Page 1
ST A TISTICS 171
13
13.1 Introduction
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives.
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
x
1
, x
2
,. . ., x
n
are observations with respective frequencies f
1
, f
2
, . . ., f
n
, then this
means observation x
1
occurs f
1
times, x
2
occurs f
2
times, and so on.
Now, the sum of the values of all the observations = f
1
x
1
+ f
2
x
2
+ . . . + f
n
x
n
, and
the number of observations = f
1
+ f
2
+ . . . + f
n
.
So, the mean x of the data is given by
x
=
11 2 2
12
nn
n
fx fx fx
ff f
?? ?
?? ?
?
?
Recall that we can write this in short form by using the Greek letter ? (capital
sigma) which means summation. That is,
ST A TISTICS
Rationalised 202324
Page 2
ST A TISTICS 171
13
13.1 Introduction
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives.
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
x
1
, x
2
,. . ., x
n
are observations with respective frequencies f
1
, f
2
, . . ., f
n
, then this
means observation x
1
occurs f
1
times, x
2
occurs f
2
times, and so on.
Now, the sum of the values of all the observations = f
1
x
1
+ f
2
x
2
+ . . . + f
n
x
n
, and
the number of observations = f
1
+ f
2
+ . . . + f
n
.
So, the mean x of the data is given by
x
=
11 2 2
12
nn
n
fx fx fx
ff f
?? ?
?? ?
?
?
Recall that we can write this in short form by using the Greek letter ? (capital
sigma) which means summation. That is,
ST A TISTICS
Rationalised 202324
172 MATHEMA TICS
x
=
1
1
n
ii
i
n
i
i
fx
f
?
?
?
?
which, more briefly, is written as x =
?
?
ii
i
fx
f
, if it is understood that i varies from
1 to n.
Let us apply this formula to find the mean in the following example.
Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students.
Marks obtained 10 20 36 40 50 56 60 70 72 80 88 92 95
(x
i
)
Number of 113 4324 4112 3 1
students ( f
i
)
Solution: Recall that to find the mean marks, we require the product of each x
i
with
the corresponding frequency f
i
. So, let us put them in a column as shown in Table 13.1.
Table 13.1
Marks obtained (x
i
) Number of students ( f
i
) f
i
x
i
10 1 10
20 1 20
. 36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Total ?f
i
= 30 ?f
i
x
i
= 1779
Rationalised 202324
Page 3
ST A TISTICS 171
13
13.1 Introduction
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives.
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
x
1
, x
2
,. . ., x
n
are observations with respective frequencies f
1
, f
2
, . . ., f
n
, then this
means observation x
1
occurs f
1
times, x
2
occurs f
2
times, and so on.
Now, the sum of the values of all the observations = f
1
x
1
+ f
2
x
2
+ . . . + f
n
x
n
, and
the number of observations = f
1
+ f
2
+ . . . + f
n
.
So, the mean x of the data is given by
x
=
11 2 2
12
nn
n
fx fx fx
ff f
?? ?
?? ?
?
?
Recall that we can write this in short form by using the Greek letter ? (capital
sigma) which means summation. That is,
ST A TISTICS
Rationalised 202324
172 MATHEMA TICS
x
=
1
1
n
ii
i
n
i
i
fx
f
?
?
?
?
which, more briefly, is written as x =
?
?
ii
i
fx
f
, if it is understood that i varies from
1 to n.
Let us apply this formula to find the mean in the following example.
Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students.
Marks obtained 10 20 36 40 50 56 60 70 72 80 88 92 95
(x
i
)
Number of 113 4324 4112 3 1
students ( f
i
)
Solution: Recall that to find the mean marks, we require the product of each x
i
with
the corresponding frequency f
i
. So, let us put them in a column as shown in Table 13.1.
Table 13.1
Marks obtained (x
i
) Number of students ( f
i
) f
i
x
i
10 1 10
20 1 20
. 36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Total ?f
i
= 30 ?f
i
x
i
= 1779
Rationalised 202324
ST A TISTICS 173
Now,
?
?
?
i i
i
f x
x
f
=
1779
30
= 59.3
Therefore, the mean marks obtained is 59.3.
In most of our real life situations, data is usually so large that to make a meaningful
study it needs to be condensed as grouped data. So, we need to convert given ungrouped
data into grouped data and devise some method to find its mean.
Let us convert the ungrouped data of Example 1 into grouped data by forming
classintervals of width, say 15. Remember that, while allocating frequencies to each
classinterval, students falling in any upper classlimit would be considered in the next
class, e.g., 4 students who have obtained 40 marks would be considered in the class
interval 4055 and not in 2540. With this convention in our mind, let us form a grouped
frequency distribution table (see Table 13.2).
Table 13.2
Class interval 10  25 25  40 40  55 55  70 70  85 85  100
Number of students 2 3 7 6 6 6
Now, for each classinterval, we require a point which would serve as the
representative of the whole class. It is assumed that the frequency of each class
interval is centred around its midpoint. So the midpoint (or class mark) of each
class can be chosen to represent the observations falling in the class. Recall that we
find the midpoint of a class (or its class mark) by finding the average of its upper and
lower limits. That is,
Class mark =
Upper class limit + Lower class limit
2
With reference to Table 13.2, for the class 1025, the class mark is
10 25
2
+
, i.e.,
17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 13.3. These class marks serve as our x
i
’s. Now, in general, for the ith
class interval, we have the frequency f
i
corresponding to the class mark x
i
. We can
now proceed to compute the mean in the same manner as in Example 1.
Rationalised 202324
Page 4
ST A TISTICS 171
13
13.1 Introduction
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives.
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
x
1
, x
2
,. . ., x
n
are observations with respective frequencies f
1
, f
2
, . . ., f
n
, then this
means observation x
1
occurs f
1
times, x
2
occurs f
2
times, and so on.
Now, the sum of the values of all the observations = f
1
x
1
+ f
2
x
2
+ . . . + f
n
x
n
, and
the number of observations = f
1
+ f
2
+ . . . + f
n
.
So, the mean x of the data is given by
x
=
11 2 2
12
nn
n
fx fx fx
ff f
?? ?
?? ?
?
?
Recall that we can write this in short form by using the Greek letter ? (capital
sigma) which means summation. That is,
ST A TISTICS
Rationalised 202324
172 MATHEMA TICS
x
=
1
1
n
ii
i
n
i
i
fx
f
?
?
?
?
which, more briefly, is written as x =
?
?
ii
i
fx
f
, if it is understood that i varies from
1 to n.
Let us apply this formula to find the mean in the following example.
Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students.
Marks obtained 10 20 36 40 50 56 60 70 72 80 88 92 95
(x
i
)
Number of 113 4324 4112 3 1
students ( f
i
)
Solution: Recall that to find the mean marks, we require the product of each x
i
with
the corresponding frequency f
i
. So, let us put them in a column as shown in Table 13.1.
Table 13.1
Marks obtained (x
i
) Number of students ( f
i
) f
i
x
i
10 1 10
20 1 20
. 36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Total ?f
i
= 30 ?f
i
x
i
= 1779
Rationalised 202324
ST A TISTICS 173
Now,
?
?
?
i i
i
f x
x
f
=
1779
30
= 59.3
Therefore, the mean marks obtained is 59.3.
In most of our real life situations, data is usually so large that to make a meaningful
study it needs to be condensed as grouped data. So, we need to convert given ungrouped
data into grouped data and devise some method to find its mean.
Let us convert the ungrouped data of Example 1 into grouped data by forming
classintervals of width, say 15. Remember that, while allocating frequencies to each
classinterval, students falling in any upper classlimit would be considered in the next
class, e.g., 4 students who have obtained 40 marks would be considered in the class
interval 4055 and not in 2540. With this convention in our mind, let us form a grouped
frequency distribution table (see Table 13.2).
Table 13.2
Class interval 10  25 25  40 40  55 55  70 70  85 85  100
Number of students 2 3 7 6 6 6
Now, for each classinterval, we require a point which would serve as the
representative of the whole class. It is assumed that the frequency of each class
interval is centred around its midpoint. So the midpoint (or class mark) of each
class can be chosen to represent the observations falling in the class. Recall that we
find the midpoint of a class (or its class mark) by finding the average of its upper and
lower limits. That is,
Class mark =
Upper class limit + Lower class limit
2
With reference to Table 13.2, for the class 1025, the class mark is
10 25
2
+
, i.e.,
17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 13.3. These class marks serve as our x
i
’s. Now, in general, for the ith
class interval, we have the frequency f
i
corresponding to the class mark x
i
. We can
now proceed to compute the mean in the same manner as in Example 1.
Rationalised 202324
174 MATHEMA TICS
Table 13.3
Class interval Number of students ( f
i
) Class mark (x
i
) f
i
x
i
10  25 2 17.5 35.0
25  40 3 32.5 97.5
40  55 7 47.5 332.5
55  70 6 62.5 375.0
70  85 6 77.5 465.0
85  100 6 92.5 555.0
Total S f
i
= 30 S f
i
x
i
= 1860.0
The sum of the values in the last column gives us S f
i
x
i
. So, the mean x of the
given data is given by
x
=
1860.0
62
30
i i
i
f x
f
S
= =
S
This new method of finding the mean is known as the Direct Method.
We observe that Tables 13.1 and 13.3 are using the same data and employing the
same formula for the calculation of the mean but the results obtained are different.
Can you think why this is so, and which one is more accurate? The difference in the
two values is because of the midpoint assumption in Table 13.3, 59.3 being the exact
mean, while 62 an approximate mean.
Sometimes when the numerical values of x
i
and f
i
are large, finding the product
of x
i
and f
i
becomes tedious and time consuming. So, for such situations, let us think of
a method of reducing these calculations.
We can do nothing with the f
i
’s, but we can change each x
i
to a smaller number
so that our calculations become easy. How do we do this? What about subtracting a
fixed number from each of these x
i
’s? Let us try this method.
The first step is to choose one among the x
i
’s as the assumed mean, and denote
it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that x
i
which lies in the centre of x
1
, x
2
, . . ., x
n
. So, we can choose a = 47.5 or a = 62.5. Let
us choose a = 47.5.
The next step is to find the difference d
i
between a and each of the x
i
’s, that is,
the deviation of ‘a’ from each of the x
i
’s.
i.e., d
i
= x
i
– a = x
i
– 47.5
Rationalised 202324
Page 5
ST A TISTICS 171
13
13.1 Introduction
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives.
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
x
1
, x
2
,. . ., x
n
are observations with respective frequencies f
1
, f
2
, . . ., f
n
, then this
means observation x
1
occurs f
1
times, x
2
occurs f
2
times, and so on.
Now, the sum of the values of all the observations = f
1
x
1
+ f
2
x
2
+ . . . + f
n
x
n
, and
the number of observations = f
1
+ f
2
+ . . . + f
n
.
So, the mean x of the data is given by
x
=
11 2 2
12
nn
n
fx fx fx
ff f
?? ?
?? ?
?
?
Recall that we can write this in short form by using the Greek letter ? (capital
sigma) which means summation. That is,
ST A TISTICS
Rationalised 202324
172 MATHEMA TICS
x
=
1
1
n
ii
i
n
i
i
fx
f
?
?
?
?
which, more briefly, is written as x =
?
?
ii
i
fx
f
, if it is understood that i varies from
1 to n.
Let us apply this formula to find the mean in the following example.
Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students.
Marks obtained 10 20 36 40 50 56 60 70 72 80 88 92 95
(x
i
)
Number of 113 4324 4112 3 1
students ( f
i
)
Solution: Recall that to find the mean marks, we require the product of each x
i
with
the corresponding frequency f
i
. So, let us put them in a column as shown in Table 13.1.
Table 13.1
Marks obtained (x
i
) Number of students ( f
i
) f
i
x
i
10 1 10
20 1 20
. 36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Total ?f
i
= 30 ?f
i
x
i
= 1779
Rationalised 202324
ST A TISTICS 173
Now,
?
?
?
i i
i
f x
x
f
=
1779
30
= 59.3
Therefore, the mean marks obtained is 59.3.
In most of our real life situations, data is usually so large that to make a meaningful
study it needs to be condensed as grouped data. So, we need to convert given ungrouped
data into grouped data and devise some method to find its mean.
Let us convert the ungrouped data of Example 1 into grouped data by forming
classintervals of width, say 15. Remember that, while allocating frequencies to each
classinterval, students falling in any upper classlimit would be considered in the next
class, e.g., 4 students who have obtained 40 marks would be considered in the class
interval 4055 and not in 2540. With this convention in our mind, let us form a grouped
frequency distribution table (see Table 13.2).
Table 13.2
Class interval 10  25 25  40 40  55 55  70 70  85 85  100
Number of students 2 3 7 6 6 6
Now, for each classinterval, we require a point which would serve as the
representative of the whole class. It is assumed that the frequency of each class
interval is centred around its midpoint. So the midpoint (or class mark) of each
class can be chosen to represent the observations falling in the class. Recall that we
find the midpoint of a class (or its class mark) by finding the average of its upper and
lower limits. That is,
Class mark =
Upper class limit + Lower class limit
2
With reference to Table 13.2, for the class 1025, the class mark is
10 25
2
+
, i.e.,
17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 13.3. These class marks serve as our x
i
’s. Now, in general, for the ith
class interval, we have the frequency f
i
corresponding to the class mark x
i
. We can
now proceed to compute the mean in the same manner as in Example 1.
Rationalised 202324
174 MATHEMA TICS
Table 13.3
Class interval Number of students ( f
i
) Class mark (x
i
) f
i
x
i
10  25 2 17.5 35.0
25  40 3 32.5 97.5
40  55 7 47.5 332.5
55  70 6 62.5 375.0
70  85 6 77.5 465.0
85  100 6 92.5 555.0
Total S f
i
= 30 S f
i
x
i
= 1860.0
The sum of the values in the last column gives us S f
i
x
i
. So, the mean x of the
given data is given by
x
=
1860.0
62
30
i i
i
f x
f
S
= =
S
This new method of finding the mean is known as the Direct Method.
We observe that Tables 13.1 and 13.3 are using the same data and employing the
same formula for the calculation of the mean but the results obtained are different.
Can you think why this is so, and which one is more accurate? The difference in the
two values is because of the midpoint assumption in Table 13.3, 59.3 being the exact
mean, while 62 an approximate mean.
Sometimes when the numerical values of x
i
and f
i
are large, finding the product
of x
i
and f
i
becomes tedious and time consuming. So, for such situations, let us think of
a method of reducing these calculations.
We can do nothing with the f
i
’s, but we can change each x
i
to a smaller number
so that our calculations become easy. How do we do this? What about subtracting a
fixed number from each of these x
i
’s? Let us try this method.
The first step is to choose one among the x
i
’s as the assumed mean, and denote
it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that x
i
which lies in the centre of x
1
, x
2
, . . ., x
n
. So, we can choose a = 47.5 or a = 62.5. Let
us choose a = 47.5.
The next step is to find the difference d
i
between a and each of the x
i
’s, that is,
the deviation of ‘a’ from each of the x
i
’s.
i.e., d
i
= x
i
– a = x
i
– 47.5
Rationalised 202324
ST A TISTICS 175
The third step is to find the product of d
i
with the corresponding f
i
, and take the sum
of all the f
i
d
i
’s. The calculations are shown in Table 13.4.
Table 13.4
Class interval Number of Class mark d
i
= x
i
– 47.5 f
i
d
i
students (f
i
) (x
i
)
10  25 2 17.5 –30 –60
25  40 3 32.5 –15 –45
40  55 7 47.5 0 0
55  70 6 62.5 15 90
70  85 6 77.5 30 180
85  100 6 92.5 45 270
Total Sf
i
= 30 Sf
i
d
i
= 435
So, from Table 13.4, the mean of the deviations, d =
i i
i
f d
f
S
S
.
Now, let us find the relation between d and x .
Since in obtaining d
i
, we subtracted ‘a’ from each x
i
, so, in order to get the mean
x , we need to add ‘a’ to d . This can be explained mathematically as:
Mean of deviations, d =
i i
i
f d
f
S
S
So,
d
=
( )
i i
i
f x a
f
S 
S
=
i i i
i i
f x f a
f f
S S

S S
=
i
i
f
x a
f
S

S
= x a 
So,
x
= a + d
i.e.,
x
=
i i
i
f d
a
f
S
+
S
Rationalised 202324
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