SM025 Self-Directed Learning

NSO/KKW/CCW

c) = 0, = 2, = 4
ℎ = − = 2−0 = 0.5

4

0 = 0.0 = 2
1 = 0.5 1.0000
2 = 1.0
3 = 1.5 1.2840
4 = 2.0 2.7183
9.4877
Total 54.5982
55.5982 13.4900

∫02 2 = 0.5 [55.5982 + 2(13.4900)]
2

= 20.645

d) = 1, = 3, ℎ = 0.5 The width is 0.5

0 = 1.0 = ln
1 = 1.5 0.0000
2 = 2.0
3 = 2.5 0.6082
4 = 3.0 1.3863
2.2907
Total 3.2958
3.2958 4.2852

3 = 0.5 [3.2958 + 2(4.2852)]
2
∫ ln

1

= 2.967 3 decimal places for final answer

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Chapter 4: Conics

4.1 Circle

1. Standard equation of circle
( − ℎ)2 + ( − )2 = 2

Centre (ℎ, )
Radius =
2. General equation of circle

2 + 2 + 2 + 2 + = 0

Centre (− , − )
Radius, = √ 2 + ℎ2 −

3. The perpendicular distance (shortest distance) from a point (ℎ, ) to a straight line
+ + = 0 is given by | √ℎ + 2 + + 2 |.

Question:
Find the equation of the circle with centre (−2, 3) and radius 5.

Solution:

[ − (−2)]2 + [ − (3)]2 = 52

2 + 4 + 4 + 2 − 6 + 9 = 25

2 + 2 + 4 − 6 − 12 = 0

Question:
Find the centre and radius of the circle 2 + 2 − 2 + 4 = 4 .

Solution:
By comparing with the general equation
2 + 2 + 2 + 2 + = 0

2 = −2 2 = 4 = −4
= −1 = 2

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Centre, = (− , − )
(1, −2)

Radius, = √ 2 + 2 −
= √(−1)2 + (2)2 − (−4)
=3

Question:
Find the equation of the radius that goes through the points (−2,0), (1, 0) and (1, −3).

Solution:

Substitute each of the point in the general equation of circle
At (−2, 0):
(−2)2 + (0)2 + 2 (−2) + 2 (0) + = 0
−4 + = −4 --------------- (1)

At (1, 0):
(1)2 + (0)2 + 2 (1) + 2 (0) + = 0
2 + = −1 --------------- (2)

At (1, −3):
(1)2 + (−3)2 + 2 (1) + 2 (−3) + = 0
2 − 6 + = −10 --------------- (3)

Solving equation (1) and (2) gives = 1 and = −2
2

Substitute values of and into equation (3),
2 (12) − 6 + (−2) = −10

= 3

2

∴ The equation of circle is

2 + 2 + 2 1 + 2 3 − 2 = 0
(2) (2)

2 + 2 + + 3 − 2 = 0
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4.2 Ellipses

Standard equation for ellipse
( − ℎ)2 ( − )2
2 + 2 = 1

Technique axby

➢ always with > >
➢ always with 2 = 2 − 2 2 = 2 − 2
refer to . refer to .
Value If > , then longer than . If > , then longer than .
Types of ellipse So, horizontal ellipse is formed. So, vertical ellipse is formed.
Only coordinate change in Only coordinate change in
Centre determining Foci and Vertices. determining Foci and Vertices.
Foci = (ℎ, ) = (ℎ, )
Vertices 1 = (ℎ + , ) 1 = (ℎ, + )
Length of major axis 2 = (ℎ − , ) 2 = (ℎ, − )
Length of minor axis 1 = (ℎ + , ) 1 = (ℎ, + )
Shape 2 = (ℎ − , ) 2 = (ℎ, − )
2 2
2 2

2 1
2 2
2 2 1 1 1
2
2

2

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Question:

Find the centre, vertices and foci of the following ellipses.

(a) ( +1)2 + ( −2)2 = 1

94

(b) ( −1)2 + ( +2)2 = 1

49

(c) 2 + 4 + 4 2 = 0

Solution:

(a) ( +1)2 + ( −2)2 = 1

94

Centre, (−1,2)

Since > , then the ellipse sits horizontally.
2 = 2 − 2
2 = 9 − 4

= ±√5

Vertices, 1(2,2) and 2(−4,2)

Foci, 1(−1 + √5, 2) and 2(−1 − √5, 2)

(b) ( −1)2 + ( +2)2 = 1

49

Centre, (1, −2)

Since > , then the ellipse sits vertically.
2 = 2 − 2
2 = 9 − 4

= ±√5

Vertices, 1(1,1) and 2(1, −5)

Foci, 1(1, −2 + √5) and 2(1, −2 − √5)
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(c) 2 + 4 + 4 2 = 0
( + 2)2 − 4 + 4 2 = 0
( + 2)2 + 4 2 = 4
( +2)2 + 2 = 1

4

Centre, (−2,0)

Since > , then the ellipse sits horizontally.
2 = 2 − 2
2 = 4 − 1

= ±√3

Vertices, 1(0,0) and 2(−4,0)

Foci, 1(−2 + √3, 0) and 2(−2 − √3, 0)

Question:

Find the equation for each ellipse as described below.
(a) Centre at (−1, −1), focus at (−4, −1), vertex at (−6, −1).
(b) Length of major axis is 10, foci at (4,9) and (4,3).
(c) Centre at (−4,3), focus at (−6,3), containing the point (−8,3).

Solution:
(a) Coordinate remain same, technique axby, > , then the ellipse sits horizontally.

Focus, (ℎ − , ) = (−1 − , −1)
∴ −1 − = −4
= 3

Vertex, (ℎ − , ) = (−1 − , −1)
∴ −1 − = −6
= 5

2 = 2 − 2
9 = 25 − 2
2 = 25 − 9

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2 = 16
∴ Equation of ellipse is ( +1)2 + ( +1)2 = 1 .

25 16

(b) Technique axby, the coordinate remain same for foci. So, > .

The length of major axis 2 = 10

= 5

Centre is the midpoint of 1 and 2, centre= (4+4 , 9+3)

2 2

(4,6)

Foci, (ℎ, ± ) = (4,6 ± )
6 + = 9
= 3

2 = 2 − 2
9 = 25 − 2
2 = 16
∴ Equation of ellipse is ( −4)2 + ( −6)2 = 1 .

16 25

(c) Technique axby, the coordinate of for centre and focus remain same. So > .
Focus, (ℎ − , ) = (−4 − , 3)
∴ −4 − = −6
= 2

2 = 2 − 2
4 = 2 − 2 ------------ (1)

Substitute (−8,3) into standard equation of ellipse,

(−8+4)2 + (3−3)2 = 1
2 2

16 = 1
2

2 = 16 ----------- (2)

Substitute (2) into (1):

4 = 16 − 2
2 = 12
∴ Equation of ellipse is ( +4)2 + ( −3)2 = 1 .

16 12

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4.3 Parabolas

1. Standard equations of parabola

( − ℎ)2 = 4 ( − )

The parabola either opens upward or downward.

Centre (ℎ, )

Focus = (ℎ, + )

Directrix = −

( − )2 = 4 ( − ℎ)

The parabola either opens to the right or to the left.

Centre (ℎ, )

Focus = (ℎ + , )

Directrix = −

Question:
Find the equation of the following parabolas
(a) Focus at (−5, 0), vertex at (0, 0)
(b) Focus at (0, −2), vertex at (0, 0)

Solution:

(a) ℎ = 0, = 0
(ℎ + , ) = (−5, 0)
∴ ℎ + = −5
(0) + = −5
= −5

( − )2 = 4 ( − ℎ)
[ − (0)]2 = 4(−5)[ − (0)]

2 = −20

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(b) ℎ = 0, = 0
(ℎ, + ) = (0, −2)
∴ + = −2
(0) + = −2
= −2

( − ℎ)2 = 4 ( − )
[ − (0)]2 = 4(−2)[ − (0)]

2 = −8

Question:
Find the vertex, focus and directrix for each parabola and sketch the graph.
(a) ( − 3)2 = 4( + 2)
(b) ( − 1)2 = −
(c) ( + 1)2 = 2( − 2)
(d) ( − 5)2 = −6( + 1)

Solution:

(a) ( − 3)2 = 4( + 2)
4 = 4
= 1

Vertex, (−2,3)
Focus, (−1,3)
Directrix, = −3

(b) ( − 1)2 = −
4 = −1
= − 1

4

Vertex, (0,1)

Focus, (− 1 , 1)
4

Directrix, = 1
4

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(c) ( + 1)2 = 2( − 2)

4 = 2

= 1

2

Vertex, (−1,2)

Focus, (−1, 52)

Directrix, = 3
2

(d) ( − 5)2 = −6( + 1)

4 = −6

= − 3

2

Vertex, (5, −1)

Focus, (5, − 5)

2

Directrix, = 1
2

Question:

Find the vertex, focus, directrix and axis of symmetry of each parabola.
(a) 2 − 4 − 4 + 16 = 0
(b) 2 + 4 − 6 + 16 = 0

Solution:
(a) 2 − 4 − 4 + 16 = 0

2 − 4 = 4 − 16

By completing the square
( − 2)2 − 4 = 4 − 16
( − 2)2 = 4( − 3)

By comparing with the standard equation
( − )2 = 4 ( − ℎ)
4 = 4
= 1

Vertex, (3,2)
Focus, (4,2)
Directrix, = 2
Axis of symmetry is the line = 2

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(b) 2 + 4 − 6 + 16 = 0
2 + 4 = 6 − 16

By completing the square
( + 2)2 − 4 = 6 − 16
( + 2)2 = 6( − 2)

By comparing with the standard equation
( − ℎ)2 = 4 ( − )
4 = 6
3
= 2

Vertex, (−2,2)

Focus, (−2, 7)

2

Directrix, = 1
2

Axis of symmetry is the line = −2

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Chapter 5: Vectors

5.1 Vectors in two and three dimensions

1. A vector is a quantity that has both magnitude and direction.

2. A scalar is a quantity with magnitude but without direction.

3. B
Notation of a vector:
⃗⃗ ⃗⃗ ⃗ , AB, , a
A



( ) or + +


4. Magnitude of vector: | ⃗⃗ ⃗⃗ ⃗ | or | |
Example:
If = + , then magnitude, | | = √ 2 + 2

5. A unit vector is vector of one unit long, ̂ = | |.

6. Parallel vector and
= , where is a scalar

7. Position vector is a vector at the origin .
Example: ⃗ ⃗ ⃗⃗ ⃗ = ⃗⃗ ⃗⃗ ⃗ − ⃗⃗ ⃗⃗ ⃗

8. Direction cosines and direction angles.
Example:
Given = + + .

Direction cosines:
cos =

cos = , cos = , | |
| | | |

Direction angles:

= −1 | |, = −1 | |, = −1
| |

2 + 2 + 2 = 1

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Question:
The position vectors of points A and B with respect to the origin are = 2 − 3 + 2 and
= + + . Find
(a) ⃗⃗ ⃗⃗ ⃗
(b) | ⃗⃗ ⃗⃗ ⃗ |
(c) A unit vector parallel to ⃗ ⃗ ⃗⃗ ⃗

Solution:
(a) ⃗ ⃗ ⃗⃗ ⃗ = ⃗ ⃗ ⃗⃗ ⃗ − ⃗ ⃗ ⃗⃗ ⃗

= ( + + ) − (2 − 3 + 2 )
= − + 4 −

(b) |⃗ ⃗ ⃗⃗ ⃗ | = √(−1)2 + (4)2 + (−1)2
= √18
= 3√2

(c) A unit vector parallel to ⃗ ⃗ ⃗⃗ ⃗ is Opposite direction of ⃗ ⃗ ⃗⃗ ⃗ .

− + 4 − or − + 4 −

3√2 −3√2

= √2 (− + 4 − ) = −√2 (− + 4 − )
6 6

Question:
Given that , and are the direction angles of the vector + − 3 .

(a) Find the direction cosines.
(b) Show that 2 + 2 + 2 = 1.

Solution:

(a) | + − 3 | = √(1)2 + (1)2 + (−3)2

= √11

∴ = 1 , = 1 and = −3
√11 √11 √11

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(b) 2 + 2 + 2

=( 1 2 1 2 + ( −3 2

) +( ) )
√11 √11 √11

=1+1+9

11 11 11

= 1 (shown)

5.2 Scalar Product (Dot Product)

1. ∙ is the scalar product of two vectors and and ∙ = | || | cos , where is
the angle between and .

2. The scalar product of two vectors, ∙ is a scalar.

3. Angle between two vectors and is given by cos = ∙ .

| || |

4. (a) If two vector and are perpendicular, then ∙ = 0.

(b) If two vector and are perpendicular, then

∙ = | || | ( and in the same direction)

Or

∙ = −| || | ( and in the opposite direction)
5. Properties of the scalar product:

(a) ∙ = ∙
(b) ( ) ∙ = ∙ ( ) = ( ∙ ), where is a scalar.
(c) ∙ ( + ) = ∙ + ∙
(d) ( + ) ∙ = ∙ + ∙
(e) ( + ) ∙ ( + ) = ∙ + ∙ + ∙ + ∙
(f) ∙ = | |2

6. If = 1 + 1 + 1 and = 2 + 2 + 2 then
∙ = 1 2 + 1 2 + 1 2

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Question:
If = 2 − + 3 , = + + and = − , find
(a) ∙
(b) ∙ ( + )
(c) ( ∙ ) −

Solution:
(a) ∙ = (2 − + 3 ) ∙ ( + + )

= (2)(1) + (−1)(1) + (3)(1)
=4

(b) ∙ ( + ) = ( + + ) ∙ [(2 − + 3 ) + ( − )]
= ( + + ) ∙ (3 − + 2 )
= (1)(3) + (1)(−1) + (1)(2)
=4

(c) ( ∙ ) = (2 − + 3 ) ∙ ( − )
= (2)(1) + (−1)(0) + (3)(−1)
= −1

( ∙ ) − = − ( + + ) − (2 − + 3 )
= −3 − 4

Question:
Given that | | = 3, | | = 4 and ∙ = 5, find
(a) ∙
(b) ( − ) ∙ ( + )

Solution:
(a) ∙ = | |2

= 32
=9

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(b) ( − ) ∙ ( + ) = ( ∙ ) + ( ∙ ) − ( ∙ ) − ( ∙ )
= ( ∙ ) − ( ∙ )
= | |2 − | |2
= 9 − 16
= −7

Question:
Given = 2 + − and = 3 − 2 − , find the angle between
(a) and
(b) ( + ) and ( − )

Solution:
(a) Angle between two vectors,

cos = ∙
| || |

= (2)(3)+(1)(−2)+(−1)(−1)

√22+12+(−1)2√32+(−2)2+(−1)2

= 5
√6√14

= 0.994 radian

(b) + = (2 + − ) + (3 − 2 − )
= 5 − − 2

− = (2 + − ) − (3 − 2 − )
= − + 3

Angle between two vectors,

( + )∙( − )

cos =

| + || − |

= (5)(−1)+(−1)(3)+(−2)(0)
√52+(−1)2+(−2)2√(−1)2+32

= −8

√30√10

= 2.05 radian

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5.3 Vector Product (Cross Product)

1. × is the scalar product of two vectors and and × = | || | sin ̂ , where
is the angle between and and ̂ is the unit vector perpendicular to both and
.

2. × is a vector perpendicular to both and .

3. Properties of vector product:
(g) × = − ×
(h) × = ( × ), where and are scalars.
(i) × ( + ) = × + ×
(j) ( + ) × ( + ) = × + × + × + ×
(k) and are parallel ⟺ × = 0
(l) and are perpendicular ⟺ × = | || | ̂

4. If = 1 + 1 + 1 and = 2 + 2 + 2 then


× = | 1 1 1|

2 2 2

= | 12 12| + | 12 12| + | 21 12|

5. 1
2
Area of triangle = | × |










Area of parallelogram = | × |

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Question:
If = + − 3 , = 2 − 3 + and = + + , find
(a) ×
(b) 3 × 5
(c) ×
(d) ×

Solution:


(a) × = |1 1 −3|
2 −3 1

= |−13 −13| − |12 −13| + |21 −13|
= −8 − 7 − 5

(b) 3 × 5 = 15( × )
= 15 (−8 − 7 − 5 )
= −120 − 105 − 75


(c) × = |1 1 −3|

11 1

= |11 −13| − |11 −13| + |11 11|

= 4 − 4

(d) × = −( × )
= −4 + 4

Question:
Given that = 2 − + 3 and = 3 + − 2 . Find a unit vector perpendicular to both
and .

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Solution:


× = |2 −1 3 |
3 1 −2

= |−11 −32| − |23 −32| + |23 −11|

= − + 13 + 5

| × | = √(−1)2 + (13)2 + (5)2
= √195

(− +13 +5 ) −(− +13 +5 )

A unit vector perpendicular to × = √195 Or √195

Question:
A plane contains points (1, 1, 0), (3, −2, 1) and (5, 7, 2). Find
(a) A vector normal to the plane
(b) The area of triangle .

Solution:
(a) ⃗⃗ ⃗⃗ ⃗ = ⃗⃗ ⃗⃗ ⃗ − ⃗⃗ ⃗⃗ ⃗

= (3 − 2 + ) − ( + + 0 )
= 2 − 3 +

⃗⃗ ⃗⃗ ⃗ = ⃗ ⃗ ⃗⃗ ⃗ − ⃗⃗ ⃗⃗ ⃗
= (5 + 7 + 2 ) − ( + + 0 )
= 4 + 6 + 2

= ⃗ ⃗ ⃗⃗ ⃗ × ⃗⃗ ⃗⃗ ⃗

12| + |42 −63|
= |2 −3 1|

462

= |−63 21| − |24

= −12 + 24

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(b) Area of ∆ = 1 | ⃗⃗ ⃗⃗ ⃗ × ⃗ ⃗ ⃗⃗ ⃗ |
2

= 1 √(−12)2 + (24)2

2

= 13.4 unit2

5.4 Application of Vectors in Geometry

1. Equation of a straight line:

Vector equation Parametric equations Cartesian equations

= + = 1 + − 1 = − 1 = − 1
= 1 +
is position vector of any = 1 +
point on the line

is position vector of a
point on the line

is the direction vector
of the line

To find the equation of a straight line, we need
(a) A given point on the line
(b) A vector parallel to the line

2. Equation of a plane: Cartesian equation
Vector equation
∙ = ∙ + + =

Normal vector,
= + +

To find the equation of a plane, we need
(a) A given point on the plane
(b) A vector normal to the plane

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Question:
Find a vector equation of the line that contains a point (1, 3, −5) and 3 + + .

Solution:
= + 3 − 5
= 3 + +

Vector equation of the line is given by
= +

∴ = + 3 − 5 + (3 + + )

Question:
Find the vector and Cartesian equation of the plane containing (1, 2, 3) and perpendicular to

3 + − 2 .

Solution:
= + 2 + 3
= 3 + − 2

Vector equation of the plane is given by
∙ = ∙
∙ (3 + − 2 ) = ( + 2 + 3 ) ∙ (3 + − 2 )
∙ (3 + − 2 ) = 3 + 2 − 6
= −1

Vector equation: ∙ (3 + − 2 ) = −1
Cartesian equation: 3 + − 2 = −1

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Angles of vectors Angle between 2 lines Angle between 2 Angle between a line
Angle between 2 planes and a plane
vectors

= −1 [ ∙ ] = −1 [ 1 ∙ 2 ] = −1 [ 1 ∙ 2 ] = −1 [ ∙ ]
| || | | 1| | 2| | 1| | 2| | || |

Question:

Calculate the angle between the planes
1: 3 + − 4 = 8
2: 4 − 3 + = −2

Solution:
Normal vector of each plane:

1 = 3 + − 4
2 = 4 − 3 +

Angle between 2 planes is given by

= −1 [ 1 ∙ 2 ]
| 1| | 2|

= −1 [ (3 + − 4 ) ∙ (4 − 3 + ) ]

√32 + 12 + (−4)2√42 + (−3)2 + 12

= −1 5
[26]

= 78.9°

Question:
Calculate the angle between the line and plane.

: = 3 + 2 , = 4 − , = 6 +
: 13 − 3 + 11 = 15

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Solution:
Direction vector of line:

= 2 − +

Normal vector of plane:
= 13 − 3 + 11

Angle between a line and a plane is given by
= −1 [ ∙ ]

| || |

= −1 [ (2 − + ) ∙ (13 − 3 + 11 ) ]

√22 + (−1)2 + (1)2√132 + (−3)2 + 112

= −1 [ 40 ]
√6√299

= 70.8°

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Chapter 6: Data Description

A. Stem-and-Leaf diagrams

Example:
The weight of 10 workers (to the nearest kg) in a company are given below:

59 63 97 83 77
75 54 66 68 84

Construct a stem-and-leaf diagram for the weight of the workers.

Solution: Leaf
Stem 4, 9
5 3, 6, 8
6 5, 7
7 3, 4
8 7
9

Key: 5|9 means 59 kg

Question 1:
The following data are collected from a group of 30 workers for their daily earning at a factory.

78 54 81 96 65 79 71 87 93 83
72 81 61 76 86 79 68 50 92 69
77 64 71 89 72 92 57 99 95 84

Solution:

Stem Leaf
5 0, 4, 7
6 1, 4, 5, 8, 9
7 1, 1, 2, 2, 6, 7, 8, 9, 9
8 1, 1, 3, 4, 6, 7, 9
9 2, 2, 3, 5, 6, 9

B. Ungrouped Data

i. Mean, median, mode, quartiles and percentiles

Mean: ̅ = 1+ 2+ 3+⋯+ = ∑


̅ = ∑
∑

( +1) ℎ value when n is an odd number
2
Median: {
( 2 ) ℎ+ ( 2 +1) ℎ value when n is an even number
2

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Mode: the value that occurs most
⌈ ⌉, when s is not an integer

Percentiles: = { + +1 , when s is an integer

2


where = 100 and ⌈ ⌉ is the least integer greater than
Quartiles: 1 = 25, 3 = 75
Interquartile Range, = 3 − 1

Question 2:
Find the mean, mode and median for the following data.
10 7 5 7 3 2 9

Solution:

Mean, ̅ = ∑

10 + 7 + 5 + 7 + 3 + 2 + 9
̅ = 7

= 43

7

= 6.1429 (4 d.p)

Mode= 7 (it appears two times)

Median:

Median is the [7+1] ℎ observation, in which the 4th observation.

2

Rearrange the data: 2 35779 10

4th
observation

∴ The median is 7.

Question 3:
The table shows the amount of money received by John after washing 35 cars.

RM 0 12345
Frequency 3 5 12 9 4 2

Find
(a) The median amount of money
(b) The interquartile range

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Solution:
(a) The cumulative frequency distribution is formed as follow:

RM 0 ≤1 ≤2 ≤3 ≤4 ≤5
Frequency 3 8 20 29 33 35

18th
observation

There are 35 observations, thus the median is the [352+1] ℎ, in which the 18th observation.
∴ The median is RM 2.00.

(b) The first quartile is the [345] ℎ = 8.75 ℎ observation, in which the 9th observation.

Thus 1 = RM 2.00
The third quartile is the [3(435)] ℎ = 26.25 ℎ observation, in which the 27th observation.

Thus 3 = RM 3.00
∴ Interquartile range is RM 3.00 − RM 2.00 = RM 1.00

ii. Box-and-whisker plots

Minimum Second Maximum Upper
value Quartile value fence

Lower X
Fence Outlier

First Third
Quartile Quartile

STEP 1: Calculate the first quartile ( 1), median ( 2), third quartile ( 3) and

interquartile range( ).

STEP 2: Calculate the lower fence and upper fence.

Lower fence: 1 − 1.5
Upper fence: 3 + 1.5
STEP 3: Find the minimum and maximum value within the lower and upper fence.

STEP 4: Draw a box from the first quartile to the third quartile and mark the second

quartile with a vertical segment.

STEP 5: Draw whiskers from 1to minimum value and 3to maximum value
STEP 6: If there are any points outside fence, plot the outlier point.

If median is nearer to 1, then the data is positively skewed.
If median is nearer to 3, then the data is negatively skewed.
If median is equal distance to 1and 3, then the data is symmetrical.

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Question 4:
Draw a box and whisker to represent the data.

Stem Leaf

4 0, 5, 7

5 1, 2, 6, 8, 8, 8, 9

6 1, 2, 4, 5, 6

7 2, 3, 7, 9

85

Key: 4|0 represents 40

Solution:

STEP 1: First Quartile, 1 = 25
= 20

4

= 5 ( )

Hence, 1 = 5+ 6

2
52+56
= 2

= 54

= 20 (even number)

Hence, median = ( 2 ) ℎ+ ( 2 +1) ℎ
2

= 10 ℎ+ 11 ℎ

2

= 59+61

2

= 60

Third Quartile, 3 = 75

= 3(20)
4

= 15 ( )

Hence,

3 = 15 + 16
2

= 66+72

2

= 69

Interquartile Range, IQR = 3 − 1

= 69 − 54
= 15

STEP 2: Lower fence = 1 − 1.5 = 54 − 1.5(15) = 31.5
Upper fence = 3 + 1.5 = 69 + 1.5(15) = 91.5

STEP 3: Minimum value: 40 Tips:
Maximum value: 85 Minimum value is the smallest value
within the fence and maximum value is
the largest value within the fence.

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31.5 60 69 91.5
40 54 85

iii. Variance and standard deviation

, 2 = 1 1 [∑ 2 − (∑ )2
− ]

, = √ 1 1 [∑ 2 − (∑ )2
− ]

Question 5:
A group of 10 married couples was asked how many children they have. The results are as follows.

Number of siblings 0 1 2 3 Ungrouped data
with frequency
Frequency 1342

Calculate the variance and standard deviation of the data.

Solution:
0, 1, 1, 1, 2, 2, 2, 2, 3, 3
∑ = 0 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 3 + 3 = 17

∑ 2 = 02 + 12 + 12 + 12 + 22 + 22 + 22 + 22 + 32 + 32 = 37

, 2 = 1 1 [∑ 2 − (∑ )2
− ]

1 (17)2
= 9 [37 − 10 ]

= 0.9

, = √

= √0.9
= 0.948

C. Grouped Data

i. Mean, median, mode, quartiles and percentiles

Mean: ̅ = ∑ −1: the cumulative frequency before class
∑ : lower boundary of class
: frequency of class
Median: + − −1)

(2

Mode: + ( 1 )

1+ 2

75

Quartiles: = + − −1) NSO/KKW/CCW

(4 1: the difference between mode class and
before mode class
Percentiles: = + − −1)
2: the difference between mode class and
(100 after mode class

: the class size

Interquartile Range, IQR: 3 − 1

Question 6:
The following table summarises the weight of 50 students (to the nearest kg). Calculate the mean,
median and mode of the following data.

Weight (kg) Number of managers
30-35 4
35-40 7
40-45 9
45-50 16
50-55 12
55-60 2

Solution:

Number of Class median
managers Model median
32.5 4 130 4
30-35 37.5 7 262.5 11
35-40 42.5 9 382.5 20
40-45 47.5 16 760 36
45-50 52.5 12 630 48
50-55 57.5 2 115 50
55-60 ∑ = 50 ∑ = 2280

Mean, ̅ = ∑
∑
= 2280

50

= 45.6

Median = + − −1)

(2

= 45 + (25−20) 5

16

= 46.56

Mode = + ( 1 )

1+ 2
= 45 + [(16−91)+6−(196−12)] 5

= 48.18

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Question 7:
The cumulative frequency table below shows the travelling times to school for 100 students in a
state. Calculate the first quartile, third quartile and the interquartile range of the following data.

Times (minutes) Cumulative frequency
≤0 0
≤7 6
≤ 14 18
≤ 21 46
≤ 28 70
≤ 35 85
≤ 42
100

Solution:

Times First quartile class
(minutes) Third quartile class
3.5 6 21 6
0-7 10.5 12 126 18
7-14 17.5 28 490 46
14-21 24.5 24 588 70
21-28 31.5 15 472.5 85
28-35 38.5 15 577.5 100
35-42 ∑ = 100 ∑ = 2275

First quartile, 1 = + − −1)

(4

25 − 18
= 14 + ( 28 ) 7
= 15.75

Third quartile, 3 = + 3 − −1)

(4

75 − 70
= 28 + ( 15 ) 7
= 30.33

Interquartile range, = 3 − 1
= 30.33 − 15.75

= 14.58

ii. Variance and standard deviation

, 2 = 1 1 [∑ 2 − (∑ )2
− ]

, = √ 1 1 [∑ 2 − (∑ )2
− ]

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Question 8:
Time taken by 100 students to answer a test was recorded and summarised in the table below.
Calculate the mean and variance of the following data.

Time (minutes) Frequency
30-35 5
35-40 12
40-45 20
45-50 32
50-55 25
55-60 6

Solution:

Times 2
(minutes)
32.5 5 162.5 5281.25
30-35 37.5 12 450 16875
35-40 42.5 20 850 36125
40-45 47.5 32 1520 72200
45-50 52.5 25 1312.5 68906.25
50-55 57.5 6 345 19837.5
55-60 ∑ = 100 ∑ = 4640 ∑ 2 = 219225

Mean, ̅ = ∑
∑

= 4640
100

= 46.40

, 2 = 1 1 [∑ 2 − (∑ )2
− ]

= 1 [219225 − (4640)2]

99 100

= 39.68

D. Pearson’s Coefficient Of Skewness

= − or = 3( − )


If < 0, then the distribution is negatively skewed or skewed to the left
If = 0, then the distribution is symmetry
If > 0, then the distribution is positively skewed or skewed to the right

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Question 9:
For a distribution, the mode is 100, the mean is 95 and the variance is 250. Calculate the Pearson’s
coefficient of skewness and interpret your answer.

Solution:

= −


95 − 100
=

√250

= −0.316

The distribution is negatively skewed.

Question 10:
Find the Pearson’s coefficient of skewness represented by the stem and leaf diagram below.

Stem Leaf
4 0, 3
5 1, 4, 6
6 2, 2, 4, 7, 8, 8, 8, 9
7 2, 5, 9
8 3, 5
96

Key: 5|3 represents 53

Solution:

∑ = 40 + 43 + 51 + 54 + 56 + 62 + 62 + 64 + 67 + 68 + 68 + 68 + 69 + 72 + 75 + 79 + 83

+ 85 + 96
= 1262

∑ 2 = 402 + 432 + 512 + 542 + 562 + 622 + 622 + 642 + 672 + 682 + 682 + 682 + 692 + 722
+ 752 + 792 + 832 + 852 + 962

= 87388

Mean, ̅ = ∑ = 1262 = 66.42
19

n = 19 (odd number)

Hence, median = ( +1) ℎ

2

= 10 ℎ

= 68

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, = √ 1 1 [∑ 2 − (∑ )2
− ]

1 (1262)2
= √18 [87388 − 19 ]
= 14.07

3( − )
=

3(66.42 − 68)
= 14.07
= −0.3369

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Chapter 7: Permutations and Combinations

A. Permutations

Permutations: when order is important, involve arrangement

Formula: = !
( − )!

Keywords: arrange, position, order, password

4 type of problems:

Type 1

different objects, use all.

Question:
How many different ways can the letters A, B, C be arranged?

Solution:

Method 1 Method 2 Method 3 Method 4
List out: Factorial: Formula: Arrangement:

ABC = The number of 3× 2× 1

ACB ways The number of ways
= × ×
BAC The number of =

BCA ways =

CAB = !

CBA =

The number of
ways =

Type 2
different objects, use partial.
Question:
How many different ways can the 2 letters from letters A, B, C be arranged?

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Solution: Method 2 Method 3 Method 4
Factorial: Formula: Arrangement:
Method 1
List out: The number of 3× 2
ways
AB The number of ways
AC = = ×
BA =
BC
CA
CB

The number of
ways =

Type 3

different objects, identical objects, use all.

Question:
How many different ways can the letters A, A, B be arranged?

Solution:

Method 1 Method 2 Method 3 Method 4
List out: Factorial: Formula: Arrangement:

AAB = , =
ABA
BAA The number of

The number of ways
ways =
! 2 identical
= ! A
=

Type 4

different objects, identical objects, use partial.
Question:
How many different ways can the 2 letters from letters A, A, B be arranged?

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Solution: Method 2 Method 3 Method 4
Factorial: Formula: Arrangement:
Method 1
List out: Case:
1. 2A
AA A
AB A !
BA
= ! =
The number of
ways =

2. 1A, 1 not A
AB
= ! =

Total number of ways
= +
=

5 techniques of solving:

Technique 1: Fixed

Example Question Solution 1
1 1 D
Given letters A, B, C, D.
How many arrangement of A
the letters with the following
conditions? 2!
1st letter: A
Last letter: D The number of ways= 1 × 2! × 1 = 2

2 Given letters A, B, B, D. 1 1
D
How many arrangement of
the letters with the following A

conditions? 2! 2 identical
1st letter: A 2! B

Last letter: D

The number of ways= 1 × 2! × 1 = 1
2!

3 Given letters A, B, B, C, D. 1 1

How many arrangement of D
the letters with the following A 3! 2 identical
2! B
conditions?

1st letter: A

Last letter: D

The number of ways= 1 × 3! × 1 = 3
2!

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4 Given letters A, B, B, C, D. 1 1
D
How many arrangement of
the letters with the following A

conditions? 3!
1st letter: B

Last letter: D The number of ways= 1 × 3! × 1 = 6

5 Given letters A, B, B, C, D. 1 1
B
How many arrangement of
the letters with the following B

conditions? 3!
1st letter: B

Last letter: B The number of ways= 1 × 3! × 1 = 6

Technique 2: Together

Example Question Solution
1 Given letters A, B, C, D. AB
How many arrangement of
the letters with AB together? 2!

3!
The number of ways= 3! × 2! = 12

2 Given letters A, A, C, D. AA
How many arrangement of

the letters with AA together? 2 identical
A
2!

2!

3!
The number of ways= 1 × 3! = 6

3 Given letters A, A, C, D. A AC
How many arrangement of

the letters with AAC 3! 2 identical
together? A

2!

2!

The number of ways= 3! × 2! = 6
2!

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4 Given letters A, A, C, D. C D AA

How many arrangement of

the letters with CD together?

2!

2 identical

3! A

2!

The number of ways= 2! × 3! = 6
2!

Technique 3: Separately

Example Question Solution
1 Given letters A, B, C, D, E.
How many arrangement of
the letters with AB CD E
separately?

Arrangement of CDE = 3!

Arrangement of AB separately, 4 spaces to
arrange AB =

The number of ways= 3! × = 72

2 Given letters A, B, C, C, D.

How many arrangement of CC D
the letters with AB

separately?

Arrangement of CCD = 3!
2!

Arrangement of AB separately, 4 spaces to

arrange AB =

The number of ways= 3! × = 36
2!

3 Given letters A, A, B, C, D.

How many arrangement of BC D
the letters with AA

separately?

Arrangement of BCD = 3!

Arrangement of AA separately, 4 spaces to

arrange AA = 2 identical A
!

The number of ways= 3! × = 36
!

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4 Given letters A, A, B, B, B.

How many arrangement of BB B
the letters with AA

separately? 3 identical B

Arrangement of BBB = 3!
3!

Arrangement of AA separately, 4 spaces to

arrange AA =

!

2 identical A

The number of ways= 3! × = 6
3! !

Technique 4: Not together

Example Question Solution
1 Given letters A, B, C, D, E, F. Arrangement ABC together:
How many arrangement of
the letters with ABC not A BC
together?
3!

4!

The number of ways= 3! × 4! = 144

Solution:
Arrangement ABC not together:

All permutation − ABC together
= 6! − (3! × 4!)
= 720 − 144
= 576

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Technique 5: Alternately

Example Question Solution
1 Given letters A, B, C and
numbers 1, 2.
How many arrangement with AB C
ABC are arranged alternately?

Arrangement of ABC = 3!
Arrangement of 1 and 2 into 2 spaces = !

The number of ways= 3! × ! = 12

2 Given letters A, B, C and

numbers 1, 1. AB C
How many arrangement with

ABC are arranged alternately?

Arrangement of ABC = 3!

Arrangement of 1 and 1 into 2 spaces = !
!

! 2 identical 1
!
The number of ways= 3! × = 6

3 Given letters A, A, B and

numbers 1, 1. AA B
How many arrangement with

AAB are arranged Arrangement of AAB = 3! 2 identical A
alternately? 2!

Arrangement of 1 and 1 into 2 spaces = !
!

3! ! 2 identical 1
2! !
The number of ways= × = 3

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B. Combinations

Combinations: order is not important

Formula: = !
!( − )!

Keyword: choose, select, pick

i) Combinations of a set of objects with condition

Question:
In how many different ways can choose 3 letters from the A, B, C, D?

Solution:

Method 1 Method 2
List out: Formula:

ABC The number of ways
ABD 4 3 = 4
ACD
BCD

The number of ways
=

ii) Combinations of r objects from n different objects

Question:
In a box, there are 4 black marbles, 3 blue marbles and 2 red marbles. In how many ways can
2 black marbles, 2 blue marbles and 1 red marble be chosen?

Solution:
Number of ways = 4 2 × 3 2 × 2 1

= 36

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Chapter 8: Probability

( ) = ℎ = ( )
( )

( ∪ ) = ( ) + ( ) − ( ∩ )

0 ≤ ( ) ≤ 1

( ′) = 1 − ( ) Complementary events
( ) = 1

Example 1:
A fair die has each face numbered 1 to 6. The die is thrown once again and the number landing
face up is recorded.
a) Find the probability of the die landing with the number 5 face up.
b) Find the probability of throwing an odd number.

Solution 1:
a) There are six faces and one has the number 5 on it, so the probability of landing with the

number 5 face up is 1/6.
b) The odd numbered faces are 1, 3 and 5, so the probability of throwing an odd number is 3/6

= 1/2 or 0.5.

Example 2:
A and B are two events and ( ) = 0.6 and ( ) = 0.7 and ( ∪ ) = 0.9. Find
a) ( ∩ )
b) ( ′)
c) ( ′ ∪ )
d) ( ′ ∩ )

Solution 2:

a) ( ∩ ) = ( ) + ( ) − ( ∪ ) Use Addition Rule.
= 0.6 + 0.7 − 0.9

= 0.4

b) ( ′) = 1 − ( ) Complementary events
= 1 − 0.6

= 0.4

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c) B
A

0.2 0.4 0.3 Draw the Venn diagram

0.1

( ′ ∪ ) = 0.3 + 0.1 + 0.4 = 0.8

d) ( ′ ∩ ) = ( ′) + ( ) − ( ′ ∪ ) Use Addition Rule. Alternative:
= (1 − 0.6) + 0.7 − 0.8 Replace A with A’ ( ′ ∩ )
= 0.3 = ( ) − ( ∩ )

Mutually Exclusive Events Both events cannot
( ∩ ) = 0 happen at the same time
( ∪ ) = ( ) + ( )

Independent Events
( ∩ ) = ( ) × ( )
( ∪ ) = ( ) + ( ) − ( ) × ( )

Example 1:

A, B and C are events in a sample space with ( ) = 7 , ( ) = 6 , ( ) = 8,
20 20
20

( ∩ ) = 1 and ( ∩ ) = 2. If B and C are mutually exclusive, find:
20
20

a) ( ∪ )

b) ( ∪ )

c) ( ∪ )

Solution 1: Use Addition Rule.

a) ( ∪ ) = ( ) + ( ) − ( ∩ )
761

= 20 + 20 − 20
3

=5

b) ( ∪ ) = ( ) + ( ) − ( ∩ ) Use Addition Rule.

782
= 20 + 20 − 20

13
= 20

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c) ( ∪ ) = ( ) + ( ) − ( ∩ )

68 B and C are mutually
= 20 + 20 − 0 exclusive

7
= 10

Example 2:

A sample space contains the events A and B satisfying ( ) = 152, ( ) = 1 and ( ∩ ′) = 13.
2

a) Find ( ∩ )

b) Are A and B independent?

Solution 2: AB
a) ( ∩ ′) = ( ) − ( ∩ )

15
3 = 12 − ( ∩ )

( ∩ ) = 1
12

b) ( ) × ( ) = 5 × 1 = 5

12 2 24

Since ( ∩ ) ≠ ( ) × ( ), then A and B are not independent

Conditional Probability

( | ) = ( ∩ )
( )

Example 1:
Let A and B be events in a sample space S with ( ) = 0.6, ( ) = 0.4 and ( ∩ ) = 0.2.
Compute the following conditional probabilities.

a) ( | )
b) ( | ′)
c) ( | )

Solution 1:

a) ( | ) = ( ∩ )

( )

0.2
= 0.4

= 0.5

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b) ( | ′) = ( ∩ ′)

( ′)

( ) − ( ∩ ) Draw Venn diagram
= 1 − ( )

0.6 − 0.2
= 1 − 0.4

2
=3

c) ( | ) = ( ∩ )
( )

0.2
= 0.6

1
=3

Example 2:
In a school, 4% of the boys and 1% of the girls are overweight. In addition, 60% of the students in
the schools are girls. If a student is selected at random and is overweight, what is the probability that
the student is a girl?

Solution 2:

0.01 Overweight

Girl Not Overweight
0.6 Overweight

0.99

0.4 0.04
Boy

0.96 Not Overweight

( | ℎ ) = ( ∩ ℎ )
( ℎ )

0.6 × 0.01 Multiply along the branch
= (0.6 × 0.01) + (0.4 × 0.04)

0.006
= 0.022

3
= 11

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Exercise
1. Two unbiased dice are tossed. Find the probability that

a) the product on the two dice is an odd number

b) the two dice show even numbers

2. Two events A and B are such that ( ) = 1 , ( ) = 1 and ( | ) = 56.
2 4

a) Find the value of ( ∩ ).

b) State whether the events are independent.

3. Given A and B are two events such that ( ) = 1 , ( | ) = 2 and ( | ) = 1.
3 5
3

a) Find ( ∩ ), ( ) and ( ∪ )

b) Determine whether events A and B are

i) independent

ii) mutually exclusive

4. The number of male and female students in a college taking three different subjects are as

follows:

Male Female

Mathematics 28 52

Biology 12 28

Physics 24 6

A student is randomly chosen. Find the probability that

a) the student is female
b) the student is female and she is taking Physics

c) the student is a female and she is taking Biology or the student is a male and he is taking
Physics

5. A box contains 10 balls of which 3 are blue and 7 are red. A ball is randomly picked from the
box and the colour is noted. The ball is not replaced. A second ball is picked. Find the
probability that
a) the first ball is red
b) the first ball is blue and the second ball is red
c) the balls are of different colours.

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Solution
1.
a) ( ) = 36
123456

1123456

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

( ℎ ) = 9
91

∴ ( ℎ ) = 36 = 4

b) 123456
1,1 2,1 3,1 4,1 5,1 6,1
1 1,2 2,2 3,2 4,2 5,2 6,2
2 1,3 2,3 3,3 4,3 5,3 6,3
3 1,4 2,4 3,4 4,4 5,4 6,4
4 1,5 2,5 3,5 4,5 5,5 6,5
5 1,6 2,6 3,6 4,6 5,6 6,6
6

( ℎ ℎ ) = 9
91

∴ ( ℎ ℎ ) = 36 = 4

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2.

a) ( | ) = ( ∩ )
( )

5 ( ∩ )
6= 1

4
5
( ∩ ) = 24

b) ( ) × ( ) = 1 × 1 = 1
2 4 8

Since ( ∩ ) ≠ ( ) × ( ), the events are not independent

3.

a) ( | ) = ( ∩ )

( )

2 ( ∩ )
5= 1

3
2
∴ ( ∩ ) = 15

( | ) = ( ∩ )
( )

1 2
3
= 15
( )

2
∴ ( ) = 5

( ∪ ) = ( ) + ( ) − ( ∩ )

21 2
= 5 + 3 − 15

3
∴ ( ∪ ) = 5

b) ( ) × ( ) = 2 × 1 = 2
i)
5 3 15
ii)
Since ( ∩ ) = ( ) × ( ), events A and B are independent.
Since ( ∩ ) ≠ 0, events A and B are not mutually exclusive.

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4. Male Female Total
28 52 80
Mathematics 12 28 40
Biology 24 6 30
Physics 64 86 150
Total

a) ( ) = 86 = 43
150 75

b) ( ∩ ℎ ) = 6 = 1

150 25

c) ( ∩ ) + ( ∩ ℎ ) = 28 + 24 = 26

150 150 75

5.
2/9 Blue

3/10 Blue Red
7/9

7/10 3/9 Blue
Red

6/9 Red

1st 2nd
=ba7ll ball
a) ( 1)
10

b) ( 1 ∩ 2) = 3 × 7 = 7
10 9 20

c) ( 1 ∩ 2) + ( 1 ∩ 2) = 3 × 7 + 7 × 3 = 7
10 9 10 9 15

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Chapter 9: Random Variable

9.1 Discrete Random Variable

0 ≤ ( = ) ≤ 1
The sum of all the probabilities must add up to one. Or in symbols, ∑ ( = ) = 1

Example 1: = 1,3
The random variable X has a probability function ( = ) = { ( − 1) = 2,4
where is a constant.
a) Find the value of .
b) Construct a table giving the probability distribution of X.

Solution 1: Substitute the value of
a) Since this is a discrete random variable

∑ ( = ) = 1

(1) + (3) + (2 − 1) + (4 − 1) = 1
+ 3 + + 3 = 1
8 = 1

1
= 8

b)

1 2 3 4

( = ) 1 3 3 1
8
888

Substitute the value of

Example 2:
A discrete random variable has the following probability distribution:
1 2 3 4 5 6
( = ) 0.1 0.2 0.3 0.25 0.1 0.05

Find
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