Vedanta Optional Math Teacher's guide

2. Compute standard deviation and its coefficients from the following method.

i) direct method ii) short – cut method iii) step – deviation

class interval 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Frequency 10 8 32 40 22 18

Solution
i) To compute S.D. by direct method

Class– interval Mid – value(m) Frequency(f) fm fm2
10 – 20 15 10 150 2250
200 5000
20 – 30 25 8 1120 39200
1800 8100
30 – 40 35 32 1210 66550
40 – 50 45 40 1170 76050
∑fm = 5650 270050
50 – 60 55 22
60 – 70 65 18

N = 130
From table, N = 130, ∑fm = 5650, ∑fm2 = 270050,

Solution ∑fm2 ∑fm 2
Standard deviation ( ) = S.D.( )= N N
–

= 270050 – 5650 2
130 130

= 2077.3076 – 1888.9053

= 188.4023

= 13.72 ∑ fm 5650
N =130x6
Mean(x) = = = 43.46

Coefficient of S.D.
13.72
= 43.46 = 0.3157

ii) To compute standard deviation by short – cut method or assumed mean method.

Let a = 45

Class– interval mid – value(m) frequency(f) d= m – a fm fd2
10 – 20 15 10 – 30 – 300 900

20 – 30 25 8 – 20 – 160 3200

30 – 40 35 32 – 10 – 320 3200
40 – 50 45
40 0 00

50 – 60 55 22 10 220 2200
60 – 70 65 18 20 360 7200

N = 130 – 200 24800

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Now, standard deviation is given by,

= ∑fd2 – ∑fd 2
N N

= 24800 – – 200 2
130 130

= 190.7652 – 2.3667

= 188.4025

= 13.72 ∑ fm
N
Mean(x) = a +

=c=oe414f330f..i74+c26ie=n–1t203o0.03f01S5=.D74. 5=.4x66

iii) To calculate standard deviation by step deviation method method

Class– interval mid – value(m) frequency(f) d1 = m –a fd1 fd1 2
i
10 – 30 90
10 – 20 15 8 –3 – 16 32
20 – 30 25 32 – 32 32
30 – 40 35 40 – 20 0
40 – 50 45 22 0 22
50 – 60 55 18 –1 22 72
60 – 70 65 N = 130 36 248
0 – 20

1

2

= ∑fd12 – ∑fd1 2 ×i
Standard deviation (S.D.) = N N

= 248 – (–20) 2 × 10
130 130

= 1.9077 – 0.0237 × 10
= 1.88 × 10
= 13.72
=CM=oe414ea330fnf..74i+(c26xi)e=–1=n32t000ao.3f×+1S51.∑7D0Nf.=d=1 4×x36.1406

402 Vedanta Optional Mathematics Teacher's Guide ~ 10

3. Calculate the standard deviation and its coefficient of variation from the following.

x 0 ≤ x ≤ 10 10 ≤ x ≤ 20 20 ≤ x ≤ 30 30 ≤ x ≤ 40 40 ≤ x ≤50

f7 10 14 12 6

Solution
Here, 0 ≤ x ≤ 10 means value of x is from 0 to 10 exclusively, we can write x belongs to
class 0 – 10 etc.

To calculate standard deviation of given data by using by step deviation method method

Let a = 25, then d1 = m–a = m – 25
i i
Mid – value
Class frequency(f) (m) d1 = m –a fd1 fd1 2
i – 14 28
0 – 10 7 5 –2

10 – 20 10 15 – 1 – 10 10

20 – 30 14 25 0 0 0

30 – 40 12 35 1 12 12

40 – 50 45 45 2 12 24

N = 49 0 70

Standard deviation = = ∑fd12 – ∑fd1 2 ×i
N N

= 74 – 02 × 10
49 49

= 1.5210 × 10
= 12.33
M=C=oe21ea25f2nf.53i+(c3xi)e=4=0n9t0a×o.4f+91S30.∑2D=Nf.d2=1 5×x6 10
4. Calculate the standard deviation and its coefficient of variation from the following.
(a)

x less than 10 less than 20 thess than 30 less than 40 less than 50
f 12 19 24 33 40

Solution

Given less than frequency table can be written in the following continuous class.
m –a
Let i = 25, then d1 = i

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Class Mid – value frequency(f) d1 = m –a fd1 fd1 2
0 – 10 (m) i
5 12 – 2 – 24 48
–7 7
10 – 20 15 19 – 12 = 7 –1 0 0
9 9
20 – 30 25 24 – 19 = 5 0 14 28
–8 92
30 – 40 35 33 – 24 = 9 1

40 – 50 45 40 – 33 = 7 2

N = 40

Standard deviation ( ) = ∑fd12 – ∑fd1 2 ×i
N N

= 92 – –8 2 × 10
40 40

= 2.3 – 0.04 × 10

= 2.26 × 10

= 15.03 ∑ fd1
N
Mean(x) =a + × i
= 25 + –8
40 × 10

= 23 15.03
23
Coefficient of variatin = x6 × 100% =

= 65.35%

b)

x above 20 above 40 above 60 above 80 above 100 and less than 120
f 50 42 30 18 7

Solution

Given more than cumulative frequency table can be written in the following continuous.
m–a =m 1–070
Let a = 70, then d1 = i

Class Mid – value frequency(f) d1 = m –a fd1 fd1 2
(m) i

20 – 40 30 50 – 48 = 8 –2 – 16 32

40 – 60 50 12 – 1 – 12 12

60 – 80 70 18 0 0 0

80 – 100 90 11 1 11 11

100 – 120 110 7 2 14 28

N = 50 – 3 83

Standard deviation ( ) = ∑fd12 – ∑fd1 2 ×i
N N

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= 83 – 83 2 × 20
50 50

= 1.66 – 0.0036 × 20

= 1.6564 × 20

= 25.74 ∑ fd1
N
Mean(x) =a + × i
= 70 + –3 × 20
50
= 68.8

Coefficient of variation (c.v) = 6 × 100%
25.74 x
= 68.8 × 100%

= 37.41%

c) 0 – 10 0 – 20 0 – 30 0 – 40 0 –50
Marks 7 18 30 42 50
No. of students

Solution

This is less than cumulative frequency table. We can change it as continuous frequency
table as follows.
m –a m – 25
Let a = 25, then d1 = i = 10

Class frequency(f) Mid – value d1 = m –a fd1 fd1 2
0 – 10 7 (m) i – 14 28
5 –2

10 – 20 18 – 7 =11 15 – 1 – 11 11

20 – 30 30 – 18 = 12 25 0 0 0

30 – 40 42 – 30 = 12 35 1 12 12

40 – 50 50 – 42 = 8 45 2 16 32

N = 50 0 3 83

Standard deviation = = ∑fd12 – ∑fd1 2 ×i
N N
83 32
= 50 – 50 × 10

= 1.66 – 0.0036 × 10

= 12.87 ∑ fd1
N
Mean(x) = a + × i
3
= 25 + 50 × 10

= 25 + 0.6

= 25.6

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c.v = x6 × 100%

= 12.87 × 100 % = 50.27 %
25.6

5. From the given data which series is more variable (inconsistent).

Variable section A 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency section B 10 18 32 40 22 18
18 22 40 32 20 10

Solution
We calculate coefficient of variation two compare variability of two givens data. More c.v.
more will be variability.

class Series A B
mdA–1=45
10 – 20 m fA –103 fAdB1 fAdA1 2 fB mdB–1=45 fBdB1 fBdB1
20 – 30 –2 –103 – 54
30 – 40 15 10 –1 – 30 90 18 162
40 – 50 25 18 0 – 36 72 22 – 2 – 44 88
50 – 60 35 32 1 – 32 32 40 40
60 – 70 45 40 2 0 32 – 1 – 40 0
55 22 0 22 20 21
65 18 22 72 10 00 40
N1A40= 36 142 =
288 NB 1 20 351
– 40
2 20

– 98

For series A,

∑fAdA1 2 ∑fAdA1 2
N N
Standard deviation ( A )= – ×i

= 288 –40 2 × 10
140 140
–

= 2.0571 – 0.0816 × 10

= 1.9755 × 10

= 14.05

Mean(xA)= a + ∑fAdA1 ×i
NA

–40 2
140
= 45 + × 10

= 42.14

6A × 100%
c.v. (A) = x

A

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= 14.05 × 100 % = 33.34 %
42.14

Again for series B,

Standard deviation ( B )= ∑fBdB1 2 ∑fBdB1 2
NB NB
– ×i

= 351 –98 2 × 10
142 142
–

= 2.4718 – 0.4763 × 10

= 1.9955 × 10

= 1.413

Mean(xB)= a + ∑fBdB1 ×i
NB
–98 2
142 × 10
= 45 +

= 38.09

6B × 100%
c.v. (B) = x

B

= 14.13 × 100 % = 37.01 %
38.09

Since c.v. (B) > c.v. (A), the series B is more variable or more inconsistent.

Questions for practice

1. If Q1 = 45, Q.D = 20, find the third quartile and coefficient of Quartile deviation.
2. If coefficient of Q.D is 0.5, third quartile is 25, find the first quartile.

3. Find the Quartile Deviation and its coefficient for the following given data.
(a)

x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90
f 2 4 6 12 15 12 6 4 3
(b)

x less than 20 less than 30 less than 40 less than 50

f 5 10 20 25

(c)

x more than 10 more than 20 more than 30 more than 40

f 40 30 20 10 5

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(d) 0 –10 0 – 20 0 – 30 0 – 40 0 – 50
5 15 25 40 50
Marks
No. of 0 – 50 0 – 40 0 – 30 0 – 20 0 – 10
students 40 30 20 10 5
(e)
Marks
No. of
students

4. Taking class interval 10 find the quartile deviation from the given data
40, 50, 60, 70, 50, 80, 70, 90, 13, 22
70, 80, 50, 60, 70, 85, 95, 65, 50, 45
22, 45, 60, 70, 85, 70, 90, 72, 55, 49

5. If Q1 = 32, find the value of x and then quartile deviation from given date.

x 0 – 20 0 – 40 0 – 60 0 – 80 0 – 100

f 10 20 30 + x 40 + x 50 + x

Mean Deviation
1. Find the coefficient of mean deviation of a continuous series having 20 samples whose

mean is 40 and ∑f|d| = 240.

2. Find the coefficient of mean deviation from median whose median is 35 and number of
item is 40 and ∑f|d| = 440.

3. Find the mean deviation from mean and its coefficients.
(a)

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of 4 10 7 6 3
students
(b)

x 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50

f 5 13 25 40 50

4. Find the mean deviation from median and its coefficient.
(a)

x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

y 5 10 15 10 5

(b)

x 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50

f 2 5 10 15 20

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(c) 10 – 20 20 – 30 20 – 40 40 – 50
5 7 15 3
Marks
No. of
students

Standard Deviation:
1. In a continuous series, N = 25, ∑fd = 480, ∑fd2 = 3240, d=(m – x ) find the coefficient

of standard deviation.

2. In a continuous series N = 40, ∑fm = 100, ∑fd2 = 4060 find the coefficient of standard
deviation.

2. In a continuous series N = 40, ∑fm = 100, ∑fd2 = 4060 find the coefficient of standard
deviation.

3. In a continuous series N = 40, ∑fd1 = 7, ∑fd1 2 = 75, i = 10, assumed mean a = 20, find
the standard deviation and its coefficient

4. Find the standard deviation and its coefficient from the following data :
(a)

x 10 – 20 20 – 30 20 – 40 40 – 50 50 – 60

y 10 15 20 10 5

(b)

x 0 –6 0 – 12 0 – 18 0 – 24 0 – 30

y 5 10 15 20 25

(c)

marks less than 10 less than 20 less than 30 less than 40 less than 50
No. of 5 9 12 20 25
students

5. Prepare a frequency distribution table taking class interval 10. Calculate the standard
deviation and its coefficient.

20, 22, 24, 25, 28, 10, 22, 70, 80, 45
33, 45, 37, 80, 75, 95, 80, 75, 78, 88
60, 66, 65, 68, 78, 90, 88, 78, 79, 90

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