Opt. Maths Book for Class-9

UNIT

fifteen Statistics: Measure of Dispersion

Estimated teaching period:7

1. Objectives Objectives
S.N. Level To define measure of dispersion
(i) Knowledge(K) To tell formula of Q.D, M.D. and S.D for individual and
discrete series.
(ii) Understanding(U) To compute Q.D., M.D. and S.D. for given individual data.
(iii) Application(A) To calculate Q.D, M.D. and S.D for discrete series.
(iv) Higher Ability (HA) To prepare frequency table from given raw data and
compute Q.D., M.D. and S.D. from it.

2. Teaching materials
Formula list in chart of paper of
Quartile Deviation (Q.D.)
Mean Deviation (M.D.)
Standard Deviation (S.D.)

3. Teaching Learning Strategies
Review the formulas to compute Quartiles, mean, median etc.
Discuss about measure of dispersion and its importance
Explain how to find Q.D., M.D. and S.D. and their coefficients for individuals and discrete
series with formula.
Discuss about the interpretation of measures of dispersion
Give examples of individuals and discrete series to calculate Q.D., M.D. and S.D.
State different methods to compute standard deviation of given data. (direct method, short-
cut method, step-deviation method).
Discuss about absolute and relative measures of dispersion.

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

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

2. Mean Deviation from mean:

i) For individual series.

M.D. = ∑|X – X|
n

ii) For Discrete Series

Vedanta Optional Mathematics Manual ~ 9 201

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

3. Mean Deviation from median

(i) For individual Series, M.D. = ∑ |X – Md|
n

For Discrete Series :

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

= M.D. from median
Median

4.Standard Deviation (S.D.)

(a) For Individual Series

i) S.D.( ) = ∑x2 – ∑x 2 , direct method
n n

ii) S.D.( )= ∑(X –x)2 , actual mean method

n

iii) S.D.( ) = ∑d2 – ∑d 2 , (short-cut method)
N N

where, d = x – a, a = assumed mean

x = a+ ∑d
N

iv) S.D.( ) = ∑d′ 2 ∑d′ 2 × i, step deviation method.
N N
–

xi) d' = x –a
i
i = common factor to each variate.

(b) For discrete Series :

i) S.D.( ) = ∑fx2 ∑fx 2 , di-rect method-
N N
–

ii) S.D.( )= ∑f(x –x)2 , actual mean method

N

iii) S.D.( ) = ∑fd2 – ∑fd 2 , short-cut method
N N

where, d = x – a,

a = assumed mean

x = a + ∑d
N

iv) S.D.( ) = ∑fd′2 – ∑fd′ 2 × i , step deviation method
N N

202 Vedanta Optional Mathematics Manual ~ 9

d' = x –a ,i = common factor to each variate value.
i

x= a + ∑ fd1 ×i
N

5. Variance = 2 (square of S.D.)

6. coefficient of S.D. = σ

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

Some solved problems

1. If Q1= 45, Q3 = 120, find the Q. D. and its coefficient.

Solution
Here Q1= 45, Q3 = 12

Quartile Deviation Q.D. = 1 (Q3 – Q1 )
2
1
= 2 (120 – 45 )

= 1 × 75
2

= 37.5 Q3 – Q1
Q3 + Q1
coefficient of Q. D. =

= 120 – 45
120 + 45

=17655
=0.4545

2. If ∑|x – x|= 220, N = 20, find M.D. from Mean.

Solution x|
Here, ∑|x – x|= 220, N = 20,

Mean Deviation from Mean (M.D) = ∑|X –
N
= 220
20
=11.

3. If M.D. from median is 20 and Md = 40, find the coefficient of M.D. from median

Solution
Here Median (Md) = 40
M.D. from median = 20

coefficient M.D. from median = M.D. from median
Median

Vedanta Optional Mathematics Manual ~ 9 203

= 20
40

= 0.5

4. If ∑fd2 = 625, ∑fd = 80, N = 40, find standard deviation. (Page 334, Q.N. 2(c) corrected)

Solution
Here, ∑fd2 = 625, ∑fd = 80, N = 40

Standard deviation ( S.D. ) = = ∑fd2 – ∑fd 2
N N

= 625 – 80 2
40 40

= 15.625 – 4

= 11.625

= 3.4095

5. Find the quartile deviation and its coefficient of the given data.
x 0 10 20 30 40 50 60
f 20 18 15 10 5 3 1

xf cf

0 20 20

10 18 38

20 15 53

30 10 63

40 5 68

50 3 71

60 1 72

N = 72

Solution
Here, N = 52

Q1 = (N + 1) th
4
item

= 73 th item
4

=(18.25)th item = 0

Q3 = 3(N + 1) th
4
item

= 3 × 18.25 th item

204 Vedanta Optional Mathematics Manual ~ 9

=54.75th item = 30

QCoueafrftiiclieenDteovfiaQti.oDn.Q=.DQQ. =33 –+QQQ3112– Q1 = 30 – 0 = 15
2
30
= 30 = 1

6. Construct a discrete frequency table and calculate the quartile deviation and its coefficient
of the following data.

Rs Daily wage(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

Solution cf
Daily wage(Rs) f

100 3 3

150 2 5

200 4 9

250 2 11

270 2 13

280 1 14

400 1 15

420 1 16

450 2 18

500 4 22

550 1 23

700 2 25

800 4 29

1000 2 31

Q1 = (N + 1) th
4
item

= 32 th item
4

= 8 th item

= Rs 200

Q3 = 3(N + 1) th
4
item

Vedanta Optional Mathematics Manual ~ 9 205

= 3 × 8 th item

= 24 th item

= Rs 700

Q.D. = Q3 – Q1 = 700 – 200 = Rs. 250
2 2

coefficient of Q. D. = Q3 – Q1 = 700 – 200 = 500 = 0.556
Q3 + Q1 700 + 200 900

7. Calculate the mean deviation and its coefficient from the given data.

x 8 10 12 16 18
f 35864

Solution
To calculate the mean deviation from mean and its coefficient

xf fx |x – x | f|x– x |

83 24 5 15

10 5 50 3 15

12 8 96 1 8

16 6 96 3 18

18 4 72 5 20

N = 26 338 9 76

x = ∑ fm = 338 = 13
N 26
x|
M.D. from mean = ∑f |x – = 76 = 2.92
N 26

coefficient of M.D. from mean = M.D. from mean
Mean

= 2.92
13

= 0.2246

8. Calculate the mean deviation from median and its coefficient.

weight in kg 10 12 15 16 20

No. of begs 6 14 20 13 7

Solution : To calculate mean deviation from median and its coefficient.

weight in kg (x) f c.f. |x – Md| f|x – Md|
10 6
12 14 65 30
15 20
20 3 42

40 0 0

206 Vedanta Optional Mathematics Manual ~ 9

16 13 53 1 13

20 7 7 5 35

N = 60 120

Median (Md) = N+1 th 60 + 1 th
2 2
item = item = 30.5 th item = 15

c.f. just greater 30.5 is 40 whose corresponding value is 15

Median (Md) = 15 kg ∑f |x – Md| = 120 =2 kg.
M.D. from median = N 60

coefficient of M.D. from median = M.D. from median
Median
2
= 15

= 0.133

9. Calculate the standard deviation and its coefficient for the following data.

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

Solution
To compute the standard deviation and its coefficient.
take a = 12 (assumed mean)

size (inch) Frequency (f) d = x – a = x – 12 fd fd2
–6 – 42 252
67 –3 – 36 108
0 0 0
9 12 3 30 90
6 18 108
12 19 ∑fd = – 30 ∑fd2 = 558

15 10

18 3

N = 51


N = 51, ∑fm = – 30, ∑fm2 = 558,

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

== 558 – – 30 2
51 51

= 10.9411 – 0.3460

= 3.255

Mean(x) = a + ∑ fd =a+ – 30
N 51

Vedanta Optional Mathematics Manual ~ 9 207

= 12 – 0.588 = 11.412 inch

coefficient of S.D. = σ
x
3.255
= 11.412

= 0.2852

10. 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

Hints :Find the mean and standard deviation of scores of the both players A and B .

Tthheenpflianydecr.wv.h=osexσco×eff1i0ci0ent of variation is less will be more consistent player.

C.V. for player A is 3.37 % and that of player B is 3.83 %

It shows that player A is more consistent.

Questions for practice

1.Find the standard deviation for given data.
a) 4, 6, 7, 8, 10, 3, 9, 5, 10, 12, 15.
b) 100, 200, 300, 400, 500, 600, 700

2. Find the mean deviation i) from mean ii) from median
(a) 10, 12, 14, 16, 18, 20, 22, 24
(b) 20, 10, 40, 60, 20, 80, 90

3. Find the standard deviation for given data
a) 10, 20, 30, 40, 50, 60, 70, 80, 90
b) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

4. Find the quartile deviation and its coefficient from given data

(a)

Marks 10 20 30 40 50
3
No. of 2 4 6 5
students 25
1
(b)

Weight in 5 10 15 18 20
kg 3542
No. of bags 2

5. Find the M.D. from mean and its coefficient

208 Vedanta Optional Mathematics Manual ~ 9

Marks 40 50 60 70 80 90 100
No. of 4 6 10 15 12 7 6
students
50 60
6. Find the M.D. from median and its coefficient. 4 2

Weights in 40 42 45 48 12 15
kg 6 3 5

No. of 5 6 9 700 800
students 3 2

7. Find the standard deviation its coefficient from given data

Height of 5 7 8 10
plants(cm)

No. of 4 6 10 7
plants

8. Find varience and coefficient of variation from given data

Dady 200 300 400 500 600
wages
(Rs)

No. of 4 10 12 8 6
labourers

Vedanta Optional Mathematics Manual ~ 9 209