Prime Optional Mathematics 8

Exercise 6.2

1.• Find•the•magnitude•of•the•following•vectors.

• i) a = e 3 o ii) b = ^ 8, 17h
4

iii) AB = 6 iv) AB for A(1, 2) and B(–3, –2)
d8n

v) PQ for P(3, 7) and Q(–1, 4)

2.• Find•the•direction•of•the•following•vectors.

• i) a = (2 3 , 2)• ii)• b =e 3 o
27

iii) AB = 4 iv) PQ for P(2, –1) and Q(6, 3)
d0 n
Vector Geometry
v) CD for C(2, 5) and D(–2, 5)

3.• Find•the•unit•vector•of•the•following•vectors.
• i) a = d12n ii) b = d–86n

iii) AB = (2 2, 2) iv) PQ for P(3, –2) & Q(–1, 1)

v) RS for R(–1, –3) & S(7, 3)

4.• If• a = 3 and• b = 1 n , ind•the•followings.
d2n d2
• i) a + b ii) a – b
iii) 2 a + b iv) 3 a – 2 b
v) 3 a + 2 b

5. PRIME more creative questions:
1 2
i) If a = d2n and• b = d3 n , ind•the•magnitude•of• a + b.

ii) If a = 2 and• b = 4 n , ind•the•magnitude•of•2 a + b.
d3 n d0

146 PRIME Opt. Maths Book - VIII

iii) If a = d24n and• b = d34n , ind•the•direction•of• a – 2 b .

iv) If A(–3, –1) and B(2, 4), ind the magnitude and direction of AB .

v) If magnitude of a = c4x m is•5•units,• ind•the•value•of•‘x’.

6. Project work
Collects the different activities in your daily life which can be
represented in vector notation. Also present such examples in your
classroom.

Answer

1.• i)• 19 units• ii)• 5•units• iii)• 10•units Vector Geometry
• iv) 4 2 units• v)• 5•units iii)• 0°
2.• i)• 30°• ii)• 60°•
v) 0°
iv) 45°

3.• i)• = KKKJKKKK 2 5 ONOOOOOO ii) = KKLKKKJKK–4355 OOOONOOO iii) KKKKKKKJ 2 5 OOOONOOO
1 5 P P L 5 P
Ua Vb 1

L

iv) KJLKKKKKK–3455 OONOOOOO v) KKKKKJKK34 5 OOOOOOON
P L 5 P

4.• i)• d44n ii) d20n iii) d76n
iv) d72n v) d1110n iii)• 45°

5.• i)• 34 units• ii)• 10•units•

• iv) 5 2 units,•45°• v)• 3•units

PRIME Opt. Maths Book - VIII 147

Vector Geometry

Unit•Test•-•1

Time•:•30•minutes

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]

Attempt all the questions:

Vector Geometry 1.• What•do•you•mean•by•unit•vector?
2.• a.• If•A(1,•2)•and•B(4,•–2)•are any two points ind the magnitude of AB

.
b. If a = e3 3 o , ind•the•direction•of• a .

c. If a = d12n and• b = d3–1n , ind•2a + b .

3.• a.• If•A(3,•2),•B(4,•–3),•P(1,•4)•and•Q(0,•9),•prove that AB = – PQ .
b. If a = d32n, b = d–41n , ind•the•magnitude•of•2a – b .

4.• If•A(3,•–2)•and•B(7,•m)• ind•the•value of m where magnitude of AB is
4 2 units.•Also• ind•the•direction•of• AB .

148 PRIME Opt. Maths Book - VIII

Unit 7 Transformation

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 – 1 2 7 10

Weight –2–5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can de!ne the transformation.
• Students can identify the isometric transformations.
• Students can !nd the image of a point under isometric transformations

geometrically.
• Students can !nd the image using co-ordinate.
• Students can plot the object and image in graph.

Materials Required:
• Chart paper.
• Graph board.
• Sample of transformations.
• List of the formula of transformations.
• Geo-board
• Chart paper.
• Shape of objects

PRIME Opt. Maths Book - VIII 149

Introduction•of•transformation

Let•us•see,

A Ai)•

Here,•the•image•of•A•is•seen•into•a•mirror, where
• A is•the•object•and•
• A into•the•mirror•is•called•the•image•of•A.

ii)•

object mirror image

Transformation Here,
• 1st ish•is•taken•as•the•object.
• Second• ish•is•taken•as•the•image.

Above examples show that one object can be transferred from one place
to another place with same shape & size or different size which is called
transformation.

The• process of changing the position or
size• of• an• object• under• any geometrical
conditions•is•called•transformation.

By using transformation patterns of objects can be drawn which is useful
to ill the pictures in clothes walls and in any kind of objects to make them
very attractive.

Here•in•grade•VIII•we•discuss•only•three•types•of•transformations

i)• Re lection ii) Rotation iii) Translation

150 PRIME Opt. Maths Book - VIII

7.1•Re•lection A P

The•image•of•person•AB•is•A‘B‘ which is formed Q A‘
at•equal•distance•from the mirror line m as the m B‘
distance• of• object• where AP = PA‘ or BQ = QB‘.
The•image•A‘B‘ so formed is laterally inverted as B
the•image•formed in the looking glass but size is
same•as•the•object.

Here,•• AB•is•called•the•object
• m is called re lection axis (mirror line)

A‘B‘ is•called•the•image.

The• transformation of an object from one place to Transformation
another•place•by•a•mirror•line•is•called•re lection.
• Object•and•image•so•formed are always congruent

in•re lection.

Re•lection using co-ordinates axes.

i)• Re•lection about x-axis (y = 0).

Y

4

3 M(2,•3)

2

1
P
X‘ -4 -3 -2 -1 O -11 2 3 4 X

-2

-3 M’(2,•–3)

-4

Y‘

PRIME Opt. Maths Book - VIII 151

Draw MP^OX and produce MP to M‘ such•that•MP•=•PM‘
Here,• M(2,•3)•is•an•object.
• x-axis (XX‘) is the re lection•axis.
• M‘(2,•–3)•is•the•image.

• It shows that: P(x, y) ® P‘(x,•–y)

ii)• Re•lection about y-axis (X = 0)

Y

P‘(–3,•4) M4 P(3,•4)

3

2

1

X’ -4 -3 -2 -1 O 1 2 3 4 X
-1

Transformation -2

-3

-4

Y’
Draw PM^OY & produce PM to MP‘ to make PM = MP‘.
Here, P(3, 4) is an object.
Y -axis (YY‘) is•the•re lection•axis.
• P‘(–3,•4)•is•the•image.

• It shows that : P(x, y) ® P‘(–x, y)

152 PRIME Opt. Maths Book - VIII

Worked out Examples

1. Find the image of the plane !igure given below under the given mirror
line M.
AB
M

C

Solution: AB Transformation

M

P

Q

C‘ B‘

R

C A‘
Here, DABC•is•the•given object, mirror line M is the re lection axis.
Draw, BP^M,•• BP•=•PB‘

AQ^M,• AQ = QA‘
CR^M,• CR•=•RC‘

Then, Join A‘, B‘ & C‘ so DA‘B‘C‘ is the image of DABC•after•re lection
under•a•mirror•line•M.

PRIME Opt. Maths Book - VIII 153

2. Find the image of a point A(–2, 3) under re!lection about x-axis.
Solution:
Under re•lection about x-axis.
P(x, y) ® P‘(x,•–y)

• A(–2, 3) ® A‘(–2,•–3)

3. Find the image of DABC having vertices A(2, 3), B(4, 6) & C(7, 1) under

re!lection about x = 0. Also plot them in a graph.

Solution :

Under re•lection about x = 0 (y-axis)
P(x, y) ® P‘(–x, y)
A(2, 3) ® A‘(–2,•3)
• B(4, 6) ® B‘(–4,•6)
• C(7, 1) ® C‘(–7,•1)
Y

B‘ B

Transformation X’ C‘ A‘ A C
O X

Y’

Here,
DABC•is•the•given object.
Y = axis (x = 0) is the mirror line.
DA‘B‘C‘ is•the•image.

154 PRIME Opt. Maths Book - VIII

Exercise 7.1

1.• Find•the•image•of•the•following plane •igures under the given re•lection
axis•‘M’.

AB
M

B

C M A
R
• i) ii)

Q M
R
Q Transformation

S
P

P

iii) M iv)

S

TR

P M
Q

v) M vi)

2.• Find• the• image• point• of• the• following under the re•lection about

x - axis.

• i) A(2, –3) ii) P(–3, 5) iii) M(–3, –2)

iv) N(3, 7) v) Q(3, –7) vi) D(–4, 6)

PRIME Opt. Maths Book - VIII 155

3.• Find•the•image•of•the•following•points•under•re•lection•about•y-axis.

• i) M(–3, –5) ii) N(–7, 2) iii) A(2, 5)

iv) B(5, –3) v) C(–8, –3) vi) D(–5, 4)

Transformation 4. PRIME more creative questions
i) Find the image of the points A(2, 2) and B(6, 7) under re•lection
about•x-axis.•Also•plot•them•in•graph.•Also•join•AB•and•A‘B‘.
ii) Find the image of DPQR having vertices P(–2, 1), Q(–5, 4) &
R(–6,•–2)•under•re•lection•about•y-axis.•Also•plot•them•in•graph.

• iii) Find the image of quadrilateral having vertices A(1, 2), B(3, 5),
C(6,•6)•and•D(7,•1)•under•re•lection about y = 0. Also plot them in
graph.

iv) Find the image of quadrilateral having vertices P(3, 1), Q(4, 5),
R(7,•4)•and•S(8,•–2)•under•re•lection about X = 0. Also plot them
in•graph.

• v) Find the coordinate of image of triangle having vertices A(2, 1),
B(4,•5)•& C(5, –3) under re•lection on x–axis to DA‘B‘C‘. Also plot
DABC,• DA‘B‘C‘ & DA‘‘B‘‘C‘‘ in graph. Also •ind the coordinates of
image•of•DABC•under•the•re•lection about y-axis and plot DABC,•
DA‘B‘C‘ and•DA‘‘B‘‘C‘‘ in graph.

Answer

1.• Discuss•with•your•subject•teacher.
2.• i)• A‘(2,•3)• ii)• P‘(–3,•–5)• iii)• M‘(–3,•2)
• iv) N‘(3,•–7)• v)• Q‘(3,•7)• vi)• D‘(–4, –6)
3.• i)• M‘(3,•–5)• ii)• N‘(7,•2)• iii)• A‘(–2,•5)
• iv) B‘(–5,•–3)• v)• C‘(8,•–3)• vi)• D‘(5,•4)
4.• i)• A‘(2 –2)•&•B‘(6, –7); graph
• ii) P‘(2,•1),•Q‘(5,•4)•&•R‘(6,•–2);•graph
• iii) A‘(1,•–2),•B‘(3,•–5),•C‘(6,•–6);•graph
• iv) P‘(–3,•1),•Q‘(–4, 5), R‘(–7, 4),•S’(–8,•–2)
• v) A‘(2,•–1),•B‘(4,•–5),•C‘(5,•3)•and•A‘‘(–2, –1),•B‘‘(–4,•–5)•&•C‘‘(–5,•3)

Note: Shot to your subject teacher after plotting in graph.

156 PRIME Opt. Maths Book - VIII

7.2•Rotation: A A‘
B‘ C‘
B
C

O Transformation

Here, O is the center of rotation, object DABC is rotated in clockwise
direction with the angle of rotation 90° where,

\ AOA’ = \ BOB’ = \ COC’ = 90°
DABC is an object.
DA’B’C’ is the image.

The transformation of an object from one
place to another place according to centre,
direction and angle is called rotation.

Note : Anticlockwise direction = Positive direction
Clock wise direction = Negative direction
+ve quarter turn = +90° or – 270°

–ve quarter turn = – 90° or + 270°
Half turn = +180° or –180°
Full turn (complete rotation) = 360°

PRIME Opt. Maths Book - VIII 157

Rotation using co-ordinate having center origin.
1. Rotation about positive quarter turn.

Y

P’(–3, 2) P(2, 3)
X’ OX

Transformation Y’

Above example shows that A(2, 3) ® A‘(–3, 2) under rotation about +90°.
It gives image of a point P(x, y) as

\ P(x, y) $ P‘(–y, x)

2. Rotation about negative quarter turn.

Y

P(4, 3)

X’ O X

P(3, –4)

Y’

Above example shows that A(4, 3) ® A’(3, –4) under rotation about, – 90°.
It gives image of a point P(x, y) as,

\ P(x, y) ® P‘(y, –x)

158 PRIME Opt. Maths Book - VIII

3. Rotation about half turn.

Y

A(4, 5)

X’ O X Transformation

A’(–4, –5)

Y’

Above example shows that A(4, 5) ® A‘(–4, –5) under rotation about
180°.
It gives image of a point P(x, y) as,

\ P(x, y) ® P‘(–x, –y)

4. Rotation about full turn.
Rotation about full turn (360°) is invariant transformation.
i.e. P(x, y) ® P‘(x, y)

A‘, A

Here, 159
Image of A is A`.
after full turn.

PRIME Opt. Maths Book - VIII

Worked out Examples

1. Rotate the given triangle under positive quarter turn with center ‘O’.
A

C B
Solution:
O
A

Transformation A‘ B‘ B
C

O

C‘

Here, DABC is an object.
O is center of rotation.
\ AOA‘ = \ BOB‘ = \ COC‘ = 90° in positive direction (Anticlockwise).
\ DA‘B‘C‘ is the image of DABC.

160 PRIME Opt. Maths Book - VIII

2. Find the image of a point A(–3, 2) under rotation about –90° with centre Transformation
origin.
Solution :
Under rotation about –90°.
P(x, y) ® P‘(y, –x)
\ A(–3,•2)•® A‘(2,•3)

3. Find the image of DPQR under rotation about positive quarter turn. Also
plot the object and image in graph. Where P(1, 2), Q(3, –2) & R(5, 4) are
the vertices.
Solution:
Under rotation about +90°
P(x, y) ® P‘(–y,•x)

• \ P(1, 2) ® P‘(–2,•1)
• Q(3, –2) ® Q‘(2,•3)
• R(5, 4) ® R‘(–4,•5)

Y

R‘
Q‘ R

P

X’ P‘ X
O

Q

Y’ 161

Here,
DPQR•is•the•object.
• under rotation about +90°.
DP‘Q‘R‘ is•the•image.

PRIME Opt. Maths Book - VIII

Exercise 7.2

1.• Draw the image of the following diagrams under the following
conditions•given in question.
i) A ii) P

B RQ
C

O (+90°) O (–90°)
iii) Q iv) E

R AD

Transformation P BC
S
O (–90°)
O (–90°) vi) A

v) G Q
E VW
A UT
S
HF
R
C
BD

O (+270°) O (–270°)

2.• Find•the•image•of•the•following points under rotation about negative

quarter•turn.

• i) (3, –5) ii) (5, 2) iii) (–2, –7)

iv) (–6, 3) v) (–7, –1) vi) (5, –3)

162 PRIME Opt. Maths Book - VIII

3.• Find•the•image•of•the•following points under rotation about positive

quarter•turn.

• i) (–7, 2) ii) (–5, –3) iii) (8, –2)

iv) (3, 7) v) (–6, 2) vi) (–2, –8)

4.• Find•the•image•of•the•following•points•under•rotation•about•half•turn.

• i) (–2, –6) ii) (5, –3) iii) (3, 6)

iv) (–5, 4) v) (–4, –5) vi) (2, –5)

5.• i)• Find•the•co-ordinate of image of DABC•having vertices A(–2, 3), Transformation
B(–3,•–1)•and•C(–5,•4)•under•rotation about –270°. Also plot the
object•and•image•in•graph.

• ii) Find the co-ordinate of the image of DPQR having vertices P(2,
1),•Q(3,•5)•& Q(5, 0) under rotation about +270°. Also plot the
DPQR•and•DP‘Q‘R‘ in•graph.

• iii) Which rotation gives the image of point A(2, –1) to A‘(1,•2)?•Also•
ind•the•image•of•the•points•B(3,•2)•and•C(5,•–4).•Also•plot•DABC•
and•DA‘B‘C‘ in•graph.

• iv) Find centre of rotation, angle of rotation and direction of rotation
which gives A(2, –3) to A‘(–3,• –2)• and• B(4,• 1)• to B‘(1,• –4)• by
plotting•the•points•in•graph.

• v) Find the co-ordinate of the image of quadrilateral having vertices
A(–3,•2),•B(–1,•5),•C(2,•4)•& D(3, –1) under rotation about positive
quarter•turn.•Also•plot•them•in•graph.

PRIME Opt. Maths Book - VIII 163

Answer

1.• Consult•to•your•subject•teacher.

2.• i)• (–5,•–3)• ii)• (2,•–5)• iii)• (–7,•2)
vi) (–3, –5)
• iv) (3, 6) v) (–1, 7)

3.• i)• (–2,•–7)• ii)• (3,•–5)• iii)• (2,•8)
• iv) (–7, 3) v) (–2, –6) vi) (8, –2)

4.• i)• (6,•2)• ii)• (3,•–5)• iii)• (–6,•–3)
• iv) (–4, 5) v) (5, 4) vi) (5, –2)

Transformation 5.• i)• A‘(–3,•–2),•B‘(1,•–3)•&•C‘(–4,•–5)
• ii) P‘(1,•–2),•Q‘(5,•–3)•&•R‘(0, –5)

iii) Rotation about +90°, B‘(–2,•3)•&•C(4,•5)
• iv) Centre origin, Angle 90° and direction is clock wise.

v) A‘(–2,•–3),•B‘(–5,•–1),•C‘(–4,•2)•&•D‘(–1,•3)

Note : Show to your subject teacher after plotting in graph.

164 PRIME Opt. Maths Book - VIII

7.3•Translation P’

P

Q’ B R’

QA

R
Here, vector AB is the magnitude & direction of translation.

DPQR is an object.
PP‘ = QQ‘ = RR‘ = AB
PP‘ ' QQ‘ ' RR‘ ' AB
DP‘Q‘R‘ is the image of DPQR

The transformation of an object from one place to Transformation
another place according to the magnitude and direction
of the given vector is called translation.

i) Translation using co-ordinate:

Y

A’(7, 5)
A(4, 3)

X’ B’(6, 1) C’(9, 1)
X
O
B(3, -1) C(6, -1)

Y’

PRIME Opt. Maths Book - VIII 165

Here,•• A(4,•3)•is•translated•to•A’(7,•5)
• B(3, –1) is translated to B’(6, 1)
C(6, –1) is translated to C’(9, 1)

i.e. All the points are translated with constant number 3 for
x-component and 2 for y - component.

i.e. A (4, 3) ® A‘(4•+•3,•3•+•2)•=•A‘(7,•5)
• i.e. Translation vector is T = <32F

i.e. under translation T = <baF
P(x, y) ® P‘(x•+•a,•y•+•b)

iii)• Translation using vector:

Transformation Let us consider a translation vector is AB where A(1, 2) and B(3, 5)
are any two points.

Then, according to the concept of column vector AB

AB = x2 – x1 = 3–1 = 2
y2 – y1 5–2 3

\ Translation vector T = AB = 2
3

Worked out Examples

1. Find the image of a point A(3, –2) under a translation vector of T = <13F .
Solution:

Under translation vector T = <13F
P(x, y) ® P‘(x•+•a,•y•+•b)
• ® P‘(x•+•1,•y•+•3)
• A(3, –2) ® A‘(3•+•1,•–2•+•3)
• ® A‘(4,•1)

• \ A(3,•–2)• T = <13F A’(4, 1)

166 PRIME Opt. Maths Book - VIII

2. If A (3, –1) and B(1, 2) are any two points, !ind the image of point A

under AB .
Solution :

Here,

Translation vector AB from•the•given points A(3, –1) & B(1, 2) is,

AB = <yx22 – xy11F = <12 – 31F = <–32F =T
– +

Then, under translation vector T = <–32F

p (x, y) ® P‘(x + a,•y•+•b)•=•P‘(x – 2,•y•+•3)
• A(3, –1) ® A‘(3 – 2,••–1•+•3)•=•A‘(1,•2)
• \ A(3,•–1)• T = <–32F A’(1, 2)

3. Find the image of DABC having vertices A(3, 1), B(–1, 5) & C(1, –3) under

a translation vector T = <32F. Also draw the object and image in graph. Transformation
Solution: Under a translation vector T = <32F
P(x, y) ® P‘(x + a,•y•+•b)
• ® P‘(x•+•3,•y•+•2)

• Then,
A(3, 1) ® A‘(3•+•3,•1•+•2)••=•A‘(6,•3)
• B(–1, 5) ® B‘(–1•+•3,•5•+•2)•=•B‘(2, 7)
C(1, –3) ® C‘(1 + 3,•–3•+•2)•=•C‘(4, –1)

Y
B’

B

X’ A A’
X
Here, DABC is• the• O
object. C’ 167
• DA‘B‘C‘ is the image of
DABC C
Y’
PRIME Opt. Maths Book - VIII

4. Draw the image of given tangle under the given vector AB .
Q

A B
P R

Solution: Q Q’

A B
P
Transformation P’
R
R’
Here, Draw PP‘, QQ‘, RR‘ || AB .
Then,
Taking PP‘ = QQ‘ = RR‘ = AB
DPQR is the object.
DP‘Q‘R‘ is the image.

168 PRIME Opt. Maths Book - VIII

Exercise 7.3

1.• Find•the•image•of•the•following••igures•under•the•given vector.
A
M

PB B

N

i) Q A R ii) C
S R BP

T B D Transformation
A C

iii) Q iv) A Q

B AG F
v) H
N
M
D
CE

2.• Find• the• image• of• the• following points under the translation vector.
T = <12F

i) A(3, 2) ii) P(–2, 5) iii) M(–3, –1)
iv) N(4, –6) v) O(–3, –4)

3.• Find•the•image•of•the•points•P(3,•–2)•under•the•translation vector AB
for the following points A and B.

PRIME Opt. Maths Book - VIII 169

i) A(3, 2) and B(1, 0) ii) A(1, –2) & B(3, 1)
iii) A(2, 1) and B(4, 5) iv) A(–1, 2) & B(2, 1)
v) A(4, 1) and B(1, –2)

Transformation 4.• Find•the•image•of•the•following triangles under the given vector. Also
plot•the•object•and•image•in•graph.

• i) A(1, –2), B(–3, 4) and C(3, –3) under a vector T= <32F

ii) P(2, 5), Q(–2, 1) and R(5, 2) under a vector T = <––32F

iii) A(–2, –4), B(–4, –1) and C(3, 0) under a vector T = <13F

iv) K(2, 3), L(4, 6) & M(6, 1) under a vector T = <–42F

v) X(–2, 5), Y(3, 1) & Z(–4, –1) under a vector T = <–34F

PRIME more creative questions
5.• i)• If•a translation T gives image of an object A(1, 3) ® A’(3, 4). Find

the•value•of•translation•vector•T.
• ii) If a point P(3, –1) is translated to P’(5, 2) under a vector T, ind

the•value•of•translation•vector•T.
• iii) If a vector T = <13F translate a point P to P’(3, 7), ind the co-

ordiante of the point P.
iv) Find the co-ordiantes of image of DABC•having vertices P(1, 2),

Q(3,• 5)• and• R(6,• –2)• under• a translation vector PQ . Also plot
DPQR•and•DP‘Q‘R‘ on graph•paper.
• v) Find the co-ordinates of the image of DABC•having A(3, 1), B(7,
3),•C(5,•–2)•under•a•translation•vector• AB .

Project work
6.• Collects• the• formula used in transformations in a chart paper and

present in your classroom as the project work.

170 PRIME Opt. Maths Book - VIII

Answer

1.• Show•to•your•subject•teacher.

2.• i)• A‘(5,•3)• ii)• P‘(0, 6)• iii)• M‘(–1,•0)
• iv) N‘(6,•–5)• v)• O‘(–1,•–3)

3.• i)• P‘(1,•–4)• ii)• P‘(5, 1)• iii)• P‘(5,•–6)
• iv) P‘(6,•–3)• v)• P‘(0, –5)

4.• i)• A‘(3,•1),•B‘(–1,•7)•&•C‘(5,•0)
• ii) P‘(0,•2),•Q‘(–4, –2)•&•R‘(3,•–1)
• iii) A‘(–1,•–1),•B‘(–3,•2),•C‘(4,•3)
• iv) K‘(0,•7),•L‘(2,•10),•M‘(4, 5)

v) X(1, 1), Y‘(6,•–3),•&•Z‘(–1, –5)

5.• i)• T•=•<12F ii) T = <32F iii) P(2, 4) Transformation

iv) P‘(3,•5),•Q‘(5,•8)•&•R‘(8,•1)
• v) AB = d24n ; A‘(7,•3),•B‘(11,•5)•&•C‘(9,•0)

Note : show to your subject teacher after plotting in graph.

PRIME Opt. Maths Book - VIII 171

Transformation

Unit Test - 1

Time•:•30•minutes

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]

Attempt all the questions:

1.• What•is•transformation?•Name•any•two•of•them.

2.• a.• Find•the•image•of•a•point•A(3,•–2)•under•re lection•about•x-axis.
• b. Find the image of a point P(–2, 5) under rotation about +90° with
center•origin.
• c. Find the translation vector ‘T’ Which transform A(2, –1) to A’(4, 2).

Transformation 3.• a.• Find• the• image• of• T ABC• having vertices A(3, 4), B(1, 0) and
• b. C(4,•–•3)•under•re lection•about•y-axis.•Also•plot•them•in•graph.
Find the co-ordinate of image of a triangle having vertices P(–2, 3),
Q(–4,•0)•and•R(0,•– 3) under rotation about –90° with centre origin.
Also•plot•them•in•graph.

4.• The• vertices of T ABC• are A(1, 2), B(3, 5)and C(4, –2). Find the co-
ordinate of image of T ABC•under•translation about AB . Also plot them
in•graph.

172 PRIME Opt. Maths Book - VIII

Unit 8 Statistics

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 – 1 27 6

Weight –2–5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can !nd the values of central tendency mean, median & mode.
• Students can !nd the partition values like quartiles and deciles.
• Students can !nd the measures of dispersion range and quartile deviation.
• Students can tabulate the collected data.
• Students can present the dat in diagrams.

Materials Required:
• Chart paper.
• Sample of data collection.
• Sample of tabulation of data.
• List of formula used in statistics.
• Graph paper.
• Scissors
• Geo-board

PRIME Opt. Maths Book - VIII 173

Introduction•of•statistics

Data•Collection
The•set•of•numerals•collected•for any fact for the purpose of investigation
or•analysis during study is called statistical data. It can be collected by the
individuals•directly for personal investigation by travelling to the related
area, through interviews and questionnaires etc, is called primary•data.

It• can• be• collected• by taking the informations from related of ices like
VDC,•Municipality, sub metropolitan city, metropolitan city, district of ice,
province of ice, central of ice etc is called secondary•data.

In• statistics,• the• collected• raw data should be tabulated, represented
diagrammatically and processed for different analysis. Among them here
we discuss about two central values called mean and median, partition
values quartiles and deciles as well as measure of dispersion range and
quartile•deviation.

Statistics Tabulation of data

The•collected•data•which is written roughly is called raw data. The categorical
presentation of data in a table using frequency is called the frequency
distribution•table. There are three types of frequency distribution table viz.
individual,•discrete and continuous frequency distribution tables. Discrete
frequency distribution table and continuous frequency distribution table.

i) Discrete frequency distribution (ungrouped) :

The•raw data collected for the investigation is tabulated individually with
respect to their repeated numbers with tally marks in it. The number of
repeatation of any variable is called the frequency of the variable.
Example : The number of family members collected in a village is as 3,
2,•4,•4,•3,•2,•3,•3,•4,•4,•4,•5,•5,•6,•5,•6,•4,•3,•2,•5,•6,•4,•3,•4,•4,•3,•4,•5,•2,•5

No.•of•members Tally Marks Frequency

2 |||| 4

3 ||||••••|| 7

4 ||||••••|||| 10

5 ||||••••| 6

6 ||| 3

Total 30

174 PRIME Opt. Maths Book - VIII

ii) Continuous frequency distribution (grouped):
When the collected raw data is too large having long range between
smallest• and• highest• items, the discrete frequency distribution is
dif•icult•to prepare and lead to a confusion. In such situation, we divide
the•distribution•in•many subgroups of suitable sizes, called the class
intervals.•Which•makes•our•presentation and study more comfortable,
understanding• and• easier. This distribution is called the continuous
frequency distribution table. There are two types of continuous
frequency distribution viz.
• inclusive class interval
• exclusive class interval
Example: The marks obtained by 40 students in •irst term examination
of•optional•maths•out•of•50•marks is as :
(No. of students)

Marks Tally marks Frequency

0 - 10 || 2

10•-•20 |||| 5 Statistics

20•-•30 ||||•••||||•••||||••| 16

30•-•40 ||||•••||| 8

40•-•50 ||||•••|||| 9

• This is an example of exclusive class interval as upper limits are
excluded in each class intervals.

8.1•Measure•of•Central•Tendency:

The• calculation• of• a central numerical value of the given statistical data
which suppose to represent the entire data is called the measure of central
tendency. They are mean, median and mode.

Arithmetic•Mean
The• average value of the statistical data which is the ratio of sum of
observations• to the total number of observations is called the arthmetic
mean•or•simply•mean.

i.e.•Arithmetic•Mean•=•

PRIME Opt. Maths Book - VIII 175

i)• WFohreinred,ixvi=duoablsoebrvsaetrivoanti(ovnaslu: Me oefanva(rxia)b=le)Rnx
Rx = sum of the observations
n = no. of observations

ii)• For discrete observation

Mean ( x ) = Rfx
Where, N

x = observations (value of variable)

f = No of observations (No. of repeatation of particular variable)

fx = product of observations and respective no. of observations

R fx•• =•sum•of•(f•×•x)

• N = R f = sum•of•number•of•observations

iii)• For continuous class interval

Mean ( x ) = Rfm or• Rfx
Where, N N

Statistics m = Mid value (mean) of each classes. (Also can be taken x)

f = No. of observations

fm = Product of mid-value and number of observation (fx)

R fm•=•sum•of•(fm)•or•(fx)

• N = R f = sum•of•number•of•observations.

Worked out Examples

1. Find the arthmetic mean of the observations 7, 9, 11, 15, 20, 29, 24, 32.
Solution :
The observations are :
x : 7, 9, 11, 15, 20, 29, 24, 32
No. of observations (n) = 8
Then,

Mean ( x ) = Rx
n

=

= 140
8
= 17.5

176 PRIME Opt. Maths Book - VIII

2. If mean of the observations 3, 7, x, 12, 15 and 14 is 10, !ind the value of

‘x’.

Solution:

The observations are :

x : 3, 7, x, 12,•15,14

• No. of obs (N) = 6

Mean ( x ) = 10
• We have,

Mean ( x ) = Rx
N

or, 10 =

or, 60 = 49 + x
\ x = 11

3. Find the mean of the marks obtained by the students given below.

Marks 12 18 24 30 36

F45894 Statistics

Solution :

x f f•×•x

12 4 48

18 5 90

24 8 192

30 9 270

36 4 144

N=30 R fx•=•744

• We have,

Mean ( x ) = Rfx
N

= 744
30

= 24.8

PRIME Opt. Maths Book - VIII 177

4. Find the arithmetic mean of the given observations.

Class 0-10 10-20 20-30 30-40 40-50

f 45894

Solution:

Class f Mid•value•(x) f•×•x

0 - 10 4 0 + 10 =5 20
2

10•-•20 5 10 + 20 = 15 75
2

20•-•30 8 20 + 30 = 25 200
2

30•-•40 9 30 + 40 = 35 315
2

40•-•50 4 40 + 50 = 45 180
2 R fx•=•790

N = 30

Statistics • We have,

Mean ( x ) = Rfx
N

= 790
30

= 26.33

5. If mean of the observations given in table is 16, !ind ‘m’.

x 5 10 15 18 20 24
5
f 4 6 m 10 7

Solution:

xf f•×•x

54 20

10 6 60

15 m 15•m

18 10 180

20 7 140

24 5 120

N = 32•+•m fx•=•520•+•15m

178 PRIME Opt. Maths Book - VIII

We have,

Mean ( x ) = Rfx
N
520 + 15m
16 = 32 + m

or, 512 + 16m = 520 + 15m

or, 16m – 15m = 520 – 512

\ m=8

Exercise 8.1 Statistics

1.• Tabulate the following statistical data in frequency distribution table.
a. Discrete frequency distribution table.
i) Age Group of 30 students of grade VIII of a school.
12, 12, 13, 14, 12, 14, 13, 15, 12, 14, 14, 14, 13, 14, 13, 15, 12,
14, 12, 13, 16, 16, 15, 15, 12, 14, 13, 14, 15, 14
ii) Number of food item used in dinner of 40 families.
2, 3, 3, 3, 4, 2, 4, 4, 4, 4, 2, 5, 6, 5, 6, 5, 4, 4, 2, 3, 3, 4, 3, 3, 2, 6,
6, 5, 4, 4, 3, 5, 3, 6, 4, 3, 5, 6, 3, 4

b. Continuous class interval taking suitable classes.
i) Marks obtained by 40 students of grade VIII is optional
maths of Enlighten school out of 50 full marks.
33, 37, 49, 48, 40, 30, 12, 7, 24, 28, 47, 49, 30, 20, 10, 40, 16,
39, 38, 50, 32, 44, 19, 29, 28, 27, 33, 35, 42, 15, 43, 36, 24, 26,
28, 30, 30, 47, 1
ii) Systolic Blood pressure of 30 people found in Pulchowk .
60, 70, 110, 100, 100, 90, 80, 80, 120, 140, 100, 70, 70, 150,
140, 80, 80, 60, 70, 70, 80, 80, 90, 90, 70, 100, 120, 90, 80, 70
iii) Number of students found in 20 government schools of
Lalitpur district.
410, 250, 210, 370, 860, 340, 890, 420, 500, 300, 400, 450,
730, 640, 670, 520, 260, 380, 360, 430

PRIME Opt. Maths Book - VIII 179

2.• Find•the•arithmetic•mean•for•the•followings.
• i) R x = 200,•N•=•25,•of•a•statistical•data.
• ii) R fx•=•450,• R f = 30•of•the•observations.
• iii) 24, 16, 12, 28, 20, 10, 24, 26

iv) 7, 9, 15, 18, 21, 24, 30, 32, 34, 40
v) 102, 112, 120, 126, 132, 136, 231

3.• Find•the•arithmetic•mean•for•the•following•observations.

• i) x 5 9 12 18 24

f 35831

• ii) Marks 36 42 46 48 50

No.•of•students 5 8 12 9 6

• iii) Age 5 9 12 18 24

No.•of•people 3 5 8 3 1

• iv) Class 0•–•8 8•–•16 16•–•24 24•–•32 32•–•40

Statistics f 7 9 13 6 5

• v) Marks 10•–•20 20•–•30 30•–•40 40•–•50 50•–•60

f 8 12 15 9 6

4. PRIME more creative questions:
i) If mean of the observations 5, 7, 9, 11, 13 and m is 10.5, ind the
value of ‘m’.
ii) If R x = 208•+•m,•N•=•5•+•m•and• x = 30,• ind•the•value•of•‘m’.
iii) If R fx•=•240•+•6p,• R f = 3p•–•10•and• x = 15,• ind•the•value•of•‘p’.
iv) If mean of the observation given below is 15, ind ‘k’.
x 5 10 15 20 25

f 265k3

• v) If mean of the observation given below is 25, ind ‘p’.

Class 0•–•10 10•–•20 20•–•30 30•–•40 40•–•50

f 34p62

180 PRIME Opt. Maths Book - VIII

5. Project work
Collect the marks obtained by the students of grade VIII of irst

terminal examination and construct frequency distribution table by

taking•suitable•class•interval.•Also• ind•the•arithmetic•mean.

Answer

1.• Show•to•your•subject•teacher

2.• i)• 8 ii) 15 iii) 20
iii)• 53.2
iv) 23 v) 137 iii) 10

3.• i)• 11.7• ii)• 45•

• iv) 18.6 v) 33.6

4.• i)• 18• ii)• 2

v) 4 v) 5

Statistics

PRIME Opt. Maths Book - VIII 181

8.2 Median,•quartiles•and•deciles•(Partition•values):

1. Median
The numerical value of a statistical data which divides the data into
two equal halves is called median.
Y

50% 50%

OX

Median
The median value divides the data 50% in left side and 50% in right
side•as•shown•in•the•given diagram.

Statistics For the calculation of median for individual data.

• Observations should be written in ascending or in descending

order. +
2
• Size of ` n 1 jth item should be calculated.
•
oMdeddinaunm(bMedr) = Corresponding observation of `n + 1 jth item. (for
of observations) 2

• For even no. of observations :

sum of two con sec utive observation of size a n + 1 th item
2 2
Md = k

Calculation of median for discrete data:

• Arrangement of the observations.

• Finding cumulative frequency column in the table.
+
• Finding ` N 2 1 jth item.
• Finding
just greater or equal to ` N + 1 jth size in c.f. column.
2
• • WMehdeiraen`(NM2+d)1=jtcholrieres•sipno•cn.fd. ing observation

182 PRIME Opt. Maths Book - VIII

Worked out Examples

` n + 1 th
2
j

a 5 + 1 th
2
k

` n + 1 th Statistics
2
j

` 6 + 1 th
2
j

24 + 30
2

PRIME Opt. Maths Book - VIII 183

3. Find median of the data.

x 18 15 10 30 24

f 95326

Solution :
The cummulative frequency table taking the observation in ascending
order.

xf c.f.

10 3 3

15 5 3+5=8

18 9 8 + 9 = 17

24 6 17 + 6 = 23

30 2 23 + 2 = 25

N = 25

Here,

Median lies in, +
2
Statistics = size•of• ` N 1 jth item

= ssiizzee••ooff•a1235th2+ite1mkth item
=

\ 17•is•just•greater•than•13•in•c.f.•column.

• \ MMed d=i1an8•(Md) = corresponding•observation•of•17•c.f.•is•18.
• \

2. Quartiles:
The•numerical•values which divide the statistical data in four equal parts

are called quartiles.

• They are Q1, Q2 and Q3.

Q1 Q2 Q3

• Q1 = 1st quartile (lower quartile)
= It divides the data 25% in left & 75% in right.

• Q2 = 2nd quartile (median)
= It divide the data 50% in left and 50% in right as the median.

• Q3 = 3rd quartile (upper quartile)
= It divide the data 75% in left & 25% in right.

184 PRIME Opt. Maths Book - VIII

• They are calculated as the median by calculating the size as follows.
+
Q1 = size of ` N 4 1 jth item

Q2 = size of 2` N + 1 jth item
4
+
Q3 = size of 3` N 4 1 jth item

3.• Deciles:
The• numerical• values of the statistical data which divide the data in ten

equal•parts•are•called•deciles.

• They are D1, D2, D3, ....... D9.

D1 D2 D3 ...................................... D9

• DTDLDih321ke===eyw231ainrsrsdtdededdeeceocacctiihlillcleeeeurwwlwadhhtheeiiiccccaihhhlsedddstiihivvvDeiiid4ddm,eeeDsses5dtt,thhhDiaeee6n,dddDbaaa7tytt,aaacD132a8000l,c%D%%u9laiiicnnntailllnneeegfffbtttteaaahncnneadddslc9i78zu000el%%%astaeiiisnnnd,.rrriiiggghhhttt... Statistics
•
•
•
•

D1 = size of a N+1 th item
10
k

D2 = size of 2^N + 1h th item
10

D3 = size of 3^N + 1h th item
10

D4 = size of 4^N + 1h th item
10

D9 = size of 9^N + 1h th item
10

PRIME Opt. Maths Book - VIII 185

Worked out Examples

1. Find the !irst quartile of the observations : 2, 5, 15, 11, 8, 21, 18

Solution: The observations in ascending order are : 2, 5, 8, 11, 15, 18,

21

No. of observations (n) = 7

Now, +
4
1st quartile•(Q1) = size of `n 1 jth item

= size of `7 +1 jth item
4
= size of 2nd item

= corresponding observation is 5

\ Q1 = 5

2. Find the third quartile of the data 14, 18, 22, 26, 32, 36, 40, 45, 48, 52.

Solution : The observations is ascending order are :

14, 18, 22, 26, 30, 36, 40, 44, 48, 52

Statistics No. of observation (n) = 10

Now, +
4
3rd quartile•(Q3) = size of 3`n 1 jth item

= size of 3 × 11 item
4
= size of 8.25th item

= 8th + 0.25(9th – 8th) obs.

• = 44 + 0.25(48 – 44)

= 44 + 0.25 × 4

= 44 + 1

\ Q3 = 45

3. Find the 9th decile of the above observations given in example 2.

Solution : The observations is ascending order are :

14, 18, 22, 26, 30, 36, 40, 44, 48, 52

No. of observation (n) = 10

Now,

9th quartile•(D9) = size of 9^n + 1h th item
10
11
= size of 9 × 10 item

186 PRIME Opt. Maths Book - VIII

= size of 9.9th item
= 9th + 0.9(10th – 9th) obs.
• = 48 + 0.9(52 – 48)
= 48 + 0.9 × 4
\ D9 = 51.6

4. Find the third decile from the observations given in descrete frequency
distribution table.

x 12 18 24 28 36 42 48

f 2 4 9 15 8 5 2
Solution :

xf c.f.

12 2 2

18 4 2•+•4•=•6

24 9 6•+•9•=•15

28 15 15•+•15•=•30

36 8 30•+•8•=•38 Statistics

42 5 38•+•5•=•43

48 2 43•+•2•=•45

N = 45

• For the third decile (D3)
item
= size of 3 a N+1 th
10
k

= size of 3 × 4.6th item

= size of 13.8th item

15 is just greater than 13.8 in c.f. column

• \ DD33 = corresponding•observation•of•15c.f.•is•24
\ = 24

PRIME Opt. Maths Book - VIII 187

Exercise 8.2

1.• Find•the•median•value•of•the•followings:
• i) 11, 15, 18, 21, 25, 30, 36

ii) 32, 8, 24, 12, 28, 20, 16, 4

x 12 16 20 25 30
iii) f 5 7 12 9 7
10
Marks 50 40 20 30 2
10 55
• iv) f 6 7 8
65 75 4
x 15 25 45 35 63
v) f 2 5 13 7 36
6
2.• Find•the• irst•quartile•of•the•followings. 20
• i) 7, 11, 15, 18, 22, 25, 30, 34, 38, 40, 45 11

Statistics ii) 48, 60, 54, 42, 36, 30, 20, 24, 52 36
6
iii) 35, 25, 15, 5, 65, 55, 45, 95, 75, 85 20
11
Marks 12 16 24 30

iv) No.•of•Students 5 8 12 9

x 7 25 18 10 35 28

• v) f 3 2 9 5 2 8

3.• Find•the•third•quartile•of•the•data•given below.
i) 7, 11, 15, 18, 22, 25, 30, 34, 38, 40, 45

ii) 48, 60, 54, 42, 36, 30, 20, 24, 52

iii) 35, 25, 15, 5, 65, 55, 45, 95, 75, 85

Marks 12 16 24 30

iv) No.•of•Students 5 8 12 9

x 7 25 18 10 35 28

• v) f 3 2 9 5 2 8

188 PRIME Opt. Maths Book - VIII

4.• Find•the•third•decile•of•the•data•given below.
i) 11, 15, 18, 21, 25, 30, 36

ii) 32, 8, 24, 12, 28, 20, 16, 4

x 12 16 20 25 30
9 7
iii) f 5 7 12
30 10
Marks 50 40 20 10 2

• iv) f 6 7 8 65 75 55
63 4
x 15 25 45 35

v) f 2 5 13 7

5.• i)• Find•the•6th decile of the observations 108, 100, 116, 112, 124,
118,•130,•142,•136.

• ii) Find the 5th decile•of•the•observations.
• 24, 36, 30, 56, 48, 42, 68, 64, 60, 72, 80

iii) Find the 4th decile•of•the•observations.

x 15 25 32 40 48 52 60 Statistics

• f 2 3 8 12 5 6 4
• iv) Find the 9th decile of the observations 10, 20, 60, 50, 40, 30, 90,

80,•70,•100.
• v) Find 8th decile•of•the•observations•given below.

Height•of•plants 80 75 72 100 86 90 99

No•of•plants 10 9 6 5 12 8 7

7. PRIME more creative questions.
i) If median of the observations taken in order of 4, 8, 12, 2x + 6, 20,
24•&•28•is•16.•Find•the•value•of•‘x’.

• ii) If ifth decile of the observation taken in order of: 11, 15, 18, 3p +
4,•5p-2,•36,•42,•48•is•29.•Find•the•value•of•‘p’.

iii) If upper quartile of the observations taken in order of: 20, 24, 30,
36,•42,•48,•2m•+•12,•56,•60•is•24,• ind•the•value•of•‘m’.

iv) Find the fourth decile of :

x 54 48 40 20 24 30 36

f 4 6 10 3 6 9 12

PRIME Opt. Maths Book - VIII 189

v) Construct the cummulative frequency distribution table of the
following data and compute lower quartile.
12, 12, 13, 14, 12, 14, 13, 15, 12, 14, 14, 14, 13, 14, 13, 15, 12, 14,
12,•13,•16,•16,•15,•15,•12,•14,•13,•14,•15,•14

1.• i)• 21• ii)• 18• Answer iv) 30 v) 45
2.• i)• 15• ii)• 27• iv) 16 v) 18
3.• i)• 38• ii)• 53• iii)• 20• iv) 30 v) 28
4)• i)• 16.8• ii)• 30• iii)• 22.5• iv) 16 v) 18
5)• i)• 124• ii)• 56• iii)• 87.5• iv) 99 v) 99
7)• i)• 5 ii) 7 iii)• 28• iv) 40 v) 13
iii)• 40•
iii) 20

Statistics

190 PRIME Opt. Maths Book - VIII

8.3•Measure•of•dispersion Statistics

Let• us• consider• the• observations• taken• from the collection of age of 5
parsons•of•two•different families.
Family A : 2, 12, 24, 30, 32.
Family B : 15, 18, 20, 22, 25.

The•arithmetic•mean•of•such•observations•is•20•in•both•families where the
observations•of•family A are very far from the mean where as of family B are
very close from mean.

• Discuss•which•type•of•data•is•better•?
• Discuss• such• type• of• scatterdness of the observations by calculating

median•also.

That•type•of•scatteredness of the observations of the data from the central
value (mean, median, mode, quartiles) is called the measure of dispersion.
The•measures•of•dispersion•are of different types which are range, quartile
deviation, mean deviation and standard deviation. The coef•icient of such
measure of dispersions are also can be taken as the measure of dispersion.

The•measurement•of•scatteredness of the observations
of• the• collected• data• from their central value is called
measure•of•dispersion.

Range:
The•measurement•of•difference between highest and lowest observations

of•the•collected•data•is•called•range.

Where,•Range•=•Highest•observation•–•lowest observation.

i.e. R = H – L

Also,

Coef•icient of range is calculated from such values as,
H – L
Coef•icient of Range = H + L .

PRIME Opt. Maths Book - VIII 191

Note : Range measures the measurement of closeness of the observations
from highest and lowest observations. [R = H – L]
• All the observations are not included in it.

Example: Find range and its coef!icient of the observations taken in order of

20, 24, 30, 36, 44, 50, 60.

Here,

Highest observation (H) = 60

Lowest observation (L) = 20

Range (R) = H – L

= 60 – 20

= 40

Again, H–L
H+L
Coef•icient of Range =

= 60 – 20
60 + 20

Statistics = 40
80
= 0.5

Quartile•deviation:

The• measurement• of• dispersion• by calculating the quartiles of the

observations•of•collected•data•is•called•quartile•deviation where only •irst

quartile•(Q1) and•third•quartile•(Q3) are•use•to••ind•it.

• Quartile deviation (Q. D.) = 1 (Q3 – Q1)
2
Q3 – Q1
Coef•icient of Q.D. = Q3 + Q1

Note: Quartile deviation is the measurement of the
observations from highest and lowest quartiles of the
observations.
• All the observations are not included in it.
• It is slightly better than range for the measurment of

dispersion.

192 PRIME Opt. Maths Book - VIII

Worked out Examples

1. Find range and its coef!icient of the observations 18, 12, 24, 28, 20.

Solution:

The given observations taken in order are, 12, 18, 20, 24, 28.

Here,

Highest observation on (H) = 28

Lowest observation (L) = 12

Range = H – L

= 28 – 12

= 16. H–L
H+L
Coef•icient of range =

= 28 – 12
28 + 12

= 16
40
= 0.4

2. Find the range and its coef!icient of the given discrete observations. Statistics

Age 12 14 15 16 18

No. of students 5 6 12 10 7

Solution:
The given frequency distribution table is,

Age• 12 14 15 16 18

No.•of•students 5 6 12 10 7

Here,

Highest observation (H) = 18

Lowest observation (L) = 12

Range = H – L

= 18 – 12

=6

Again, H–L
H+L
Coef•icient of range =

= 18 – 12
18 + 12
6
= 30

= 0.2

PRIME Opt. Maths Book - VIII 193

3. Find quartile deviation and its coef!icient of 20, 14, 12, 36, 24, 30, 42.

Solution:

The given observations taken in order are: 12, 14, 20, 24, 30, 36, 42.

Number of obs. (N) = 7

Then, N+ 1
4
First quartile lies in = Size of ` j th item

= ` 7 + 1 jth item
4
= 2nd item

= 14

Third quartile lies in = size of 3 ` N + 1 jth item
4
= size of (3 × 2)th item

= size of 6th item
= 36

Statistics Then, 1
2
Quartile deviation (Q.D.) = (Q3 – Q1)

= 1 (36•–•14)
2
• = 11

Coef icient of Q.D. = Q3 – Q1
Q3 + Q1

= 36 – 14
36 + 14

= 22
50
= 0.44.

194 PRIME Opt. Maths Book - VIII

4. Find quartile deviation and its coef!icient of:

In come in thousands 25 27 30 32 37 50

Family No. 3 4 762 1

Solution:
Cumulative frequency table:

In•come Family No. (f) C.F.

x in•thousand

25 3 3

27 4 3+4=7

30 7 7 + 7 = 14

32 6 14 + 6 = 20

37 2 20 + 2 = 22

50 1 22 + 1 = 23

N = 23

• Here, N + 1 Statistics
4
First quartile (Q1) = Size•of• ` jth item

= Size o ` 23 + 1 jth item
4
= Size of 6th item

7 is just greater than 6 in c. f.

\ Q1 = 27•(thousands)

• Third quartile (Q3) = size•of•3 ` N + 1 jth item
4
= Size of 3(6)th item

= Size of 18th item

20 is just greater than 18 in c.f.

\ Q3 = 32•(Thousands)

• Then, 1
2
Quartile deviation (Q.D.) = (Q3 – Q1)

= 1 ×5
2
= 2.5

PRIME Opt. Maths Book - VIII 195