Mathematics_Anyflip Sem 2

PENERBIT
ILMU
BAKTI SDN. BHD.

Contents

PENERBITChapter 1 Numerical Solution 1
ILMU 3
BAKTI SDN. BHD.1.1 Numerical Solution of Equations 7
1.2 Newton-Raphson Method
Summative Practice 1 8
12
Chapter 2 Integration 15
22
2.1 Integration of Functions 29
2.2 Integration of Trigonometric Functions 33
2.3 Techniques of Integration
2.4 Definite Integrals 34
2.5 Trapezoidal Rule
Summative Practice 2

Summative Assessment Test (UPS) 1

Chapter 3 First Order Differential Equations

3.1 Separable Variables 36

3.2 First Order Linear Differential Equations 40

3.3 Applications 44

Summative Practice 3 50

Chapter 4 Conics

4.1 Circles 51

4.2 Ellipses 65

4.3 Parabolas 72

Summative Practice 4 78

Summative Assessment Test (UPS) 2 79

Chapter 5 Vectors 81
87
5.1 Vectors in Three Dimensions 91
5.2 Scalar Product 96
5.3 Vector Product 112
5.4 Applications of Vectors in Geometry
Summative Practice 5

Chapter 6 Data Description

6.1 Introduction to Data 114
6.2 Measures of Location 120
6.3 Measures of Dispersion 134
Summative Practice 6 143
PENERBIT
ILMUChapter 7 Probability
BAKTI SDN. BHD.
7.1 Probability 145

7.2 Probability Involving Permutations and Combinations 164

Summative Practice 7 180

Summative Assessment Test (UPS) 3 182

Chapter 8 Random Variables

8.1 Discrete Random Variables 185
8.2 Continuous Random Variables 199
Summative Practice 8 211

Chapter 9 Special Probability Distributions 213
219
9.1 Binomial Distribution 225
9.2 Poisson Distribution 235
9.3 Normal Distribution 240
9.4 Distribution Approximation
Summative Practice 9 243
245
Semester Examination
Answers

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Solutions

1CHAPTERPENERBIT
ILMU
Numerical Solution BAKTI SDN. BHD.

1.1 Numerical Solution of Equations

 Learning Outcomes

● Locate approximately a root of an equation, by graphical considerations or
searching for a sign change.

Locate Root of an Equation

 1 If f is a continuous function on the closed interval [a, b], where f(a)  0
 f(b) or f(b)  0  f(a), then the function contains at least one root in the
interval.

 2 There are two methods to determine the interval where the root lies. It can be
done by using the graphical method or the algebraic method.

Graphical method

 1 Sketch the graph of y = f (x). The exact root is the point where the graph f (x)
intercepts the x-axis as shown in Figure 1.1.

y

y = f(x)

0a b x
Root

Figure 1.1

1

 Numerical Solution

 2 Sketch the graph of y = f(x) and y = g(x). The real root is the x-coordinate of
the intersection points between the two graphs as shown in Figure 1.2.

y

y = f(x)

BAKTI SDN. BHD. y = g(x)

0a b x
Root
Chapter 1
Figure 1.2

 3 In both cases, the root lies in the interval [a, b].

Algebraic method

 1 Find two values of a and b such that f (a) and f(b) have different signs by trial
and error. The root of f (x) = 0 lies in the interval [a, b] if f(a) • f(b)  0.

Example 1

Show that there is a real root for x3 – 2x – 10 = 0 between x = 2 and x = 3.

Solution
f (2) = (2)3 – 2(2) – 10 = 8 – 4 – 10 = –6  0
f (3) = (3)3 – 2(3) – 10 = 27 – 6 – 10 = 11  0
ILMU
Since f (2) and f (3) have opposite signs, x3 – 2x – 10 = 0 has a root between
x = 2 and x = 3.

Example 2
PENERBIT
Given the equation ln 2x = 0, show that there exists a real root between x = 0
and x = 1 by using graphical consideration.

Smart Tips! Solution y

Step 1: y = In 2x
Sketch the graph of
y = ln 2x 0 0.5 1 x
Step 2:
Make conclusions based The graph crosses the x-axis between x = 0 and x = 1, so there is a root between
on the graph. x = 0 and x = 1.

2

Quick Check 1.1 Numerical Solution 

 1 (a) Given the equation ex – 1 = 2 , show that there is a real root between 1 3
and x = 2. x

(b) Find the real root of e–x = 2x that lies between x = 0 and x = 1.PENERBIT
ILMU
(c) Show that 2x = 2 – ex has a root in the interval [0, 1].BAKTI SDN. BHD.

 2 (a) Show that there is a real root for cos x = x between x = 0.5 radian and
x = 1 radian.

(b) Show that the equation sin x = x + 0.5 has a root lying in the interval
[–2, –1].

 3 (a) Sketch the graph of y = 4 – x and y = x2 for 0  x  2 on the same axes. Chapter 1
Hence, find an approximate value (x0) of the root of the equation
x2 + x – 4 = 0.

(b) Show that the equation e x = 3 – x has only one real root.

 4 (a) Using the graphical method, show that the equation –ex = x + 1 has only
one real root. Hence, determine the interval where the root lies.

(b) Sketch the graphs of y = –2 – x and y = –e–x. Hence, find the interval
a < x < b, where a and b are consecutive negative integers, in which the
point of intersection lies.

1.2 Newton-Raphson Method

 Learning Outcomes

● Find the root using the Newton-Raphson method.

 1 The Newton-Raphson method is based on the use of the tangent to the curve
f(x) to obtain a better approximation to the root of f (x) = 0.

 2 Steps for using the Newton-Raphson method:
Step 1: Find the initial approximate value of the root, x1, for f (x) = 0.
Step 2: Use the first approximation to get the second approximation, x2, by

using the formula

xn + 1 = xn – f (xn) , where n = 1, 2, 3, …
f '(xn)

Step 3: Repeat this process until the desired accuracy is obtained.

Example 1

Use the Newton-Raphson method to obtain the root of the equation
x3 + 3x + 2 = 0 with an initial approximation x1 = 0. Give your answer correct
to four decimal places.

Solution
Let f (x) = x3 + 3x + 2
f '(x) = 3x2 + 3

 Numerical Solution

xn + 1 = xn – f (xn)
f '(xn)

= xn – xn3 + 3xn + 2
3xn2 + 3
x1 = 0 BAKTI SDN. BHD.
x13 + 3x1 + 2
x2 = x1 – 3x12 + 3

= 0 – (0)3 + 3(0) + 2
3(0)2 + 3

= –0.66667

Chapter 1 x3 = x2 – x23 + 3x2 + 2
3x22 + 3
(–0.66667)3 + 3(–0.66667) + 2
= –0.66667 – 3(–0.66667)2 + 3

= –0.59829

x4 = x3 – x33 + 3x3 + 2
3x32 + 3

= –0.59829 – (–0.59829)3 + 3(–0.59829) + 2
3(–0.59829)2 + 3

= –0.59607

x5 = x4 – x43 + 3x4 + 2
3x42 + 3

= –0.59607 – (–0.59607)3 + 3(–0.59607) + 2
3(–0.59607)2 + 3

= –0.59607

ILMU
\ The approximate real root is –0.5961 (correct to four decimal places).

PENERBIT Smart Tips! Example 2

Make sure to change the Show that there is a real root for sin x + 2x = 1 lies between x = 0 and x = 1
calculator mode to radian. Hence, determine the real root correct to four decimal places using the
R (radian). Newton-Raphson method.

Solution
sin x + 2x = 1
sin x + 2x – 1 = 0

f(x) = sin x + 2x – 1
f(0) = sin (0) + 2(0) – 1 = –1
f(1) = sin (1) + 2(1) – 1 = 1.8415

Since f(0) and f(1) has opposite sign, therefore f(x) has a root lies between x = 0
and x = 1.

Let f(x) = sin x + 2x – 1
f'(x) = cos x + 2

4

Numerical Solution 

xn + 1 = xn – f (xn)
f '(xn)
sin xn + 2xn – 1
= xn – cos xn + 2

x1 = 0.5 sin x1 + 2x1 – 1
x2 = x1 – cos x1 + 2
PENERBIT
ILMU =0.5–sin 0.5 + 2(0.5) – 1
BAKTI SDN. BHD.cos 0.5 + 2

= 0.33339

x3 = x2 – sin x2 + 2x2 – 1
cos x2 + 2
Chapter 1
= 0.33339 – sin 0.33339 + 2(0.33339) – 1
cos 0.33339 + 2

= 0.33542

x4 = x3 – sin x3 + 2x3 – 1 Smart Tips!
cos x3 + 2

= 0.33542 – sin 0.33542 + 2(0.33542) – 1 We stop the process at x4
cos 0.33542 + 2 because x3 and x4 have

= 0.33542 the same answer correct

to four decimal places.

\ The approximate real root is 0.3354 (correct to four decimal places).

Example 3

Given the equation ln x = – x + 3.
(a) Show that there is a real root lies between x = 2 and x = 3.
(b) Hence, use the Newton-Raphson method to solve the equation. Give your

answer correct to three decimal places.

Solution
(a) f(x) = ln x + x – 3
f(2) = ln (2) + (2) – 3 = –0.3069
f(3) = ln (3) + (3) – 3 = 1.0986

Since f(2) and f(3) has opposite sign, therefore f(x) has a real root lies
between x = 2 and x = 3.

(b) Let f(x) = ln x + x – 3

f' (x) = 1 +1
x

xn + 1 = xn – f (xn)
f '(xn)

= xn – ln xn + xn – 3

1 +1
xn

5

 Numerical Solution x1 = 2.5 ln x1 + x1 – 3
x2 = x1 –
6 1
x1 +1

ln 2.5 + 2.5 – 3
= 2.5 – BAKTI SDN. BHD.
1
2.5 + 1

= 2.20265

x3 = x2 – ln x2 + x2 – 3

1 +1
x2
Chapter 1
= 2.20265 – ln 2.20265 + 2.20265 – 3

1 +1
2.20265

= 2.20794

x4 = x3 – ln x3 + x3 – 3

1 +1
x3

= 2.20794 – ln 2.20794 + 2.20794 – 3

1 +1
2.20794

= 2.20794

\ The approximate real root is 2.208 (correct to three decimal places).

ILMU Quick Check 1.2

 1 (a) Show that ex = 2 – x has a root between x = 0 and x = 1.
(b) By taking x = 0.5 as the first approximation, evaluate this root to three

significant figures using the Newton-Raphson method.

PENERBIT  2 (a) Show that there is a root of x + ex = 0 between x = –1 and x = 0.
(b) Use the Newton-Raphson method to find this root correct to four

decimal places.

 3 (a) Show that the equation sin x + x + 1 = 0 has a root lying in the interval

[–1, 0].

(b) By using the Newton-Raphson method and the initial value x = –0.5,

determine the real root that lies between x = –1 and x = 0, correct to

two decimal places.

 4 (a) Show that the equation 2x2 – 1 = 0 has a positive root in between
x = 0.5 and x = 1. x

(b) Use the Newton-Raphson method to find this root correct to two

decimal places, taking the middle value in the interval as the initial

approximation to the root.

 5 Use the Newton-Raphson method to solve the equation cos x = 2 – 2x
correct to four decimal places. Take x = 0.5 as the first approximation.

PENERBIT Summative Practice 1 Chapter 1
ILMU
BAKTI SDN. BHD. 1 Show that the equation 2x4 = 5 + x has a root between x = 1 and x = 2. By taking x = 1.4 as the
first approximation, evaluate this root to three significant figures using the Newton-Raphson
method.

 2 Given ex = 4 – x.
(a) Show that there is a real root between x = 1 and x = 2.
(b) Hence, by using the Newton-Raphson method, find the root of the equation correct to four

decimal places. Take x0 = 1.2 as the first approximation.
 3 Show that the equation 2x + ex – 2 = 0 has a root between x = 0 and x = 1. Using the Newton-

Raphson method and taking x0 = 0.7, find the root correct to four decimal places.
 4 Use the Newton-Raphson method with the initial approximation x1 = 1 to find 6 2 in the

interval [0, 2] correct to three decimal places.
 5 Given f1(x) = 2x and f2(x) = –ln x.

(a) Without using curve sketching, show that y = f1(x) and y = f2(x) intersect in the interval
[0.1, 1].

(b) Use the Newton-Raphson method to estimate the intersection point of y = f1(x) and
y = f2(x) with the initial value x1 = 1. Iterate until | f (xn + 1) – f (xn ) |  0.005. Give your
answer correct to three decimal places.

 6 Show that the equation –4x2 + 5x + 7 = 0 has a root in the interval [–2, 0]. Use the Newton-
Raphson method to find the root of the equation correct to four decimal places.

7

Summative Assessment Test (UPS) 1

Time: 20 minutes
[19 marks]

Answer all questions.
BAKTI SDN. BHD.
1 Integrate ∫ 6 ln x dx with respect to x. 5 ∫If 3 f (x) dx = 8 and ∫1 g(x) dx = 5, evaluate
6 1 3
x
A + c ∫3 [2f (x) + g(x)] dx.
1
B 6ex + c
A 21

C 6xex – 6x + c B 16

D 6x ln x – 6x + c C 12

   2 1 m+4 x2 – 1 D 11
x x2
∫x + dx is equal to 6 Integrate 1 with respect to x.
5x ln
 A 1 m+5 x
x
x + A 1 ln x + c
5
m+5 +c

 B m+5 B 1 ln | ln x| + c
x2 + 1 5
x2 (m + 5) + c

 C m+5 C 5 ln x + c
x
x2 + 1 (m + 5) + c

UPS 1 D None of the above D 5 ln |ln x | + c

3 By using the trapezoidal rule with 5 ordinates, the 7 Evaluate ∫5 | x| dx.
∫1 ILMU –5
approximate value of 0 tan x dx is 0.628. What is
25
the value of h? A 2

A 0.02 50
3
B 0.15 B

C 0.20 C 25

D 0.25 D 50

AnswersPENERBIT
4 Find the volume of the solid formed by rotating ∫8 Given that h(x)e 2x3 + 5x dx = e 2x3 + 5x + c, what is the

the region bounded by y = x2, x = 0 and x = 2 function of h(x)?

through 360° about the x-axis. A 6x2 + 5

A 2π B 2x3 + 5x

B 8 π C x4 + 5x2
3 2 2

C 4π D 6x3

D 32 π
5

34

Summative Assessment Test (UPS) 1 |

 9 1 2 14 If ∫ cosn x sin x dx = – cos6 x + c, what is the
2x 6
The integration of ∫ – 4x3 ( x – x4)3 dx = value of n?

1 ( x – x4)n + c. A 0
n
B 1
What is the value of n?
C 2
PENERBIT 3
ILMUA 5 D 5
BAKTI SDN. BHD.
B 2 15 ∫ x n dx = x n +1 + c is not possible when
5 n+1

C 5 A n = –1
3
B n = 0

D 2 C n = 1
3
D n = 2

10 Integrate ∫ 7x + 1 ln 7 dx. 16 ∫ 1 dx = ln |x | + c is not possible when
x

A ln 7(7x + 1) + c A x = –2

B 7x + 1 + c B x = –1

7x + 1 C x = 0
ln 7
C +c D x = 1

D x+1 +c 17 Given the equation ln x = –2x + 5, determine the
ln 7 f ʹ of the function.

11 Integrate ∫ 12 sin (3x + 1) dx. A 1 UPS 1
x
A 6 cos (3x + 1) + c
B 4 cos (3x + 1) + c B 1 + 2
C –6 cos (3x + 1) + c x
D –4 cos (3x + 1) + c
C ln x + 2x
D ln x + 2x – 5

12 What is the lower limit and upper limit for the 18 Determine the first approximation of the equation
area of the region bounded by the curves ex = 2 – x.
y = x2 – 3x, y = 9 – x2 and the x-axis? A 1.5
B 0.5
A – 2 and 3 C – 0.5
3 D – 1.5
Answers
B – 3 and 3
2

C 0 and 27 19 Use the Newton-Raphson method to solve the
4 equation e–x = 2x correct to four decimal places.
Take x0 = 0.5 as the first approximation.
D 0 and 4 A 0.3490
27 B 0.3491
C 0.3517
13 Evaluate ∫2 |x| dx. D 0.3518
A –1 –1 x

B 0

C 1

D None of the above

35

PENERBIT 6CHAPTER
ILMU
BAKTI SDN. BHD.Data Description

6.1 Introduction to Data

 Learning Outcomes

● Identify the discrete and continuous data.
● Identify ungrouped and grouped data.
● Construct and interpret stem-and-leaf diagrams.

Types of Data

  1 Qualitative data is data which can be observed but not measured.
Example: colour, gender, name, smells, religion, country
 2 Qualitative data can be nominal or ordinal.
(a) Nominal data is data which is not in an ordered sequence.
Example: red, green, yellow, blue, white, black
(b) Ordinal data is data which is in an ordered sequence.
Example: small, medium, large, extra large
  3 Quantitative data is data which can be measured and can be expressed

numerically.
Example: length, weight, age, area, volume, cost
 4 Quantitative data can be discrete or continuous.
(a) Discrete data takes only certain values.
Example: Number of cars sold in a day
(b) Continuous data takes any value within an interval.
Example: Time taken to answer a question
  5 Ungrouped data is data which are not sorted in any way, while grouped data

is data which are sorted into different classes.

114

 6 The types of data can be summarised as shown in Figure 6.1. Data Description 
Types of data
115
PENERBITQualitative Quantitative
ILMU
BAKTI SDN. BHD.NominalOrdinalDiscrete Continuous

Figure 6.1 Types of data Chapter 6

Example 1

Determine whether the following are qualitative or quantitative data.

(a) Weight of an individual (d) Breed of cat

(b) Level of education (e) Duration of green lights

(c) Age of an individual

Solution (d) Qualitative
(a) Quantitative (e) Quantitative
(b) Qualitative
(c) Quantitative

Example 2

Determine whether the following are discrete or continuous data.

(a) Number of fish in a basket (d) Head circumference

(b) Speed of buses (e) Duration of a movie

(c) Number of goals in a football match

Solution (d) Continuous
(a) Discrete (e) Continuous
(b) Continuous
(c) Discrete

Stem-and-Leaf Diagrams

 1 A stem-and-leaf diagram is used to display quantitative data visually.
 2 Each value in the data is split into two parts. The stem represents the first digit

or digits and the leaf represents the last digits.
Example:

Stem Leaf
1 2466

 Data Description  3 A key is used to guide others on how to read the values.
Example:
Smart Tips! Key: 1 | 4 means 14 cm

Arrange the data in  4 A stem-and-leaf diagram illustrates the shape and skewness of the distribution.
ascending order before
constructing the stem- (a) Symmetrical skewed
ChapterPENERBIT 6and-leaf diagram.
ILMU Stem Leaf 5
BAKTI SDN. BHD.11610
3 23
5 34
7 01
97

(b) Skewed to the left or negatively skewed

Stem Leaf
10
3 23
5 345
7 015678
9 78

(c) Skewed to the right or positively skewed

Stem Leaf
1 01
3 235799
5 345
7 08
91

Example 3

The mass of 15 students (to the nearest kg) in a class are given as follows:

40 52 63 71 70 72 49 48
53 56 59 58 64 55 65
Construct a stem-and-leaf diagram for the mass of the students.

Solution

Stem Leaf
4 089
5 235689
6 345
7 012

Key: 4 | 8 means 48 kg

Example 4 Data Description 

The pH levels for 20 freshwater aquariums are recorded. The results are as 117
follows:

4.1 4.2 4.3 4.5 4.6 4.8 5.0 5.2 5.3 5.3 5.4
5.7 5.8 5.9 5.9 6.2 6.3 6.7 7.2 8.4
Construct a stem-and-leaf diagram for the given data.
PENERBIT
ILMUSolution Repeat 9 two times since Chapter 6
BAKTI SDN. BHD. the value 5.9 appears twice.
Stem Leaf
4 123568
5 023347899
6 237
72
84

Key: 7 | 2 means 7.2

Example 5

Interpret the following stem-and-leaf diagrams.
(a) Stem Leaf

01
1 02
2 237
3 1457
4 445678
5 132
6 13
71
81

(b) Stem Leaf
0 12
1 024556788
2 23756
3 1457
4 445
5 13
62
71
81

 Data Description (c)
Stem Leaf
118
05
10
28
3 12
4 44
5 257
6 44556
7 124556788
8 13
ChapterPENERBIT 6
ILMU Solution
BAKTI SDN. BHD.
(a) Stem Leaf Symmetrical.

01
1 02
2 237
3 1457
4 445678
5 132
6 13
71
81

(b) Stem Leaf Skewed to the right or
0 12 positively skewed.
1 024556788
2 23756
3 1457
4 445
5 13
62
71
81

(c) Stem Leaf Skewed to the left or
05 negatively skewed.
10
28
3 12
4 44
5 257
6 44556
7 124556788
8 13

PENERBIT Quick Check 6.1 Data Description  Chapter 6
ILMU
BAKTI SDN. BHD. 1 Determine whether the following are qualitative or quantitative data. 119
(a) Household income
(b) Types of pasta
(c) Number of districts in each state
(d) Musical genres
(e) Shapes
(f) Places of interest in Perak

 2 Determine whether the following are discrete or continuous data.
(a) Beds in private hospitals by state
(b) Train braking distances
(c) Number of books sold in a day
(d) Height of tomato plants
(e) Marathon finish times
(f) Volume of milk cartons

 3 The following shows the high jump records, in centimetres of 22 students at a
school competition. Construct a stem-and-leaf diagram to represent the data.
200 202 203 204 206 211 212 215 216 217 228
225 223 228 230 240 235 240 244 243 245 244

 4 What is the skew of the following stem-and-leaf diagrams?

(a) Stem Leaf
1 112

3 023459
5 2377
7 145
9 44
11 1

(b) Stem Leaf
1 1125
3 0234
5 2377
7 145679
94
11 1 3 6

(c) Stem Leaf
0 37
1 125
2 02379
3 237
4 48

 Data Description 6.2 Measures of Location

120  Learning Outcomes

● Find and interpret the mean, mode, median, quartiles and percentiles for
ungrouped data.

● Construct and interpret box-and-whisker plots for ungrouped data.
● Find and interpret the mean, mode, median, quartiles and percentiles for

grouped data.
ChapterPENERBIT 6
ILMU Mean, Mode, Median, Quartiles and Percentiles for
BAKTI SDN. BHD.Ungrouped Data

 1 Measures of location describe the central tendency of the data. The three most
common measures of central tendency are the mean, mode and median.

 2 The mean for a set of data for n observations is simply the average of the
values and is denoted by x̅ .

x̅ = x1 + x2 + x3 + ... + xn = ∑x
n n

 3 The mode is the value that occurs most frequently.

 4 The median is the value in the middle when the data are arranged in ascending
order.

(a) For an odd number of observations, the median is the value in the
middle.

n + 1 th
2
∙ ∙Median =

(b) For an even number of observations, the median is the average value of
two numbers in the middle.

∙ ∙ ∙ ∙Median =n th n th
2 2
+ + 1

2

 5 Percentiles indicate the percentage of scores that fall below a particular
value. Percentiles divide ordered data into hundredths. Recall that a percent
means one hundredth. Therefore, percentiles mean the data is divided into 100
sections.

∙x⌈s⌉, when s is not an integer
Pk = xs + xs + 1 , when s is an integer
2

where s = kn and ⌈s⌉ is the least integer greater than s.
100

Data Description 

 6 Quartiles divide a list of data into four equal parts. The first quartile is equal
to the 25th percentile, the second quartile is equal to 50th percentile (median)
and the third quartile is equal to the 75th percentile.

Q1 = P25, Q2 = P50, Q3 = P75

  7 Interquartile range, IQR is the difference between the upper and lower
quartile of a given data.

IQR = Q3 – Q1

For a given data set, if
(a) mean = median = mode, then the data is symmetrical,
(b) mean < median < mode, then the data is skewed to the left,
(c) mean > median > mode, then the data is skewed to the right.
PENERBIT Chapter 6
ILMU
BAKTI SDN. BHD.Example 1

The following data shows the Air Pollutant Index (API) readings recorded in
Selangor.

65 64 63 61 62 58 58 57 58 51
52 53 50 75 74 49 48 46 49 73

Find the (b) median, (c) mode.
(a) mean,

Solution
46, 48, 49, 49, 50, 51, 52, 53, 57, 58, 58, 58, 61, 62, 63, 64, 65, 73, 74, 75

Arrange the data in
ascending order.

46 + 48 + 49 + 49 + 50 + 51 + 52 + 53 + 57 + 58 + 58 + 58 + 61

(a) Mean = + 62 + 63 + 64 + 65 + 73 + 74 +75
= 58.3 20

(b) n = 20 (even number)

20 th 20 th Use this formula when n
2 2 is an even number.
+ +1
∙ ∙ ∙ ∙
Hence, median = 2

= 10th + 11th
2

= 58 + 58
2
= 58

(c) Mode = 58 58 occurs most frequently.

121

 Data Description Example 2

122 The following observations are arranged in ascending order.
23, 31, 34, 39, 40, x – y, 51, 2x + y, 58, 58

The median of the data is 43.5 and the mean of the data is 43.3. Find the values
of x and y.
ChapterPENERBIT 6
ILMU Solution
BAKTI SDN. BHD.
n = 10 (even number) th th

∙ ∙ ∙ ∙H ence, median =10+ 10 +1
2 2

5th + 6th 2
2
43.5 =

43.5 = 40 + x – y
2
87 = 40 + x – y

x – y = 47 ……… 1

M ean = ∑x
n

43.3 = 23 + 31 + 34 + 39 + 40 + x – y + 51 + 2x + y + 58 + 58
10
334 + 3x
43.3 = 10

334 + 3x = 433

3x = 99

x = 33

From 1 , 33 – y = 47

y = –14

Example 3

The marks for five students in three Quiz Marks
different trigonometry quizzes are shown A 58, 66, 86, 90, 90
in the table on the right. For each quiz, B 90, 90, 90, 90, 90
calculate its mean, median and mode. C 70, 70, 75, 85, 100
Then, analyse the shape of its distribution.

Solution

For Quiz A, 5 + 1 th
2
Mean = 58 + 66 + 86 + 90 + 90 ∙ ∙Median =
5
= 78 = 3rd observation

n = 5 (odd number) = 86

Mode = 90

Mean < Median < Mode. Hence, this

distribution is skewed to the left.

For Quiz B, For Quiz C, Data Description 

Mean = 90 + 90 + 90 + 90 + 90 Mean = 70 + 70 + 75 + 85 + 100 123
5 5
= 90 = 80

n = 5 (odd number) n = 5 (odd number)
PENERBIT
ILMU5 + 1 th 5 + 1 th
BAKTI SDN. BHD.2 2
∙ ∙Median = ∙ ∙Median =

= 3rd observation = 3rd observation

= 90 = 75

Mode = 90 Mode = 70

Mean = Median = Mode. Mean > Median > Mode.

Hence, this distribution is Hence, this distribution is Chapter 6
symmetrical. skewed to the right.

Example 4

The following data shows the height (to the nearest cm) of 15 students in a class.

Find the Stem Leaf
(a) first quartile 14 0 3 5 5
15 2 3 5 6
16 1 3 4 5 6
17 0 1

Key: 14 | 3 means 143 cm

(b) third quartile (c) interquartile range.

Solution

(a) First quartile, Q1 = P25

s = 15 = 3.75 (not an integer) The least integer greater than 3.75 is 4.
4
Hence, Q1 = x4

= 145 cm

(b) Third quartile, Q3 = P75

s= 3(15) = 11.25 (not an integer) The least integer greater than 11.25 is 12.
4
Hence, Q3 = x12

= 165 cm

(c) Interquartile range, IQR = Q3 – Q1
= 165 cm – 145 cm

= 20 cm

 Data Description Example 5

124 Find the value of the first quartile, the median and the 80th percentile of the
following data.

8, 12, 15, 24, 35, 44, 47, 49, 52, 58, 61, 65
ChapterPENERBIT 6
ILMU Solution
BAKTI SDN. BHD.
First quartile, Q1 = P25

s= 12 = 3 (integer)
4
x3 + x4
Hence, Q1 = 2

= 15 + 24
2

= 19.5

n = 12 (even) th th

∙ ∙ ∙ ∙Hence, median = 12 + 12 + 1
2 2

2

= 6th + 7th
2

= 44 + 47
2
= 45.5

For 80th percentile, s = 80(12) = 9.6 (not an integer) The least integer
100 greater than 9.6 is 10.

P80 = x10 = 58

Box-and-whisker Plots

 1 A box-and-whisker plot is a graphical method used to illustrate the spread
(dispersion) and skewness of a set of data.

 2 A box-and-whiskers plot displays the five number summary of a set of data.
(a) Minimum value: the smallest value in the data set excluding any outliers.
(b) Maximum value: the biggest value in the data set excluding any outliers.
(c) Median: the middle value in the data set.
(d) First quartile: the middle value of the first term and the median.
(e) Third quartile: the middle value of the median and the last term.

Minimum Median or Maximum Upper
value second quartile value fence

Lower • Outlier
fence

First quartile Third quartile
Figure 6.2 Box-and-whisker plot

 3 The following are the steps to construct a box-and-whisker plot: Data Description 
Step 1: Calculate the first quartile (Q1), median (Q2), third quartile (Q3) and
125
interquartile range (IQR).
Step 2: Calculate the lower fence and upper fence.

Lower fence = Q1 – 1.5(IQR)
Upper fence = Q3 + 1.5(IQR)
Step 3: Find the minimum and maximum values within the lower and upper
fences.
Step 4: C onstruct a box from the first quartile to the third quartile and mark
the second quartile with a vertical segment.
Step 5: Draw whiskers from Q1 to the minimum value and Q3 to the maximum
value.
Step 6: If there are any points outside the fence, plot the outlier point.
 4 The comparison between the shape and skewness of the distribution are as the
following.
(a) If the median is of equal distance to Q1 and Q3, then the data distribution is
symmetrical.

Q1 Q2 Q3

(b) If the median is nearer to Q1, then the data distribution is positively skewed.
PENERBIT Chapter 6
ILMU
BAKTI SDN. BHD.Q1 Q2Q3

(c) If the median is nearer to Q3, then the data distribution is negatively skewed.

Q1 Q2 Q3

Example 6

Draw a box-and-whisker plot to represent the following data.

Stem Leaf
4 578
5 0245789
6 12456
7 0368
81

Key: 7 | 0 means 70

SolutionPENERBIT f fx fx2 Data Description 
ILMU 1 3 9
ScoreBAKTI SDN. BHD.2 8 32 137
3 16 80 400
4 11 77 539
5 6 48 384
7 9 81 729
8 5 50 500
9 ∑ f = 50 ∑ fx = 347 ∑ fx2 = 2 593
10

Mean, x ̅ = 347 = 6.94 Chapter 6
50

∙ ∙ ∙Standard deviation, s =1
49 2 593 – (347)2
50
= 1.942

Example 4

A random sample of 100 students was asked how long it took them to
complete their assignment the previous night. The time was recorded and
summarised in the following table.

Time (minutes) Frequency
30 – 35 6
35 – 40 10
40 – 45 19
45 – 50 30
50 – 55 20
55 – 60 15

Calculate the mean and variance of the data.

Solution

Time (minutes) x f fx fx2
6 337.5
30 – 35 32.5 6 195 14 062.5
34 318.75
35 – 40 37.5 10 375 67 687.5
55 125
40 – 45 42.5 19 807.5 49 593.75
∑ fx2 = 227 125
45 – 50 47.5 30 1 425

50 – 55 52.5 20 1 050

55 – 60 57.5 15 862.5

∑ f = 100 ∑ fx = 4 715

 Data Description Mean, x̅ = 4 715 = 47.15
100
138
∙ ∙Variance, 1
s2 = 99 227 125 – (4 715)2
100
= 48.61
ChapterPENERBIT 6
ILMU Example 5
BAKTI SDN. BHD.
A group of 70 students took a statistics test. Their marks are summarised in the
following table. Calculate the mean and standard deviation of the data.

Marks Number of students
30 ⩽ x ⩽ 39 2
40 ⩽ x ⩽ 49 3
50 ⩽ x ⩽ 59 11
60 ⩽ x ⩽ 69 14
70 ⩽ x ⩽ 79 20
80 ⩽ x ⩽ 89 15
90 ⩽ x ⩽ 99 5

Solution

Marks xf fx fx2
30 – 39
40 – 49 34.5 2 69 2 380.5
50 – 59
60 – 69 44.5 3 133.5 5 940.75
70 – 79
80 – 89 54.5 11 599.5 32 672.75
90 – 99
64.5 14 903 58 243.5

74.5 20 1 490 111 005

84.5 15 1 267.5 107 103.8

94.5 5 472.5 44 651.25

∑ f = 70 ∑ fx = 4 935 ∑ fx2 = 361 997.55

Mean, x̅ = 4 935
70
= 70.5

∙ ∙ ∙Standard deviation, s = 1
69 361 997.55 – (4 935)2
70
= 14.28

Example 6 Data Description 

The values of daily sales made at a restaurant last year are summarised in the 139
following table. Calculate the variance of the data.

PENERBIT Sales Number of days Chapter 6
ILMU 1 – 250 170
BAKTI SDN. BHD.251 – 500 90
501 – 750 76
751 – 1 000 15
1 001 – 1 250 10
1 251 – 1 500 4

Solution

Sales x f fx fx2

1 – 250 125.5 170 21 335 2 677 542.5

251 – 500 375.5 90 33 795 12 690 022.5

501 – 750 625.5 76 47 538 29 735 019

751 – 1 000 875.5 15 13 132.5 11 497 503.75

1 001 – 1 250 1 125.5 10 11 255 12 667 502.5

1 251 – 1 500 1 375.5 4 5 502 7 568 001

∑ f = 365 ∑ fx = 132 557.5 ∑ fx2 = 76 835 591.25

∙ ∙Variance, 1
s2 = 364 76 835 591.25 – (132 557.5)2
365

= 78 831.1

Pearson’s Coefficient of Skewness

 1 The skewness of a set of data can be determined using the Pearson’s coefficient
of skewness.

 2 There are two ways to calculate the value of Pearson’s coefficient of skewness.

Sk = Mean – Mode
Standard deviation

or

3(Mean – Median)
Sk = Standard deviation

 Data Description 3 The value for Pearson’s coefficient of skewness can be negative, zero, or
positive.
140 (a) If Sk < 0, then the distribution is negatively skewed or skewed to the left.
(b) If Sk = 0, then the distribution is symmetrical.
(c) If Sk > 0, then the distribution is positively skewed or skewed to the right.

Example 7
ChapterPENERBIT 6
ILMU For a distribution, the mode is 50, the mean is 45 and the variance is 150.
BAKTI SDN. BHD.Calculate the Pearson’s coefficient of skewness and interpret your answer.

Solution

Sk = Mean – Mode Use the formula that has
Standard deviation both the mode and mean.

= 45 – 50
√150

= – 0.408
The distribution is negatively skewed.

Example 8

Find the Pearson’s coefficient of skewness for the data represented by the
following stem-and-leaf diagram.

Stem Leaf
4 24
5 355
6 23666899
7 357
8 11
97

Key: 5 | 3 means 53

Solution

∑x = 4 2 + 44 + 53 + 55 + 55 + 62 + 63 + 66 + 66 + 66 + 68 + 69 + 69 + 73
+ 75 + 77 + 81 + 81 + 97

= 1 262
∑x2 = 422 + 442 + 532 + 552 + 552 + 622 + 632 + 662 + 662 + 662 + 682 + 692

+ 692 + 732 + 752 + 772 + 812 + 812 + 972
= 87 000

Mean, x̅ = 1 262 = 66.42
19

n = 19 (odd number)

PENERBIT8CHAPTER
ILMU
BAKTI SDN. BHD.Random Variables

8.1 Discrete Random Variables

 Learning Outcomes

● Construct probability distribution tables and probability distribution functions.
● Find the cumulative distribution function.
● Find the probabilities from:
(i) probability distribution tables or functions
(ii) cumulative distribution functions
● Find the expectation and variance of discrete random variables.

Random Variables

1 A random variable is a variable whose value is determined by the outcome of
a random experiment.

2 There are two types of random variables:
(a) discrete random variables
(b) continuous random variables

Random variables

Discrete random variable Continuous random variable

A random variable that assumes A random variable that can assume

countable values. any real value in an interval.

Example: The number of students Example: The mass of students in

participating in the National a particular art class.

Statistics Competition.


Figure 8.1 Types of random variables

185

 Random Variables Example 1

186 Two fair coins are tossed. Let x be the random variable which represents the
number of tails. Find the possible values of x.

Solution
Let T represents the event of obtaining a tail
H represents the event of obtaining a head

Sample space, S = {TT, TH, HT, HH}
The possible outcomes will be 2 tails, 1 tail and 1 head or 2 heads.
The possible values of x are 0, 1 and 2.
BAKTI SDN. BHD.
Chapter 8 Example 2

Indicate whether the following random variables are discrete or continuous.
(a) The number of vehicles owned by 100 families living in a particular

housing area.
(b) The time taken to finish a game.
(c) The number of cars in a particular shopping mall car park.
(d) The lifespan of a particular smartphone battery.

Solution The number of vehicles can be counted.
(a) Discrete random variable
(b) Continuous random variable The time taken is considered as an
(c) Discrete random variable interval of time.
(d) Continuous random variable
ILMU The number of cars can be counted.

The lifespan is considered as an
interval of time.

Probability Distribution Function (PDF) for a Discrete
Random Variable

PENERBIT 1 The probability distribution of a discrete random variable lists all the
possible values that the random variable can assume and their corresponding
probabilities.

2 If X is a discrete random variable which takes values x1, x2, x3, …, xn, then its
probability distribution function P(X = x) is the respective probability of X,
that is

P(X = xi) = f (xi), i = 1, 2, 3, …, n

3 There are two characteristics of a probability distribution for a discrete random
variable.
(a) Each probability is between zero and one, inclusive.

0  P(X = x )  1, where P(X = x) is the probability distribution function.

(b) The sum of all probabilities is 1.

n
∑ P(X = xi) = 1

i=1

4 The mode of a discrete random variable is the value of x when the probability Random Variables 

P(X = x) is the highest. ∑x 187
x0
5 F(x) is the cumulative distribution function where F(x) = P(X  x) = P(X = x).

Example 3
PENERBIT
A coin is tossed twice. x represents the number of heads obtained.ILMU
(a) Show that x is a discrete random variable.BAKTI SDN. BHD.
(b) Construct a probability distribution table.
(c) Sketch a graph for the probability distribution of x.
(d) Determine the mode.

Solution Chapter 8
(a) Let T represents the event of obtaining a tail
H represents the event of obtaining a head

Sample space, S = {TT, TH, HT, HH}

P(X = 0) = P(HH) = 1 × 1 = 1
2 2 4

P(X = 1) = P(TH) + P(HT)
   
= 1 × 1 + 1 × 1
2 2 2 2

= 1 + 1
4 4
1
= 2

P(X = 2) = P(TT)
1 1
= 2 × 2

= 1
4

∑2x = 0 P(X = x) = P(X = 0) + P(X = 1) + P(X = 2)
= 1 1 + 1
4 + 2 4

=1

Therefore, x is a discrete random variable.

(b) x 0 1 2

P(X = x) 1 1 1
4 2 4

(c) P(x = x)

1 x
2

1
4

0 0 1 2

(d) Mode = 1

 Random Variables Example 4

188 The random variable X has the following probability distribution.

 k, x = 1, 3, 5
P(X = x) = kx2, x = 2, 4

(a) Determine the value of k.BAKTI SDN. BHD.
(b) Construct a probability distribution table for X.
(c) Find P(X = 3).
(d) Find P(X = 2 or X = 4).

Solution

Chapter 8 (a) ∑5x = 1 P(X = x) = 1

P (X = 1) + PX = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

k + 4k + k + 16k + k = 1
1
k = 23

(b) x 123 4 5

P(X = x) 141 16 1
23 23 23 23 23

1
23
(c) P(X = 3) =

(d) P(X = 2 or X = 4) ‘Or’ means plus (+)

= P(X = 2) + P(X = 4)

= 4 + 16
23 23
ILMU
= 20
23

Example 5

The probability distribution function of a discrete random variable W is given
by
PENERBIT
P(W = w) = h(3w + 1), w = 0, 1, 2, 3, 4

(a) Determine the value of h.

(b) Construct a probability distribution table for W.

(c) Find

(i) P(W  3) (iv) P(W  3) (vii) P(2  W  4)
(viii) P(2  W  4)
(ii) P(W  3) (v) P(W  3) (ix) P(2  W  4)

(iii) P(W = 3) (vi) P(2  W  4)

Solution

(a) ∑w4 = 0 P(W = w) = 1

P(W = 0) + P(W = 1) + P(W = 2) + P(W = 3) + P(W = 4) = 1

h + 4h + 7h + 10h + 13h = 1

35h = 1

h = 1
35

(b) w 0123 4 Random Variables 

P(W = w) 1 4 7 10 13 189
35 35 35 35 35

(c) (i) P(W  3) = P(W = 0) + P(W = 1) + P(W = 2)
PENERBIT
ILMU = 1 + 4 + 7
BAKTI SDN. BHD.353535

= 12
35

(ii) P(W  3) = P(W = 0) + P(W = 1) + P(W = 2) + P(W = 3)

= 1 + 4 + 7 + 10 Chapter 8
35 35 35 35

= 22
35

(iii) P(W = 3) = 10
35

= 2
7

(iv) P(W  3) = P(W = 4)

= 13
35

(v) P(W  3) = P(W = 3) + P(W = 4)

= 10 + 13
35 35

= 23
35

(vi) P(2  W  4) = P(W = 3)

= 10
35

= 2
7

(vii) P(2  W  4) = P(W = 2) + P(W = 3)

= 7 + 10
35 35

= 17
35

(viii) P(2  W  4) = P(W = 3) + P(W = 4)

= 10 + 13
35 35

= 23
35

(ix) P(2  W  4) = P(W = 2) + P(W = 3) + P(W = 4)

= 7 + 10 + 13
35 35 35

= 30
35

= 6
7

 Random Variables Example 6

190 Let x be the number of students participating in a mathematics competition.
x
The probability distribution function for x is given by P(X = x) = 45 for x = 1,

2, 3, 4, 5, 6, 7, 8. What is the probability of the number of participants being as
BAKTI SDN. BHD.
follows?

(a) Fewer than 5 students (f) Between 5 and 7 students

(b) At most 5 students (g) At least 5 but fewer than 7 students

(c) Exactly 5 students (h) More than 5 but not more 7 students

(d) More than 5 students (i) At least 5 and at most 7 students

(e) At least 5 students

Chapter 8 Solution

(a) P(fewer than 5) = P(X  5)

= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 1 + 2 + 3 + 4
45 45 45 45

= 10
45

= 2
9
‘At most 5’ is the same as 'not more than 5’
(b) P(at most 5) = P(X  5)

= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

= 1 + 2 + 3 + 4 + 5
45 45 45 45 45

= 15
45
ILMU
= 1
3

(c) P(exactly 5) = P(X = 5)

= 5
45

= 1
9
PENERBIT
(d) P(more than 5) = P(X > 5)

= P(X = 6) + P(X = 7) + P(X = 8)

= 6 + 7 + 8
45 45 45

= 21
45

= 7
15

(e) P(at least 5) = P(X  5)

= P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= 5 + 6 + 7 + 8
45 45 45 45

= 26
45

Random Variables 

(f) P(between 5 and 7) (h) P(more than 5 but not more than 7)

= P(5  x 7 ) = P(5  X  7)

= P(X = 6) = P(X = 6) + P(X = 7)

= 465 = 6 + 7
45 45
PENERBIT
ILMU =2 = 13
BAKTI SDN. BHD.15 45

(g) P(at least 5 but fewer than 7) (i) P(at least 5 and at most 7)

= P(5  X  7) = P(5  X  7)

= P(X = 5) + P(X = 6) = P(X = 5) + P(X = 6) + P(X = 7)

= 5 + 465 = 5 + 6 + 7 Chapter 8
45 45 45 45

= 11 = 18
45 45

= 2
5

Cumulative Distribution Function (CDF) for a Discrete
Random Variable

1 Given a discrete random variable X, its cumulative distribution function
(CDF) is the probability that X is less than or equal to a given value x.

2 The cumulative distribution function, F(x) is given by

x

F(x) = P(X  x) = ∑ P(X = x)
x0

P(a  x  b) = P(x  b) – P(x  a)
= F(b) – F(a)

3 The median, m, of a discrete probability distribution function is the value
of m when P(X  m) = 0.5.

4 The median, m, of a cumulative distribution function is the value of m when
F(m) = 0.5.

Example 7

The cumulative distribution for a discrete random variable X is given by the
following table.

x123456

F(x) 1 1 5 2 3 1
12 3 12 3 4

Find

(a) P(X  3) (d) P(X  3) (g) P(3  X  5)

(b) P(X  3) (e) P(X  3) (h) P(3  X  5)

(c) P(X = 3) (f) P(3  X  5) (i) P(3  X  5)

191

 Random Variables Solution

192 (a) P(X  3) = F(2) (f) P(3  X  5) = F(4) – F(3)

= 1 = 2 – 5
3 3 12

(b) P(X  3) = F(3) BAKTI SDN. BHD. = 1
4
= 5
12 (g) P(3  X  5) = F(4) – F(2)

(c) P(X = 3) = F(3) – F(2) = 2 – 1
3 3
= 5 – 1
12 3 1
= 3
= 1
Chapter 8 12

(d) P(X  3) = 1 – F(3) (h) P(3  X  5) = F(5) – F(3)

= 1 – 5 = 3 – 5
12 4 12

= 7 = 1
12 3

(e) P(X  3) = 1 – F(2) (i) P(3  X  5) = F(5) – F(2)

= 1 – 1 = 3 – 1
3 4 3

= 2 = 5
3 12

Conversion between Probability Distribution Functions
(PDF) and Cumulative Distribution Functions (CDF) for
Discrete Random Variables

1 A probability distribution function can be converted into a cumulative
distribution function and vice versa.
ILMU
Example 8

The probability distribution function of a discrete random variable X is given as
follows.
PENERBIT
x0123

P(X = x) 0.2 0.1 0.4 0.3

(a) Find the cumulative distribution function of X and draw the graph of F(x).
(b) Find the value of the mode and median.

Solution
(a) F(0) = P(X  0)

= P(X = 0)
= 0.2
F(1) = P(X  1)
= P(X = 0) + P(X = 1)
= 0.2 + 0.1
= 0.3

F(2) = P(X  2) Random Variables 
= P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.1 + 0.4 193
= 0.7

F(3) = P(X  3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.2 + 0.1 + 0.4 + 0.3
= 1.0

Therefore, the cumulative distribution function is

0, x  0
0.2, 0  x  1
F(x) = 0.3, 1  x  2
0.7, 2  x  3
1.0, x  3

The graph of F(x)

F(x)

1.0

0.7
PENERBIT Chapter 8
ILMU
BAKTI SDN. BHD.0.3
0.2

0 1 2 3 x



(b) Mode = 2 The mode is the value of X which has the highest
probability. P(X = 2) = 0.4 has the highest probability.

Median, m = 2 P(X  m) = 0.5
P(X  1) = 0.3  0.5
P(X  2) = 0.7  0.5
Thus, the median is 2.

Example 9

The probability distribution function of a discrete random variable Y is as
follows.

y 0123

P(Y = y) 0.3 0.2 0.2 0.3

(a) Find the cumulative distribution function of Y and draw the graph of F(y).
(b) Find P(1  Y  3).

Solution F(1) = P(Y  1)
(a) F(0) = P(Y  0) = P(Y = 0) + P(Y = 1)
= 0.3 + 0.2
= P(Y = 0)
= 0.3 = 0.5

 Random Variables F(2) = P(Y  2)
= P(Y = 0) + P(Y = 1) + P(Y = 2)
194 = 0.3 + 0.2 + 0.2
= 0.7

F(3) = P(Y  3)
= P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3)
= 0.3 + 0.2 + 0.2 + 0.3
= 1.0

Thus, the cumulative distribution function is

0, y  0
0.3, 0  y  1
F(y) = 0.5, 1  y  2
0.7, 2  y  3
1.0, y  3
Chapter 8 BAKTI SDN. BHD.

The graph of F(y)

F(y)

1.0

0.7
0.5
0.3

0 1 2 3 y

ILMU
(b) P(1  Y  3) = P(Y  3) – P(Y  1)

= F(3) – F(1)

= 1.0 – 0.5

= 0.5

PENERBIT Example 10
A discrete random variable X has the following cumulative distribution.

2 , 0x1
21

8 , 1x2
21
F(x) =
12 , 2x3
21

17 , 3x4
21

1.0, x  4



(a) Construct the probability distribution function of X.

(b) Sketch the graph of the probability distribution of X.

Summative Practice 8

1 Let X be the random variable representing the number obtained when a biased dice is rolled.
The probability of the biased dice to give odd numbers is three times higher than even numbers
when it is rolled.
(a) If the dice is rolled once,
(i) construct a probability distribution table for X,
(ii) find the probability of getting a number less than 3,
(iii) find the mean and variance of X.
(b) If the dice is rolled 100 times, find the expected value of getting the number 6.
PENERBIT Chapter 8
ILMU2 The cumulative distribution function of a continuous random variable X is given as follows.
BAKTI SDN. BHD.
0, x  0

F(x) = 1 x(x + 4), 0x4
32

1, x  4



(a) Calculate P(|X – 1|  1).

(b) Find the median.

(c) Determine the probability density function of X. Hence, evaluate E(3X2 – 1).

3 The continuous random variable X has the cumulative distribution function F(x) given by

0, x  0

x2 , 0x2
6 2x3
F(x) =
– x2 + 2x – 2,
3

1, x  3


Find

(a) P(1  X  2.2),

(b) the value of the median,

(c) the probability density function of X,

(d) the expected value of X,

(e) the variance of X, given that E(X2) = 19 .
6

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

xy, x = y
W = f(x) = |x – y|, x ≠ y



(a) Write all the outcomes for W = 4 and show that P(W = 4) = 5 .
36

211

Random Variables 

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

(c) Find P(W  9).
(d) Find the mode of W.
(e) Find E(W) and calculate E(3 – 4W).

5 The probability distribution for a discrete random variable X is as shown in the following table.BAKTI SDN. BHD.

x –2 –1 0 1 2 3

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

Chapter 8 Given that p and q are constants. If E(X) = 0.65, determine the values of p and q. Hence,
calculate the standard deviation of X.


6 The probability distribution function of a discrete random variable X is given by

x , x = 1, 2, 3
17

f(x) = x , x = 4, 5, 6, 7
34

0, otherwise

(a) Calculate P(2  X  5).

(b) Determine the value of Var(X).
(c) Hence, calculate the standard deviation of Y = ( 5 X – 1).



7 A continuous random variable X has a probability density function given by

ax , 0  x  1

ILMUf(x) = a (4 – x), 1  x  4
3

0, otherwise

where a is a constant.

(a) Find the value of a.

(b) Find E(X) and Var(2 – 3X).

(c) Evaluate P(X – E(X)  a).
PENERBIT
(d) Estimate the median.

212

Answers

y

y=x+1
1
PENERBIT1 Numerical Solution 4 (a) 4 (a) Since f(0.5) and f(1) have
ILMU
BAKTI SDN. BHD. opposite signs, 2x2 – 1 = 0
x
Quick Check 1.1 has a root between x = 0.5

1 (a) Since f(1) and f(2) have and x = 1.

opposite signs, ex – 1 – 2 = 0 –2.5 –2 –1.5 –1 –0.5 0 x (b) 0.79
x
has a root between x = 1 and 5 0.5824

x = 2. –1 y = –ex Summative Practice 1

(b) Since f(0) and f(1) have 1 Since f (1) and f (2) have different
signs, 2x4 = 5 + x has a root
opposite signs, e –x – 2x = 0 From the graph, there is between x = 1 and x = 2.
only one real root lies
has a root between x = 0 and between x = –1.5 and x = –1. Approximate root: 1.33
2 (a) Since there is a change in
x = 1. The root lies in the interval
[–1.5, –1]. sign, the root lies between
(c) Since f(0) and f(1) have x = 1 and x = 2.
(b) 1.0737
opposite signs, 2x – 2 + ex = 0 3 (a) Since f(0)  0 and f(1)  0,
f(x) has a root between x = 0
has a root between x = 0 and (b) y and x = 1.
(b) 0.3149
x = 1. 4 1.123
5 (a) Since f(0.1)  0 and
2 (a) Since f(0.5) and f(1) have x f(1)  0, from the
–2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 intermediate value theorem,
opposite signs, x – cos x = 0 there is at least one root in
the interval [0.1, 1].
has a root between x = 0.5 rad y = –2 – x –0.5 y = –e–x (b) Intersection point:
–1 (0.426, 0.852)
and x = 1 rad. 6 –0.8381

(b) Since f(–2) and f(–1) have

opposite signs, x + 0.5 – sin x –1.5

= 0 has a root between x = –2 –2

and x = –1. From the graph, there is only
one real root between x = –1
3 (a) y and x = 0.

4 The root lies in the interval
3 y = x2 –1  x  0.

2 y=4–x Quick Check 1.2
1
1 (a) Since f(0) and f (1) have 2 Integration Answers
x opposite signs, ex – 2 + x = 0
0 1 2 3 4 has a root between x = 0 and
x = 1.
From the graph, the Quick Check 2.1
(b) 0.443
approximate value of the root 2 (a) Since f(–1) and f (0) have

is x = 1.5. opposite signs, x + ex = 0 has 1 (a) x2 –x+c
a root between x = –1 and 2
(b) y x = 0.
(b) –0.5671
3 (a) Since f(–1) and f (0) have (b) ln |x – 1| + c
opposite signs, sin x + x + 1 = 0
3 has a root between x = –1 and
x = 0.
2 (b) –0.51 (c) x – ln |x| + c

y = ex 1 y=3–x (((abd))) 2––e5x2x3e–1+5x1c+ +c c (c) ln188x + c
2
–2 –1 0 1 2 3 x

From the graph, there is only 3 (a) 1 ln |7x + 4| + c
one real root between x = 0 7
and x = 1.
(b) 2 ln |5x + 2| + c

(c) 3 ln |3 – 4x| + c

245

| Answers

4 (a) 1 x2 + c (f) 1 (2 + ln 3 + c Summative Practice 2
2 6
x)2 1 4.876

(b) e3x + 1 e5x + c 2 (a) (3x – 2)(2 + 5x)3 – (2 + 5x)4 + c 2 3
5 15 100 16

(c) 5 x2 + 2x + c (b) 2x (3 + 3 – 4 (3 + 5 + c 3 ln 32
2 3 15 27
x)2 x)2
AnswersPENERBIT
ILMU5 (a) 1x5+ 1 + 5 x2 + c (c) x6 ln x – x6 + c 4 2π units3
BAKTI SDN. BHD.52x2 2 6 36 3

(b) – 2 – 2 +c (d) – 2 (1 + 2x)(1 – 1 – 5 (a) (1 – x)10 – (1 – x)9 +c
3x3 3 10 9
5 3x)2

5x2 8 3 π
27 2
(c) 2x2 + 8 3 + x + c (1 – 3x)2 + c (b) – 1 unit2
3
x2

6 (a) e2x + 2x – e–2x +c (e) x5 ln 4x – x5 +c 6 2.925
2 2 5
7 0.6478

(b) – 1 + 1 +c (f) – 31x3ln x – 1 + c 8 2.1061
e2x ex 9x3
9 1.1039

(c) – 6 + c 3 (a) 1 ln |x2 – 1| + c Summative Assessment Test
 ex 2
(UPS) 1
| | (b) 1 ln x–1 +c 1 D 2 A 3 D 4 D 5 D
2 x+1 6 B 7 C 8 A 9 C 10 B
Quick Check 2.2 11 D 12 B 13 C 14 D 15 A
| | (c) 1 ln x–9 +c 16 C 17 B 18 B 19 C
 1 1 18 x+9
(a) 2 x – sin 2x +c
2 (d) ln |x| + 2 + c
x + 2
(b) – 1 cos 2x + c
4 4 (a) –x cos x + sin x + c 3 First Order Differential
Equations
  (c) 1 x – 1 sin 4x +c (b) –x2 cos x + 2x sin x + 2 cos x
8 4
+c

2 (a) – 1 cos 8x + c 5 (a) – 1 cos x2 + c Quick Check 3.1
4 2
1 (a) y = cx
(b) 1 sin 10x + c (b) sin3 x + c (b) y = c(x – 1) – 1
2 3
3
(c) 2 tan 2x + c (c) – cos3 x +c (c) y = 3 2 x2 + c
3
3 (a) – 1 cos 2x + 1 sin 3x + c   1
2 3 (d) –ln |cos x| + c (d) y = –ln – 2 e2x – c

(b) 2 tan 3x + 1 cos 5x + c Quick Check 2.4 (e) P = Aekt
5
1 1 (f) R = Ae–kt
(c) 8 sin 2x + 5 cos 3x + c 1 (a) –2
(g) θ = Ae–kt + 10
(b) 10.208
1 1 2
4 (a) 2 x – 8 sin 4x + c 2 9 units2 2 y =
2 1
(1 + x2)2
1 1
(b) 4 sin 2x + 2 x+c 3 4π units2 3 y = 1
– 3x2
(c) – 1 cos 8x + c 4 13 π units2 28
16 3
4 y4 – 4 = 5x – 1 – 2
5 136.53π units3 4 y x

Quick Check 2.3 Quick Check 2.5 Quick Check 3.2

1 (a) x – 1 + ln |x – 1| + c 1 (a) 70.376 1 (a) y = e5x + ce4x
1 (b) 22.500
(c) 0.911 1 c
(b) (x2 – 1)2 + c (d) 1.475 3 ex3
(e) 1.056
(c) 1 (x2 – 3 + c 2 10.582 (b) y = +
3 3 1.032
1)2

(d) 2 (x – 5 + 2 (x – 3 + c (c) y= x – 1 + c
5 3 2 4 e2x
1)2 1)2

(e) 6 – 3x – ln |2 – x| + c (d) y = ex cos (e–x) + cex

(e) y = x2 + cx

246

PENERBIT
ILMU
BAKTI SDN. BHD.