Optional Math 9

Answer

1. Show to your teacher.

2. Show to your teacher.

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

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

5. i) A’(3, 1), B’(–1, 7) & C’(5, 0); graph
ii) P’(0, 2), Q’(–4, –2) & R’(3, –1); graph
iii) A’(–1, –1), B’(–3, 2), C’(4, 3); graph
iv) K’(0, 7), L’(2, 10), M’(4, 5); graph
v) X’(1, 1), Y’(6, –3), & Z’(–1, –5); graph

6. i) T = <2F ii) T = <2F iii) P(2, 4)
1 3

iv) P’(3, 5), Q’(5, 8) & R(8, 1)

v) AB = d4 n ; A’(7, 3), B’(11, 5) & C’(9, 0)
2

7. iii) A’(7, 2), B’(3, 5), C’(2, 3); A’’(3, 5), B’’(–1, 8), C’’(–2, 0); graph
iv) P(x, y) → P’(2 – x, y) → P’’(x + 4, y); both are equal
v) Discuss with your teacher.

PRIME Opt. Maths Book - IX 297

7.4 Enlargement B’
B
Let us consider DABC is an object. It’s image DA’B’C’
A’
is drawn taking o as the constant point and OA’ = A

20A, OB’ = 20B and OC’ = 20C. The image so formed C’

is 2 times bigger than the object DABC where the
fixed point O is called the centre of enlargement

and constant 2 (double) is called the Scale factor O C’

where k= OA` = OB` = OC` =2.
OA OB OC

The size of image depends upon the constant value (k) (which is used to draw the image
by taking less or more than the length of OA, OB and OC from the centre) which is called
the scale factor.

The transformation where the image is enlarge or contract
in a certain proportion according to the centre and scale
factor is called the enlargement.

• In the enlargement E [0, k] where, the centre of enlargement is origin and ‘k’ is the
scale factor.

• In the enlargement E[A, K] where A is the centre of enlargement other than origin
and ‘k’ is the scale factor.

If k = 1, the object and image are invariant.
If k > 1, the image is enlarged than object.
If k < 1, the image is reduced than object.
If k < 0, the image is in opposite direction than the object.
If k = –1, the image and object are equal in size but in opposite direction.

Example:
Enlarge the given arrow about the centre of enlargement at ‘O’ and scale factor –2.
Here,

Join OA and produced to OA’ where B

OA’ = 20A similarly D’ A’ A C
OB’ = 20B C’ O D’
OC’ = 20C

OD’ = 20D B’

Then,
Arrow A’B’C’D’ is the image of arrow ABCD.

298 PRIME Opt. Maths Book - IX

7.4.1 Enlargement using Co-ordinates

1. Having centre origin and scale factor ‘k’. Y

i.e. E[o, k]. Let p1(x1, y1) be the image of an O (0, 0)
Y’
object p(x, y) under an enlargement having, P’(x’, y’)
k
origin 0(0, 0) is the centre of enlargement and
P(x, y)
‘k’ is the scale factor where,
X
scale factor (k) = OP' =k
OP
i.e. OP’ = k . OP

OP = 1

∴ PP’ = k – 1. X’

Using internal section formula for OP’.
m1 y2 + m2 y1
(x , y) = a m1 x2 + m2 x1 , m1 + m2 k
m1 + m2

P(x , y) = a 1×x` + (k–1) 0 , 1×y` + (k–1) 0 k
1 + k–1 1 + k–1

P(x , y) = a x' , y' k
k k
Equating the corresponding elements,

x' = x and y' =y
k k

∴ x’ = kx and y’ = ky

∴ p(x , y) → p’(kx , ky)

ii. Having center origin (a, b) and scale factor ‘k’.

Let, A’ (x’, y’) be the image of a point A(x, y) under,

Center of enlargement p(a,b) k–1 A’(x’, y’)
A(x, y)
and scale factor PA' =K 1
PA
where, PA’ = K. PA

PA = 1, P(a, b)
AA’ = K–1

Then, Using section formula internally for PA’.
m1 y2 + m2 y1
(x,y) = a m1 x2 + m2 x1 , m1 + m2 k
m1 + m2

A(x.y) = a 1×x` + (k–1) a , 1×y` + (k–1) b k
1 + k–1 1 + k–1

or, (x,y) = a x` + ka – a , y` + kb – b k
k k
By equating corresponding elements,
y' + kb–b
or, x= x' + ka–a , y= k
k

or, x’ = kx – ka + a , y’ = ky – kb + b ,

or, x’ = k(x–ka)+a , y’ = k(y–b)+b

∴ p(x , y) → p’[k(x–a)+a , k(y–b)+b]

PRIME Opt. Maths Book - IX 299

Worked out Examples

1. Draw the image of a quadrilateral given in diagram under E[0 , 2].
B

O A C
D’

Solution:

B’
B

A’ C’
OA

C
D’

D’

Here, O is the center of enlargement quad. ABCD is an object.

Then, scale factor (k) = 2.

Join OA and produce to A’ where OA’ = 2OA

Similarly, OB’ = 2OB
OC’ = 2OC and
OD = 2OD.
\ Quad. A’B’C’D’ is formed which is the image of quad. ABCD

2. Find the image of a point p(–2, 3) under an enlargement (i) E1[0, –2] (ii) E2[(1,2), 2].
Solution:

Under an enlargement of an quadrilateral given in diagram under E[0,-2].

p(x, y) → p’ (kx, ky)

→ p’ (–2x, – 2y)

\ p(-2 , 3) → p’[–2 (–2), (–2) 3]

→ p’ (4, –6)

Again,

Under an enlargement E2[(1, 2), 2]
p(x,y) → p’[k(x – a)+a, k(y – b) + b]

→ p’[2(x – 1) + 1, 2(y – 2) + 2]

→ p’ (2x – 1, 2y – 2)

\ p(–2 , 3) → p’[2(–2) – 1, 2(3) – 2]

→ p’ (–5, 4)

300 PRIME Opt. Maths Book - IX

3. Find the image of DABC having vertices A(–2, 3), B(1,–3) and C(2, 4) under an

enlargement E[0, –2]. Also plot the object and image in graph.

Solution:

Under enlargement about E[0,–2]

p(x, y) → p’(kx , ky)

→ p’(–2x , –2y)

\ A(–1, 3) → A’ (–2(–1), (–2) 3) = A’(2, –6)

B(1, –3) → B’ (–2(1), (–2) (–3) = B’(–2, 6)

C(2, 4) → C’ (–2(2), (–2) 4) = C’(–4, –8)

Y

A’(2, –6)

A(–1, 3) C(2, 4)

X’ O X

B(1, –3)

B’(–2, 6)

C’(–4, –8) Y’

Here, DABC is an object.
DA’B’C’ is the image under an enlargement E[0, –2].

4. Find the image of quadrilateral having vertices A(2, 2), B(–1, 3), C(–2, –3) and D(3,

–2) under an enlargement E(1, –1), 2). Also plot the object and image in graph.

Solution:

Under an enlargement E[(1, –1), 2]

p(x, y) → P’[k(x – a) +a, k(y – b) + b)

→ P’[2(x – 1) +1, 2(y + 1) – 1)

→ P’(2x – 1, 2y + 1)

\ A(2, 2) → A’(2 × 2 – 1, 2 × 2 + 1) =A’(3, 5)
B(–1, 3) → B’(2(–1) – 1, 2 × 3 + 1) =B’(–3, 7)
C(–2, –3) → C’(2(–2) – 1, 2(–3) + 1) =C’(–5, –5)
D(3, –2) → D’(2(3) – 1, 2(–2) + 1) =D’(5, –3)

PRIME Opt. Maths Book - IX 301

Y
B’(–3, 7)

A’(3, 5)
B(–1, 3)

A(2, 2)

X’ O X

C(–2, –3) D(3, –2)
D’(5, –3)

C’(–5, –5)
Y’

Here, Quad. ABCD is an object.
Quad. A’B’C’D’ is the image under an enlargement E[(1, –1), 2].

5. Find the center of enlargement and scale factor where it maps DABC having vertices
A(1, 2), B(3, 5) and C(5, –2) to DA’B’C’ having vertices A’(4, 3), B(8, 9) and C’(12, –5).

Solution :
Plotting the object and image given under an elargement in graph having
A(1, 2) → A’(4, 3)
B(3, 5) → B(8, 9)
C(5, –2) → C’(12, –5), we get,

Y
B’(8, 9)

B(3, 5)

A(1, 2) A’(4, 3)

X’ P(–2, 1) X
O

C(5, –2)

Y’ C’(12, –5)
PRIME Opt. Maths Book - IX
302

Now, join AA’, BB’, and CC’, which are intersected at point p(–2, 1)

Again,

Scale factor (k) = PA' = PB' = PC'
PA PB PC

= (x2 – x1)2 + (y2 – y1)2 for PA`
(x2 – x1)2 + (y2 – y1)2 for PA

(4 + 2)2 + (3 – 1)2
=

(1 + 2)2 + (2 – 1)2

= 36 + 4
10

= 2 10
10

=2

\ Center (a, b) = (–2, 1)

Scale factor (k) = 2.

Alternative method.
Let An enlargement be E[a, b), k]
Where,
P(x, y) → P’[k(x – a) + a, k(y – b) + b]
\ A (1, 2) → A’[k(1 – a) + a, k(2 – b)+b]
B (3, 5) → B’[k(3 – a) + a, k(5 – b)+b]
C(5, –2) → C’[k(5 – a) + a, k(–2 – b)+b]
From given,
A(1, 2) →A’(4, 3)
B (3, 5) →B’(8, 9)
C (5, –2) →C’(12, –5)

By equating the image A’.

k(1 – a) + a = 4 and k(2 – b) + b = 3

\ k= 4–a ..........................(i) k = 3–b ..........................(ii)
1–a 2–b

By equating the image B’,

k(3 – a) + a = 8 and k(5 – b) + b = 9

\ k= 8–a .......................... (iii) k= 9–b ..........................(iv)
3–a 2–b


Solving equation (i) and (ii)

4–a = 8–a
1–a 3–b

or, 12 – 4a – 3a + a2 = 8 – a – 8a + a2

or, 2a = –4

\ a = –2

PRIME Opt. Maths Book - IX 303

Solving equation (ii) and (iv)

3–b = 9–b
2–b 5–b

or, 15 – 3b –5b + b2 = 18 – 2b – 9b + b2

or, 3b = 3

\ b = 1.

Putting the value of ‘a’ in equation (i),
4+2
k= 1+2 =2

\ Center = (–2, 1)

Scale factor (k) = 2

7.4.2 Properties of enlargement:

i. Object and image are similar to each other.

ii. If scale factor k >1, the object is said to be enlarged in same side of the object.

iii. If 0< k <1, the object is said to be reduced in same side of the object.

iv. If k <0, the image will be in opposite side to the object and it will be inverted
also.

v. If k = 1, object and image are coincided.

vi. If k = –1, object and image are equivalent but remain in opposite from the center
of enlargement.

vii. The scale factor is the ratio of distance of image and object from the center.

i.e. k = PA' where p is the center of enlargement.
PA

viii. Sides of object and image are also parallel.

ix. A and A’ are always lie in a same straight line passes through the center of
enlargement.

304 PRIME Opt. Maths Book - IX

Exercise 7.4

1. i) What do you mean by enlargement in transformation?
ii) Write down the formula to find the image of A(x, y) under enlargement about

E[(a, b), k].

iii) Find the image of a point p(a, b) under enlargement about E[0, 2], followed by
3
Enlargement about E2[0, 2 ].

iv) In what condition in enlargement the object will be invariant?

v) In what condition the object is reduced in enlargement?

2. Draw the image of the given objects under the center of enlargement ‘o’ and scale

factor ‘k’. B

i) ii) Q

A CO PR
(k = 2) O E(0, 3) S
iii)
B O (k = 2) iv) O (k = 32 ) P
D
v)
C

R Q

A AB
HG

EF
DC

O (k = –1)

3. Find the image of a point p(–2, 4) under the following enlargements.

i) E1[0, 2] ii) E2[0, – 3 ]
2

iii) E3[0, –2] iv) E4[0, 1 ]
v) E5[0, –3] 2

4. Find the image of an object point A(3, –5) under the following enlargements.

i) E1[(1, 1), 2] ii) E2[(1, 2), –2]
iii) E3[(0, 2), –3] iv) E4[(3, 0), –1]
v) E5[(–2, 1), 3]
PRIME Opt. Maths Book - IX 305

5. Find the image of the following under the given enlargements. Also plot the object

and image in graph.

i) Triangle having vertices A(2, 1), B(4, 5) and C(5, – 3) under the enlargement

E[0, 2]

ii) Triangle having vertices P(–3, 0), Q(4, 0) and R(2, 4) under the enlargement E[0, 3]

iii) Triangle having vertices K(2, 4), L(–4, 8) and O(0, 0) under the enlargement

E[0, 0), – 3 ].
2
iv) Quadrilateral having vertices A(2, 4), B(–4, 2), C(–2, –4) and (4, –2) under the

enlargement E[0, 3 ].
2
v) Quadrilateral having vertices P(2, 1) Q(3, 4), R(5, 3) and S(6, – 2) under an

enlargement E[0, –2).

6. Find the image of the followings under the given enlargements. Also plot the object

and image in graph.

i) Triangle PQR having vertices P(6, 2) Q(6, 6) and R(2, 4) under the enlargement

E[(2, 4), 1 ].
2
ii) Triangle having vertices A(3, 4), B(–2, 6) and C(1, 6) under an enlargement

E[(–1, 2), –2].

iii) Triangle having vertices P(–3, 2) Q(0, 4) and R(–1, 5) under the enlargement

E[(1, 2), 2].

iv) Quadrilateral having vertices A(2, 3) B(–3, 2), C(–2, –2) and D(3, –3) under an

enlargement E[(1, –1), 3].

v) Triangle having vertices A(3, 4), B(–2, 6) and C(1, –2) under an enlargement

having centre (2, 2) and Scale factor –2.

7. PRIME more creative questions:
i) The vertices image of a triangle are A’(8, –2) B’(6, 2) and C’(16, –6) after

enlargement about centre origin and scale factor 2. Find the vertices of DABC.
Also plot the object and image in graph.
ii) The vertices of image of triangle PQR are P’(3, –5), Q’(5, – 1) and R’(–1, –1)
under an enlargement E [(3, –1), –2]. Find the coordinate of the vertices for
DPQR. Also plot the object and image in graph.
iii) An enlargement gives image of an object at A(3, 2) → A’(5, 7), B(4, –2) → B’(7, 1)
and C(5, 4) → C’(9, 11). Find centre of enlargement and scale factor.
iv) Draw the DPQR having vertices P(2, 4), Q(4, 2) and R(6, 6) and image of DP’Q’R’
having vertices P’(–4, –8), Q’ (–8, –4) and R’ (–12, –12). Then find the centre of
enlargement and Scale factor.
v) If an enlargement transferred a triangle having vertices A(3, 2), B(4, –5) and
C(5, 0) to the triangle having vertices A’(–3, –7), B’(–5, 7) and C’(–7, –3). Find the
centre of enlargement and scale factor.
vi) What is enlargement? Write down its properties. Where is it applicable in our
daily life? Discuss and prepare a report in group.

306 PRIME Opt. Maths Book - IX

8. Project work
Prepare a list of formula in a chart paper and paste at the project board kept in your

classroom.

Answer

1. Show to your teacher.

2. Show to your teacher.

3. i. (–4, 8) ii. (3, –6) iii. (4, –8)
iv. (–1, 2) v. (6, –12)

4. i. (5, –11) ii. (–3, 16) iii. (–9, 23)
iv. (3, 5) v. (13, –17)

5. i. A’(4, 2), B’(8, 10), C’(10, –6) ; graph.
ii. P’(–9, 0), Q’(12, 0), R’(6, 12) ; graph.
iii. K’(–3, –6), L’(6, –12), O’(0, 0) ; graph.
iv. A’(3, 6), B’(–6, 3), C’(–3, –6), D’(6, –3) ; graph.
v. P’(–4, –2), Q’(–6, –8), R’(–10, –6), S’(–12, 4) ; graph.

6. i. P’(4, 3), Q’(4, 5), R’(2, 4) ; graph.
ii. A’(–9, –2), B’(1, –6), C’(–5, –6) ; graph.
iii. P’(–7, 6), Q’(–1, 10), R’(–3, 12) ; graph.
iv. A’(4, 11), B’(–11, 8), C’(–8, –4), D’(7, –7) ; graph.
v. A’(0, –2), B’(10, –6), C’(4, 10) ; graph.

7. i. A(4, –1), B’(3, 1), C’(8, –3) ; graph.
ii. P(3, 1), Q(2, –1), R(5, –1) ; graph.
iii. Centre = (1, –3), Scale factor = 2
iv. Centre = O(0, 0), Scale factor = 2 ; graph.
v. Centre = (1, –1), Scale factor = –2

vi. Show to your teacher.

PRIME Opt. Maths Book - IX 307

Tramsformation

Unit Test - 1 Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. Which transformations are isometric and which are non- isometric? Name them.

2. a. Find the image a point A(2, –3) under reflection about x-axis followed by

translation about T = <1F .
2

b. If A(3, –2) is the image of a point ‘A’ under rotation about – 90° with centre
origin followed by reflection about y – 2 = 0. Find the co-ordinate of point A.

c. If a point P(2, 4) is translated to P’(4, –2) under a translation ‘T’. Find the

translation vector.

3. a. Find the image of a triangle having vertices A(2, 4), B(–1, 1) and C(3, –3). Find
the image of it under reflection about x + 2 = 0 followed by rotation about –270°
with centre origin. Also plot the object and images in graph.

4. a. Find the image of quadrilateral having vertices (–3, 3), (–2, –1), (1, –3) and (2,
4) under translation about AB followed by enlargement about E[(3, 2), 2]. Also
plot the object and image in graph.

b. Find the centre of enlargement and scale factor where the vertices of triangle are
transferred as A(2,1) → A’(3,4), B(–2,–1) → B’(–5,0) and C(–1,3)→C’(–3,8).

308 PRIME Opt. Maths Book - IX

Unit 8 Statistics

Specification Grid Table

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

No. of Questions – 1 2 – 3 10 12
Weight – 2 4 –

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

Objectives : At the end of the lesson
• Students are able to understand and find the central tendency, its types and

procedure.
• Students are able to find the disparsion.
• Student are able to find the partition values like median, quartiles, decides

and percentiles.
• Students are able to find the value of dispersion like mean deviation,

quartile deviation and standard deviation for individual and discrete
observations.

Materials Required:
• Chart paper.
• Formula chart of central tendency.
• Secondary data of different field like marks, population, cost etc
• Formula chart and model of measure of dispersion.

PRIME Opt. Maths Book - IX 309

8.1 Statistics

Enjoy the recall

The collection of information like marks, ages, incomes, productions etc in a certain area

to analyses the informations is called data.
• The number of the members involve in the collected data is called frequency.
• Distribution of the information in a table to analyses is called tabulation.
• Diagrammatic representation of the tabulated data called diagrammatic

interpretation.

• The collected data should be taken in ascending or in descending order for systemic

analysis.
• The observation lies at the middle of the data is called median.
+ th
i.e. median = size of a N 2 1 item.
k

• The average value of the collected data is called arithmetic mean
Σx Σfx
i.e. AM (X) = N = N .

• The central value of the observations like mean, median and mode are called the

measures of central tendency. The quartiles, deciles and percentiles of observation

are called the partitions.

8.1.1 Partition Values :

Let us take an example of the marks obtained by 7 students of grade 9:

8, 12, 24, 18, 14, 20, 16.

Here, Taking the marks in ascending order as 8, 12, 14, 16, 18, 20, 24. Where 16 divides
the observations in two equal parts which is called median.

Also, 12, 16 and 20 divides the observations in four equal parts which are called quartiles.

Here, the observations 12, 16 and 20 divides the observations in different equal parts
which are called the partition values.

Mean, mode and median are the central values of a
frequency distribution where as median, quartiles, deciles
and percentiles are the values of partitions.

8.1.2 Quartiles:

The partition values which divide the arranged (ascending or decending) data into four
equal parts are called quartiles.
• There are three quartiles which divide the data into four equal parts called first

quartile (Q1), median (M2) and third quartile (Q3)

310 PRIME Opt. Maths Book - IX

• First quartile (Q1) cuts the data in two parts, making 25% in the left and 75% in the
right, which is called lower quartile too.

• Median (Md) cuts the data making in two parts, 50% in the left and 50% in the right
which is called median too.
• Third quartile Q3 cuts the data in two parts making 75% in the left and 25% in the
right which is called upper quartile.

• Size of ` N + 1 th item is the first quartile, 2` N + 1 th is the second quartile and
• 3` N 4 is the third quartile. 4
j j item

+ 1 th item
4
j

Items for different quartiles to be determined from cumulative frequency table.

Examples

1. Find the 3 quartiles of the marks obtained by 10 students in mathematics 44, 68, 95,

98, 86, 70, 32, 50, 58, 72.

Solution:

The marks obtained by 10 students in ascending order are: 32, 44, 50, 58, 68, 70, 72,

72, 86, 95, 98.

No. of students (n) = 10

First quartile (Q1) = Size of ` n + 1 th item
4
j

= Size of ` 10 + 1 th item
4
j

= Size of 2.75th item

=2nd + 0.75 (3rd – 2nd)

= 44 + 0.75 (50 – 44)

= 44 + 4.5

= 48.5

Second quartile (Q2) = Size of 2` n + 1 th item
4
j

= Size of 5.5th item.

= 5th + 0.5 (6th – 5th)

= 68 + 0.5(70 – 68)

= 68 + 1

= 69

Third quartile (Q3) = Size of 3 ` n + 1 th item
4
j

= Size of 8.25th item

= 8th + 0.25 (9th – 8th)

= 86 + 0.25 (95 – 86)

= 86 + 2.25

= 88. 25

PRIME Opt. Maths Book - IX 311

2. Find the first and third quartile from the given observations.

x 24 18 12 48 40 36 30
10 6
f 9732 8
Solution:
The observations in ascending order with c.f. table.

xf c.f

12 3 3

18 7 3 + 7 = 10

24 9 10 + 9 = 19

30 6 19 + 6 = 25

36 10 25 + 10 = 35

40 8 35 + 8 = 43

48 2 43 + 2 = 45

N = 45

First quartile lies in = Size of N + 1 th item
4

= Size of 45 + 1 th item
4

= Size of 11.5th item

19 is the just greater than 11.5 in c.f.

\ First quartile (Q1) = 24.

Again

Third quartile lies in, = Size of 3 ` N + 1 th item
4
j

= Size of 3 × 11.5th item

= Size of 34.5th item

35 is the just greater than 34.5 in c.f.

\ Third quartile (Q3) = 36.

8.1.3 Deciles:

The partition values which divide the given data into ten equal parts are called deciles.

• There are 9 deciles D1, D2, D3, .................... D9 which divide the data into ten equal parts.
• 5th decile cuts the data in two equal half 50% in the left and 50% in the right which is

called median of the data.

• Size of a N+1 th item gives the first decile (D1) and 9 a N+1 th item gives the last
10 10
k k

decile (D9)
• Cumulative frequency distribution table to be constructed for discrete observations.

312 PRIME Opt. Maths Book - IX

Example: 3

3. Find 4th and 9th deciles from the observations 15, 21, 24, 30, 36, 40, 44, 48, 50, 54.

Solution:

The observations in ascending order are 15, 21, 24, 30, 36, 40, 44, 48, 50, 54.

No. of observations (n) = 10.

Then,

4th decile (D4) = Size of 4 a n+1 th item
10
k

= Size of 4 a 10 + 1 th item
10
k

= Size of 4.4th item

= 4th + 0.4 (5th – 4th)

= 30 + 0.4 (36 – 30)

= 30 + 0.4 × 6

= 32.4.

9th decile (D9) = Size of 9 a n+1 th item
10
k

= Size of 9.9th item

= 9th item + 0.9 (10th – 9th)

= 50+ 0.9 (54 – 50)

= 50 + 3.6

= 53.6

4. Find the 6th and 8th deciles of the marks obtained by the students of grade IX.

x 10 18 28 34 40 48 52

f 5 7 9 15 8 6 4
Solution :
The observations in ascending order with c.f. table.

xf c.f

10 5 5

18 7 5 + 7 = 12

28 9 12 + 9 = 21

34 15 21 + 15 = 36

40 8 36 + 8 = 44

48 6 44 + 6 = 50

52 4 50 + 4 = 54

N = 54

Now, 6th deciles lies in, = Size of 6 a N+1 th item
10
k

= Size of 6 × 55 th item
10

= Size of 33rd item

36 is just greater than 33 in c.f.

\ D6 = 34

PRIME Opt. Maths Book - IX 313

Again,

8th decile lies in, = Size of 8 a N + 1 th item
2
k

= Size of 44th item

44 is equal to he 44 in c.f.

\ D8 = 40

8.1.4 Percentiles:

The partition values which divide the data in 100 equal parts are called percentiles.
• There are 99 percentiles from P1, P2, P3, ................, P99 which divide the data into 100

equal parts.
• P50 divides the data 50% in the left and 50% in the right which is called median of

the data.

• Size of a N+1 th item given the first percentile (P1) and 99 a N+1 th item gives the
100 100
k k

• last percentile (P99) of the data.
Cumulative frequency table have to be used to calculate percentiles for discrete

frequency distribution.

5. Find 25th percentile and 90th percentile of the observations 54, 58, 62, 70, 76, 84, 90,

94, 98, 106, 110 and 116.

Solution:

The given observations in ascending order are,

54, 58, 62, 70, 76, 84, 90, 94, 98, 106, 110, 116.

No. of observation (N) = 11

Now,

25th percentiles (P25) = Size of 25 a n+1 th item
100
k

= Size of 25 a 11 + 1 th item
100
k

= Size of 3rd item

= 62

Again,

90th percentile (P90) = Size of 90 a n+1 th item
100
k

= Size of 10.8th item

= 10th + 0.8(11th – 10th)

= 110 + 0.8(116 – 110)

= 110 + 4.8

= 114.8

314 PRIME Opt. Maths Book - IX

6. Find the 32nd percentile and 80th percentile of the observation given below.

x 18 28 32 40 46 52 60
3
f 2 6 9 13 8 5
Solution:
The observation in ascending order are with c.f table.

xf c.f

18 2 6+2=8

32 6 8 + 9 = 17

40 13 17 + 13 =30

46 8 30 + 8 = 38

52 5 38 + 5 = 43

60 2 43 + 5 = 45

N = 45

Now, = Size of 32 a N+1 th item.
32nd percentile lies in, 100
k

= Size of 14.72th item.

17 is just greater than 4.72 in c.f,

\ P32 = 32

Again,

80th percentile lies in, = Size of 80 a N+1 th item.
100
k

= Size of 36.8th item.

38 is just greater then 36.8 in c.f.

\ P80 = 46

PRIME Opt. Maths Book - IX 315

Exercise 8.1

1. Answer the following questions.
i) What is median? What step have to be done to calculate median of the individual

observation?
ii) What is quartile? What steps have to be done to calculate 3rd quartile from

discrete observations?
iii) What is decile? What steps have to be done to calculate 6th decides of the

individuals observations?
iv) What is percentiles? What steps have to be done to calculate 75th percentile from

the discrete observation?
v) What is partition value? Write down the types of such values.

2. Find the followings from the given observations.
i) All the quartiles from the marks obtained by the students of grade IX
18, 12, 16, 8, 20, 14, 10, 24, 22, 26, 28
ii) First and third quartiles from:
112, 108, 104, 100, 124, 120, 116, 96
iii) Find quartiles from the given observations
60, 64, 68, 70, 75, 79, 82, 84, 92, 96
iv) Find first and last quartiles of the observations given below.
210, 200, 140, 150, 180, 170, 160, 130, 120, 110, 100, 190, 220
v) Find the lower and upper quartile of the following production of crops given in

metric tonn.
220, 230, 240, 260, 270, 210, 200, 290, 280, 310, 300, 250

3. i. Find the first quartile from the given frequency distribution table.

x 28 36 42 48 52 60 64

f 4 5 7 12 9 6 3

ii) Find the third quartile of data given in table.

x 55 65 50 40 45

f 4 5 7 12 9

iii) Find the lower quartile of the observations given below.

marks 120 110 100 160 140 150 130

f 13 8 4 3 8 9 15

iv) Find the second quartile of the marks obtained by the students out of 20.

Age 12 14 16 18 20

f 25963

316 PRIME Opt. Maths Book - IX

v) Find the upper quartile of the weight of children given below.

Height 20 10 25 45 15 40 35 30

f 12 4 9 3 7 6 8 5

4. Find the following deciles from the observation given below.
i) Find the 4th and 9th deciles of the production of cropes in metric tonn of a place.
124, 136, 130, 120, 140, 160, 154, 150, 170, 164, 146
ii) Find 6th decile of the weight of students given below.
21, 25, 30, 35, 40, 44, 50, 54, 60
iii) Find the 7th decile of the observations:
60, 21, 27, 35, 42, 46, 39, 12, 15, 18, 50, 54, 24, 30
iv) Find the 6th decile of the observation given below:

x 44 34 24 14 74 64 54

f 15 12 10 7 12 8 6

v) Find the 3thdeciles of the marks obtained by the grade IX student in mathematics
in the third terminal examination given below.

Marks 210 250 290 370 330

f 31 37 45 32 34

vi) Find the 8th deciles of the observations given in table.

Age 15 10 5 25 20 30 35

f 9 7 3 6 11 5 2

5. Find the following percentiles from the given observations.
i) Find the 15th percentiles and 80th percentiles of the observations.
12, 20, 30, 36, 44, 52, 60, 68, 76, 84, 90, 98, 100
ii) Find the 20th and 90th percentiles of the observations given below.
105, 120, 100, 115, 110, 140, 135, 130, 125
iii) Find 28th and 67th percentiles of the number of students of different schools

given below in hilly region.
230, 240, 250, 220, 210, 220, 300, 290, 280, 270, 260
iv) Find the 8th percentile of the observations given below.

x 25 35 45 55 65 75 85

f 90 100 110 120 105 95 79

iv) Find the 49th percentiles of the production of crops of districts given in metric tonn:

x 200 300 400 500 600 700 800

f 3 7 9 13 20 12 8

v) Find the 85th percentiles of the observations given in table.

x 120 110 100 160 140 130 85

f 24 18 12 15 24 30 16

PRIME Opt. Maths Book - IX 317

6. PRIME more creative questions:

i) If 3rd quartile of the observations taken in order 24, 30, 36, 40, 44, 48, 2x + 3, 3x
– 20, 60 is 54, find the value of ‘x’.

ii) If 7th decile of the observations taken in order 52, 60, 68, 76, 80, 84, 88, 4x + 12,
5x – 4, 100, 106 is 93.6, find the value of ‘x’.

iii) If 60th percentile of the observations taken in order 114, 116, 118, 120, 122, 2x
+ 24, 126, 128, 130, 132 is 124, find the value of ‘x’.

iv) If 75th percentiles of the observations taken in order 12, 16, 20, 24, 28, 32, 36,
40, 44, 48, x +2, 2x – 44, 60, 64 is 53, find the value of ‘x’.

v) If lower quartile of the observations taken in order 13, 2x – 12, 24, 28, 32, 36, 40
is 18, find the value of ‘x’.

Answer

1. Discuss with your subject teacher.

2. i) 12, 18, 24 ii) 101, 110, 119
iii) 67, 77, 86 iv) 125, 160, 195
v) 222.5, 255, 287.5

3. i) 42 ii) 55
iii) 120 iv) 16
v) 35

4. i) 139.2, 168.8 ii) 44
iii) 44 iv) 15
v) 250

5. i) 21,91.6 ii) 105, 140
iii) 223.6, 270.4 iv) 25
v) 140 vi) 600

6. i) 25 ii) 20 iii) 50 iv) 50 v) 15

318 PRIME Opt. Maths Book - IX

8.2 Measure of dispersion:

The measure of central values mean, median and mode are learned in previous classes
as well as in compulsory mathematics in the topic of measure of central tendency. But
central values fail to represent the characteristic of data when the values are scattered
widely and deviated more irregularly from central values. The study of scatteredness of
the values of variable from its central value is called the measurement of dispersion.

Let us take two types of observations having same mean and median.

Data A 30 36 40 44 50 40 40
40
Data B 10 25 40 45 60 40

Here the mean and median of both types of observations are same which are 40 for each.
But observations are dispersed closely in ‘A’ and dispersed far in ‘B’. It results distribution
of data in example A is better than in example B. This type of analysis is called dispersion.

The measurement of scatteredness of the observation from
their central values mean, median and mode to know the
detail of distribution is called the measure of dispersion.

In grade IX, we discuss the following measure of dispersions for individual and discrete
series.

• Range (It is given in compulsory mathematics)
• Quartile deviation
• Mean deviation
• Standard deviation

8.2.1 Quartile deviation:

It is defined as the half of the range of quartiles of the given
observations out of three quartiles Q1, Q2 and Q3 which is
also called semi-interquartile range (quartile deviation).

Range of quartiles out of Q1, Q2 and Q3 is Q3 – Q1 (which is inter-quartile range) and half
of its value is called semi-inter-quartile range (Quartile deviation).

i.e. Q.D. = Q3 – Q2 where,
2
Q1 = First quartile
Q3 = Third quartile
Range o=fQq3u–arQtl1efodrevqiuaatriotinleiss.
Coefficient calculated by using the formula,

PRIME Opt. Maths Book - IX 319

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

Note : Only first and third quartiles are involved in it.

Steps for the calculation of Quartile deviation
• Arranging the observation in ascending order.
• Finding of total number of observations N = ∑f.
• Finding of Q1 and Q3 of the observations.

Q1 = size of ` N + 1 th item
4
j

Q3 = size of 3 ` N + 1 th item
4
j

• Find of corresponding observations for such sizes which are Q1 and Q3
respectively.

• Find quartile deviation as,

Q.D. = 1 (Q3 – Q1)
2
Q3 – Q1
Coefficient of Q.D. = Q3 + Q1

• Corresponding observations have to be taken in discrete observations

from cumulative frequency table.

Table :

Xf C.f.

Q1 ` N + 1 th
4
j

Q3 3 ` N + 1 th
N = ∑f 4
j

Q1 lies in size of ` N + 1 th item.
4
j

Q3 lies in size of 3 ` N + 1 th item.
4
j

Then,

Q.D. = 1 (Q3 – Q1) [Semi-interquartile range]
2

Interquartile range = Q3 – Q1

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

320 PRIME Opt. Maths Book - IX

Worked out Examples

1. If first quartile and third quartile of the observations are 10 and 30 respectively. Find
the quartile deviation and its coefficient.

Solution :

First quartile (Q1) = 10
Third quartile (Q3) = 30
Then,

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

= 1 (30 – 10)
2

= 10

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

= 30 – 10
30 + 10

= 0.5

2. If quartile deviation and its coefficient are respectively 30 and 0.6. Find th first and

third quartiles.

Solution :

Quartile deviation (Q.D.) = 30
Coefficient of Q. D. = 0.6

We have,

Q.D. = 1 (Q3 – Q1)
2

or, 30 × 2 = Q3 – Q1
or, Q1 = Q3 – 60 .................... (i)
Again,

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

or, 0.6 = Q3 – Q1
Q3 + Q1

or, 6 = Q3 – Q1
10 Q3 + Q1

or, 3Q3 + 3Q1 = 5Q3 – 5Q1
or, 8(Q3 – 60) = 2Q3
or, 6Q3 = 480
\ Q3 = 80

Putting the value of Q3 in equation (i) 321
Q1 = 80 – 60 = 20
\ Q1 = 20
Q2 = 60
PRIME Opt. Maths Book - IX

3. Find the quartile deviation of the observations 15, 24, 18, 12, 36, 33, 30 27, 21.

Solution :

The given observations in ascending order are:

12, 15, 18, 21, 24, 27, 30, 33, 36

Number of observations (N) = 9

Now,

First quartile (Q1) = size of ` N + 1 th item
4
j

= size of ` 10 th item
4
j

= size of 2.5th item

= 2nd + 0.5 (3rd – 2nd)

= 15 + 0.5(18 – 15)

= 15 + 1.5

= 16.5

Third quartile (Q3) = size of 3 ` N + 1 th item
4
j

= size if 3 × 2.5th item

= size of 7.5th item

= 7th + 0.5(8th – 7th)

= 30 + 0.5(33 – 30)

= 31.5

Then,

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

= 1 (31.5 16.5)
2

= Q3 – Q1 × 15
Q3 + Q1

= 7.5

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

= 31.5 – 16.5
31.5 + 16.5

= 15
48

= 0.3125

322 PRIME Opt. Maths Book - IX

4. Find quartile deviation and its coefficient of the marks obtained by students in grade
IX.

Marks 35 30 24 12 18 40 48

f 15 12 7 3 5 4 4

Solution :

The observations in ascending order:

Marks f c.f.

12 3 3

18 5 3+5=8

24 7 8 + 7 = 15

30 12 15 + 12 = 27

35 15 27 + 15 = 42

40 4 42 + 4 = 46

48 4 46 + 4 = 50

N = 50

Here, = size of ` N + 1 th item
First quartile lies in, 4
j

= size of a 50 + 1 th item
4
k

= size of 12.75th item

15 is just greater than 12.75 in c.f.

\ Q1 = 24

Third quartile lies in, = size of 3 ` N + 1 th item
4
j

= size of 38.25th item

42 is just greater than 38.25 in c.f.

\ Q3 = 35
Then,

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

= 1 (35 – 24)
2

= 1 ×1
2

= 5.5

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

= 35 – 24
35 + 24

= 11
59

= 0.1864

PRIME Opt. Maths Book - IX 323

Exercise 8.2

1. Find the followings.
i) If first and third quartiles of the observations are 20 and 60. Find the quartile

deviation and its coefficient.

ii) The upper and lower quartiles of the observations are 25 and 55 respectively.
Find the semi-inter-quartile range and its coefficient.

iii) The first quartile of the observations having inter-quartile range 20 is 10, find
the coefficient of quartile deviation.

iv) The quartile deviation and its coefficient are respectively 30 and 0.6. Find the
first and third quartiles of the observations.

v) The inter-quartile range and coefficient of quartile deviation are 100 and 0.5
respectively find the two extremes quartiles.

2. Find the quartile deviation and its coefficient of the following observations.
i) 28, 24, 18, 12, 6, 52, 48, 42, 38, 34, 56
ii) 42, 48, 44, 46, 58, 56, 52, 50, 38
iii) 120, 110, 100, 140, 130, 160, 150,170
iv) 200, 180, 160, 260, 240, 220, 300, 280
v) 10, 20, 30, 100, 90, 80, 70, 60, 50, 40, 120, 130, 110

3. Find semi-interquartile range and its coefficient of the followings.

i) x 12 18 24 30 36 42
4
f 3 7 12 9 6

ii) Marks 44 50 25 30 36 40

f 6 4 5 8 10 7

iii) Age 15 20 28 32 40 46 50
f 2478631

iv) Height 100 140 120 180 160

f 18 42 60 45 34

v) Weight 34 42 22 30 60 56 50

f 16 20 10 14 9 11 15

324 PRIME Opt. Maths Book - IX

4. PRIME more creative questions
i) What is quartile deviation? Also define coefficient of quartile deviation with

their calculation formula.

ii) I1f6Q, 1fionfdtthheeoqbusaerrtvialetidoenvsiatatikoennainndoritdsecro1e2ff,imcie–n5t ,bmy f–in3d,i2n1g,‘m24’., 27, 31, 35, 36 is

iii) If third quartile of the observations taken in order 100, 110, 120, 130, 140, P +
50, 2P – 40 and 170 is 157.5, find the ‘P’. Also find Q.D. and its coefficient.

iv) If 3rd quartile is three times the first quartile and Q.D. is 10, find the value of
quartiles and coefficient of quartile deviation.

v) collect the marks obtained by the students of your class in first term examination.
Construct discrete frequency distribution table and find the value of inter
quartile range and coefficient of quartile deviation.

1. i) 20, 0.5 Answer iii) 0.5
iv) 20, 80 iii) 22.5, 0.1667
ii) 15, 0.375 iii) 6, 0.176
v) 50, 150
iv) 10, 30, 0.5
2. i) 15, 0.454 ii) 5.5, 0.113
iv) 45., 0.195 v) 35, 0.5

3. i) 6, 0.2 ii) 7, 0.189
iv) 20, 0.142 v) 10, 0.25

4. i) Show to your teacher iii) 100, 22.5, 0.1667
ii) 20, 8.5, 0.346
v) Show to your teacher

PRIME Opt. Maths Book - IX 325

8.3 Mean deviation (Average deviation)

It is defined as the average of positive deviation
of all the observations from their central values
mean and median.

It is calculated by taking the positive difference of the observations from mean and
median and their arithmetic mean.

• For individual observations:

Mean deviation from mean (M.D. from X) = Σ X–X
N

Mean deviation from median (M.D. from, Md) = Σ X–Md
N

• For discrete observations:

M.D. (X) = Σf X–X
N

M.D. (Md) = Σf X–Md
N

• Where X is the arithmetic mean acnaldcuMlad ties the median of the observations.
• Coefficient of mean deviation is by using,
M.D.
Coefficient of M.D. = X (from mean)

Coefficient of M.D. = M.D. (from median)
Md

Worked out Examples

1. Find mean deviation and its coefficient of 18, 12, 24, 20, 36 from mean.
Solution :

The observations taken in ascending order,

XX |X – X|

12 10

18 4

20 22 2

24 2

36 14

∑X = 110 ∑|X – X| = 32

No. of observations (N) = 5
ΣX
Mean (X) = N = 110 = 22
5

326 PRIME Opt. Maths Book - IX

Here,

M.D. ^Xh = R X–X
N
32
= 5

= 6.4

Coefficient of M.D. ^Xh = M.D.
X

= 6.4
22

= 0.29

2. Find mean deviation from median and its coefficient of: 108, 112, 116, 122, 126, 130,
136

Solution :
The observation in ascending order,

X Md X – Md
108 14

112 10

116 6

122 122 0

126 4

130 8

136 14

R X – Md = 56

Here,

No of observations (N) = 7

Median (Md) = size of a N + 1 th item
2
k

= size of a 7 + 1 th item
2
k

= size of 4th item

= 122

Then,

M.D. (Md) = R X–Md
N

= 56
7

Coefficient of M.D. (Md) = M.D.
Md

= 8
122

= 0.065

PRIME Opt. Maths Book - IX 327

3. Find mean deviation from mean and its coefficient of:

x 15 22 30 35 40
84
f 6 10 12
Solution :

X f f×x X X–X f X–X
12.75 76.5
15 6 90 5.75 57.5
2.25 27
22 10 220 7.25 58
12.25 49
30 12 360
35 8 280 27.75 Rf X – X = 268

40 4 160

N = 40, ∑fx = 1110

Here, Rfx
N
Mean ( X ) = = 1110 = 27.75
40
Then,

Mean deviation (M.D.) = Rf X–X
N

= 268
40

= 6.7

Coefficient of M.D. = MD
X

= 6.7
27.75

= 0.245

4. Find mean deviation from median and its coefficient of:

Marks 18 24 30 36 42 50
1
f 6 8 15 7 5
f X – Md
Solution : 72
48
X f c.f. Md X – Md 0
42
18 6 6 12 60
20
24 8 14 6
Rf X – Md =
30 13 27 0

36 7 34 30 6

42 1 40 12

50 1 40 20

N = 40

328 PRIME Opt. Maths Book - IX

Here,

Median lies in = size of a N + 1 th item
2
k

= size of a 40 + 1 th item
2
k

= size of 20.5th item

= 27 is just greater than 20.5

\ Median (Md) = 30
Then,

Mean deviation (M.D.) = Rf X–Md
N

= 242
6.5

= 6.5

Coefficient of M.D. = M.D. (Md)
Md

= 6.5
30

= 0.216

4. If mean of the observations taken in order 90, 100, 110, P, 125, is 108, find mean
deviation from mean and its coefficient.

Solution :

X X X–X

90 18

100 8

110 108 2

P7

125 17

∑x = 425 + P R X – X = 52

Here,

No of observations (N) = 5
Rx
Mean ( X ) = N

or, 108 = 425 + P
5

or, 540 = 425 + P
\ P = 115

Again, = R X–X = 52 = 10.4
M.D. from mean (M.D.) N 5 = 0.096
Coefficient of M.D.
= M.D. (X) = 10.4
PRIME Opt. Maths Book - IX X 108

329

Exercise 8.3

1. Find mean deviation from mean and its coefficient of the followings.
i) Weight kg : 24, 30, 36, 40, 50
ii) Rainfall (mm) : 85, 115, 120, 140, 135, 125
iii) Temperature (°C) : 41, 25, 30, 18, 26, 45, 32, 35, 31, 27
iv) Income (In Rs. 1000) : 24, 28, 29, 23, 36, 35, 25, 18
v) Marks (Out of 20) : 19, 17, 20, 18, 15, 12, 9, 16, 14, 10

2. Find mean deviation from median and its coefficient of the followings:
i) Temperature (°C) : 17, 10, 15, 7, 13, 9, 6, 18, 11, 14, 1
ii) Weight (kg) : 32, 35, 42, 30, 33, 37, 40, 43
iii) Height (ft) : 5.4, 5.2, 5.6, 5.8, 8, 4.8
iv) Marks (out of 25) : 24, 23, 25, 18, 20, 16, 21
v) In come (in Rs. 1000) : 41, 25, 30, 18, 20, 26, 43, 32, 35, 31, 27

3. Find the mean deviation from mean and its coefficient of the followings.

i) Marks 5 8 15 18 20 22
5
f 6 5 8 10 6

ii) Age 12 18 24 30 36

f4 5 8 9 4

iii) x 14 10 5 25 20

f5 6 2 4 3

iv) Cost (Rs.) 50 45 60 70 75 55

No fo shops 12 4 7 3 6 8

v) Weight 65 70 48 50 60 45 55
2
f 415436
40
4. Find mean deviation from median and its coefficient of the followings. 8

i) Marks 48 44 38 25 36

No of Students 2 4 15 9 12

330 PRIME Opt. Maths Book - IX

ii) Rainfall (mm) 15 35 25 65 55 45 75
No of Places 17 564 14 3

iii) Height (cm) 50 60 75 82 90 95

Numbers 8 5 10 12 3 2

iv) x 10 18 14 16 12

f 6 10 16 13 5

v) x 12 16 20 24 28 32 36
f 7 8 10 13 6 4 2

5. PRIME more creative questions:

i) If mean of the observations taken in order 2, 4, 6, m, 10, 12 is 7, find the mean
deviation from mean of the observations:

ii) If median of the observations taken in order is 110, find mean deviation from
median of the observations 90, 100, P + 5, P + 15, 125, 130

iii) If mean of the given observations is 12, find the value of ‘m’, mean deviation
from mean and its coefficient of the observations.

Marks 6 8 10 12 14 16 18 20

f 1 14 25 27 m 94 2

iv) Differentiate between mean deviation from mean and mean deviation from
median. Also write down their calculating formula.

v) Collect the marks obtained by the students of grade IX in first terminal
examination of your school in optional mathematics and calculate mean
deviation from mean and from median by constructing discrete frequency
distribution table.

PRIME Opt. Maths Book - IX 331

1. i) 7.2, 0.2 Answer iii) 5.8, 0.187
iv) 4.75, 0.1773 iii) 0.3, 0.0566
ii) 13.33, 0.111 iii) 5.5, 0.3667
v) 3, 0.2 iii) 12.15, 0.148
iii) 18, 2.24, 0.187
2. i) 4, 0.3636 ii) 4, 0.111
iv) 2.57, 0.1224 v) 6, 0.2

3. i) 4.75, 0.3167 ii) 6.106, 0.246
iv) 8, 0.139 v) 7.04, 0.132

4. i) 4.02, 0.1057 ii) 10, 0.222
iv) 2, 0.142 v) 5.36, 0.268

5. i) 3 ii) 12.5

iv) Show to your teacher.

v) Show to your teacher.

332 PRIME Opt. Maths Book - IX

8.4 Standard deviation (Root mean square deviation)

It is defined as the positive square root of the mean of square

of deviation taken from the mean.

i.e. S.D.(σ) = Σ (X – X)2

N

We calculate it as the positive square root the mean of the square of the deviation from
the mean (or assume mean). So it is called the root mean square deviation too.

For standard deviation,

S.D.(σ) = Σ (X – X)2

N

coefficient of standard deviation = S.D.
X

variance(σ2) = (S.D.)2

Coefficient of variation (C.V.) = σ × 100%
X

Where ‘σ’ is the Greek letter sigma.

i) For individual observations:
• Direct method (using mean)

Σ (X – X)2
S.D. (σ) = N , where the table is,

X X X–X (X – X)2

∑X = ∑(X – X)2 =

• Direct method (without mean)

S.D. (σ) = ΣX2 – ` ΣX 2 , where the table is,
N N
j

X X2

∑X = ∑X2 =

• Indirect method (using assume mean) [shortcut method]

PRIME Opt. Maths Book - IX 333

S.D. (σ) = Rd2 – a Rd 2 , where the table is,
N N
k

X d=X–A d2

∑d = ∑d2 =

A = assume mean taken from X.

ii) Discrete observations:
• Direct method (using mean)

S.D. (σ) = Rf (X – X)2
N , where the table is

X f fx X X – X(X – X)2 f(X – X)2

N = ∑fx = ∑f(X – X)2

• Direct method (without mean)

S.D. (σ) = Σfx2 – a Σfx 2 , where the table
N N
k

X f fx x2 fx2

N = ∑fx = ∑fx2

• Indirect method (using assume mean) [shortcut method]

S.D. (σ) = Σfd2 – a Σfd 2 , where the table is
N N
k

X f d=X–A fd d2 fd2

N = ∑fx = ∑fx2
A = assume mean
PRIME Opt. Maths Book - IX
334

Worked out Examples

1. Find the standard deviation of the observations by using mean of 32, 35, 40, 45, 48.
Also find the coefficient of variation.

Solution :

XX X–X (X – X)2

32 –12 144

35 –9 81

40 44 –4 16

45 1 1

48 4 16

∑x = 220 ∑(X – X)2 = 258

Here,

No. of observations (N) = 5

Mean (X) = Rx = 220 = 44
N 5

(σ) = R (X – X)2

N

= 258
5

= 51.6

= 7.18

Again, σ
X
Coefficient of variation (C.V.) = × 100% = 7.18 × 100% = 16.32%
44

2. Find C.V. of the given observations by using assume mean (direct method).
12, 15, 18, 20, 24, 18, 32, 35
Solution :

X A d = X – A d2

12 –8 64

15 –5 25

18 –2 4

20 0 0
20

24 4 16

28 8 64

32 12 144

35 15 225

∑d = 24, ∑d2 = 542

PRIME Opt. Maths Book - IX 335

No. of observations (N) = 8

Here,

S.D. (σ) = Rd2 – a Rd 2
N N
k

= 542 – a 24 2
8 8
k

= 67.75 – 9

= 7.665

Mean (X) = A + Rd = 20 + 24 = 23
N 8

\ C.V. = σ × 100% = 7.665 × 100% = 33.32%
X 23

3. Find standard deviation and its coefficient of the observation given below.

Height (cm) 20 25 35 44 50

No. of persons 8 6 12 5 9
Solution :

x f fx X (X – X) (X – X)2 f(X – X)2

20 8 160 –15 225 1800

25 6 150 –10 100 600

35 12 420 35 0 0 0

44 5 220 9 81 405

50 9 450 15 225 2025

N = 40, ∑fx = 1400 ∑f(X – X)2 = 4830

Here, Rfx
N
Mean (X) = = 1400 = 35
40

S.D. (σ) = Rf (X – X)2

N

= 4830
40

= 120.75

= 10.9886

Coefficient of S.D. = σ
X

= 10.9886
35

= 0.3139

336 PRIME Opt. Maths Book - IX

4. Find standard deviation and its coefficient of :

x 85 90 100 124 130 142
5
f 6 5 9 15 10
Solution :

X f fx x2 fx2

85 6 510 7225 43350

90 5 450 8100 40500

100 9 900 10000 90000

124 15 1860 15376 230640

130 10 1300 16900 169000

142 5 710 20164 100820

N = 50, ∑fx = 5730, ∑fx2 = 674310

Here, Σfx2 – a Σfx 2
S.D. (σ) = N N
k

= 13486.20 – (114.6)²

= 13486.20 – 13133.16

= 353.04
= 18.789

Mean (X) = Rfx = 5730 = 114.60
N 50 = 0.1639

\ Coefficient of S.D. = σ = 18.789
X 114.60

5. If X = 12, ∑fx2 = 4880 and standard deviation of discrete series is 10, find the total

number of participates.

Solution :
Rfx
X = 12 = N

∑fx2 = 4880

S.D. = 10

We have,

S.D. = Σfx2 – a Σfx 2
N N
k

or, 10 = 4880 – ^12h2
N

or, 100 = 4880 – 144 [ a SBS]
N

or, 244N = 4880

or N= 4880
244

\ N = 20

PRIME Opt. Maths Book - IX 337

Exercise 8.4

1. Find the followings:
i) If ∑(X – X)2 = 5000 and N = 50 of individual observations, find the S.D. and

coefficient of variation where x = 4000.
ii) If standard deviation of discrete observations is 4 where ∑ f(X – X)2 = 480, find

the total number of participants.
iii) If ∑fx = 200, ∑fx2 = 3280 and ∑f = 20, find the standard deviation and its

coefficient.
iv) If standard deviation of the observations having ∑fx2 = 22250 out of 40 students

is 12.5, find the value of arithmetic mean of the discrete observations. Also find
its C.V.
v) What is standard deviation? Write down about the coefficient of variation with
calculating formula of them.

2. Find the standard deviation and its coefficient following individual observations:
i) Temperature (°C) : 30, 28, 35, 25, 42, 20
ii) Weight (kg) : 50, 30, 40, 20, 10
iii) Marks (out of 100) : 80, 75, 85, 95, 90, 70, 65
iv) Rainfall (mm) : 28, 34, 38, 42, 48, 52, 56, 60, 64, 68
v) Height (cm) : 40, 45, 48, 52, 56, 62, 64, 33

3. Find the standard deviation and C.V. of the following observations.

i) Height (cm) 10 20 25 30 35 40
4
f 1 5 10 12 8
40
ii) Marks 15 20 24 28 3

No. of students 4 8 10 5 320
2
iii) Wages 270 250 260 240 300
6 8 12 15 7 72
No of workers 5

iv) Age 12 24 36 48 60 40
6
f 6 4 6 12 7

v) X 8 16 24 32
f 2345

4. PRIME more creative questions:
i) Construct discrete frequency distribution table and find standard deviation of

the observations : 20, 20, 20, 10, 10, 30, 40, 30, 30, 30, 30, 20, 40, 40, 50, 30, 40,
50, 10, 40.

338 PRIME Opt. Maths Book - IX

ii) Construct discrete frequency distribution table and find C.V. of the observations

: 15, 15, 10, 20, 20, 20, 25, 25, 25, 30, 30, 30, 10, 15, 20, 25, 25, ,25, 25, 30, 25, 30,

15, 25, 30, 20, 20.
iii) Construct discrete frequency distribution table and find varience of the

observations : 130, 140, 120, 120, 140, 140, 140, 130, 160 ,150, 140, 150, 160,

170, 170, 130, 120, 140, 140, 150, 160, 150, 140, 130, 150.
iv) If mean of the observations is 15, find the varience of the observations: 5, 10, x,

x + 5 & 25.
v) If mean of the observations is 22, find the standard deviation of the data.

Marks 10 15 20 25 30

f 24p85

5. Project work
Collects the marks obtained by the students of classes V, VI, VII, VIII and IX in first

terminal examination in mathematics of your school and constructs the discrete
frequency distribution of each classes. Also compare the standard deviation and C.V.
of each classes.

Answer

1. i) 10, 12.5% ii) 30 iii) 8, 0.8
iv) 20, 62.5%
v) Show to your teacher.
2. i) 7.047, 0.2349
iv) 12.657, 0.258 ii) 14.14, 0.47 iii) 10, 0.125
v) 9.987, 0.1997
3. i) 6.538, 22.54%
iv) 18.72, 43.04% ii) 6.582, 27.42% iii) 22.659, 8.66%
v) 10.58, 37.79%
4. i) 12.03
iv) 50 ii) 27.27% iv) 196.16
v) 6
5. Show to your teacher.

PRIME Opt. Maths Book - IX 339

Statistics

Unit Test Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. What do you mean by measure of dispersion?

2. a. If Σ x – x = 80 of the 8 observations where ∑x = 200, find the coefficient of
mean deviation.

b. If ∑fx2 = 4000 and ∑fx = 1000 of 50 students, find the standard deviation.
c. If first quartile is 12 and coefficient of quartile deviation is 0.4, find the third

quartile.


3. a. Find mean deviation from median and its coefficient of:

Marks 24 18 12 40 32 36

No. of students 12 10 8 4 9 7

b. If quartile deviation and its coefficient are respectively 8 and 0.4 of a data. Find

the lower and upper quartiles of the data.

4. Find the root mean square deviation of the observations given below. Also find the
coefficient of variation.

Age 14 18 25 36 40

No. of people 5 10 14 5 6

340 PRIME Opt. Maths Book - IX

Proposed Syllabus with Grid for
First Terminal Examination

S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Relation 2 2 2 1 7 18 8
ii. function

2 Matrices i. Introduction 2 2 1 – 5 10 6
ii. Addition

3 Co-ordinate i. Distance Formula 2 2 2 1 7 19 12
Geometry ii. Locus
iii. Section Formula

4 Trigonometry i. Measurement of A 24 3 1 9 27 18
angle
ii. Trigonometric
Ratios
iii. Conversion of TR

5 Transformation i. Reflection 1 2 1 1 5 14 6
ii. Rotation

6 Statistics Partition Values 1 1 2 – 4 11 6

Total Questions 10 13 11 4 38

Total Marks 10 26 44 20 100 56

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

Model Question Set for First Terminal Examination

Group : A [10 × 1 = 10]

1. a. What do you mean by antecedent and consequence is an ordered pair?

b. Find element of image of a function f(x) = 2x – 1 where pre-image is 1.

2. a. What do you mean by scalar matrix?

b. If A = <3 2 1F , find a13 + a22.
4 –2 –3

3. a. What do you mean be points of trisection of a line segment AB?

b. Is (1, 2) a point in the locus of a point having equation 3x – y = 1?

4. a. Express 12° 15’ 5’’ into seconds. A
b. Write down the trigonometric ratios of Tanq and Secq
q
from the given right angled DABC.
5. a. What is the image of a point A(2, 1) under reflection

about x = 1?

b. Write down the calculating formula of quartile deviation
and its coefficient for the individual observations.
B C

Group : B [13 × 2 = 26]
6. a. If ordered pairs (2x – y, 4 – y) and (4, y), are the equal ordered pairs, find the

value of ‘x’ and ‘y’.

PRIME Opt. Maths Book - IX 341

b. If a relation R = {(x, y) : x + y = 8, x, y∈N}, find R in ordered Pairs form.
c.
7. a. If aij = 2i + j is the general element of a matrix, find the matrix of order 2 × 2.
b.
If A = <2 –1F and B = <1 –2F , find the matrix 3A – 2B.
8. a. 3 1 –1 3
b.
9. a. Find the co-ordinate of a point on y-axis which is 5 units distant from a point
b.
c. (4, – 1)

10. a. In what ratio does x-axis cuts the line joining the points (1, 4) and (5, –6)?
b.
Convert 45g 24’ 42’’ into sesagesimal measurement.
c.
If nSinA = m, prove that Sec2A – Tan2A = 1

Prove that 1 – Sin4 A = 1 + 2Tan2A.
Cos4 A

Find the length of an arc of a circle which subtends an angle of 60° at the centre

of the circle of radius 4.2 cm.

What is transformation? Write down the name of geometrical transformations.

Find the image of a point A(3, –2) under rotation about –90° with centre origin
followed by reflection about x + y = 0.

If 42 is the 7th decile of the observations taken in order 12, 16, 20, 24, 28, 32, 36,
2m, m + 25, 48, 52. Find the value of ‘m’.

Group : C [11 × 4 = 44]
11. If a relation R = {(x, y) : 2x + y ≤ 6} domain A = {1, 2, 3, 4}, find R from A × A. Also show

in arrow diagram.

12. If f(x + 2) = f(x) + f(2), prove that f(0) = 0 and f(–2) = –f(2). Also prove that f(4) = 2f(2)

13. If A + B = <3 –3F and A – B = <1 1F , find the matrices A and B.
2 4 4 –2

14. Find the equation of locus of a point which moves so that its distance from (1, 2) is

equidistance from the point (2, 3).

15. Find the co-ordinate of fourth vertex of a parallelogram where three of the vertices

are A(1, 2), B(3, 5) and C(5, 4). [Ans : (3, 1)] kc ,

16. Find the angles of a triangle in degrees where the angles are a 2x a 8x g and ` rx jC
5 9 75
k

respectively. [Ans : 18°, 36°, 108°]

17. Prove that : 1 + Secq – Tanq = 1 – Sini
1 + Secq + Tanq Cosi

18. If 2Sin2q + Cosq – 1 = 0, find the value of Cosq.

19. Find the image of DABC having vertices A(1, –2), B(3, 3) and C(5, 0) under an

enlargement about E[(0, 1), 2]. Also plot the object and images in graph.

20. Find the 60th percentile of the observations. [Ans : 40]

Class 10 20 30 40 50

f 9 13 24 20 14
21. Find quartile deviation and its coefficient of the observations 15, 35, 25, 20, 30, 10.

342 PRIME Opt. Maths Book - IX

Group : D [4 × 5 = 20]
22. If f(x) = 3x – 2, range = {1, 4, 7, 10}, find the domain. Also show the function in arrow

diagram.

23. Find the co-ordinate of a point which is equidistant from the points (–2, 0), (5, –1)

and (–1, 7). [Ans: (2, 3)]

24. Prove that : Tani + Coti = Secq.Cosecq + 1
1 – Coti 1 – Tani

25. Find the co-ordinate of image of DABC having vertices A(–1, 1), B(2, 4) and C(3, –3)
under reflection about x + 2 = 0 followed by rotation about 180° with centre (0, 0).

Also plot the object and images in graph.

Proposed Syllabus with Grid for
Second Terminal Examination

S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Polynomial 22 2 1 7 19 6

ii. Sequence & Series

2 Matrices i. Multiplication 2 2 1 – 5 10 12
ii. Transpose

3 Co-ordinate i. Equation of Straight 2 2 2 1 7 19 20
Geometry line
ii. Perpendicular dis.
iii. Area of Triangle

4 Trigonometry i. Standard Angles 24 4 – 10 26 16
ii. Certain Angles
iii. Conversion of TR

5 Transformation i. Translation 1 2 1 1 5 14 10
ii. Enlargement

6 Statistics i. Quartile Deviation 11 1 1 4 12 12
ii. Mean Deviation
iii. Standard
Deviation

7. First Term Review 6

Total Questions 10 13 11 4 38

Total Marks 10 26 44 20 100 80

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

PRIME Opt. Maths Book - IX 343

Model Question Set for Second Terminal Examination

Group : A [10 × 1 = 10]
1. a. Define the term cartesian product A × B.
b. Evaluate /3 n (n + 1)
n=1
2. a. Find A = <a bF
(AT)T where c d
b.
3. a. If A = <1 2F and I is an identity matrix of order 2 × 2, find IA.
3 –2

Write down the formula of finding perpendicular distance from a point to the

straight line.
b. If (1, 2) lies in the locus of equation 2x + my = 8, find the value of ‘m’.

4. a. Find the value of Tan(675)°
b. If 5SinA = 3, find the value of CosA.

5. a. What is the image of A(–2, 1) under translation about T = <3F ?
1
b. Write down the calculation formula of standard deviation and its C.V. by using

mean of the observations.

Group : B [13 × 2 = 26]
6. a. If (32x + 1, 16) = (27, 23x – 8), find the value of ‘x’ and ‘y’.

b. If Sn = 3n + 2, find the value of t5. [Ans: 3]

c. If A + B = <2 4F and B = <–1 2F , find the matrix A and its transpose.
3 –2 1 –4
<2 0F
7. a. Which matrix pre-multiplies to 3 1 Y
results the matrix [7 1]?

b. Find the equation of straight line AB B X
from the given diagram where OC = 2 C
units.
30°
8. a. Prove that the points (3, 4), (7, 7) and AO
(11, 10) are collinear points.
Y’
b. Find the central angle of a circle in X’
degrees having radius 8.4cm which
subtends by an arc of length 8.8 cm.

9. a. Prove that : Sec²q + Cosec²q = Tanq + Cotq.
b.
c. Prove that : 1 – Tan30° =2– 3
10. a. 1 + Cot60°
If Secq – Cosecq = 0, find the value of Cosecq.
b.
Find the image of a point A(2, –3) under reflection about y = 0 followed by

rotation about +90° with centre (0, 0).

Find the reflection axis which transferred a point A(3, 5) to A’(1, 5)

c. If 25th percentile of the observations 24, 30, 2p+6, 42, 48, 54, 60, 66, 72
respectively taken in order is 33, find the value of ‘p’.

344 PRIME Opt. Maths Book - IX

Group : C [11 × 4 = 44]
11. If f(x) = 2Sinx + 1, domain = {0°, 30°, 60°, 90°} find range. Also show the function in

arrow diagram.

12. If p(x) = 2x3 – 3x + 5 – x2 and q(x) = 3x3 + 5x – 2 + 3x2, answer the questions given

below.

i. Find p(x) + q(x)

ii. Write down the degree of the polynomial obtained after addition.

iii. Write down the types of the polynomial according to degree.

13. If A = <3 2F and B = <2 –1F , prove that (AB)T = BT AT.
1 –1 3 1

14. Prove that the points P(2, 3), Q(2, –4) and R(9, –4) are the vertices of an isosceles

right angled triangle.

15. Find the points of trisection of the line joining the points (–5, –5) and (25, 10).
r
16. Prove that : Sin2 16 + Sin2 3r + Sin2 5r + Sin2 7r =2
16 16 16

17. Prove that : (3 – 2Sin2q)(3Tan2q – 1) = (3 + Tan2q)(4Sin2q – 1)
18. If 8Cosq + 15Sinq = 17, find the value of Cotq.
[Ans: 8/15]

19. The angles of a triangle in degrees, grades and radians respectively are in the ratio

288:280:p. Find the angles of the triangle in degrees. [Ans: 72°, 63°, 45°]

20. Find the centre of enlargement and scale factor which transform the points A(3, 3)

→ A’(5, 4), B(4, –1) → B’(7, –4) and C(6, 1) → C’(11, 0). [Ans: 6.16, 0.205]
21. Find the mean deviation from median and its coefficient of:

x 12 18 24 30 36 40

f 3 7 12 15 8 5

Group : D [4 × 5 = 20]
22. Answer the questions given below by the study of given diagrams.



i) Add one more diagram in the given pattern.

ii) Calculate the nth term of the sequence of number of dots used there in the

diagrams.

iii) Write down in sigma notation.

23. Find the equation of locus of a point P which moves in such a way that from the

points A(3, 0) and B(–3, 0) as AP2 + BP2 = AB2.

24. Find the image of a triangle having vetices A(1, –2), B(3, 1) and C(5, –4) under a

translation AB followed by enlargement about E[(1, 2), –2]. Also plot them in graph.

25. Find the standard deviation and C.V. of : [Ans: 12.107, 30.8%]

x 15 24 33 42 51 60

f 3 5 8 12 9 3

PRIME Opt. Maths Book - IX 345

Specification Grid for Final Examination
referred by CDC Nepal

S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Relation 23 2 1 8 21 33

ii. function

iii. Polynomial

iv. Sequence & Series

2 Limits & Limit & Continuity 1 – 1 – 2 5 10
Continuity

3 Matrices i. Introduction 1 2 1 – 4 9 20
ii. Addition
iii. Multiplication

4 Co-ordinate i. Locus 2 2 1 1 6 15 30
Geometry ii. Section Formula
iii. Equation of
Straight line
iv. Area of Triangle

5 Trigonometry i. Measurement of 2 3 3 – 8 20 35

Angle (4)

ii. Trigonometric

Ratios

iii. Conversion of TR

iv. Standard Angles

v. Certain Angles

vi. Compound Angle

6 Vector i. Introduction 12 – 1 4 10 12
1– 1 1 3 10 18
ii. Vector Geometry

7 Transformation i. Reflection

ii. Rotation

iii. Translation

iv. Enlargement

8 Statistics i. Partition Values –12 – 3 10 12

ii. Quartile Deviation

ii. Mean Deviation

iii. Standard

Deviation

First Term Review 4

Second Term Review 4

Total Questions 10 13 11 4 38 34

Total Marks 10 26 44 20 100 170

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

346 PRIME Opt. Maths Book - IX