v“Statistics may be rightly called the science of averages and their
estimates.” – A.L.BOWLEY & A.L. BODDINGTON v
15.1 Introduction
We know that statistics deals with data collected for specific
purposes. We can make decisions about the data by
analysing and interpreting it. In earlier classes, we have
studied methods of representing data graphically and in
tabular form. This representation reveals certain salient
features or characteristics of the data. We have also studied
the methods of finding a representative value for the given
data. This value is called the measure of central tendency.
Recall mean (arithmetic mean), median and mode are three
measures of central tendency. A measure of central
tendency gives us a rough idea where data points are
centred. But, in order to make better interpretation from the
data, we should also have an idea how the data are scattered or how much they are
bunched around a measure of central tendency.
Consider now the runs scored by two batsmen in their last ten matches as follows:
Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71
Batsman B : 53, 46, 48, 50, 53, 53, 58, 60, 57, 52
Clearly, the mean and median of the data are
Batsman A Batsman B
Mean 53 53
Median 53 53
Recall that, we calculate the mean of a data (denoted by
x
) by dividing the sum
of the observations by the number of observations, i.e.,
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Karl Pearson
(1857-1936)
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348 MATHEMATICS
1
1
n
i
i
x x
n
=
=
∑
Also, the median is obtained by first arranging the data in ascending or descending
order and applying the following rule.
If the number of observations is odd, then the median is
th
1
2
n +
observation.
If the number of observations is even, then median is the mean of
th
2
n
and
th
1
2
n
+
observations.
We find that the mean and median of the runs scored by both the batsmen A and
B are same i.e., 53. Can we say that the performance of two players is same? Clearly
No, because the variability in the scores of batsman A is from 0 (minimum) to 117
(maximum). Whereas, the range of the runs scored by batsman B is from 46 to 60.
Let us now plot the above scores as dots on a number line. We find the following
diagrams:
For batsman A
For batsman B
We can see that the dots corresponding to batsman B are close to each other and
are clustering around the measure of central tendency (mean and median), while those
corresponding to batsman A are scattered or more spread out.
Thus, the measures of central tendency are not sufficient to give complete
information about a given data. Variability is another factor which is required to be
studied under statistics. Like ‘measures of central tendency’ we want to have a
single number to describe variability. This single number is called a ‘measure of
dispersion’. In this Chapter, we shall learn some of the important measures of dispersion
and their methods of calculation for ungrouped and grouped data.
Fig 15.1
Fig 15.2
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15.2 Measures of Dispersion
The dispersion or scatter in a data is measured on the basis of the observations and the
types of the measure of central tendency, used there. There are following measures of
dispersion:
(i) Range, (ii) Quartile deviation, (iii) Mean deviation, (iv) Standard deviation.
In this Chapter, we shall study all of these measures of dispersion except the
quartile deviation.
15.3 Range
Recall that, in the example of runs scored by two batsmen A and B, we had some idea
of variability in the scores on the basis of minimum and maximum runs in each series.
To obtain a single number for this, we find the difference of maximum and minimum
values of each series. This difference is called the ‘Range’ of the data.
In case of batsman A, Range = 117 – 0 = 117 and for batsman B, Range = 60 – 46 = 14.
Clearly, Range of A > Range of B.
Therefore, the scores are scattered or dispersed in
case of A while for B these are close to each other.
Thus, Range of a series = Maximum value – Minimum value.
The range of data gives us a rough idea of variability or scatter but does not tell
about the dispersion of the data from a measure of central tendency. For this purpose,
we need some other measure of variability. Clearly, such measure must depend upon
the difference (or deviation) of the values from the central tendency.
The important measures of dispersion, which depend upon the deviations of the
observations from a central tendency are mean deviation and standard deviation. Let
us discuss them in detail.
15.4 Mean Deviation
Recall that the deviation of an observation x from a fixed value ‘a’ is the difference
x – a. In order to find the dispersion of values of x from a central value ‘a’ , we find the
deviations about a. An absolute measure of dispersion is the mean of these deviations.
To find the mean, we must obtain the sum of the deviations. But, we know that a
measure of central tendency lies between the maximum and the minimum values of
the set of observations. Therefore, some of the deviations will be negative and some
positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations
from mean (
x
) is zero.
Also Mean of deviations
Sum of deviations 0
0
Number of observations n
= = =
Thus, finding the mean of deviations about mean is not of any use for us, as far
as the measure of dispersion is concerned.
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Remember that, in finding a suitable measure of dispersion, we require the distance
of each value from a central tendency or a fixed number ‘a’. Recall, that the absolute
value of the difference of two numbers gives the distance between the numbers when
represented on a number line. Thus, to find the measure of dispersion from a fixed
number ‘a’ we may take the mean of the absolute values of the deviations from the
central value. This mean is called the ‘mean deviation’. Thus mean deviation about a
central value ‘a’ is the mean of the absolute values of the deviations of the observations
from ‘a’. The mean deviation from ‘a’ is denoted as M.D. (a). Therefore,
M.D.(
a) =
Sum of absolute values of deviations from ' '
Number of observations
a
.
Remark Mean deviation may be obtained from any measure of central tendency.
However, mean deviation from mean and median are commonly used in statistical
studies.
Let us now learn how to calculate mean deviation about mean and mean deviation
about median for various types of data
15.4.1 Mean deviation for ungrouped data Let n observations be x
1
, x
2
, x
3
, ...., x
n
.
The following steps are involved in the calculation of mean deviation about mean or
median:
Step 1 Calculate the measure of central tendency about which we are to find the mean
deviation. Let it be ‘a’.
Step 2 Find the deviation of each x
i
from a, i.e., x
1
– a, x
2
– a, x
3
– a,. . . , x
n
– a
Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is
there, i.e.,
axaxaxax
n
−−−− ....,,,,
321
Step 4 Find the mean of the absolute values of the deviations. This mean is the mean
deviation about a, i.e.,
1
( )M.D.
n
i
i
x a
a
n
=
−
=
∑
Thus M.D. (
x
) =
1
1
n
i
i
x x
n
=
−
∑
, where
x
= Mean
and M.D. (M) =
1
1
M
n
i
i
x
n
=
−
∑
, where M = Median
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A
Note
In this Chapter, we shall use the symbol M to denote median unless stated
otherwise.Let us now illustrate the steps of the above method in following examples.
Example 1 Find the mean deviation about the mean for the following data:
6, 7, 10, 12, 13, 4, 8, 12
Solution
We proceed step-wise and get the following:
Step 1 Mean of the given data is
6 7 10 12 13 4 8 12 72
9
8 8
x
+ + + + + + +
= = =
Step 2 The deviations of the respective observations from the mean
,
x
i.e., x
i
–
x
are
6 – 9, 7 – 9, 10 – 9, 12 – 9, 13 – 9, 4 – 9, 8 – 9, 12 – 9,
or –3, –2, 1, 3, 4, –5, –1, 3
Step 3 The absolute values of the deviations, i.e.,
i
x x
−
are
3, 2, 1, 3, 4, 5, 1, 3
Step 4 The required mean deviation about the mean is
M.D.
(
)
x
=
8
1
8
i
i
x x
=
−
∑
=
3 2 1 3 4 5 1 3 22
2 75
8 8
.
+ + + + + + +
= =
A
Note Instead of carrying out the steps every time, we can carry on calculation,
step-wise without referring to steps.
Example 2 Find the mean deviation about the mean for the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5
Solution We have to first find the mean (
x
) of the given data
20
1
1
20
i
i
x x
=
=
∑
=
20
200
= 10
The respective absolute values of the deviations from mean, i.e.,
xx
i
−
are
2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5
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Therefore
20
1
124
i
i
x x
=
− =
∑
and M.D. (
x
) =
124
20
= 6.2
Example 3 Find the mean deviation about the median for the following data:
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Solution Here the number of observations is 11 which is odd. Arranging the data into
ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
Now Median =
th
11 1
2
+
or 6
th
observation = 9
The absolute values of the respective deviations from the median, i.e.,
M
i
x −
are
6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
Therefore
11
1
M 58
i
i
x
=
− =
∑
and
( )
11
1
1 1
M.D. M M 58 5.27
11 11
i
i
x
=
= − = × =
∑
15.4.2 Mean deviation for grouped data We know that data can be grouped into
two ways :
(a) Discrete frequency distribution,
(b) Continuous frequency distribution.
Let us discuss the method of finding mean deviation for both types of the data.
(a) Discrete frequency distribution Let the given data consist of n distinct values
x
1
, x
2
, ..., x
n
occurring with frequencies f
1
, f
2
, ..., f
n
respectively. This data can be
represented in the tabular form as given below, and is called discrete frequency
distribution:
x : x
1
x
2
x
3
... x
n
f : f
1
f
2
f
3
... f
n
(i) Mean deviation about mean
First of all we find the mean
x
of the given data by using the formula
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1
1
1
1
N
n
i i
n
i
i i
n
i
i
i
x f
x x f
f
=
=
=
= =
∑
∑
∑
,
where
∑
=
n
i
ii
fx
1
denotes the sum of the products of observations x
i
with their respective
frequencies f
i
and
∑
=
=
n
i
i
f
1
N
is the sum of the frequencies.
Then, we find the deviations of observations x
i
from the mean
x
and take their
absolute values, i.e.,
xx
i
−
for all i =1, 2,..., n.
After this, find the mean of the absolute values of the deviations, which is the
required mean deviation about the mean. Thus
1
1
M.D. ( )
n
i i
i
n
i
i
f x x
x
f
=
=
−
=
∑
∑
=
xxf
i
n
i
i
−
∑
=1
N
1
(ii) Mean deviation about median To find mean deviation about median, we find the
median of the given discrete frequency distribution. For this the observations are arranged
in ascending order. After this the cumulative frequencies are obtained. Then, we identify
the observation whose cumulative frequency is equal to or just greater than
N
2
, where
N is the sum of frequencies. This value of the observation lies in the middle of the data,
therefore, it is the required median. After finding median, we obtain the mean of the
absolute values of the deviations from median.Thus,
1
1
M.D.(M) M
N
n
i i
i
f x
=
= −
∑
Example 4 Find mean deviation about the mean for the following data :
x
i
2 5 6 8 10 12
f
i
2 8 10 7 8
5
Solution Let us make a Table 15.1 of the given data and append other columns after
calculations.
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Table 15.1
x
i
f
i
f
i
x
i
xx
i
−
f
i
xx
i
−
2 2 4 5.5 11
5 8 40 2.5 20
6 10 60 1.5 15
8 7 56 0.5
3.5
10 8 80 2.5 20
12 5 60 4.5 22.5
40 300 92
40N
6
1
==
∑
=i
i
f
,
300
6
1
=
∑
=i
ii
xf
,
92
6
1
=−
∑
=
xxf
i
i
i
Therefore
6
1
1 1
300 7.5
N 40
i i
i
x f x
=
= = × =
∑
and
6
1
1 1
M. D. ( ) 92 2.3
N 40
i i
i
x f x x
=
= − = × =
∑
Example 5 Find the mean deviation about the median for the following data:
x
i
3 6 9 12 13 15
21 22
f
i
3 4 5 2 4 5 4 3
Solution The given observations are already in ascending order. Adding a row
corresponding to cumulative frequencies to the given data, we get (Table 15.2).
Table 15.2
x
i
3 6 9 12 13 15 21 22
f
i
3 4
5 2 4 5 4 3
c.f. 3 7 12 14 18 23 27 30
Now, N=30 which is even.
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Median is the mean of the 15
th
and 16
th
observations. Both of these observations
lie in the cumulative frequency 18, for which the corresponding observation is 13.
th th
15 observation 16 observation 13 13
Therefore, Median M 13
2 2
+ +
= = =
Now, absolute values of the deviations from median, i.e.,
M
i
x −
are shown in
Table 15.3.
Table 15.3
M
i
x −
10 7 4 1 0 2
8 9
f
i
3 4 5 2 4 5 4 3
f
i
M
i
x −
30 28 20 2 0 10 32 27
We have
8 8
1 1
30 and M 149
i i i
i i
f f x
= =
= − =
∑ ∑
Therefore
8
1
1
M. D. (M) M
N
i i
i
f x
=
= −
∑
=
1
149
30
×
= 4.97.
(b) Continuous frequency distribution A continuous frequency distribution is a series
in which the data are classified into different class-intervals without gaps alongwith
their respective frequencies.
For example, marks obtained by 100 students are presented in a continuous
frequency distribution as follows :
Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60
Number of Students 12 18
27 20 17 6
(i) Mean deviation about mean While calculating the mean of a continuous frequency
distribution, we had made the assumption that the frequency in each class is centred at
its mid-point. Here also, we write the mid-point of each given class and proceed further
as for a discrete frequency distribution to find the mean deviation.
Let us take the following example.
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Example 6 Find the mean deviation about the mean for the following data.
Marks obtained 10-20 20-30
30-40
40-50
50-60
60-70
70-80
Number of students 2 3 8 14 8 3 2
Solution We make the following Table 15.4 from the given data :
Table 15.4
Marks Number of Mid-points f
i
x
i
xx
i
−
f
i
xx
i
−
obtained students
f
i
x
i
10-20 2 15 30 30 60
20-30 3 25 75 20
60
30-40 8 35 280 10 80
40-50 14 45 630 0 0
50-60 8 55 440 10 80
60-70 3 65 195 20 60
70-80 2 75 150 30 60
40 1800 400
Here
7 7 7
1 1 1
N 40, 1800, 400
i i i i i
i i i
f f x f x x
= = =
= = = − =
∑ ∑ ∑
Therefore
7
1
1 1800
45
N 40
i i
i
x f x
=
= = =
∑
and
( )
7
1
1 1
M.D. 400 10
N 40
i i
i
x f x x
=
= − = × =
∑
Shortcut method for calculating mean deviation about mean We can avoid the
tedious calculations of computing
x
by following step-deviation method. Recall that in
this method, we take an assumed mean which is in the middle or just close to it in the
data. Then deviations of the observations (or mid-points of classes) are taken from the
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assumed mean. This is nothing but the shifting of origin from zero to the assumed mean
on the number line, as shown in Fig 15.3
If there is a common factor of all the deviations, we divide them by this common
factor to further simplify the deviations. These are known as step-deviations. The
process of taking step-deviations is the change of scale on the number line as shown in
Fig 15.4
The deviations and step-deviations reduce the size of the observations, so that the
computations viz. multiplication, etc., become simpler. Let, the new variable be denoted
by
h
ax
d
i
i
−
=
, where ‘a’ is the assumed mean and h is the common factor. Then, the
mean
x
by step-deviation method is given by
1
N
n
f d
i i
i
x a h
∑
=
= + ×
Let us take the data of Example 6 and find the mean deviation by using step-
deviation method.
Fig 15.3
Fig 15.4
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358 MATHEMATICS
Number of
students
Marks
obtained
Take the assumed mean a = 45 and h = 10, and form the following Table 15.5.
Table 15.5
Mid-points
45
10
i
i
x
d
−
=
i i
f d
xx
i
−
f
i
xx
i
−
f
i
x
i
10-20 2 15 – 3 – 6 30 60
20-30 3 25 – 2 – 6 20 60
30-40 8 35 – 1 – 8 10
80
40-50 14 45 0 0 0 0
50-60 8 55 1 8 10 80
60-70 3 65 2 6 20 60
70-80 2 75 3 6 30 60
40 0 400
Therefore
7
1
N
f d
i i
i
x a h
∑
=
= + ×
=
0
45 10 45
40
+ × =
and
7
1
1 400
M D ( ) 10
N 40
i i
i
x f x x
=
= − = =
∑
. .
A
Note The step deviation method is applied to compute
x
. Rest of the procedure
is same.
(ii) Mean deviation about median The process of finding the mean deviation about
median for a continuous frequency distribution is similar as we did for mean deviation
about the mean. The only difference lies in the replacement of the mean by median
while taking deviations.
Let us recall the process of finding median for a continuous frequency distribution.
The data is first arranged in ascending order. Then, the median of continuous
frequency distribution is obtained by first identifying the class in which median lies
(median class) and then applying the formula
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frequency
N
C
2
Median
l h
f
−
= + ×
where median class is the class interval whose cumulative frequency is just greater
than or equal to
N
2
, N is the sum of frequencies, l, f, h and C are, respectively the lower
limit , the frequency, the width of the median class and C the cumulative frequency of
the class just preceding the median class. After finding the median, the absolute values
of the deviations of mid-point x
i
of each class from the median i.e.,
M
i
x −
are obtained.
Then
1
M.D. (M) M
1
N
n
f x
i i
i
= −
∑
=
The process is illustrated in the following example:
Example 7 Calculate the mean deviation about median for the following data :
Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 6 7 15
16 4 2
Solution Form the following Table 15.6 from the given data :
Table 15.6
Class Frequency Cumulative Mid-points
Med.
x
i
−
f
i
Med.
x
i
−
f
i
(c.f.) x
i
0-10 6 6 5 23 138
10-20 7 13 15 13 91
20-30 15 28 25 3 45
30-40 16 44 35 7 112
40-50 4 48 45 17 68
50-60 2 50 55 27 54
50 508
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The class interval containing
th
N
2
or 25
th
item is 20-30. Therefore, 20–30 is the median
class. We know that
Median =
N
C
2
l h
f
−
+ ×
Here l = 20, C = 13, f = 15, h = 10 and N = 50
Therefore, Median
25 13
20 10
15
−
= + ×
= 20 + 8 = 28
Thus, Mean deviation about median is given by
M.D. (M) =
6
1
1
M
N
i i
i
f x
=
−
∑
=
1
508
50
×
= 10.16
EXERCISE 15.1
Find the mean deviation about the mean for the data in Exercises 1 and 2.
1. 4, 7, 8, 9, 10, 12, 13, 17
2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Find the mean deviation about the mean for the data in Exercises 5 and 6.
5. x
i
5 10 15 20 25
f
i
7 4 6
3 5
6. x
i
10 30 50 70 90
f
i
4 24 28 16 8
Find the mean deviation about the median for the data in Exercises 7 and 8.
7. x
i
5 7 9 10 12 15
f
i
8 6 2 2
2 6
8. x
i
15 21 27 30 35
f
i
3 5 6 7 8
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Find the mean deviation about the mean for the data in Exercises 9 and 10.
9. Income per 0-100 100-200 200-300
300-400
400-500
500-600
600-700
700-800
day in `
Number 4 8 9 10 7 5 4 3
of persons
10. Height 95-105 105-115 115-125 125-135
135-145 145-155
in cms
Number of 9 13 26 30 12 10
boys
11. Find the mean deviation about median for the following data :
Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of 6 8 14 16 4 2
Girls
12. Calculate the mean deviation about median age for the age distribution of 100
persons given below:
Age 16-20 21-25 26-30 31-35 36-40 41-45
46-50 51-55
(in years)
Number 5 6 12 14 26 12 16 9
[Hint Convert the given data into continuous frequency distribution by subtracting 0.5
from the lower limit and adding 0.5 to the upper limit of each class interval]
15.4.3 Limitations of mean deviation In a series, where the degree of variability is
very high, the median is not a representative central tendency. Thus, the mean deviation
about median calculated for such series can not be fully relied.
The sum of the deviations from the mean (minus signs ignored) is more than the
sum of the deviations from median. Therefore, the mean deviation about the mean is
not very scientific.Thus, in many cases, mean deviation may give unsatisfactory results.
Also mean deviation is calculated on the basis of absolute values of the deviations and
therefore, cannot be subjected to further algebraic treatment. This implies that we
must have some other measure of dispersion. Standard deviation is such a measure of
dispersion.
15.5 Variance and Standard Deviation
Recall that while calculating mean deviation about mean or median, the absolute values
of the deviations were taken. The absolute values were taken to give meaning to the
mean deviation, otherwise the deviations may cancel among themselves.
Another way to overcome this difficulty which arose due to the signs of deviations,
is to take squares of all the deviations. Obviously all these squares of deviations are
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non-negative. Let x
1
, x
2
, x
3
, ..., x
n
be n observations and
x
be their mean. Then
2
2 2 2 2
1
1
( ) ( ) ....... ( ) ( )
n
n i
i
x x x x x x x x
=
− + − + + − = −
.
If this sum is zero, then each
)( xx
i
−
has to be zero. This implies that there is no
dispersion at all as all observations are equal to the mean
x
.
If
∑
=
−
n
i
i
xx
1
2
)(
is small , this indicates that the observations x
1
, x
2
, x
3
,...,x
n
are
close to the mean
x
and therefore, there is a lower degree of dispersion. On the
contrary, if this sum is large, there is a higher degree of dispersion of the observations
from the mean
x
. Can we thus say that the sum
∑
=
−
n
i
i
xx
1
2
)(
is a reasonable indicator
of the degree of dispersion or scatter?
Let us take the set A of six observations 5, 15, 25, 35, 45, 55. The mean of the
observations is
x
= 30. The sum of squares of deviations from
x
for this set is
∑
=
−
6
1
2
)(
i
i
xx
= (5–30)
2
+ (15–30)
2
+ (25–30)
2
+ (35–30)
2
+ (45–30)
2
+(55–30)
2
= 625 + 225 + 25 + 25 + 225 + 625 = 1750
Let us now take another set B of 31 observations 15, 16, 17, 18, 19, 20, 21, 22, 23,
24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45. The
mean of these observations is
y
= 30
Note that both the sets A and B of observations have a mean of 30.
Now, the sum of squares of deviations of observations for set B from the mean
y
is
given by
∑
=
−
31
1
2
)(
i
i
yy
= (15–30)
2
+(16–30)
2
+ (17–30)
2
+ ...+ (44–30)
2
+(45–30)
2
=
(–15)
2
+(–14)
2
+ ...+ (–1)
2
+ 0
2
+ 1
2
+ 2
2
+ 3
2
+ ...+ 14
2
+ 15
2
= 2 [15
2
+ 14
2
+ ... + 1
2
]
=
15 (15 1) (30 1)
2
6
× + +
×
= 5 × 16 × 31 = 2480
(Because sum of squares of first n natural numbers =
( 1) (2 1)
6
n n n
+ +
. Here n = 15)
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If
∑
=
−
n
i
i
xx
1
2
)(
is simply our measure of dispersion or scatter about mean, we
will tend to say that the set A of six observations has a lesser dispersion about the mean
than the set B of 31 observations, even though the observations in set A are more
scattered from the mean (the range of deviations being from –25 to 25) than in the set
B (where the range of deviations is from –15 to 15).
This is also clear from the following diagrams.
For the set A, we have
For the set B, we have
Thus, we can say that the sum of squares of deviations from the mean is not a proper
measure of dispersion. To overcome this difficulty we take the mean of the squares of
the deviations, i.e., we take
∑
=
−
n
i
i
xx
n
1
2
)(
1
. In case of the set A, we have
1
Mean
6
=
× 1750 = 291.67 and in case of the set B, it is
1
31
× 2480 = 80.
This indicates that the scatter or dispersion is more in set A than the scatter or dispersion
in set B, which confirms with the geometrical representation of the two sets.
Thus, we can take
∑
−
2
)(
1
xx
n
i
as a quantity which leads to a proper measure
of dispersion. This number, i.e., mean of the squares of the deviations from mean is
called the variance and is denoted by
2
σ
(read as sigma square). Therefore, the
variance of n observations x
1
, x
2
,..., x
n
is given by
Fig 15.5
Fig 15.6
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364 MATHEMATICS
Deviations from mean
(x
i
–
x
)
∑
=
−=
n
i
i
xx
n
1
22
)(
1
σ
15.5.1 Standard Deviation In the calculation of variance, we find that the units of
individual observations x
i
and the unit of their mean
x
are different from that of variance,
since variance involves the sum of squares of (x
i
–
x
). For this reason, the proper
measure of dispersion about the mean of a set of observations is expressed as positive
square-root of the variance and is called standard deviation. Therefore, the standard
deviation, usually denoted by
σ
, is given by
∑
=
−=
n
i
i
xx
n
1
2
)(
1
σ
... (1)
Let us take the following example to illustrate the calculation of variance and
hence, standard deviation of ungrouped data.
Example 8 Find the variance of the following data:
6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Solution From the given data we can form the following Table 15.7. The mean is
calculated by step-deviation method taking 14 as assumed mean. The number of
observations is n = 10
Table 15.7
x
i
14
2
i
i
x
d
−
=
(x
i
–
x
)
6 –4 –9 81
8 –3 –7 49
10 –2 –5 25
12 –1 –3 9
14 0 –1 1
16 1 1 1
18 2 3 9
20 3 5 25
22 4 7 49
24 5 9 81
5 330
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Therefore Mean
x
= assumed mean +
h
n
d
n
i
i
×
∑
=1
=
5
14 2 15
10
+ × =
and Variance (
2
σ
) =
10
2
1
1
)
i
i
( x x
n
=
−
∑
=
1
330
10
×
= 33
Thus Standard deviation (
σ
) =
33 5 74
.
=
15.5.2 Standard deviation of a discrete frequency distribution Let the given discrete
frequency distribution be
x : x
1
, x
2
, x
3
,. . . , x
n
f : f
1
, f
2
, f
3
,. . . , f
n
In this case standard deviation
( )
2
1
1
( )
N
n
i i
i
f x x
σ
=
= −
∑
... (2)
where
1
N
n
i
i
f
=
=
∑
.
Let us take up following example.
Example 9 Find the variance and standard deviation for the following data:
x
i
4 8 11 17 20 24 32
f
i
3 5 9 5
4 3 1
Solution Presenting the data in tabular form (Table 15.8), we get
Table 15.8
x
i
f
i
f
i
x
i
x
i
–
x
2
)( xx
i
−
f
i
2
)( xx
i
−
4 3 12 –10 100 300
8 5 40 –6 36 180
11 9 99 –3 9 81
17 5 85 3 9 45
20 4 80 6 36 144
24 3 72
10 100 300
32 1 32 18 324 324
30 420 1374
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N = 30,
( )
7 7
2
1 1
420, 1374
i i i i
i i
f x f x x
= =
= − =
∑ ∑
Therefore
7
1
1
420 14
N 30
i i
i
f x
x
=
= = × =
∑
Hence variance
2
( )
σ
=
7
2
1
1
( )
N
i i
i
f x x
=
−
∑
=
1
30
× 1374 = 45.8
and Standard deviation
8.45)( =
σ
= 6.77
15.5.3 Standard deviation of a continuous frequency distribution The given
continuous frequency distribution can be represented as a discrete frequency distribution
by replacing each class by its mid-point. Then, the standard deviation is calculated by
the technique adopted in the case of a discrete frequency distribution.
If there is a frequency distribution of n classes each class defined by its mid-point
x
i
with frequency f
i
, the standard deviation will be obtained by the formula
2
1
1
( )
N
n
i i
i
f x x
σ
=
= −
∑
,
where
x
is the mean of the distribution and
1
N
n
i
i
f
=
=
∑
.
Another formula for standard deviation We know that
Variance
2
( )
σ
=
2
1
1
( )
N
n
i i
i
f x x
=
−
∑
=
2
2
1
1
( 2 )
N
n
i i i
i
f x x x x
=
+ −
∑
=
2
2
1 1 1
1
2
N
n n n
i i i i i
i i i
f x x f x f x
= = =
+ −
∑ ∑ ∑
=
2
2
1 1 1
1
2
N
n n n
i i i i i
i i i
f x x f x x f
= = =
+ −
∑ ∑ ∑
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STATISTICS 367
=
2 2
1
1
N 2 . N
N
=
+ −
n
i i
i
f x x x x
1 1
1
Here or N
N
n n
i i i i
i i
x f x x f x
= =
= =
∑ ∑
=
2
2
2
1
2
1
N
n
i i
i
x x
f x
=
+ −
∑
2
2
1
1
N
n
i i
i
x
f x
=
= −
∑
or
2
σ
=
2
2
2 2
=1
2
1 1 =1
1 1
N
N N
N
n
i i
n n n
i
i i i i i i
i i i
f x
f x f x f x
− =
− = −
∑
∑ ∑ ∑
Thus, standard deviation
( )
2
2
1 =1
1
N
N
n n
i i i i
i i
f x f x
σ
=
= −
∑ ∑
... (3)
Example 10 Calculate the mean, variance and standard deviation for the following
distribution :
Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency 3
7 12 15 8 3 2
Solution From the given data, we construct the following Table 15.9.
Table 15.9
Class Frequency Mid-point f
i
x
i
(x
i
–
x
)
2
f
i
(x
i
–
x
)
2
(f
i
) (x
i
)
30-40 3 35 105 729 2187
40-50 7 45 315 289 2023
50-60 12 55 660 49 588
60-70 15 65 975 9 135
70-80 8 75 600 169 1352
80-90 3 85 255
529 1587
90-100 2 95 190 1089 2178
50 3100 10050
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Thus
7
1
1 3100
Mean 62
N 50
i i
i
x f x
=
= = =
∑
Variance
(
)
2
σ
=
7
2
1
1
( )
N
i i
i
f x x
=
−
∑
=
1
10050
50
×
= 201
and Standard deviation
(
)
201 14 18
.
σ
= =
Example 11 Find the standard deviation for the following data :
x
i
3 8 13 18 23
f
i
7 10 15 10 6
Solution Let us form the following Table 15.10:
Table 15.10
x
i
f
i
f
i
x
i
x
i
2
f
i
x
i
2
3
7 21 9 63
8 10 80 64 640
13 15 195 169 2535
18 10 180 324 3240
23 6 138 529 3174
48 614 9652
Now, by formula (3), we have
σ
=
( )
2
2
1
N
N
i i i i
f x f x
−
∑ ∑
=
2
1
48 9652 (614)
48
× −
=
1
463296 376996
48
−
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STATISTICS 369
=
1
293 77
48
.
×
= 6.12
Therefore, Standard deviation (
σ
) = 6.12
15.5.4. Shortcut method to find variance and standard deviation Sometimes the
values of x
i
in a discrete distribution or the mid points x
i
of different classes in a
continuous distribution are large and so the calculation of mean and variance becomes
tedious and time consuming. By using step-deviation method, it is possible to simplify
the procedure.
Let the assumed mean be ‘A’ and the scale be reduced to
h
1
times (h being the
width of class-intervals). Let the step-deviations or the new values be y
i
.
i.e.
A
i
i
x
y
h
−
=
or x
i
= A + hy
i
... (1)
We know that
1
N
n
i i
i
f x
x
=
=
∑
... (2)
Replacing x
i
from (1) in (2), we get
x
=
1
A )
N
n
i i
i
f ( hy
=
+
∑
=
1 1
1
A
N
n n
i i i
i i
f h f y
= =
+
∑ ∑
=
1 1
1
A
N
i
n n
i i
i i
f h f y
= =
+
∑ ∑
=
1
N
A
N N
n
i i
i
f y
. h
=
+
∑
1
because N
n
i
i
f
=
=
∑
Thus
x
= A + h
y
... (3)
Now Variance of the variable x,
2
2
1
1
)
N
n
x i i
i
f ( x x
σ
=
= −
∑
=
2
1
1
(A A )
N
n
i i
i
f hy h y
=
+ − −
∑
(Using (1) and (3))
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=
2 2
1
1
( )
N
n
i i
i
f h y y
=
−
∑
=
2
2
1
( )
N
n
i i
i
h
f y y
=
−
∑
= h
2
× variance of the variable y
i
i.e.
2
x
σ
=
2
2
y
h
σ
or
x
σ
=
y
h
σ
... (4)
From (3) and (4), we have
x
σ
=
2
2
1 1
N
N
n n
i i i i
i i
h
f y f y
= =
−
∑ ∑
... (5)
Let us solve Example 11 by the short-cut method and using formula (5)
Examples 12 Calculate mean, variance and standard deviation for the following
distribution.
Classes 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency 3
7 12 15 8 3 2
Solution Let the assumed mean A = 65. Here h = 10
We obtain the following Table 15.11 from the given data :
Table 15.11
Class Frequency Mid-point y
i
=
65
10
i
x
−
y
i
2
f
i
y
i
f
i
y
i
2
f
i
x
i
30-40 3 35 – 3 9 – 9 27
40-50 7 45 – 2 4 – 14 28
50-60 12 55 – 1 1 – 12 12
60-70 15 65 0 0 0 0
70-80 8 75 1
1 8 8
80-90 3 85 2 4 6 12
9 0-100 2 95 3 9 6 18
N=50 – 15 105
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Therefore
x
=
15
A 65 10 62
50 50
i i
f y
h
+ × = − × =
∑
Variance
2
σ
=
( )
2
2
2
N
2
N
i i
h
f y f y
i i
−
∑ ∑
=
( )
2
10
2
50 105 (–15)
2
(50)
× −
=
1
[5250 225] 201
25
− =
and standard deviation
(
)
201
σ
=
= 14.18
EXERCISE 15.2
Find the mean and variance for each of the data in Exercies 1 to 5.
1. 6, 7, 10, 12, 13, 4, 8, 12
2. First n natural numbers
3. First 10 multiples of 3
4. x
i
6 10 14 18 24 28 30
f
i
2
4 7 12 8 4 3
5. x
i
92 93 97 98 102 104 109
f
i
3 2 3 2 6 3 3
6. Find the mean and standard deviation using short-cut method.
x
i
60 61 62 63 64 65 66 67 68
f
i
2 1 12 29 25 12 10 4 5
Find the mean and variance for the following frequency distributions in Exercises
7 and 8.
7. Classes 0-30 30-60
60-90 90-120 120-150 150-180 180-210
Frequencies 2 3 5 10 3 5 2
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372 MATHEMATICS
8.
Classes 0-10 10-20 20-30
30-40
40-50
Frequencies 5
8 15
16 6
9. Find the mean, variance and standard deviation using short-cut method
Height 70-75
75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115
in cms
No. of 3 4
7 7 15 9 6 6 3
children
10. The diameters of circles (in mm) drawn in a design are given below:
Diameters 33-36 37-40 41-44 45-48 49-52
No. of circles 15 17 21 22 25
Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5,
40.5-44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.]
15.6 Analysis of Frequency Distributions
In earlier sections, we have studied about some types of measures of dispersion. The
mean deviation and the standard deviation have the same units in which the data are
given. Whenever we want to compare the variability of two series with same mean,
which are measured in different units, we do not merely calculate the measures of
dispersion but we require such measures which are independent of the units. The
measure of variability which is independent of units is called coefficient of variation
(denoted as C.V.)
The coefficient of variation is defined as
100
C.V.
x
σ
= ×
,
0
≠
x
,
where σ and
x
are the standard deviation and mean of the data.
For comparing the variability or dispersion of two series, we calculate the coefficient
of variance for each series. The series having greater C.V. is said to be more variable
than the other. The series having lesser C.V. is said to be more consistent than the
other.
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15.6.1 Comparison of two frequency distributions with same mean Let
1
x
and σ
1
be the mean and standard deviation of the first distribution, and
2
x
and σ
2
be the
mean and standard deviation of the second distribution.
Then C.V. (1st distribution) =
1
1
100
x
σ
×
and C.V. (2nd distribution) =
2
2
100
x
σ
×
Given
1
x
=
2
x
=
x
(say)
Therefore C.V. (1st distribution) =
1
100
x
σ
×
... (1)
and C.V. (2nd distribution) =
2
100
x
σ
×
... (2)
It is clear from (1) and (2) that the two C.Vs. can be compared on the basis of values
of
1
σ
and
2
σ
only.
Thus, we say that for two series with equal means, the series with greater standard
deviation (or variance) is called more variable or dispersed than the other. Also, the
series with lesser value of standard deviation (or variance) is said to be more consistent
than the other.
Let us now take following examples:
Example 13 Two plants A and B of a factory show following results about the number
of workers and the wages paid to them.
A B
No. of workers 5000 6000
Average monthly wages Rs 2500 Rs 2500
Variance of distribution 81 100
of wages
In which plant, A or B is there greater variability in individual wages?
Solution
The variance of the distribution of wages in plant A (
2
1
σ
) = 81
Therefore, standard deviation of the distribution of wages in plant A (
1
σ
) = 9
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Also, the variance of the distribution of wages in plant B (
2
2
σ
) = 100
Therefore, standard deviation of the distribution of wages in plant B (
2
σ
) = 10
Since the average monthly wages in both the plants is same, i.e., Rs.2500, therefore,
the plant with greater standard deviation will have more variability.
Thus, the plant B has greater variability in the individual wages.
Example 14
Coefficient of variation of two distributions are 60 and 70, and their
standard deviations are 21 and 16, respectively. What are their arithmetic means.
Solution
Given C.V. (1st distribution) = 60,
1
σ
= 21
C.V. (2nd distribution) = 70,
2
σ
= 16
Let
1
x
and
2
x
be the means of 1st and 2nd distribution, respectively. Then
C.V. (1st distribution) =
1
1
x
σ
× 100
Therefore 60 =
1
1
21 21
100 or 100 35
60
x
x
× = × =
and C.V. (2nd distribution) =
2
2
x
σ
×100
i.e. 70 =
2
2
16 16
100 or 100 22 85
70
x .
x
× = × =
Example 15 The following values are calculated in respect of heights and weights of
the students of a section of Class XI :
Height Weight
Mean 162.6 cm 52.36 kg
Variance 127.69 cm
2
23.1361 kg
2
Can we say that the weights show greater variation than the heights?
Solution To compare the variability, we have to calculate their coefficients of variation.
Given Variance of height = 127.69cm
2
Therefore Standard deviation of height =
127.69cm
= 11.3 cm
Also Variance of weight = 23.1361 kg
2
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STATISTICS 375
Therefore Standard deviation of weight =
23 1361 kg
.
= 4.81 kg
Now, the coef
ficient of variations (C.V.) are given by
(C.V.) in heights =
Standard Deviation
100
Mean
×
=
11 3
100
162 6
.
.
×
= 6.95
and (C.V.) in weights =
4 81
100
52 36
.
.
×
= 9.18
Clearly C.V. in weights is greater than the C.V. in heights
Therefore, we can say that weights show more variability than heights.
EXERCISE 15.3
1. From the data given below state which group is more variable, A or B?
Marks 10-20 20-30 30-40 40-50 50-60 60-70
70-80
Group A 9 17 32 33 40 10 9
Group B 10 20 30 25 43 15 7
2. From the prices of shares X and Y below, find out which is more stable in value:
X 35 54 52 53 56 58
52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101
3. An analysis of monthly wages paid to workers in two firms A and B, belonging to
the same industry, gives the following results:
Firm A Firm B
No. of wage earners 586 648
Mean of monthly wages Rs 5253 Rs 5253
Variance of the distribution 100 121
of wages
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
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4. The following is the record of goals scored by team A in a football session:
No. of goals scored 0 1
2
3
4
No. of matches 1
9
7 5 3
For the team B, mean number of goals scored per match was 2 with a standard
deviation 1.25 goals. Find which team may be considered more consistent?
5. The sum and sum of squares corresponding to length x (in cm) and weight y
(in gm) of 50 plant products are given below:
50
1
212
i
i
x
=
=
∑
,
50
2
1
902 8
i
i
x .
=
=
∑
,
50
1
261
i
i
y
=
=
∑
,
50
2
1
1457 6
i
i
y .
=
=
∑
Which is more varying, the length or weight?
Miscellaneous Examples
Example 16 The variance of 20 observations is 5. If each observation is multiplied by
2, find the new variance of the resulting observations.
Solution Let the observations be x
1
, x
2
, ..., x
20
and
x
be their mean. Given that
variance = 5 and n = 20. We know that
Variance
( )
2
20
2
1
1
( )
i
i
x x
n
σ
=
= −
∑
, i.e.,
20
2
1
1
5 ( )
20
i
i
x x
=
= −
∑
or
20
2
1
( )
i
i
x x
=
−
∑
= 100 ... (1)
If each observation is multiplied by 2, and the new resulting observations are y
i
, then
y
i
= 2x
i
i.e., x
i
=
i
y
2
1
Therefore
20 20
1 1
1 1
2
20
i i
i i
y y x
n
= =
= =
∑ ∑
=
20
1
1
2
20
i
i
. x
=
∑
i.e.
y
= 2
x
or
x
=
y
2
1
Substituting the values of x
i
and
x
in (1), we get
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STATISTICS 377
2
20
1
1 1
100
2 2
i
i
y y
=
− =
∑
, i.e.,
∑
=
=−
20
1
2
400)(
i
i
yy
Thus the variance of new observations =
2
1
400 20 2 5
20
× = = ×
A
Note The reader may note that if each observation is multiplied by a constant
k, the variance of the resulting observations becomes k
2
times the original variance.
Example17 The mean of 5 observations is 4.4 and their variance is 8.24. If three of
the observations are 1, 2 and 6, find the other two observations.
Solution Let the other two observations be x and y.
Therefore, the series is 1, 2, 6, x, y.
Now Mean
x
= 4.4 =
1 2 6
5
x y
+ + + +
or 22 = 9 + x + y
Therefore x + y = 13 ... (1)
Also variance = 8.24 =
2
5
1
)(
1
xx
n
i
i∑
=
−
i.e. 8.24 =
( ) ( ) ( ) ( )
2 2 2 2
2 2
1
3 4 2 4 1 6 2 4 4 ( ) 2 4 4
5
. . . x y . x y .
+ + + + − × + + ×
or 41.20 = 11.56 + 5.76 + 2.56 + x
2
+ y
2
–8.8 × 13 + 38.72
Therefore x
2
+ y
2
= 97 ... (2)
But from (1), we have
x
2
+ y
2
+ 2xy = 169 ... (3)
From (2) and (3), we have
2xy = 72 ... (4)
Subtracting (4) from (2), we get
x
2
+ y
2
– 2xy = 97 – 72 i.e. (x – y)
2
= 25
or x – y =
±
5
... (5)
So, from (1) and (5), we get
x = 9, y = 4 when x – y = 5
or x = 4, y = 9 when x – y = – 5
Thus, the remaining observations are 4 and 9.
Example 18 If each of the observation x
1
, x
2
, ...,x
n
is increased by ‘a’, where a is a
negative or positive number, show that the variance remains unchanged.
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378 MATHEMATICS
Solution Let
x
be the mean of x
1
, x
2
, ...,x
n
. Then the variance is given by
2
1
σ
=
2
1
1
( )
n
i
i
x x
n
=
−
∑
If ‘a is added to each observation, the new observations will be
y
i
= x
i
+ a ... (1)
Let the mean of the new observations be
y
. Then
y
=
1 1
1 1
( )
n n
i i
i i
y x a
n n
= =
= +
∑ ∑
=
1 1
1
n n
i
i i
x a
n
= =
+
∑ ∑
=
ax
n
na
x
n
n
i
i
+=+
∑
=1
1
i.e.
y
=
x
+ a ... (2)
Thus, the variance of the new observations
2
2
σ
=
2
1
1
( )
n
i
i
y y
n
=
−
∑
=
2
)(
1
1
axax
n
n
i
i
−−+
∑
=
[Using (1) and (2)]
=
2
1
1
( )
n
i
i
x x
n
=
−
∑
=
2
1
σ
Thus, the variance of the new observations is same as that of the original observations.
A
Note We may note that adding (or subtracting) a positive number to (or from)
each observation of a group does not affect the variance.
Example 19 The mean and standard deviation of 100 observations were calculated as
40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one
observation. What are the correct mean and standard deviation?
Solution Given that number of observations (n) = 100
Incorrect mean (
x
) = 40,
Incorrect standard deviation (σ) = 5.1
We know that
∑
=
=
n
i
i
x
n
x
1
1
i.e.
100
1
1
40
100
i
i
x
=
=
∑
or
100
1
i
i
x
=
∑
= 4000
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STATISTICS 379
i.e. Incorrect sum of observations = 4000
Thus the correct sum of observations = Incorrect sum – 50 + 40
= 4000 – 50 + 40 = 3990
Hence Correct mean =
correct sum 3990
100 100
=
= 39.9
Also Standard deviation
σ
=
2
2
2
1 1
1 1
n n
i i
i i
x x
n
n
= =
−
∑ ∑
=
( )
2
1
2
1
xx
n
n
i
i
−
∑
=
i.e. 5.1 =
2
2
1
1
Incorrect (40)
100
n
i
i
x
=
× −
∑
or 26.01 =
2
1
1
Incorrect
100
n
i
i
x
=
×
∑
– 1600
Therefore Incorrect
2
1
n
i
i
x
=
∑
= 100 (26.01 + 1600) = 162601
Now Correct
2
1
n
i
i
x
=
∑
= Incorrect
∑
=
n
i
i
x
1
2
– (50)
2
+ (40)
2
= 162601 – 2500 + 1600 = 161701
Therefore Correct standard deviation
=
2
2
Correct
(Correct mean)
i
x
n
−
∑
=
2
161701
(39 9)
100
.
−
=
1617 01 1592 01
. .
−
=
25
= 5
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380 MATHEMATICS
Miscellaneous Exercise On Chapter 15
1. The mean and variance of eight observations are 9 and 9.25, respectively. If six
of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the
observations are 2, 4, 10, 12, 14. Find the remaining two observations.
3. The mean and standard deviation of six observations are 8 and 4, respectively. If
each observation is multiplied by 3, find the new mean and new standard deviation
of the resulting observations.
4. Given that
x
is the mean and σ
2
is the variance of n observations x
1
, x
2
, ...,x
n
.
Prove that the mean and variance of the observations ax
1
, ax
2
, ax
3
, ...., ax
n
are
a
x
and a
2
σ
2
, respectively, (a ≠ 0).
5. The mean and standard deviation of 20 observations are found to be 10 and 2,
respectively. On rechecking, it was found that an observation 8 was incorrect.
Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted. (ii) If it is replaced by 12.
6. The mean and standard deviation of marks obtained by 50 students of a class in
three subjects, Mathematics, Physics and Chemistry are given below:
Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard 12 15 20
deviation
Which of the three subjects shows the highest variability in marks and which
shows the lowest?
7. The mean and standard deviation of a group of 100 observations were found to
be 20 and 3, respectively. Later on it was found that three observations were
incorrect, which were recorded as 21, 21 and 18. Find the mean and standard
deviation if the incorrect observations are omitted.
Summary
®
Measures of dispersion Range, Quartile deviation, mean deviation, variance,
standard deviation are measures of dispersion.
Range = Maximum Value – Minimum Value
®
Mean deviation for ungrouped data
M
M.D. ( ) M.D. (M)
i i
x – x x –
x ,
n n
= =
∑ ∑
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STATISTICS 381
®
Mean deviation for grouped data
M.D.
N
M.D. M
M
N
where N( )
, ( )
,x
f x x f x
f
i i i i
i
= = =
∑ ∑
∑
®
Variance and standard deviation for ungrouped data
2 2
1
( )
i
x – x
n
σ
=
∑
,
2
1
( – )
i
x x
n
σ
=
∑
®
Variance and standard deviation of a discrete frequency distribution
( ) ( )
2 2
2
1 1
,
N N
i i i i
f x x f x x
σ σ
= − = −
∑ ∑
®
Variance and standard deviation of a continuous frequency distribution
( )
( )
2
2
2 2
1 1
, N
N N
i i i i i i
f x x f x f x
σ σ
= − = −
∑ ∑ ∑
®
Shortcut method to find variance and standard deviation.
( )
2
2
2 2
2
N
N
i i i i
h
f y f y
σ
= −
∑ ∑
,
( )
2
2
N
N
i i i i
h
f y f y
σ
= −
∑ ∑
,
where
A
i
i
x
y
h
−
=
®
Coefficient of variation (C.V.)
100, 0.
x
x
= × ≠
σ
For series with equal means, the series with lesser standard deviation is more consistent
or less scattered.
Historical Note
‘Statistics’ is derived from the Latin word ‘status’ which means a political
state. This suggests that statistics is as old as human civilisation. In the year 3050
B.C., perhaps the first census was held in Egypt. In India also, about 2000 years
ago, we had an efficient system of collecting administrative statistics, particularly,
during the regime of Chandra Gupta Maurya (324-300 B.C.). The system of
collecting data related to births and deaths is mentioned in Kautilya’s Arthshastra
(around 300 B.C.) A detailed account of administrative surveys conducted during
Akbar’s regime is given in Ain-I-Akbari written by Abul Fazl.
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382 MATHEMATICS
—
vv
vv
v
—
Captain John Graunt of London (1620-1674) is known as father of vital
statistics due to his studies on statistics of births and deaths. Jacob Bernoulli
(1654-1705) stated the Law of Large numbers in his book “Ars Conjectandi’,
published in 1713.
The theoretical development of statistics came during the mid seventeenth
century and continued after that with the introduction of theory of games and
chance (i.e., probability). Francis Galton (1822-1921), an Englishman, pioneered
the use of statistical methods, in the field of Biometry. Karl Pearson (1857-1936)
contributed a lot to the development of statistical studies with his discovery
of Chi square test and foundation of statistical laboratory in England (1911).
Sir Ronald A. Fisher (1890-1962), known as the Father of modern statistics,
applied it to various diversified fields such as Genetics, Biometry, Education,
Agriculture, etc.
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