CHOW
~ ( , 2 ) −
~ (52, 2.52 )
=
(46 < < 56)
-2.4 1.6
46 − 52 56 − 52
= ( 2.5 < < 2.5 )

= (−2.4 < < 1.6)

= 1 − ( > 1.6) − ( > 2.4)

= 1 − 0.0548 − 0.0082

= 0.937

Thus, the expected number of normal length babies =
= 100 0.937
= 93.7
≈ 94

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3. A car rental company has 7 cars available for rental each day. Assuming that each rental is for the
whole day and that the number of demands has a mean of 3 cars per day. Find the probability that
a) The company cannot meet the demand in any one day.
b) Less than 5 cars are rented in a period of 3 days.

SOLUTION CHOW
(a) X = Number of cars rented in one day

= 3
~ (3)
( > 7) = ( ≥ 8) = 0.0119

(b) Y = number of cars rented in 3 days
= 9
~ (9)
( < 5) = 1 − ( ≥ 5)
= 1 − 0.9450
= 0.055

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4. Given ( ) = 0.37, ( | ) = 0.13 ( ′ ∩ ) = 0.47. Find
a) ( ∩ ).
b) ( ) and hence calculate ( ∪ )′

SOLUTION CHOW
(a) ( ) = 0.37

( | ) = 0.13
( ∩ )

( | ) = ( )
( ∩ ) = ( ). ( | )

= ( 0.37). (0.13)
= 0.0481

(b) ( ∪ )′ = 1 − [ ( ) + ( ) − ( ∩ )]
( ′ ∩ ) = ( ) − ( ∩ )
( ) = ( ′ ∩ ) + ( ∩ )
= 0.47 + 0.0481
= 0.5181

( ∪ )′ = 1 − [ ( ) + ( ) − ( ∩ )]
= 1 − [0.37 + 0.5181 − 0.0481]
= 0.16

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5. Given a set of digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
a) Find the number of different ways to choose two prime digits from the set.
b) Four-digit numbers are to be formed from the set and the numbers do not start with
digit 0. Find the possible number of ways of getting
i. Even numbers between 6000 and 7000 if every digit can be repeated.
ii. Numbers greater than 6000 that end with digit 5 and the digits can only be used
once.
iii. Numbers that contain exactly two odd digits and they must be next to each other
with no repetitions of digits allowed.

SOLUTION CHOW
(a) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
= { 2, 3, 5, 7 }
The number of different ways to choose two prime digits from the set = 4C2 = 6
(bi) Possible number of ways of getting Four-digit even numbers between 6000 and 7000 if
every digit can be repeated.

1 x 10 x 10 x 5 - 1

{ 6} {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} {0, 2, 4, 6, 8} 6000



Possible number of ways = (1 10 10 5) – 1

= 499

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(bii) Possible number of ways of getting Four-digit Numbers greater than 6000 that end with
digit 5 and the digits can only be used once.

4x 8 x 7 x1

{ 6, 7, 8, 9} {5}



Possible number of ways = (4 8 7 1)

= 224

(biii) Possible number of ways of getting Four-digit Numbers Numbers that contain exactly two CHOW
odd digits and they must be next to each other with no repetitions of digits allowed.

O OE E 5 X 4 X 5 X 4 = 400
4 X 5 X 4 X 4 = 320
E OO E

4 X 4 X 5 X 4 = 320

E EOO

Total 1040

Possible number of ways = (5 4 5 4) + (4 5 4 4) + (4 4 5 4)
= 1040

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6. The time taken for 70 students to walk from the hostel to class in a certain college are shown in the
following table.

Time (minute) Number of student
2–4 5
5–7 9
8 – 10 19
21
11 – 13 12
14 – 16 4
17 - 19

a) Find the mean and mode. CHOW
b) Determine the 40th percentile.
c) Find the standard deviation.
d) Calculate the Pearson’s coefficient of skewness. Interpret your answer.

SOLUTION

Time (minute) Class Boundary

2–4 1.5 – 4.5 35 15 45
5–7 4.5 – 7.5 69 54 324
8 – 10 7.5 – 10.5 9 19 171 1539
11 – 13 10.5 – 13.5 12 21 252 3024
14 – 16 13.5 – 16.5 15 12 180 2700
17 - 19 16.5 – 19.5 18 4 72 1296
Total
∑ = 70 ∑ = 744 ∑ 2 = 8928

(a) Mean, ̅ = ∑
∑

744
= 70 = 10.63

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PSPM I QS 025/2 Session 2015/2016

= + [ 1 1 2]
+

21 − 19
= 10.5 + [(21 − 19) + (21 − 12)] (13.5 − 10.5)

2
= 10.5 + [2 + 9] 3

= 11.05

Time (minute) Class Boundary

2–4 1.5 – 4.5 35 15 45 5 CHOW
5–7 4.5 – 7.5 69 54 324 14
8 – 10 7.5 – 10.5 9 19 171 1539 33
11 – 13 10.5 – 13.5 12 21 252 3024 54
14 – 16 13.5 – 16.5 15 12 180 2700 66
17 - 19 16.5 – 19.5 18 4 72 1296 70
Total
∑ = 70 ∑ = 744 ∑ 2 = 8928

(b) Persentile: = + [(1 0 0) − −1]



40 = 40 + [(14000) 70 − 40−1
40 ]

28 − 14
= 7.5 + [ 19 ] (10.5 − 7.5)
= 9.71

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(c) Variance:
2 = ∑ 2− 1 (∑ )2

−1

2 = 8928 − 1 (744)2
70 1

70 −

= 14.788

Standard deviation: = √ 2
= √14.788
= 3.85

(d) Pearson’s coefficient of skewness = − CHOW


= 10.63−11.05
3.85

= −0.109

∴

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7. A car insurance company offers two types of insurance plan for privately-owned cars, namely Plan
X and Plan Y. For a random sample of 60 clients for each insurance plan, the number of claims is
given in the following table.

Plan Claim
Yes
No
X 38 22
15
Y 45

Let
A = the event that no claim is made by the client
B = the event that the customer takes Plan X.

a) Find ( ∩ ). CHOW
b) Find ( ′ ∩ )
c) Given that the chosen client did not make any claim, find the probability that the insurance

plan taken was Plan X.
d) Determine whether the events “make a claim” and “the type of each insurance plan taken” are

independent. Give reason for your answer.

SOLUTION

Plan Claim Total

X / (B) Yes / (A’) No / (A) 60
Y / (B’) 60
Total 38 22 120

45 15

83 37

(a) ( ∩ ) = 22

120

11
= 60
(b) ( ′ ∪ ) = ( ′) + ( ) − ( ′ ∩ )

83 60 38
= 120 + 120 − 120

105 7
= 120 = 8

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(c) ( | ) = ( ∩ ) = 22

( ) 37

(d) ( ) = 1
2

( | ′) = 38
83

( | ′) ≠ ( )

∴ ℎ .

CHOW

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8. It is known that 37% of the students at a college do not take breakfast regularly. A random sample
of 20 students is chosen.

a) Find the probability that there are at least two students who do not take breakfast regularly.
b) Use normal approximation to calculate the probability that there are more than 10 students

who do not take breakfast regularly. Verify that the distribution can be approximated by a
normal distribution.

SOLUTION

(a) X – Students who do not take breakfast regularly

~ (20, 0.37)

( ≥ 2) = 1 − ( ≤ 1)

= 1 − ( ≤ 0) − ( = 1)

= 1 − 20C0(0.37)0 (0.63)20- 20C1(0.37)1 (0.63)19 CHOW

= 0.9988

(b) = = 20 0.37 = 7.4
2 = = 20 0.37 0.63 = 4.662

~ (7.4, 4.662) Continuity Correction
−
( > 10)
=
= ( > 10.5)

= ( > 10.5 − 7.4 )

√4.662

= ( > 10.5 − 7.4 )

√4.662

= ( > 1.44)

= 0.0749

The distribution can be approximated by a normal distribution as
= 20 0.37 = 7.4 > 5

= 20 0.63 = 12.6 > 5

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9. The continuous random variable X has cumulative distribution function F(x) given by

0 , ≤ 0
2 , 0 ≤ ≤ 2

( ) = 6 , 2 ≤ ≤ 3
2 , ≥ 3
− 3 + 2 − 2
{1


a) (1 < < 2.2)
b) ℎ
c) The probability density function of X.
d) The expected value of X.
e) The variance of X, given that ( 2) = 169.

SOLUTION

(a) (1 < < 2.2) CHOW

= (2.2) − (1)

(2.2)2 12
= [− 3 + 2(2.2) − 2] − ( 6 )

= 0.62

(b) ( ) = 0.5
2
6 = 0.5
2 = 0.5 6
2 = 3

= √3 since > 0

(c) ( )
( ) = ( )
≤ 0 ( ) = 0
0 ≤ ≤ 2 2
2 ≤ ≤ 3 ( ) = (0) = 0
( ) = 6
≥ 3 2 2
( ) = ( 6 ) = 3
( ) = − 3 + 2 − 2
( ) = 1 2 2
( ) = (− 3 + 2 − 2) = − 3 + 2

( ) = (0) = 0


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PSPM I QS 025/2 Session 2015/2016



, 0 ≤ ≤ 2

3
2
( ) = − 3 + 2, 2 ≤ ≤ 3

{0, ℎ

(d) ( ) = ∫−∞∞ ( )

0 2 3 2 ∞

= ∫ (0) + ∫ ( ) + ∫ (− 3 + 2) + ∫ (0)

−∞ 0 32 3

2 2 3 2 2
=∫ + ∫ − + 2
3 3
0 2

2 2 3 2 2
= ∫ − ∫ 3 − 2
03
2

= 3 2 − 2 3 − 3 CHOW
[9
[] 2]

9 2

0

= 23 − 03 − 2(3)3 − (3)2) − 2(2)3 − (2)2)]
[( 9 (9
[( ) ( )]

9 9

8 16
= 9 − [(6 − 9) − ( 9 − 4)]

= 8 20 )]
− [−3 − (−
9
9

5

=

3

(e) ( ) = ( 2) − [ ( )]2
19 5 2

= 6 − [3]
7

= 18

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PSPM I QS 025/2 Session 2015/2016

10. Two dice are thrown and the numbers x and y obtained from each dice are noted. The discrete
random variable W is defined as

= ( ) = {| , =
− |, ≠

a) Write all the outcomes for W=4 and hence show that

( = 4) = 5
36

b) Construct a table of the probability distribution of the random variable W. Hence, show that W CHOW
is a discrete random variable.

c) Find ( > 9).
d) Find the mode of W.
e) Find ( ) and hence, calculate (3 – 4 ).

SOLUTION

X1 2 3 4 5 6

y

1112345
2141234
3219123
4 3 2 1 16 1 2
5 4 3 2 1 25 1
6 5 4 3 2 1 36

a) W=4

{(2,2), (5,1), (1,5), (2,6), (6,2)}

( = 4) = 5
36

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PSPM I QS 025/2 Session 2015/2016

b)

W 1 2 3 4 5 9 16 25 36

P(W=w) 11 8 6 5 2 1 1 1 1

36 36 36 36 36 36 36 36 36

∑ ( = ) = 11 + 8 + 6 + 5 + 2 + 1 + 1 + 1 + 1 = 1
36 36 36 36 36 36 36 36 36

c) ( > 9) = ( = 16) + ( = 25) + ( = 36)

111 CHOW
= 36 + 36 + 36

3
= 36

1
= 12

d) Mode = 1
e) ( ) = ∑ [P(W = w)]

11 8 6 5 2 1 1 1 1
= 1 (36) + 2 (36) + 3 (36) + 4 (36) + 5 (36) + 9 (36) + 16 (36) + 25 (36) + 36 (36)

161
= 36

f) (3 − 4 ) = 3 − 4 ( )

161

= 3−4( )

36

134
=− 9

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PAPER 1 CHOW
SEMESTER 2
2016/2017

86

PSPM I QS 025/1 Session 2016/2017

1. Find the angle between the line : 〈 , , 〉 = 〈1,3, −1〉 + 〈2,1,0〉 and the plane Π: 3 −
2 + = 5.

SOLUTION

= 〈2, 1, 0〉

= 〈3, −2,1〉

. = 〈2, 1, 0〉. 〈3, −2,1〉

= (2)(3) + (1)(−2) + (0)(1)

=6−2+0

=4

| | = √22 + 12 + 02 CHOW

= √5

| | = √32 + (−2)2 + 12

= √14
.

cos =
| |. | |
4

=
√5. √14

= 0.4781
= −10.4781

= 61.4° (Angle between the line and the normal vector of plane Π)
Angle between the line and the vector of plane Π = 90° − 61.4° =28.6°

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PSPM I QS 025/1 Session 2016/2017

2. Solve ∫ 2 . Page 3
1− 2
88
SOLUTION

2
∫ 1 − 2

= 1 − 2

= −2 2


= −2 2

2 = −
2

2 1 1 CHOW
∫ 1 − 2 = − 2 ∫

1
= − 2 ln| | +

= − 1 ln|1 − 2 | +
2

Alternative

2
∫ 1 − 2

= 2

= 2 2


= 2 2

2 = − 1
2

2 1 1
∫ 1 − 2 = ∫ 1 − ( 2 )

PSPM I QS 025/1 Session 2016/2017

11
= 2 ∫ 1 −

= − 1 ln|1 − | +
2

= − 1 ln|1 − 2 | +
2

CHOW

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PSPM I QS 025/1 Session 2016/2017

3. Given four points = (−2, −8, 4), (2, − , −1), = (0, −9, 0) = (−4, −3, 7).
Determine the value of if ⃗ ⃗ ⃗⃗ ⃗ . (⃗ ⃗ ⃗⃗ ⃗ x ⃗⃗ ⃗⃗ ⃗ ) = 64.

SOLUTION

= (−2, −8, 4), (2, − , −1), = (0, −9, 0) = (−4, −3, 7)

⃗⃗⃗ ⃗⃗ ⃗ = ⃗ ⃗⃗ ⃗⃗ ⃗ − ⃗ ⃗⃗ ⃗⃗ ⃗

= 〈2, − , −1〉 − 〈−2, −8, 4〉

= 〈4, 8 − , −5〉

⃗⃗ ⃗⃗ ⃗ = ⃗ ⃗⃗ ⃗⃗ ⃗ − ⃗ ⃗⃗ ⃗⃗ ⃗

= 〈0, −9, 0〉 − 〈−2, −8, 4〉 CHOW

= 〈2, −1, −4〉

⃗ ⃗ ⃗⃗ ⃗⃗ = ⃗⃗ ⃗⃗ ⃗⃗ − ⃗ ⃗⃗ ⃗⃗ ⃗

= 〈−4, −3, 7〉 − 〈−2, −8, 4〉

= 〈−2,5, 3〉


⃗⃗ ⃗⃗ ⃗ ⃗⃗ ⃗⃗ ⃗⃗ = | 2 −1 −4|

−2 5 3

= 17 + 2 + 8

= 〈17,2,8〉

⃗ ⃗ ⃗⃗ ⃗ . (⃗ ⃗ ⃗⃗ ⃗ x ⃗⃗ ⃗⃗ ⃗ ) = 64

〈4, 8 − , −5〉. 〈17,2,8〉 = 64

(4)(17) + ( 8 − )(2) + (−5)(8) = 64

68 + 16 − 2 − 40 = 64

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PSPM I QS 025/1 Session 2016/2017
44 − 2 = 64
2 = −20
= −10

CHOW

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PSPM I QS 025/1 Session 2016/2017

4. Find the vertex, focus and directrix for the parabola 2 + 64 = 8 − 16 . Hence, sketch
and label the vertex, focus and directrix for the curve.

SOLUTION

2 + 64 = 8 − 16
2 − 8 = −16 − 64

2 − 8 + −8 2 = −16 − 64 + −8 2
(2) (2)

( − 4)2 = −16 − 64 + 16

( − 4)2 = −16 − 48 Compare
( − 4)2 = −16( + 3)
( − 4)2 = −16( + 3) ( − )2 = 4 ( − ℎ) CHOW

ℎ = −3, = 4, 4 = −16 ==> = −4



(ℎ, ) = (−3, 4)

,

(ℎ + , ) = (−3 − 4, 4)

= (−7, 4)

,

= ℎ −

= −3 + 4

= 1

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PSPM I QS 025/1 Session 2016/2017
(− , )

=

(− , )



CHOW

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PSPM I QS 025/1 Session 2016/2017

5. The end points of the diameter of a circle are (0, 1) (3, −3).
a) Determine an equation of the circle.
b) Find an equation of the tangent line to the circle at the point (0, 1).

SOLUTION

a) Diameter: (0, 1) (3, −3)

(0, 1)

(3, −3)

, (ℎ, ) = ( + , − ) CHOW



3
= (2 , −1)
= √(3 − 0)2 + (−3 − 1)2

= √25

=5

1
, = 2 ( )

5
=2


( − ) + ( − ) =

( − 3 + ( + 1) = 5
2) (2)

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PSPM I QS 025/1 Session 2016/2017

( − 3 + ( + 1) = 25
2) 4

b)
(0, 1)

(3, −3)

CHOW

−3 − 1
= 3 − 0

4
= −3



1
= −

= − 1

− 4
3

3
=4

( , )
− 1 = ( − 1)

Page 10

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PSPM I QS 025/1 Session 2016/2017

3
− 1 = 4 ( − 0)

3
= 4 + 1

CHOW

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PSPM I QS 025/1 Session 2016/2017

6. In a Chemistry experiment, sodium hydroxide, NaOH, reacts with hydrochloric acid,
HCl, to form sodium chloride salt, NaCl, and water. Before the reaction starts, no NaCl
salt is formed. At time (minute), the mass of NaCl salt formed is grams and the rate
of change of is given by = (50 − ), where is a positive constant.



a) Find the general solution for the above equation.
b) Find the particular solution if 35 grams of NaCl salt has formed in the first 30

minutes.
c) Hence, find

i. The mass of NaCl salt formed in 60 minutes.
ii. The time taken to form 40 grams of NaCl salt.

SOLUTION

a) = (50 − ) CHOW



1
(50 − ) =

1
∫ (50 − ) = ∫

− ln(50 − ) = +

ln(50 − ) = −( + )

50 − = −( + )

50 − = − . −

50 − = − ; = −

= , =

50 − 0 = −0

= 50

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PSPM I QS 025/1 Session 2016/2017

Page 13
50 − = 50 −
50 − 50 − = 98
= 50(1 − − )

b) ℎ = 30, = 35

35 = 50(1 − −30 )

1 − −30 = 35
50

1 − −30 = 0.7 CHOW

−30 = 1 − 0.7

−30 = 0.3

ln −30 = ln 0.3

−30 = −1.2040

−1.2040
= −30
= 0.0401

Particular
= 50(1 − −0.0401 )

PSPM I QS 025/1 Session 2016/2017

d) i) ℎ = 60

= 50[1 − −0.0401(60)]

= 50[1 − −0.0401(60)]

= 45.5

) ℎ = 40

40 = 50(1 − −0.0401 )

40 = 1 − −0.0401
50

0.8 = 1 − −0.0401

−0.0401 = 1 − 0.8 CHOW

−0.0401 = 0.2

ln −0.0401 = ln 0.2

−0.0401 = ln 0.2
ln 0.2

= −0.0401
= 40.1

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