Vedanta Opt. Math 9 Final (2078)

vedanta Excel In Opt. Mathematics - Book 9 Transformation

scale factor (k) = 2 P'((k(x – a) + a, k(y – b) + b)
Then, P(x, y) E[(a, b), k] P'(2(2 – 1) + 1, 2(3 – 1) + 1) = P'(3, 5)
Now, P(2, 3) E[(1, 1), 2] Q'(2(4 – 1) + 1, 2(5 – 1) + 1) = Q'(7, 9)
R'(2(7 – 1) + 1, 2(0 – 1) + 1) = R'(13, –1)
Q(4, 5) E[(1, 1), 2]
R(7, 0) E[(1, 1), 2]

'PQR and 'P'Q'R' are plotted in the adjoint graph.

Y
9 Q'

8

7

6 P' Q
5

4 P
3

2

X' 1R X
-2 -1-1O
-2 1 234 5 6 7 8 9 10 11 12 13
R'

-3

Y'

Example 5. Find centre and scale factor of enlargement if two points P(4, 2) and Q(3, 1)
Solution: are mapped to P'(–8, –2) and R'(–6, –2).

Let (a, b) be the centre of enlargement and k the scale factor of enlargement.
Then we have,

P(x, y) E[(a, b), k] P'(k(x – a) + a, k(y – b) + b)

Now, for point P(4, 2)

P(4, 2) P'[k(4 – a) + a, k(2 – b) + b] = P'(4k – ka + a, 2k – kb + b)

But, P'(–8, –2),

? 4k – ka + a = –8 ...................... (i)

2k – kb + b = –4 ...................... (ii)

Again, for point Q(3, 1),

Q(3, 1) Q'[k(3 – a) + a, k(1 – b) + b] = Q'(3k – ka + a, k – kb + b)

But, Q'(–6, –2),

? 3k – ka + a = – 6 ...................... (iii)

k – kb + b = – 2 ........................ (iv)

Subtracting equation (iii) from (i), we get,
4k – ka + a = – 8

–3k+– ka +– a = +– 6
k = –2

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vedanta Excel In Opt. Mathematics - Book 9 Transformation

Putting the value of k in equation (i), we get
4 (–2) – (–2) a + a = –8

or, a = 0
Putting the value of k in equation (ii), we get,

2 (–2) – (–2)b + b = –4
? b = 0.
Hence, centre of enlargement is (0, 0) and scale factor –2.

Exercise 13.4

Very Short Questions:

1. (a) Define enlargement.

(b) Under which conditions the object and image are

(i) in the same direction of centre of enlargement.

(ii) in the opposite direction of centre of enlargement.

(c) What is invariant point in enlargement?

(d) Write the meaning of E[(0, 0), k].

(e) Write the formula to find image of P(x, y) under E[(a, b), k].

2. Find the image of the following points under E[0, 2]:

(a) A(7, 8) (b) B(4, 5) (c) C(2, 7)

3. Find the image of the following points under E (0, 0), 1 :
2

(a) P(2, 4) (b) Q(4, 6) (c) R(10, 8)

4. Find the image of the following points under E[(1, 2), 3]:

(a) P(4, 5) (b) Q(6, 5) (c) R(2, 4)

Short Questions:

5. Sketch the image of the following geometrical objects with the centre of enlargement O
and the given scale factor.

(a) C (b) P S

A

O B QR
k=2 O
k = –2
302
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vedanta Excel In Opt. Mathematics - Book 9 Transformation

(c) P S (d) U

O
W

QR V X

k = 1 k = 2 1
2 2

6. Find the centre of enlargement and scale factor from the given object and its image:

(a) D C (b) P'

D' C' P

R R'
Q
A' B'
A B Q'

(a) P R'
Q'

Q P'

R

Long Questions:

7. (a) A(4, –4), B(4, 0) and C(2, 2) are the vertices of 'ABC. Find the image of 'ABC
under the enlargement E[0, 2]. Present both object and image triangles on the
same graph.

(b) P(2, –1), Q(2, 3), and R(2, 5) are the vertices of 'PQR. Find the image of 'PQR
under enlargement E[O, 2]. Present both the object and the image triangle on the
same graph.

(c) A triangle with vertices P(4, 0), Q(4, 4), and R(1, 4) is enlarged about centre O and
scale factor –1. Find the vertices of the image triangle so obtained. Plot both the
object triangle and image triangle on the same graph.

(d) A triangle PQR with vertices P(3, 0), Q(0, 2), and R(3, 2) is enlarged about centre O
and scale factor –2. Find the vertices of image triangle P'Q'R'. Plot both the object
triangle and the image triangle on the same graph.

8. (a) Enlarge 'PQR having the vertices P(4, 0), Q(4, 4), and R(7, 2) under enlargement
E[(2, 2), 2] so that the image 'P'Q'R' is formed. Find the coordinates of vertices of
image triangle.

(b) An enlargement maps 'ABC with vertices A(3, 2), B(4, 2), and C(6, –5) into 'A'B'C'
about centre (0, –2) and scale factor 2. Find the coordinates of the image vertices
of the triangle.

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vedanta Excel In Opt. Mathematics - Book 9 Transformation

9. (a) P(2, 1), Q(2, 0), and R(4, 2) are the vertices of 'PQR. Find the image of 'PQR under
the enlargement with centre (1, –1) and scale factor 2. Plot the 'PQR and 'P'Q'R'
on the same graph.

(b) P(4, 0), Q(3, –3), and R(7, 4) are the vertices of 'PQR under the enlargement with
scale factor 2 at the centre (2, 1). Plot the 'PQR and 'P'Q'R' on the same graph.

10. (a) A(2, 2), B(6, 2), C(6, 6), and D(2, 6) are the vertices of square ABCD. Find the image
under E[O, –2]. Show them in the same graph.

(b) P(2, 2), Q(10, 2), R(13, 6), and S(5, 6) are the vertices of parallelogram PQRS. Find
1
the image under E O, 2 . Plot both of them in the same graph.

11. (a) A triangle ABC with vertices A(4, 2), B(–3, 1), and C(0, 8) is enlarged to 'A'B'C'

with A'(–8, –4), B'(6, –2), C'(0, –16). Find the centre and scale factor enlargement.

(b) A(2, 3) and Q(1, 4) are mapped to P'(6, 9) and Q'(3, 12) under an enlargement with
a scale factor. Find the centre and scale factor of the enlargement.

(c) P triangle PQR with vertices P(–2, –1), Q(2, 3), and R(1, –1) enlarged to 'P'Q'R'
with vertices P'(–6, –3), Q'(6, 9), and B'(3, –3). Find the centre and scale factor of
the enlargement.

12. An enlargement mapped A to A' and B to B'. Find the centre and the scale factor of the
enlargement.

(a) A(–1, 2) A'(–4, 3) (b) A(2, 0) A'(2, –6)

B(2, –3) B'(2, –7) B(4, 3) B'(8, 3)

Project Work

13. Prepare a report with group discussion stating the following:
(a) Definition and properties of enlargement.
(b) Use of enlargement in our daily life.

2. (a) A'(14, 6) (b) B'(8, 10) (c) C'(4, 14) 3. (a) P'(1, 2)
(b) Q'(2, 3)
(c) R'(4, 8) (c) R'(5, 4) 4. (a) P'(10, 11) (b) Q'(16, 11)

5. Show to your teacher. 6. Show to your teacher.

7. (a) A'(8, –8), B'(8, 0), C'(4, 4) (b) P'(4, –2), Q'(4, 6), R'(4, 10)

(c) P(–4, 0), Q'(–4, –4), R'(–1, –4) (d) P'(–6, 0), Q'(0, –4), R'(–6, –4)

8. (a) P'(6, –2), Q'(6, 6), R'(12, 2) (b) A'(6, 6), B'(8, 6), C'(12, –8)

9. (a) P'(3, 3), Q'(3, 1), R'(7, 5) (b) P'(6, –1), Q'(4, –7), R'(12, 7)

10. (a) A'(–4, –4), B'(–12, –4), C'(–12, –12), D'(–4, –12)

(b) P'(1, 1), Q'(5, 1), R' 13 , 3 , S' 5 , 3 11.(a) E[(0, 0), –2] (b) E[(0, 0), 3]
2 2 (b) E[(2, 3), 3]
(c) E[(0, 0), 3] 12.(a) E[(2, 1), 2)

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Statistics : 14
Partition Values

14.0 Review

The following are the marks obtained by 11 students in optional mathematics:
65, 45, 15, 25, 75, 85, 95, 20, 30, 90, 30. Then,

(i) write above marks in ascending and descending orders.
(ii) find the middle value of the marks.
(iii) find the difference between the maximum and minimum marks.
(iv) is there any difference between mean and median of the data ?
In the above data when the marks are arranged in ascending or descending order, the middle
mark is called median. It divides the given data into two equal parts. The difference of the
maximum and minimum value is called range.

Discuss the answer of the following questions in groups.
(a) What is statistics ?
(b) What is frequency ?
(c) What do cummulative frequencies represent ?
(d) Why is arithmetic mean or average used ?
(e) Define individual, discrete and continuous series.

Partition Values

Partition values are those variate values which divide the whole distribution of data into a
number of equal parts.

For example; Median is a partition value which divides a given distribution of data into two
equal parts.

The following are the types of partition values :

(a) Median (b) Quartiles (c) Deciles (d) Percentiles.

In this section, we discuss about quartiles, deciles, and percentiles only.

We learn to calculate partition values of individual and discrete series.

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

14.1 Quartiles

Quartiles are those partition values which divide the whole distribution of data into four
equal parts. There are three quartiles Q1, Q2 and Q3. The first quartile Q1 is called the lower
quartile and the third quartile Q3 is called the upper quartile. The second quartile Q2 is called
the median.

50% 50%

25% 75% Q2 75% 25%
Q1 Md Q3

The variate value Q1 divides the lower half into two equal parts. It means that Q1 divides the
whole data into lower 25% and upper 75%. Q2 divides the given data 50% and 50%. Q2 is
also called median it divides the whole data into two equal halves. Q3 divides the data into
lower 75% and upper 25%.

Calculation of Quartiles for individual and discrete series:

Individual Series

Let n be the total number of observations.

[ ]Q1 = value ofi(n + 1) th
4 item, i = 1, 2, 3.

( )Then, the first Quartile (Q1) = value of th
n+1 item.
4
( )The second Quartile (Q2) = Median (Md) = value of
( )The third Quartile (Q3) = Value of n + 1 th
2 item.

3(n+1) th item.

4

Worked out Examples

Example 1. Find the three quartiles for each of the following set of observations:
Solution:
(a) 15, 16, 10, 13, 17, 22, 21, 25, 18, 23, 19

(b) 70, 60, 80, 50, 90, 100, 40, 85

(a) Arranging the given data in ascending order, we have 10, 13, 15, 16, 17,

18, 19, 21, 22, 23, 25 number of observations (n) = 11

Lower Quartile (Q1) = value of ( )n + 1 th4 item

( )11+ 1 th item = 3rd item = 15
=4
( )n + 1 th
The second Quartile (Q2) = Median (Md) = value of 2 item

= value of ( )11+ 1 th2 item = 6th item = 18

Upper Quartile (Q3) = value of ( )3(n+1) th4 item

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

= value of ( )3 × 12 th4 item

= Value of 9th item = 22

(b) Arranging the given data in ascending order, 40, 50, 60, 70, 80, 85, 90, 100

Number of observations (n) = 8

( )Now, the first Quartile (Q1) = value of th item
n+1
( )= 4
th item = 2.25th item
8+1
4
= 2nd item + (3rd item – 2nd item) × 0.25

= 50 + (60 – 50) × 0.25

= 50 + 10 × 0.25 = 50 + 2.5 = 52.5

Median (Md) = Second Quartile (Q2)
Upper Quartile (Q3) ( )= Value of
( )= Value ofn+1 th item
2
th
8+1
2 item = 4.5th item

4th item + 5th item = 70 + 80 = 75
2 2
( ) ( )=
3(n+1) th item = 3.9 th
=4 4
item

= (6. 75)th item

= 6th item + (7th item – 6th item) × 0.75

= 85 + (90 – 85) × 0.75

= 85 + 5 × 0.75 = 88.75

Example 2. Calculate lower quartile (Q1) and upper quartile (Q3) from the given data:
Solution:
x 10 15 20 25 30 35 40 45 50
f 4 7 9 15 12 10 7 2 1
To calculate the lower quartile (Q1) and upper quartile (Q3).

x f c.f (cumulative frequency)

10 4 4

15 6 10

20 8 18

25 15 33

30 12 45

35 10 55

40 6 61

45 2 63

50 1 64

∑f= N = 64

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

( )N+1 th
Now, lower quartile (Q1) = Value of 4 item

= Value of 4( )64 +1 th item

= Value of (16.25)th term

From the table, c.f. just greater then 16.25 is 18

whose corresponding value is 20.

? Q1 = 20

Upper Quartile (Q3) = Value of 4( )3(N+1) th item

( )3(64 + 1) th

= Value of 4 item

= Value of 48.75th item

c.f. just greater than 48.75 is 55, whose corresponding value is 35.

? Q3 = 35

Exercise 14.1

Very Short Questions
1. (a) Define partition values.

(b) Define the lower quartile and upper quartile.
(c) Which quartile is equal to median ?
(d) Write formula to calculate Q1 and Q3 for discrete series.
(e) What are the positions of the lower quartile and the upper quartile for a discrete

series with 119 terms?
Short Questions
2. Find the lower quartile, median and the upper quartile of the following data:

(a) 4, 6, 8, 12, 14, 16, 20
(b) 10, 15, 20, 25, 30, 35, 40
(c) 2, 5, 7, 8, 9, 10, 12, 20, 22, 24, 26
(d) 5, 6, 8, 10, 12, 14, 20, 24, 26, 29, 34, 35, 38, 40, 45
3. Find Q1, Q2 and Q3 of the following data:
(a) 11, 15, 8, 9, 12, 13, 14, 18
(b) 20, 25, 15, 10, 30, 45, 22, 35
(c) 45, 50, 15, 20, 25, 35, 40, 42, 41, 18, 17, 44
(d) 11, 21, 14, 16, 18, 20, 24, 27, 30, 35, 40, 25

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Long Questions

4. Find Q1 and Q3 of the following data :

(a) x 10 20 25 30 35 40 50
f 4 6 10 15 13 10 2

(b) x 16 18 20 10 25 30
f 3 4 10 2 12 8

(c) x 40 20 30 10 60 80
f 12 2 8 1 4 3

5. Find Q1 and Q3 of the following data :

(a) Size 11 13 15 17 19 21 22
5
No. of boxes 2 9 20 25 24 15

(b) Class (marks) 5 15 25 35 45
No. of students 5 8 15 10 6

(c) Age (in years) 10 11 12 13 14 15 16 17
No. of students 4 6 8 10 8 6 4 2

(d) Income (Rs.) 100 150 200 250 300 400

No. of labourers 5 8 10 13 10 4

6. Construct discrete frequency distribution table and find Q1 and Q3:
(a) 50, 20, 60, 30, 48, 60, 70, 20, 30, 50, 40, 46, 30, 40, 60, 70, 70, 37, 40, 50

(b) 15, 12, 23, 35, 46, 57, 18, 12, 39, 41, 32, 43, 25, 49, 18, 38, 45, 40, 32, 33.

7. (a) If 2x + 1, 3x – 1, 3x + 5, 5x – 7, 51, 63 and 70 are in ascending order and Q1 = 20,
find the value of x and Q3.

(b) If 10, x + 4, 20, 25, 30, 35, 40 are in ascending order and Q1 = 15, find the value
of x and Q3.

Project Work

8. List the first 100 natural numbers. Find Q1, Q2 and Q3 from the data.

1. (e) 30, 90 2. (a) 6, 12, 16 (b) 15, 25, 35 (c) 7, 10, 22 (d) 10, 24, 35

3. (a) 9.5, 12.5, 14.75 (b) 16.25, 23.5, 33.75 (c) 18.5, 37.5, 43.5

(d) 16.5, 22.5, 29.25 4.(a) 25, 35 (b) 20, 25 (c) 30, 60

5. (a) 15, 19 (b) 15, 35 (c) 12, 15 (d) 150, 300

6. (a) 37, 60 (b) 23, 43

7. (a) x = 7, Q3 = 63 (b) x = 11, Q3 = 35

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

14.2 Deciles

The variate values which divide the total number of observations (arranging in ascending
or descending order) into ten equal parts are called deciles. There are nine deciles. D1, D2,
D3,....... D9.

The first decile (D1)
The second decile (D2)
The third decile (D3)
.............................
The nineth decile (D9)

10% 20% 30% 40% 50% 60% 70% 80% 90%

D1 D2 D3 D4 D5 D6 D7 D8 D9

Note : 5th decile is median.

Calculation of Deciles for Individual and Discrete Series:

Let n = total number of observation

[ ]For Individual Series : Di = value of
i(n+1) th item i = 1, 2, 3...., 9

10
( )The first decile(D1) = Value of
(n+1) th item

10
( )The second decile(D2) = Value of
2(n+1) th item

10
( )The fifth decile (D5) = value of
5(n+1) th item.

10
( )The nineth decile (D9) = value of
9(n+1) th item.

10

For Discrete series: Let N = total number of observations.

( )The first decile (D1) = value of 10
N+1) th item

( )The second decile (D2) = value of
2(N+1) th item.

10
( )The fifth decile(D5) = Median value
5(N+1) th item.

10
( )The nineth decile (D9) = value of
9(N+1) th item.

10

Worked out Examples

Example 1. Find D3, D6, and D9 from the following data :
Solution: (a) 2, 4, 6, 20, 18, 12, 14, 21, 24
(b) 5, 10, 20, 15, 25, 22, 30, 35, 12, 14, 11
(a) Arranging the given data in ascending order we get,

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

2, 4, 6, 12, 14, 18, 20, 21, 24

Here, total number of observation (n) = 9

The third decile (D3) = value of ( )3(n+1) th10 item

( )3(9+1) th item
= value of 10

= value of 3rd item = 6

The sixth decile(D6) = value of ( )6(n+1) th10 item

= 6th item = 18

The nineth decile (D9) = value of ( )9(n+1) th10item = 9th item = 24

(b) Arranging the given data in ascending order, we get,
5, 10, 11, 12, 14, 15, 20, 22, 25, 30, 35

Here, total number of observation (n) = 11

The third decile (D3) = value of ( )3(n+1) th10 item

= value of 3.6th item

= 3rd item + (4th item – 3rd item) × 0.6

= 11 + (12 – 11) × 0.6

= 11 + 0.6

= 11.6

The sixth decile (D6) = value of ( )6(n+1) th10 item

( )= value of
6(11+1) th item
10

= 7.2th item

= 7th item + (8th item – 7th item) × 0.2

= 20 + (22 – 20) × 0.2

= 20 + 0.4

= 20.4

The nineth decile (D9) = value of ( )9(n+1) th10item

( )= value of
9(11+1) th item
10

= 10.8th item

= 10th item + (11th item – 10th item) × 0.8

= 30 + (35 – 30) × 0.8

= 30 + 4 = 34

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Example 2. Find the 4th and the 7th decile from given data :

x 40 30 20 10 60 80 100
f 12 10 7 5 7 3 2

Solution: To calculate the 4th and the 7th decile, writing the given data in ascending order :

xf c.f
10 5 5
20 7 12
30 10 22
40 12 34
60 7 41
80 3 44
100 2 46

N = 46

Total observation (N) = 46

( )4(N+1) th
The fourth decile (D4) = value of 10 item

= value ( )4(46+1) th10 item

= 18.8th item

Then, 18.8

The c.f., just greater is 22 whose corresponding value is 30.

? D4 = 30 ( )7(N+1) th item
The seventh decile (D7) = value of 10

( )7(46+1) th item
= 10

= 32.9th

c.f., just greater than 32.9 is 41, whose corresponding value is 60.

? D7 = 60

Exercise 14.2

Very Short Questions.

1. (a) Define decile.

(b) Which decile is equal to median ?

(c) If N is the total number of observations in a descrete series, write the formulas to
calculate D3 and D9.

(d) If N = 199, what are the positions of 4th and 7th deciles?

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Short Questions

2. Find D4, D6, and D9 from the following data:
(a) 20, 25, 5, 10, 15, 8, 7, 25, 28.

(b) 22, 42, 12, 32, 48, 13, 16, 52, 45.

(c) 10, 20, 80, 70, 60, 50, 40, 90, 100.

(d) 1, 2, 6, 7, 8, 12, 14, 5, 7, 9, 22, 11, 13, 19, 25, 26, 23, 29, 28.

3. Find D3, D7, and D8 from the following data :
(a) 80, 90, 70, 50, 40

(b) 11, 21, 12, 18, 20, 24, 28, 30, 32, 19.

(c) 15, 5, 18, 25, 40,

(d) 22, 45, 23, 67, 18, 70, 80, 75.

Long Questions

4. Find D2 , D5 and D7 from the following data :

(a) x 5 10 15 20 25 30
f 2 4 6 10 6 2

(b) x 22 24 26 28 30 34
f 459842

(c) x 20 30 40 50 60 70
f 2 6 8 10 6 4

5. Find D4 , D6, and D9 from the following data: 70
6
(a) Marks 10 20 30 40 60 80
No. of students 4 6 8 16 8 4

(b) Age (year) 12 13 14 15 16 17 18

No. of students 4 5 9 12 10 6 4

(c) Weight (kg) 15 18 20 22 24 26

No. of children 2 3 5 8 4 2

6. Construct discrete frequency distribution table and find D4 and D6:

(a) 12, 14, 18, 20, 16, 22, 18, 14, 12, 18, 13, 25, 22, 23, 22, 16, 20, 18, 23, 14, 16, 22,
28, 23, 12

(b) 40, 45, 42, 41, 46, 75, 80, 75, 40, 45, 42, 43, 41, 45, 48, 50, 52, 54, 70, 75, 80, 60,
75, 60

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Project Work

7. List the marks obtained by students in your class in compulsory mathematics and find
D2, D4, D6 and D8 with their meanings..

1. (a) 80th, 140th 2. (a) 10, 20, 28 (b) 22, 42, 52 (c) 50, 70, 100
(d) 9, 14, 28 3. (a) 48, 82, 88 (b) 18.3, 26.8, 29.6 (c) 13, 28, 37
(d) 22.7, 71.5, 76 4. (a) 15, 20, 20 (b) 24, 26, 28 (c) 30, 50, 50

5. (a) 40, 40, 70 (b) 15, 16, 17 (c) 20, 22, 26

6. (a) 18, 22, 25 (b) 45, 54, 80

14.3 Percentiles

Percentiles are those partition values which divide a given distribution of data into 100
equal parts. There are 99 percentiles which are denoted by P1, P2 P3, ........... P99.

Note :
50th percentile = Median = 5th decile = 2nd Quartile

Calculation of Percentiles for Individual and Discrete Series.

For Individual Series : Let n = total number of observations.

( )i(n+1) th
Pi = value of 100 item, i = 1, 2, 3, ........., 99.

( )First percentile (P1) = value of 100
n+1 th item.

( )Sixtieth percentile (P5) = value of 100
5(n+1) th item.

( )Sixth percentile (P60) = value of th item.
60(n+1)
10
( )Ninety ninth Percentile (P99) = value of
99(n+1) th item.
100

For Discerte Series:

Let N = total number of observations

( )First Percentile (P1) = value of th
N+1 item
100
( )Fortyth percentile (P40) value of
40(N+1) th item.
100
( )Ninetith percentile (P90) = value of
90(N+1) th item.
100

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Worked out Examples

Example 1. Find the 20th and the 70th percentiles for the following data:
Solution:
(a) 20, 40, 50, 30, 60, 70, 75, 35, 25.

(b) 80, 75, 200, 125, 225, 90, 120, 110, 140, 150, 260.

(a) Writing the given data in ascending order, we get,

20, 25, 30, 35, 40, 50, 60, 70, 75

number of observations (n) = 9

30th percentile (P30) ( )= value of30(n+1) th item.
( )= value of 10 th item.

30(9+1)
100
= value of 3rd item. = 30

? P30 = 30

70th percentile (P70) ( )= value of70(n+1) th item.
( )= value of 100 th item.

70(9+1)
100
= value of 7th item = 60

? P70 = 60

(b) Writing the given data in ascending order, we get,
75, 80, 90, 110, 120, 125, 140, 150, 200, 225, 260

Number of observations = 11

( )= Value of
( )= value of
30th percentile 30(n+1) th item.
100
30×(11+1) th
100
item.

= value of (3.6)th item.

= 3rd item + (4th value – 3rd value) × 0.6

= 90 + (110 – 90) × 0.6

= 90 + 12 = 102

? P30 = 102
(( ))70th percentile (P70)
= value of 70(n+1) th item.
= value of 100
70(11+1) th item.
10
= value of 8.4th item

= 8th item + (9th item – 8th item) × 0.4

= 150 + (200 – 150) × 0.4

= 150 + 20 = 170

? P70 = 170

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Example 2. Find the 25th and the 85th percentiles of the given data:

x 5 10 15 20 25 30 40 50
f 2 7 11 15 12 10 8 5

Solution: To calculate P25 and P85.

x f cf

522

10 7 9

15 11 20

20 15 35

25 12 47

30 10 57

40 8 65

50 5 70

N = 70

From table, we have, N = 70

25th percentle (P25) ( )= value of25(N+1) th item.
( )= value of100
25(70+1) th
100 item

= 17.75th item

c.f., just greater then 17.75 is 20 whose corresponding value is 15.

? P25 = 15 ( )= value of85(N+1) th item.
Similarly, 85th percentile ( )= value of100

85 × (70 + 1) th item
100
= value of 60.35th item

c.f., just greater than 60.35 is 65 whose corresponding value is 40.

? P85 = 40

Exercise 14.3

Very short Questions

1. (a) Define percentile.

(b) What is meaning of the 30th percentile ?

(c) Which percentile is also called median ?

(d) If N is the total number of observations in a discrete series, write the formula to
calculate P25 and P75.

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Short Questions

2. Find P10, P30, and P80 of the following data :
(a) 5, 10, 15, 20, 25, 30, 35, 46, 45.

(b) 85, 10, 75, 80, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 5, 78, 42, 43, 32.

(c) 10, 20, 30, 40, 50, 80, 100, 90, 100, 110, 120, 140, 130, 150, 45, 23, 45, 55, 65.

3. Find P25, P50, and P75, from the following data.
(a) 10, 20, 30, 40, 50, 60, 70, 80, 90, 100

(b) 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40

(c) 100, 120, 130, 150, 140, 170, 180, 200, 210, 165

Long Questions

4. Find P25, P60, and P80 of the following data :

(a) x 10 20 30 40 50 60 70
f 2 4 8 12 10 8 6

(b) x 15 20 25 30 35 40
f 8 12 20 15 10 4

5. Calculate P25, P45, and P90 of the following data. 90
10
(a) Marks 40 50 60 70 80 100
5
No. of students 5 10 15 20 15

(b) Age (years) 10 11 12 13 14 15 16
No. of students 4 6 10 13 12 10 6

(c) Size (inch) 12 13 14 15 16 17 18
Frequency 2 4 6 8 6 4 2

6. Construct discrete frequency table and find the 40th and the 80th percentiles.

(a) 2, 1, 0, 3, 2, 2, 4, 1, 2, 3, 4, 2, 5, 0, 2, 3, 2, 4, 0, 3, 2, 3, 2, 4, 1, 1, 2, 3, 4, 5, 1.

(b) 200, 100, 150, 300, 350, 250, 100, 300, 150, 100, 200, 300, 200, 250, 200, 300, 350,
150, 200, 400, 200, 350, 400

2. (a) 5, 15, 45 (b) 10, 30, 75 (c) 20, 45, 120
3. (a) 27.5, 55, 82.5 (b) 18, 26, 34 (c) 127.5, 157.5, 185
4. (a) 30, 50, 60 (b) 20, 30, 35
5. (a) 60, 70, 90 (b) 12, 13, 16 (c) 14, 15, 17
6. (a) 2, 4 (b) 200, 350

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Statistics :

15Measure of Dispersion

15.0 Introduction

Let us study the marks of two group of students each group containing 11 students.

Marks of Group A : 40, 42, 46, 48, 54, 60, 61, 62, 67, 80, 100.

Marks of Group B : 15, 25, 32, 40, 55, 60, 70, 80, 93. 94, 96

Now, group discuss the answers of the following questions.

(a) Calculate the mean and median marks of each group of students.

(b) Find the range marks of each group of students.

(c) Which group has better students and why?

(d) Do mean and median represent all the properties of set of data ?

Each of above set of data has mean 60 and median 60.

Central tendency gives only an idea about concentration of the average values in the central
part of the given data. It tells nothing about how each variate value of a data is scattered
from an average value of those items. Two or more than two data set may have equal mean,
median and mode but with one or more scattered than the other. It may be clear from above
two sets of data.

To have an idea how given set of data are deviated from the central value (or average), we
need another type of measure which is called measure of dispersion. The main objective to
study measure dispersion is to test whether there is homogeneity or heterogeneity of data.

Meaning of Dispersion

The literal meaning of dispersion is scatteredness. The word scatteredness means the
scatteredness of indivisual items from its central value. It gives an idea about the homogeneity
or heterogeneity of the distribution of data.

Methods of Measuring Dispersion
The following are the measure of dispersions :

(a) Range

(b) Quartile Deviation (Q.D.) (or semi–interquartile range)

(c) Mean Deviation (M.D.) (or Average Deviation)

(d) Standard Deviation (S.D.)

Here, we study about Q.D., M.D., and S.D. only for individual and discrete data.

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Absolute and Relative Measure of Dispersion.

When the units of measures of dispersion have the same units as that of the given series,
they are called absolute measure of dispersion. These types of dispersion can be used only
in comparing the variability of two series having the same unit and are not suitable for
comparing the variability of two series expressed in different units.

Relative measure of dispersion is a pure number carrying no unit and is obtained by taking
a ratio of an absolute measure of dispersion to suitable average. It is used to compare two or
more series having different units.

15.1 Quartile Deviation

Quartile deviation is a measure of dispersion based on the upper quartile and the lower

quartile of a set of data. The difference between the upper quartile (Q3) and the lower quartile
(Q1) is called the interquartile range.

? Interquartile Range = Q3 – Q1

Half of the interquartile range is called semi–interquartile range or quartile deviation. It is
denoted by Q.D. and is given by

Q.D. = 1 (Q3 – Q1)
2

It is an absolute measure of dispersion. It has the same unit of original data.

For the comparative study of two or more data, we calculate coefficient of quartile deviation.

It is the relative measure of dispersion, based on the quartiles Q3 and Q1.

? Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

The coefficient of Q.D. is unitless. It is used to compare variability of two or more than two

data sets. The data having less value of coefficient of quartile deviation is less variable or

less heterogenous or more homogeneous or more consistent or more uniform or more stable

than other and vice versa.

Calculation of Quartile Deviation for Individual and Discrete Series.

( )Q1 = value ofN+1 th item
( )Q3 = value of 4

3(N+1) th item.
4

where, N denotes total number of observations items.

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

Note :

N is used for total frequency for discrete or continuous series and n is used for individual

series.

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Worked out Examples

Example 1. Calculate the quartile deviation and its coefficient for given data:
Solution:
(a) 6, 10, 5, 7, 11, 12, 8

(b) 15, 14, 18, 20, 12, 24, 28, 26, 25, 30

(a) Writing the given data in ascending order, we get, 5, 6, 7, 8, 10, 11, 12.

Number of terms (n) = 7

( )Q1 = value of n+1 th item.
4
( )= value of 7+1 th
4 item.

= value of 2nd item = 6

( )Q3 = value of 3(n+1) th item.
( )= value of 4 th item.

3(7+1)
4
= value of 6th item = 11

? Quartile Deviation (Q.D.) = 1 (Q3 – Q1) = 1 (11 – 6) = 2.5
2 2
Q3 – Q1
Coefficient of Quartile Deviation = Q3 + Q1

= 11 – 6 = 5 = 0.294
11 + 6 17

(b) Writing the given data in ascending order, we have 12, 14, 15, 18, 20,
24, 25, 26, 28, 30

Number of terms (n) = 10

( )Q1 n+1 th item
= value of 4

( )= value of
10+1 th item.

4

= value of 2.75th item

= value of 2nd item + (3rd item – 2nd item) × 0.75

= 14 + (15 – 14) × 0.75 = 14.75

( )Q3 th item.
= value of 3(n+1)
4
( )= value of th item
3(10+1)
4
= value 8.25th item

= value of item 8th item + (9th item – 8th item) × 0.25

= 26 + (28 – 26) × 0.25

= 26 + 0.5 = 26.5

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Now, Quartile Deviation = 1 (Q3 – Q1)
2
1
= 2 (26.5 – 14.75) = 11.75

Coefficient of Quartile Deviation = Q3 – Q1
Q3 + Q1
26.5 – 14.75
= 26.5 + 14.75 = 0.285

Example 2. Calculate the quartile deviation and its coefficient for the following data :

x 5 15 25 35 45 55 65
f 11 18 25 28 30 33 22

Solution: To calculate the quartile deviation and its coefficients

xf cf
11
5 11 29
54
15 18 82
112
25 25 145
167
35 28

45 30

55 33

65 22

N = 167

( )Q1 N+1 th item.
= value of 4

( )4
= value of 167+1) th item.

= value of 42th item.

c.f. just greater than 42 is 54. Hence, value of Q1 is 25.

? Q1 = 25

( )Q3
= value of 3(N+1) th item
4
( )= value of
3(167+1) th
4
= value of 126th item

c.f. just greater than 126 is 145, Hence, value of Q3 is 55.

Q3 = 55

? Quartile Deviation (Q.D) = 1 (Q3 – Q1)
2
1
= 2 (55 – 25)

= 15

Coefficient of Q.D. = Q3 – Q1 = 55 – 25 = 30 = 0. 375
Q3 + Q1 55 + 25 80

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Exercise 15.1

Short Questions

1. (a) Define measure of dispersions.

(b) List the methods measure of dispersions.

(c) Define quartile deviation with its formula.

(d) If N is the number of terms of discrete series, write the formula for the following.

(i) Q1 (ii) Q3

(e) Write the formula of coefficient of Q.D.

2. (a) If N = 239, find the positions of Q1 and Q3.
(b) If Q1 = 25, Q3 = 75, find Q.D.
(c) If Q1 = 45, Q3 = 120, find the coefficient of Q.D.

Short Questions

3. Find the quartile deviations of the following data :

(a) 1, 2, 3, 4, 5, 6, 7

(b) 6, 4, 7, 9, 10, 12, 14

(c) 30, 50, 20, 70, 80, 70, 60

4. Calculate the quartile deviations and coefficients of the following data.

(a) 100, 50, 80, 70, 40, 30, 85, 90, 35, 45

(b) 200, 500, 400, 250, 450, 550, 600, 150, 280

(c) Price (Rs) : 20, 40, 60, 80, 70, 22, 90, 85, 100, 50

(d) Height of plants (cm) : 10, 20, 15, 12, 8, 4, 6, 20, 18, 22

(e) Rainfall (mm) : 20, 40, 50, 30, 17, 22, 24, 12, 15, 44

5. (a) The quartile deviation of a data is 10. If the upper quartile is 40, find the first
quartile.

(b) The quartile deviation of a data is 20.5. If the lower quartile is 40, find the upper
quartile.

(c) The coefficient of quartile deviation is 0.50 and the lower quartile is 20. Find the
upper quartile.

(d) The coefficient of quartile deviation is 0.75. If the upper quartile is 175, find the
lower quartile.

(e) In a discrete data, if the first quartile is p and quartile deviation P, find the third
quartile and the coefficient of quartile deviation.

Long Questions

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6. Find the quartile deviation and its coefficient of the following data :

(a) x 0 10 20 30 40 50 60
f
20 18 15 10 5 3 1

(b) Size 4 6 8 10 12 14 16

Frequency 1 3 3 6 3 3 1

(c) Height (in cm) 10 20 30 40
No. of plants 2 3 9 21

(d) Marks 30 50 70 90 95 100

No. of students 6 10 12 18 20 4

(e) Wages (Rs.) 400 500 600 700 800 1000
No. of workers 30 10 30 40 20 10

7. Construct discrete frequency distribution table and calculate quartile deviation and its
coefficients of the following data :

(a) 10, 12, 24, 12, 16, 24, 25, 30, 22, 25, 10, 12, 15, 24, 18, 24, 18, 16, 25, 35, 22, 35,
15, 18

(b) Daily Wages (Rs) : 100, 200, 400, 500, 200, 700, 800,

200, 250, 420, 550, 100, 200, 800, 500

100, 800, 700, 500, 450, 250, 800, 1000

150, 270, 500, 450, 150, 270, 280, 1000

2. (a) Q1 = value of 60th item, Q3 = value of 180th item (b) 25 (c) 0.4545

3. (a) 2 (b) 3 (c) 20

4. (a) 23.75, 0.38 (b) 150, 0.4 (c) Rs 26.625, 0.428 (d) 6.25 cm, 0.455

(e) 12.25 mm, 0.426 5.(a) 20 (b) 81 (c) 60

(d) 25 (e) 3p, 0.5 6.(a) 15, 1 (b) 2, 0.2

(c) 5, 0.143 (d) 12.5, 0.152 (e) Rs. 100, 0.167

7. (a) 5, 0.25 (b) Rs. 250, 0.556

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15.2 Mean Deviation

The arithmetic mean of the deviations of the items taken from the mean or median or mode
is called mean deviation where all the deviations are taken positive. Mean deviation is also
called average deviation. Mean deviation can be calculated from mean, median and mode.
Here, we learn to calculate mean deviation from mean and median only. Mean deviation is
denoted by M.D.

Mean deviation calculated from median is considered better than the deviation from
arithmetic mean because the sum of deviations from the median is less than the sum of
deviations from the mean. Hence, the value of the mean deviation from median is always
less than that calculated from mean.

In calculation of mean deviations from mean, some deviations are positive and some are
negative. The sum of deviations from the mean is zero, ie., ∑ (x – x– ) = 0; hence, their
arithmetic mean is always zero. We are interested in magnitude of the deviations and not
whether they are positive or negative. We ignore the signs and define a measure of variation
in terms of the absolute values of the deviations from the mean. i.e. |x – –x|

Calculation of Mean Deviation

(I) For Individual Series :
Let X1, X2, X3, ..... Xn be the n variate values and X and Md be the arithmetic mean and
median of the series respectively. Then,

M. D. from mean = ∑|X – X|
n
∑|X –Md|
M.D. from median = n

(II) For Discrete Series :

Let X1, X2, X3 ....... XN be the N variate values and f1, f2, f3, ........ fN be their corresponding
frequencies. Then

M.D. from mean = ∑f|x – –x | = ∑f|D|
N N

M.D. from median = ∑f|x – Md| = ∑f|D|
N N

Coefficient of M.D. from mean = M.D. from mean
mean

Coefficient of M.D. from median = M.D. from median
Median

Worked out Examples

Example 1. Find the mean deviation and its coefficient from mean and median:
Solution :
20, 30, 40, 50, 60, 70, 80.

Mean Deviation from mean:

Mean (x) = ∑x = 350 = 50
n 7

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X |X – X| = |X – 50|

20 30
30 20

40 10
50 0

60 10
70 20

80 30

∑X = 350 ∑|X – X| = 120

M.D. from mean = ∑|X – X| = 120 = 17. 14
n 7
M.D. from mean 17. 14
Coefficient of M.D. from mean = mean = 50 = 0.3428

( ) ( )For mean deviation from median
Median (Md) = n+1 th item = 7+1 th item = 4th item = 50
2 2

X |X – Md| = |X – 50|
20 30
30 20

40 10

50 0
60 10

70 20
80 30

∑|X – Md| = 120

Mean Deviation from median = ∑|X – Md| = 120 = 17. 14
n 7
M.D. from mean 17. 14
Coefficient of M.D. from Median = median = 50 = 0.3428

Example 2. Calculate the mean deviation from the mean and median and its coefficient.
Solution :
Marks 10 15 20 25 35 40 45

No. of student 3 4 6 10 8 5 4
Computation of M.D. from mean

Marks(x) No. of student (f) fx |x – x| f|x – x|

10 3 30 18 54
15 4 60 13 52
20 6 120 8 48
25 10 250 3 30
35 8 280 7 56
40 5 200 12 60
45 4 180 17 68

N = 40 1120 ∑f|x – x| = 368

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Mean ( x– ) = ∑fx = 1120 = 28 marks
N 40
x– |
Mean Deviation from mean = ∑f|x –
N
368
= 40 = 9.2 marks

Coefficient of M.D. from mean = M.D. from mean
mean
9.2
= 28 = 0.3286

Computation of M.D. from median

Marks(x) No. of student (f) c.f. |x – Md| f|x – Md|
10 3 3 15 45

15 4 7 10 40

20 6 13 5 30

25 10 23 0 0

35 8 31 10 80

40 5 36 15 75

45 4 40 20 80

N = 40 ∑f|x –Md| = 350

( ) ( )Median (Md) = th th item = (20.5)th item
N+1 item = 40+1
2 2
C.f. just greater than 20.5 is 23 whose corresponding value is 25.

? Median (Md) = 25

M.D. from Median = ∑f|x– Md|
N

= 350 = 8.75
40

Coefficient of M.D. from median = M.D. from median
Median

= 8.75 = 0.35
25

Exercise 15.2

Very Short Questions
1. (a) What is mean deviation ?

(b) What are importances of mean deviation ? Write any two importances.
(c) If sum of deviation from mean of a data with n observation is ∑|D|, write formula

of M.D. from mean.
(d) If ∑f|x – Md| and N is the total frequency of a data, write formula of M.D. from

median.
(e) Write the formula for coefficient of M.D. from mean and median.

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2. (a) If ∑ |x– –x| = 220, n = 20, find M.D. from mean.
(b) If ∑f|x – –x| = 450, N = 20, find the mean deviation from mean.
(c) If ∑f |x – Md| = 240, N = 40, find the mean deviation from median.
(d) If M.D. from median is 20 and median (Md) = 40, find the coefficient M.D. from
median.
(e) If M.D. from mean is 12 and mean = 25, find the coefficient of M.D. from mean.

Short Questions
3. Calculate the mean deviation from (i) Mean (ii) median from the following data. Also

find their coefficient.
(a) 11, 18, 19, 20, 24, 25, 30,
(b) 20, 22, 25, 27, 28, 30, 35, 40, 45, 50, 55, 70, 73
(c) 90, 100, 110, 115, 125
(d) Height : 24, 28, 30, 32, 36, 38, 40, 52
(e) Marks : 25, 40, 55, 60, 65, 70, 75, 80, 90, 95
Long Questions
4. Calculate the mean deviation from mean and its coefficient from the following data:

(a) x 12 16 20 24 28 32 36 40
f 1 14 25 27 18 9 4 2

(b) x 8 10 12 16 18
f 35864

(c) Wages (R1) 50 55 70 80

No. of workers 4 3 5 4

5. Calculate the M.D. from median and its coefficient from the following data :

(a) Size (inch) 4 6 8 10 12 14 16

Frequency 1 3 3 6 3 3 1

(b) Weight (kg) 10 12 15 16 20

No. of bags 6 14 20 13 7

(c) x 6 8 10 12 14 16 18 20
f 1 14 25 28 18 9 4 2

6. Construct a discrete frequency table and find the mean deviation from median:

19, 23, 30, 29, 11, 21, 26, 36, 41, 42, 49, 52, 56, 58, 53, 27, 20, 34, 62, 22, 23, 25, 27, 36,
42, 42, 52, 50, 58, 53, 30, 11, 29, 26, 41, 36, 58, 53, 19, 23, 30.

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2. (a) 11 (b) 22.5 (c) 6 (d) 0.5 (e) 0.48

3. (a) M.D. from mean = 4.571, coeff. = 0.28, M.D. from median = 4.429, coeff. = 0.221

(b) M.D. from mean = 14.308, coeff. = 0.358, M.D. from median = 13.923, coeff. = 0.398

(c) M.D. from mean = 10.4, coeff. = 0.096, M.D. from median = 10, coeff. = 0.091

(d) M.D. from mean = 6.5, coeff. = 0.186, M.D. from median = 6.5, coeff. = 0.191

(e) M.D. from mean = 16.5, coeff. = 0.248, M.D. from median = 15.5, coeff. = 0.230

4. (a) 4.48, 0.187 (b) 2.92, 0.22 (c) 10.976, 0.170

5. (a) 2.4, 0.24 (b) 2, 0.133 (c) 2.202, 0.183 6. 12.171, 0.358

15.3 Standard Deviation

Standard deviation is the best measure of dispersion. It is non-negative value. It is denoted
by the Greek letter V (sigma). It is defined as the positive square root of the arithmetic mean of
the squared deviation of a set of value of observations from their arithmetic mean. Standard
deviation is denoted by S.D. in abbreviated form.

Calculation of standard Deviation

I. For Individual Series

Let X1, X2, X3, ........, Xn be the value of n observations of variable X.

( )(i) ∑X2 – ∑X 2
Direct Method : S.D. (V) = n n

(ii) Actual Mean Method S.D. (V) = ∑(X – X)2
n

(iii) Short–cut Method (or assumed mean method) :

Let d = X – A, A = assumed mean.

( )S.D. (V) = ∑d2 – ∑d 2
n n

II. For Discrete Series :

Let x1, x2, x3....... xn be values of N variate values with f1, f2, f3 ........ fn as their corresponding
frequencies.

( )(i) ∑fx2 ∑fx 2
N N
Direct Method : S.D. (V) = –

(ii) Actual Mean Method S.D. (V) = ∑f(x – x )2
(iii) Short cut Method N

Let d = x – A

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Where A = assumed mean

S.D. (V) = ( )∑fd2 – ∑fd 2
N N

(iv) Step – deviation method

This method is specially used for continuous series or data with common factor in

every variate value.
x–A
Let d'= h

Let d' (V) = ( )∑fd'2∑fd' 2 ×h
N N
–

where, h = class difference or common factor.

Coefficient of Standard Deviation.

Coefficient of S.D. = S.D = V–x
mean

Coefficient of variation (C.V) = V–x × 100%
The square of standard deviation is called variance. Variance = V 2

Worked out Examples

Example 1. Find the standard deviation and its coefficient by (i) direct method
Solution: (ii) actual mean method (iii) short–cut method (iv) step deviation method of
the following data : 60, 65, 70, 75, 80

(i) Calculation of S.D.(V) by direct method

x x.x = x2 Mean ( x– ) = ∑x
60 60.60 = 3600 n
65
70 4225 = 350
75 4900 5
80 5625
∑x = 350 6400 = 70
∑x2 = 24750

( )S.D. (V) = ∑x2 – ∑x 2
n n

( )=
24750 – 350 2
5 5

= 4950 – 4900

= 50 = 7.07

Coefficient of S.D. = S.D.
x

= 7.07 = 0.101
70

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

(ii) Calculation of S.D. by actual mean method

x x = 70, (x – x) (x – x)2
60 60– 70 = – 10 100

65 65 – 70 = – 5 25

70 70 – 70 = 0 0

75 75 – 70 = 5 25

80 80 – 70 = 10 100

∑(x – x)2 = 250

Standard Deviation (V) = ∑(x – x– )2
n

= 250
5

= 50 = 7.07

Coefficient of S.D. = V–x

= 7.07
70

= 0.101

(iii) Calculation of S.D. (V) by short-cut method

Let d = x – A, Where A = assumed mean = 70

x d = x – A = x – 70 d2
60 60– 70 = – 10 100
25
65 65 – 70 = – 5
0
70 70 – 70 = 0 25
100
75 75 – 70 = 5 ∑d2 = 250

80 80 – 70 = 10

Mean ∑d = 0
S.D. (V)
= A + ∑d = 70 + 0 = 70
n 5

( )= ∑d2 – ∑d 2
n n

( )= 250 – 02
5 5

= 50 = 7.07

Coefficient of S.D. = V–x

= 7.07
70

= 0.101

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

(iv) Calculation of S.D. by step-deviation method

Common factor (h) = 5 d' = x–A = x – 70
h 5

x d'= x–A = x – 70 d'2
h 5 4

60 – 2

65 –1 1

70 –0 0

75 1 1

80 2 4
∑d' = 0 ∑d'2 = 10

( )= ∑d'2 ∑d' 2
n n
S.D. (V) – ×h

( )= 10 – 0 2
5 5
×5

= 2 . 5 = 7.07

Mean ( x– ) = a+ ∑d' × h = 70 + 0 ×5
= 70 n 5

Coefficient of S.D. = V–x = 7.07 = 0.101
70

Example 2. Calculate the standard deviation for the given data:

x 45 55 65 75 85 95

f 6 10 20 30 20 14

use (i) direct method (ii) actual mean method

(iii) short-cut method (iv) step-deviation method

Solution: (i) Calculation of S.D. (V) by direct method.

x f fx fx2 = fx.x
45 270 270 × 45 = 12150
55 6
65
75 10 550 30250
85
95 20 1300 84500

S.D. (V) = 30 2250 168750

20 1700 144500

14 1330 126350
∑fx = 7400 ∑fx2 = 566500
N = 100

( )∑fx2 ∑fx 2 ( )=
N – N 566500 – 7400 2
100 100

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

( )=
S.D. (V) ∑fx2 – ∑fx 2
N N
( )=
566500 – 7400 2
100 100

= 5665 – (74)2

= 5665 – 5476 = 189

= 13.75

(ii) Calculation of S.D. (V) from actual mean method.

S.D. (V) = ∑f(x – x– )2
n

xf fx x – x (x – x)2 f(x – x)2
45 6 270 –29 841 5046

55 10 550 –19 361 3610

65 20 1300 –9 81 1620

75 30 2250 1 1 30

85 20 1700 11 121 2420

95 14 1330 21 441 6174
∑f(x – x)2 = 18900
N = 100 ∑fx = 7400

Arithmetic mean ( x– ) = ∑fx
N

= 7400 = 74
100

S.D. (V) = ∑f(x – x– )2
N

= 18900 = 13.745
100

(iii) Calculation of S.D.(V) by short cut method

S.D. (V) = ( )∑fd2– ∑fd 2
N N

Let A = assumed mean = 75, d = x – A = x – 75

xf d=x–A fd fd2
–180 5400
45 6 45–75 = – 30
–200 4000
55 10 – 20
–200 2000
65 20 – 10
0 0
75 30 0
200 2000
85 20 10
280 5600
95 14 20 ∑fd = –100 ∑fd2 = 19000
N = 100

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

( )=
Standard deviation (V) ∑fd2 – ∑fd 2
N N

( )=
19000 – –100 2
100 100

= 13.748

(iv) Calculation of S.D. (V) by step – deviation method.

Let A = 75, d' = x –A = x – 75
h 10

xf d'= x–A fd' fd'2
h – 18
45 6 54
55 10 –3 40
65 20 20
75 30 – 2 – 20 0
85 20 20
95 14 – 1 – 20 56
∑fd'2 = 190
N = 100 00

1 20

2 28
∑fd' = 10

( )∑fd'2
N
S.D. (V) = – ∑fd' 2 ×h
N × 10

= ( )190––10 2
= 100
100

1.9 – (0.1)2 × 10

= 1.9 – 0.12 × 10

= 1.89 × 10

= 13.75

Exercise 15.3

Very Short Question

1. (a) Define standard deviation.
(b) If S.D. = V, mean = x– write formula for-

(i) Coefficient of S.D. (ii) Coefficient of variation

(iii) Variance

(c) Write a difference between S.D. (V) and M.D.

(d) Write an importance of coefficient of S.D.

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

(e) Write list of formulas of standard deviation of individual series.

(f) Write list of formulas of standard deviation of discrete series.

2. Find standard deviation under given conditions:

(a) ∑x2 = 240, ∑x = 16, n = 10.

(b) ∑fx2 = 425, ∑fx = –20, N = 25.

(c) ∑fd2 = 625, ∑fd = 80, N = 40.

(d) ∑fd'2 = 225, ∑fd' = –8, N = 25, h = 10

Short Questions :

3. Find S.D. and its coefficient of given data by using :

(i) Direct Method (ii) Actual mean method

(iii) Short-cut method

(a) 2, 4, 6, 8, 10, 12, 23

(b) Income (R) : 40, 100, 90, 80, 60

(c) Temperature (°C) = 5, 6, 7, 8, 9

(d) Wages (Rs) : 300, 325, 350, 375, 425

Long Questions :

4. Calculate standard deviation and its coefficient for the following data :

(a) x 10 11 12 13 14
f 3 12 19 12 3

(b) Size (inch) 6 9 12 15 18
Frequency 7
12 19 10 3

(c) x 6 20 25 30 35 40
f 1 5 10 12 8 4

(d) Weight 20 25 30 35 40

No. of persons 5 7 10 5 9

(e) x 45 55 65 75 85
f 35872

(f) Height of plants 2 4 6 8 10 12
2
No. of plants 1 2 3 4 0

5. (a) Find the standard deviation of the first ten natural numbers.

(b) Find the standard deviation of the first ten even natural numbers.

(c) Find the standard deviation of the cube of first five natural numbers.

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

6. Scores of two golfers for 9 rural were as follows:
Golfer A : 74, 75, 78, 72, 77, 79, 78, 81, 76

Golfer B : 86, 84, 80, 88, 89, 85, 86, 82, 79.

Find which golfer may be considered to be a more consistent player.
7. Construct discrete frequency table and find S.D. and its coefficient.

10, 20, 40, 50, 60, 70, 80, 90, 100, 20, 10, 40, 50, 60, 70, 80, 40, 50, 70, 20, 30, 40, 70,
80, 50, 90, 50, 60, 40, 20
Project Work

8. Collect the marks obtained by your friends in optional mathematics in the first terminal
exam. Find the range mean, S.D. and C.V. Discuss about the result obtained.

9. List the prices of daily commodities in your area and find standard deviation and its
coefficient. Interpret the result..

2. (a) 4.63 (b) 4.04 (c) 3.41 (d) 29.83
(c) 1.414, 0.202 (d) 43.01, 0.121
3. (a) 6.43, 0.692 (b) 21.541, 0.291 (c) 6.851, 0.237 (d) 6.851, 0.371

4. (a) 0.9897, 0.082 (b) 3.255, 0.285

(e) 11.314, 0.174 (f) 2.887, 0.412

5. (a) 2.872, 0.522 (b) 5.745, 0.522 (c) 45.585, 1.013
S.D. (B) = 3.232, C.V. (B) = 3.83%
6. S.D. (A) = 2.582, C.V. (A) = 3.37%

A is more consistent.

7. 24.413, 0.469

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

Curriculum Development Centre, Sanothimi, Bhaktapur

Syllabus of Optional Mathematics - Grade 9

1. Algebra 33 periods

(a) Relation and function

(i) Ordered pair

(ii) Cartesian Product

(iii) Introduction to relation, types, representation of relation, domain and range
of relation

(iv) Introduction and notation of function, domain, range and co domain, image
and pre-image

(v) Representation of function, vertical line test of function Types of function
(onto, into, one to-one, many to one)

(b) Polynomials

(i) Introduction and classification (on the basis of function)

(ii) Degree of polynomial, standard form of the polynomial, equal polynomials

(iii) Basic operation on polynomials (Addition, subtraction and multiplication)

(c) Sequence and Series

(i) Introduction to sequence and general term

(ii) Introduction to series and use of sigma (6) notation

2. Limit 10 periods

(a) Simple concept of limit
(i) By the sequence of numbers
(ii) By the sequence of diagrams
(iii) By the sum of infinite series

(b) Value of function
(c) Concept of limiting value of function
(d) Notational Representation of limit
(e) Meaning and introduction to x o a

3. Matrix

(a) Introduction to matrix and its components

(b) Types of matrix (row, column, zero, rectangular, square, diagonal, scalar, unit,
equal, symmetric and triangular matrices)

(c) Addition and subtraction of matrices

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

(d) Properties of matrix addition
(e) Transpose of a matrix
(f) Multiplication of matrices
(g) Multiplication by a scalar
(h) Properties of matrix multiplication (closure, associative and distributive)

4. Coordinate Geometry 30 periods

(a) Locus and its general introduction

(b) Division of line segment in the given ratio

(c) Equations of straight lines

(i) (Parallel to the axes, slope intercept form, intercept form, perpendicular form)

(ii) Reduction to the standard forms

(iii) Point slope form, two points form, distance between a point and a straight line

(d) Area of triangle and quadrilateral using coordinates

5. Trigonometry 35 periods

(a) Measurement of angles

(b) Introduction to Trigonometrical ratios

(c) Problems on Trigonometrical indentities (sin2T + cos2T = 1, sinT = tanT, etc.) and
their conversion cosT

(d) Values of trigonometrical ratios of the angles 0°, 30°, 45°, 60°, 90° (Using the
concept of unit circle)

(e) Trigonometrical ratios of the angles (90° ± T), (180° ± T), (–T)

(f) Trigonometrical ratios of compound angles

6. Vector 12 periods

(a) Introduction to vector

(b) Types of vectors (row, column, position, unit, null, equal and negative vector)

(i) Like and unlike vectors

(ii) Magnitude and direction of vectors

(c) Operations on vectors

(i) Multiplication of a vector by a scalar (condition of vectors being parallel)

(ii) Addition and subtraction of vectors

(iii) Laws of vector addition

7. Transformations 18 periods

(a) Introduction to transformation and types (Isometric and non isometric)

(b) Rotation of ± 90°, ± 180°, 360° about the point (a, b) (Graphically and using formula)

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

(c) Reflection in (X-axis, Y-axis, y = x, y = –x, x = a, y = b) (Graphically and using formula)

(d) Translation

(e) Enlargement or reduction (centre at (0, 0) and (a, b) and scale factor k, using
formula and graphical method)

8. Statistics 12 periods

(a) Partition values

(i) Quartiles (Individual and discrete series only)

(ii) Deciles (Individual and discrete series only)

(iii) Percentiles (Individual and discrete series only)

(b) Dispersion

(i) Quartile deviation and its coefficient (Individual and discrete series only)

(c) Mean deviation (from mean and median) and its coefficient (Individual and
discrete series only)

(d) Standard deviation and its coefficient (Individual and discrete series only)

Specification Grid 2076

Issued by CDC

SN Contents K U A HA

1. Algebra Each of 1 Each of 2 Each of 4 Each of 5 TQ TM TTH
2. Limit and Continuity Mark Marks Marks Marks
3. Matrix
4. Coordinate Geometry 2 3 2 1 8 21 33
5. Trigonometry
6. Vectors 1 - 1 - 2 5 10
7. Transformation
8. Statistics 1 2 1 - 4 9 20

Total 2 2 1 1 6 15 30
Marks
2 3 3 - 8 20 35

1 2 - 1 4 10 12

1 - 1 1 3 10 18

- 1 2 - 3 10 12

10 13 11 4 38 170
10 26 44 20 100 100

INDEX

K = Knowledge U = Understanding A = Application
TM = Total Marks
HA = Higher Ability TQ = Total no. of Questions

TTH = Total Teaching Hours

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

New Model Questions 2076

Group A [5×(1+1) = 10]

1. (a) If (x + 5, 4) = (10, y + 2), find the value of x and y.

(b) Write the following polynomial in ascending order.

2x3 + 5x2 + 7x + 9x4 + 1 + x5

2. (a) Find the limiting value of the sequence

3, 3 , 3 , 3 , ................
2 4 8
(b) In matrix A = (aij)3×3, where aij = i + 2j then find the value of a33.

3. (a) Define locus.

(b) Write the equation of straight line in normal form.

4. (a) If sinA = 1 , find the value of cosecA.
3
(b) Find the value of cosT from the given figure. 2

5. (a) If oa = (3, 5) and ob = (–1, 3), writeoa + ob in column vector. T
1
(b) Find the image of a point A(4, –1) under the reflection on line y = 4.

Group B [3(2+2+2) + 2(2 + 2) = 26)

6. (a) If f : A o B is defined by f(x) = 3x – 2 and A = {1, 2, 3}, find the range of the
function f.

(b) If the polynomials x3 + (a + 2)x2 + x + 8 and x3 + 4x2 + (b – 1)x + 8 are equal,
find the values of a and b.

(c) In a sequence, if Sn = n(n + 1) , find the value of S5.
4

7. (a) If two matrices x–1 5 and 6 5 are equal, find the values of x and y.
–3 –2 –3 y–5

(b) If P = 1 –1 and Q = 2 –5 show that PQ is a null matrix.
3 –3 2 –5
8. (a) If A(6, 4), B(4, 2) and C(1, –1) are three collinear points, find the ratio that B

divides AC.

(b) Find the equation of a straight line making intercepts equal in magnitude but
opposite in sign and passing through the point (4, 5).

9. (a) Prove that sinA 1 + cot2A + cosA 1 + tan2A = 2

(b) Prove that 2 cosT + 1 sinT = sinT + cosT
– tanT – cotT

(c) Prove that tan(180° – A).sec(90° + A).sin(90° – A) = 1

10. (a) If AoB displaces A(3, 3) to B(6, 7) and CoD displaces C(2, 0) to D(5, 4), prove that
AoB = CoD.

(b) If vectors 2oi – 3oj and moi – 12oj are parallel to each others, find the value of m.

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Measure of Dispersion

(c) Find the standard deviation and coefficient of standard deviation from following data.

10, 20, 30, 40

Group C [11 × 4 = 44]

11. If f(x) = x3 + 1 and g(x) = x + 1 , find f(2) .
3 + 2g(2)
f(2)
12. Find the nth term of the sequence 1, 3, 6, 10, 15, ...... and use it to find the value of
105th term.

13. Define limit of a function. Evaluate lim x2 – 25
xo5 x–5

14. If A + B = 7 0 and A – B = 3 0 , find the matrices A and B.
2 5 0 3
15. The length of perpendicular drawn from the origin on a straight line is 4 units and the
perpendicular is inclined at 60° to X-axis. Find the equation of the straight line and
prove that it passes through the point (5, 3).

16. Prove that tan(A + B) = sin2A – sin2B
sinA.cosA – sinB.cosB
1 + (cosecA.tanA)2 1 + (cotA.sinB)2
17. Prove that 1 + (cosecC.tanB)2 = 1 + (cotC.sinB)2

18. Solve that: xcotA.tan(90° + A) = tan(90° + A).cot(180° – A) + xsec(90° + A).cosecA

19. Points A(6, 8), B(5, 6) and C(–1, –5) are the vertices of ABC. Find the images of vertices
–2
of 'ABC under translations through translation vector 3 . Also draw the object and
image in the graph paper.

20. Find the quartile deviation and its coefficient of the following data:

Marks obtained 80 70 60 50 40 30 20

No. of students 10 8 9 6 5 7 3

21. Find the mean deviation and its coefficient from mean:

Weight (kg) 6 8 10 12 14 16 18 20

No. of children 1 14 25 27 18 9 4 2

Group D [4 × 5 = 20]

22. If A = {1, 4), B = {2, 3, 6}, and C = {2, 3, 7}, then verify:

(a) A × (B – C) = (A × B) – (A × C)

(b) A × (B ‰ C) = (A × B) ‰ (A × C)

23. If the point A(a, b) lies on the line 3x – 4y = 26 and B(b, a) lies on the line 5x – 3y + 31 = 0,
find the equation and length of AB.
ED

24. In the given figure, ABCDEF is a regular polygon.

Prove that AoB + AoC + AoD + EoA + FoA = 4AoB FC

25. A(1, 4), B(3, 2), and C(2, 3) are the vertices of 'ABC. Enlarge 'ABC A B
with centre at (4, 1) and scale factor 2, and get the first image
'A'B'C'. Again rotate 'A'B'C' through +90° about the origin and get the second image
'A"B"C". Find the co-ordinates of final image and show all the figures in graph.

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