Class X-STATISTICS- PPT.pptx new new new

MATHEMATICS
CLASS-X
TOPIC:STATISTICS

LINK FOR REFERENCE OF TEXTS
NCERT TEXT BOOK CHAPTER-14
NCERT EXEMPLAR BOOK CHAPTER-14
STATISTICS -X EXEMPLAR.pdf
STATISTICS -X NCERT.pdf

LEARNING OBJECTIVES:
To find mean for group data by direct method, assumed
mean method and step deviation method.
To learn to find the mode for grouped data.
To learn to calculate cumulative frequency of a class.
To find median for grouped data using formula.
To represent cumulative frequency distribution graphically
as cumulative frequency curve (ogive) of less than type and
more than type.
To apply the knowledge of ogives to find median of grouped
data graphically.

PREVIOUS KNOWLEDGE
Concept of Exclusive form of group(Continuous Form) and Inclusive
form of group (Discontinuous Form);
Finding Class size ;
Finding mid value of a class or Class Mark;
Conversion of a Discontinuous Class to Continuous.
Measure of central tendency.
1. Calculating Mean of Raw data & Ungrouped Data.
2. Finding median of ungrouped Data;
3. Finding mode of Ungrouped data;
Graphical representation of Continuous and Discontinuous data.
(Such as Bar Graphs, Histograms of uniform width and of varying
widths & Frequency polygons)

INTRODUCTION
In the previous year we learnt about the ungrouped data , its
presentation and central tendencies ,now this year we will proceed
to find the central tendencies of grouped data with various methods.
In this regard we will discuss about various methods of finding mean
of a grouped data such as Direct method, Assumed Mean
method , and Step Deviation method;
Various methods of finding median of a grouped data such as from
formula and from cumulative frequency curves or Ogives.
Finding mode of a grouped frequency distribution;
Finding missing frequency of a class when the median or mean of
that data is given.

Class X-STATISTICS- PPT.pptx new new new

Marks obtained ) 10 20 36 40 50 56 60 70 72 80 88 92 95
Number of students ) 1 1 3 4 3 2 4 4 1 1 2 3 1
Marks obtained Number of
students )
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

MEAN OF GROUPED DATA (DIRECT METHOD)
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 class-intervals of width, say 15. Remember that, while
allocating frequencies to each class-interval, students falling in any
upper class-limit would be considered in the next class, e.g., 4 students
who have obtained 40 marks would be considered in the class interval
40-55 and not in 25-40.With this convention in our mind, let us form a
grouped frequency distribution table.

Example-1. Find the mean marks of the following distribution by Direct method.
Now, for each class-interval, 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 mid-point. So the
mid-point (or class mark) of each class can be chosen to represent the observations
falling in the class.
Recall that we find the mid-point of a class (or its class mark) by finding the average
of its upper and lower limits.That is,
Class mark =
With reference toTable given above , for the class 10 -25, the class mark is ,
i.e.,17.5. Similarly, we can find the class marks of the remaining class intervals.We
put them inTable given below.These class marks serve as our xi’s. Now, in general,
for the ith class interval, we have the frequency fi corresponding to the class mark xi.
We can now proceed to compute the mean in the same manner as in Example 1
Class
interval.
10-25 25-40 40-55 55-70 70-85 85-100
Number of
students
2 3 7 6 6 6

Table.
So, the mean of the given data is given by
= = = 62.
This new method of finding Mean is called as Direct Method for grouped data.
NOTE :The difference in the two values first mean and second is because of
the mid-point assumption in second table,59.3 being the exact mean,while
62 an approximate mean.
Class interval Number of
students( )
Class mark( )
10-25 2 17.5 35
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 = 30

Example-2:Find the mean of the following data by Direct
method :
Solution:
.
Table: (2 marks)
Mean === 20.5 (1 mark )
Class Interval 0-10 10-20 20-30 30-40 40-50
Frequency 5 6 4 3 2

ACTIVITY BY STUDENTSTO FIND MEAN OF A
GROUPED DATA BY USING DIRECT METHOD.
STATISTICS.mp4

(SHORTCUT METHOD OR ASSUMED MEAN METHOD)
Sometimes when the numerical values of xi and fi are large, finding the product of xi and fi
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 fi’s, but we can change each xi 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 xi’s? Let us try this method.
The first step is to choose one among the xi’ s as the assumed mean, and denote it by‘a’.Also,
to further reduce our calculation work, we may take‘a’ to be that xi which lies in the centre of x1, x2 ,
. . ., xn. 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 di between a and each of the xi’s, that is, the
deviation of‘a’ from each of the xi’s .
i.e., di = xi – a = xi – 47.5
The third step is to find the product of di with the corresponding fi, and take the sum of all
the fi di‘s.The calculations are shown inTable given below.

Example-3. Find the mean marks of the following distribution by shortcut method or assumed
mean method.
Solution: From the table Let a = Assumed Mean= 47.5 ,= .
a = 47.5, ,
So, mean ( ) = a +
= 47.5 + = 47.5 + 14.5 = 62
Therefore, the mean marks obtained by the students is 62.
Class
interval.
10-25 25-40 40-55 55-70 70-85 85-100
Number of
students
2 3 7 6 6 6
Class interval. Number of
students()
Class mark) =
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

Activity 1 :
From the aboveTable if the students will be asked to find the mean by taking each of xi(i.e.,
17.5, 32.5, and so on) as‘a’.What do they observe?
They will find that the mean determined in each case is the same, i.e., 62.
Because as we find the difference of assumed mean a from each of xi and finally when we add the a in
the formula it balances the differences occurred at the first step.
DERIVATION OF FORMULA OF ASSUMED MEAN METHOD.
Proof: Let x1,x2,...,xn are observations with respective frequencies f1,f2,...,fn, respectively.Taking
deviations about an arbitrary point, 'a' we have
= - a, i = 1,2,3,...n.
Or, ,
i = 1,2,3,...n.
Or,
Or, =
Or, []= []
Or[]= = - a
Or, =a + (proved)
So, we can say that the value of the mean obtained does not depend on the choice of ‘a’

Example-4. Find the average cost of the following data by assumed
mean method.
Solution: Let, assumed mean (a) = 12
a = 12, (2 marks)
fi = 130,
fidi = 560
= 16.307
Therefore, average cost = 16.307 (1 mark)
Cost 2 - 6 6 - 10 10 - 14 14 - 18 18 - 22 Total
Frequency 1 9 21 47 52 130
Cost Frequency
fi
Midvalue
xi
di = xi - a fidi
2-6 1 4 - 8 - 8
6-10 9 8 - 4 - 36
10-14 21 12 0 0
14-18 47 16 4 188
18-22 52 20 8 416
Total fi = 130 60 fidi = 560
( ) = a +

Example:5-The data of 30 CORONA patients attending a hospital in a month are
given below. Find the average number of patients attending the hospital in a day,
using assumed mean method.
Solution: Let assumed mean (a) = 25
A = 25, , (table: 2marks)
So, mean ( ) = a +
= 25 + =25 + = 28.666= 28.67. (1 mark)
Number of CORONA
patients
0-10 10-20 20-30 30-40 40-50 50-60
Number of days 2 6 9 7 4 2
Class interval Midvalue
xi
fi di = xi - A fidi
0-10 5 2 -20 -40
10-20 15 6 -10 -60
20-30 25 9 0 0
30-40 35 7 10 70
40-50 45 4 20 80
50-60 55 2 30 60
Total

MEAN OF GROUPED DATA
BY STEP DEVIATION METHOD
Sometimes ,during the application of the short-cut method for finding
mean , the deviations are divisible by a common number h (class size of
each class interval.). In such case the calculation is reduced to great
extent by taking
= ; i=1,2,3,... ,n
or, = a + h , (i =1,2,3,..., n)
or, = , (multiplied on both sides)
or,
or, =
or, = a + h (proved)
Following algorithm may be used to find the Mean by Step Deviation
method:

ALGORITHM FOR STEP DEVIATION METHOD
STEP-I Obtain the frequency distribution and prepare the frequency table in
such a way that its first column consists of the values of the variable and the
second column corresponding frequencies.
STEP- II Choose a number‘a' (generally known as the assumed mean)
and take deviations di= xi-a about a .Write these deviations against the
corresponding frequencies in the third column.
STEP- III Choose a number h, generally common factor for all di's in III column,
divide deviations di by h to get ui. Write these ui's against the corresponding di's
in the IV column.
STEP-IV Multiply the frequencies in the II column with the
corresponding ui's in IV column to prepareV column of fiui.
STEP -V Find the sum of all entries inV column to obtain and the sum
of all frequencies in II column to obtain N=
STEPVI Use the formula: = a + h

Following Example-6 will illustrate the above algorithm to find mean :
Solution: Let assumed mean = 47.5
Therefore using the formula:
= + h
Now substituting the values of a, h, and from theTable we get
= 47.5 + 15 =47.5+14.5 = 62 .
so, the mean marks obtained by a student is 62.
Class
interval.
10-25 25-40 40-55 55-70 70-85 85-100
Number of
students
2 3 7 6 6 6
Class interval
10-25 2 17.5 -30 -2 -4
25-40 3 32.5 -15 -1 -3
40-55 7 47.5 0 0 0
55-70 6 62.5 15 1 6
70-85 6 77.5 30 2 12
85-100 6 92.5 45 3 18
Total

Example :7 The table below gives the percentage distribution of female teachers in
the primary schools of rural areas of various states and union territories (U.T.) of India. Find
the mean percentage of female teachers by all the three methods discussed in this section.
Solution: Let us solve this problem by using all the three methods.
From the table above, we obtain , , ,.
Using the direct method = = = 39.71
Assuming the mean from the class mark as a=50,
then using the assumed mean method = = 50 + = 39.71
Using the step deviation method = a + = 50 + = 39.71
Percentage of female teachers 15-25 25-35 35-45 45-55 55-65 65-75 75-85
Number of states/U.T. 6 11 7 4 4 2 1
%of female
teachers
Number of
states/U.T.()
15-25 6 20 -30 -3 120 -180 -18
25-35 11 30 -20 -2 330 -220 -22
35-45 7 40 -10 -1 280 -70 -7
45-55 4 50 0 0 200 0 0
55-65 4 60 10 1 240 40 4
65-75 2 70 20 2 140 40 4
75-85 1 80 30 3 80 30 3
Total 35 1390 -360 -36

REMARK
The result obtained by all the three methods is the same. So the
choice of method to be used depends on the numerical values of xi
and fi. If xi and fi are sufficiently small, then the direct method is an
appropriate choice. If xi and fi are numerically large numbers, then
we can go for the assumed mean method or step-deviation method.
If the class sizes are unequal, and xi are large numerically, we can
still apply the step-deviation method by taking h to be a suitable
divisor of all the di’s.

Example-8:The mean of the following distribution is 18.The frequency f in
the class interval 19-21 is missing. Determine f (3 Marks)
Solution :
Let us take assumed mean (a) = 18. Here h = 2 (2 Marks)
Using the step deviation method = a + = 18 +2
or, 18 = 18 +2 ; = 18(given)
f= 8 (1Mark)
Hence, the frequency of the class interval 19-21 is 8.
Class interval 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Frequency 3 6 9 13 f 5 4
Class interval Mid-point Frequency ( =
11-13 12 3 -3 -9
13-15 14 6 -2 -12
15-17 16 9 -1 -9
17-19 18 13 0 0
19-21 20 F 1 F
21-23 22 5 2 10
23-25 24 4 3 12
= 40+ f

MODE OF GROUPED DATA.
A mode is that value among the observations which occurs most often, that is, the value of the
observation having the maximum frequency.
Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of
obtaining a mode of grouped data. It is possible that more than one value may have the same
maximum frequency. In such situations, the data is said to be multimodal.Though grouped data
can also be multimodal, we shall restrict ourselves to problems having a single mode only.
In a grouped frequency distribution, it is not possible to determine the mode by looking at the
frequencies. Here, we can only locate a class with the maximum frequency, called the modal
class.The mode is a value inside the modal class, and is given by the formula:
Mode = l+
where l= lower limit of the modal class,
h= size of the class interval (assuming all class sizes to be equal),
f1= frequency of the modal class,
f0= frequency of the class preceding the modal class,
f2= frequency of the class succeeding the modal class.
Let us consider the following examples to illustrate the use of this formula.

Example 1 : A survey conducted on 20 households in a locality by a
group of students resulted in the following frequency table for the number
of family members in a house hold.Findthe mode of this data.
Solution :Here the maximum class frequency is 8, and the class
corresponding to this frequency is (3 – 5).
So, the modal class is 3 – 5.Now
lower limit (l ) of modal class = 3, class size (h) = 2
frequency (f1) of the modal class = 8,
frequency (f0 ) of class preceding the modal class = 7,
frequency (f2 ) of class succeeding the modal class = 2.
Now, let us substitute these values in the formula:
Mode = l +
Mode = 3 + = 3+ = 3.286.
Therefore, the mode of the data above is 3.286.
Family size 1-3 3-5 5-7 7- 9 9- 11
Number of families 7 8 2 2 1

Example 2.The marks distribution of 30 students in a mathematics
examination are given below . Find the mode of this data. (2Marks)
Solution: From the aboveTable ,since the maximum number of students
(i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 -
55.Therefore, the lower limit ( l ) of the modal class = 40,
the class size (h) = 15,
the frequency (f1 ) of modal class = 7,
the frequency ( f0 ) of the class preceding the modal class = 3,
the frequency ( f2) of the class succeeding the modal class = 6.
Now, let us substitute these values in the formula: (1 Mark)
Mode = l+
Mode = 40 + = 40+ 12= 52.
So, the mode marks is 52 (1 Mark)
Class interval. 10-25 25-40 40-55 55-70 70-85 85-100
Number of
students
2 3 7 6 6 6

Example 3.The frequency distribution table of agricultural holdings
in a village is given below. Find the modal agricultural holdings of the
village.
Solution :Here the maximum class frequency is 80, and the class
corresponding to this frequency is 5-7.
So, the modal class is (5-7).
l ( lower limit of modal class) = 5
f1(frequency of the modal class) = 80
f0(frequency of the class preceding the modal class) = 45
f2(frequency of the class succeeding the modal class) = 55
h (class size) = 2
Mode = l +
Mode = 5 + = 5 + = 5+ 1.2 =6.2.
Hence, the modal agricultural holdings of the village is 6.2 hectares.
Area of land (in hectares) 1-3 3-5 5-7 7-9 9-11 11-13
Number of families 20 45 80 55 40 12

MEDIAN OF GROUPED DATA:
Median:The median is a measure of central tendency which gives the
value of the middle-most observation in the data. It is the value of the
variable such that the number of observations above it is equal to the
number of observations below it. Recall that for finding the median of
ungrouped data, we first arrange the data values of the observations in
ascending order.
Then, if n is odd, the median is the observation.
And, if n is even, then
the median will be the average of the th and the(thobservations.
Let us first recall how we found the median for ungrouped data through
the following

Example 1: Suppose, we have to find the median of the following
data, which gives the marks, out of 50, obtained by 100 students in
a test :
First, we arrange the marks in ascending order and prepare a frequency table as
follows :
Here n = 100, which is even.The median will be the average of the th and the
(th observations, i.e., the 50th and 51st observations.To find these observations,
we proceed as follows:
Marks
obtained
20 29 28 33 42 38 43 25
Number of
students
6 28 24 15 2 4 1 20
Marks obtained Number of studets(Frequency)
20 6
25 20
28 24
29 28
33 15
38 4
42 2
43 1
Total 100

Now we add another column depicting this information to the
frequency table above and name it as cumulative frequency column.
From the table above, we see that:
50th observation is 28 (Why?)
(Because out of 100 observations 50th
and 51st
are the middle places)
51st observation is 29
So, Median = = 28.5.
The median marks 28.5 conveys the information that about 50% students obtained
marks less than 28.5 and another 50% students obtained marks more than 28.5.
Marks obtained Number of students Cumulative frequency (cf)
20 6 6
25 20 26
28 24 50
29 28 78
33 15 93
38 4 97
42 2 99
43 1 100

Now, let us see how to obtain the median of grouped data, through the
following situation.
Example 2: Consider a grouped frequency distribution of marks obtained, out
of 100, by 53 students, in a certain examination, as follows:
From the table above, try to answer the following questions:
How many students have scored marks less than 10? The answer is clearly 5.
How many students have scored less than 20 marks? Observe that the number
of students who have scored less than 20 include the number of students who have
scored marks from 0 - 10 as well as the number of students who have scored marks
from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, i.e., 8.
We say that the cumulative frequency of the class 10 -20 is 8.
Similarly, we can compute the cumulative frequencies of the other classes, i.e.,
the number of students with marks less than 30, less than 40, . . ., less than 100.
We give them inTable given below:
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Number of
students
5 3 4 3 3 4 7 9 7 8

Cumulative frequency distributionTable of less than type:
Marks obtained Number of students
(Cumulative freqency)
Less than 10 5
Less than 20 5+3=8
Less than 30 8+4=12
Less than 40 12+3= 15
Less than 50 15+3 =18
Less than 60 18+4 = 22
Less than 70 22+7=29
Less than 80 29+ 9= 38
Less than 100 45+8=53

We can similarly make the table for the number of students with scores, more than or
equal to 0, more than or equal to 10, more than or equal to 20, and so on. From the
Frequency DistributionTable , we observe that all 53 students have scored marks
more than or equal to 0. Since there are 5 students scoring marks in the interval 0 - 10, this
means that there are 53 – 5 = 48 students getting more than or equal to 10 marks. Continuing
in the same manner, we get the number of students scoring 20 or above as 48 – 3 = 45, 30 or
above as 45 – 4 = 41, and so on, as shown inTable.
Cumulative Frequency DistributionTable of the more than type.
Marks Number of students (Cumulative freqency)
More than or equal to 0 53
More than or equal to 10 53-5= 48
More than or equal to 20 48-3=45
More than or equal to 30 45-4 =41

The table above is called a cumulative frequency distribution of the more
than type. Here 0, 10, 20, . . ., 90 give the lower limits of the respective
class intervals.
Now, to find the median of grouped data, we can make use of any of these
cumulative frequency distributions.
By combining QuestionTable & Less than typeTable we can get aTable
given below.Marks Number of students(f) Cumulative frequency(cf)
0-10 5 5
10-20 3 8
20-30 4 12
30-40 3 15
40-50 3 18
50-60 4 22
60-70 7 29
70-80 9 38
80-90 7 45
90-100 8 53

To find this class, we find the cumulative frequencies of all the classes and .We now
locate the class whose cumulative frequency is greater than (and nearest to) .This is called
the median class. In the distribution above, n = 53. So, = 26.5.
Now 60 – 70 is the class whose cumulative frequency 29 is greater than (and nearest
to) = 26.5 .Therefore, 60 – 70 is the median class.
After finding the median class, we use the following formula for calculating the median.
Median = + , Where = lower limit of the median class.
n = number of observations.
cf= cumulative frequency of class preceding the median class.
f= frequency of the median class.
h= class size(assuming class size to be equal)
Substituting the values = 26.5, l = 60, cf = 22, f = 7, h = 10 in the formula above, we
get
Median = 60 +
=60+
=66.4
So, about half the students have scored marks less than 66.4, and the other half have
scored marks more than 66.4.

Sometime the median of a data is given but one or two of the frequencies
from the table are found missing . In such cases how to find the missing
frequencies by using the formula for median lets learn.
Example 3 : The median of the following data is 525. Find the values of x and y, if the
total frequency is 100.
Class interval Frequency
0 - 100 2
100 - 200 5
200 - 300 x
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 y
700 - 800 9
800 - 900 7
900 - 1000 4

Solution :
Class interval Frequency Cumulative frequency
0 - 100 2 2
100 - 200 5 7
200 - 300 x 7 + x
300 - 400 12 19 + x
400 - 500 17 36 + x
500 - 600 20 56 + x
600 - 700 y 56 + x + y
700 - 800 9 65 + x + y
800 - 900 7 72 + x + y
900 - 1000 4 76 + x + y
It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24 (1)
The median is 525, which lies in the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula : Median = 𝑙 + ൤
𝑛
2
−𝑐𝑓
𝑓
൨
× ℎ
525 = 500 +ቂ
50−36−𝑥
20
ቃ× 100
i.e., 525 – 500 = (14 – x) × 5
i.e., 25 = 70 – 5x
i.e., 5x = 70 – 25 = 45
So, x = 9
Therefore, from (1), we get 9 + y = 24
i.e., y = 15

A study on the Nature of the Central tendencies
Now, that we have studied about all the three measures of central tendency, let us discuss
which measure would be best suited for a particular requirement.
 The mean is the most frequently used measure of central tendency because it takes
into account all the observations, and lies between the extremes, i.e., the largest and
the smallest observations of the entire data. It also enables us to compare two or
more distributions.
For example, by comparing the average (mean) results of students of different
schools of a particular examination, we can conclude which school has a better
performance.
However, extreme values in the data affect the mean. For example, the mean of
classes having frequencies more or less the same is a good representative of the data.
But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20,
21, 18, then the mean will certainly not reflect the way the data behaves. So, in such
cases, the mean is not a good representative of the data.
 In problems where individual observations are not important, and we wish to find
out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical
productivity rate of workers, average wage in a country, etc. These are situations
where extreme values may be there. So, rather than the mean, we take the median as
a better measure of central tendency.
In situations which require establishing the most frequent value or most popular
item, the mode is the best choice, e.g., to find the most popular T.V. programme
being watched, the consumer item in greatest demand, the colour of the vehicle used
by most of the people, etc.
There is an empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean

GRAPHICAL REPRESENTATION OF CUMULATIVE FREQUENCY
DISTRIBUTION
In previous class IX we have studied simple frequency curves which are drawn by plotting
frequencies against class marks of the class intervals.
In this class we will learn the technique of drawing Cumulative Frequencies plotted against
corresponding upper or lower limits of the classes depending upon the manner in which the
series has been cumulated.
Such type of curves are called Ogives.
An Ogive is a graphical representation of cumulative frequency distribution, in the form of a
smooth continuous, free hand curve, which is either ever rising upwards or ever falling
downwards.
As per the data is cumulated there are two types of Ogives are possible according to it i.e
less than type and more than type.
Less than type Ogive is drawn by taking upper limits with corresponding increasing
cumulative frequencies ,while More than type Ogive is drawn by taking lower limits with
corresponding decreasing cumulative frequencies .
Median can be found from these graphs by finding corresponding value of observation
from class interval axis by joining middle place (n/2)from the cumulative
frequency axis.
Lets learn through an ACTIVITY:
ACTIVITY FOR OGIVE.mp4

CUMULATIVE FREQUENCY DISTRIBUTIONTABLE OF LESS
THANTYPE AND GRAPH:
Marks obtained Number of students
(Cumulative frequency)
Less than 10 5
Less than 20 5+3=8
Less than 30 8+4=12
Less than 40 12+3= 15
Less than 50 15+3 =18
Less than 60 18+4 = 22
Less than 80 29+ 9= 38
Less than 100 45+8=53

CUMULATIVE FREQUENCY DISTRIBUTIONTABLE OFTHE
MORE THAN TYPE AND GRAPH.
Marks Number of students
(Cumulative freqency)
More than or equal to 0 53
More than or equal to 10 53-5= 48
More than or equal to 30 45-4 =41

Example4.A student draws a cumulative frequency curve for the marks obtained
by 40 students of a class, as shown below.Find the median obtained by the students
of the class.
Sol: Here n = 40 ,
𝑛
2
= 20
Take a point A(0, 20) on y-axis and draw AP || x-axis, meeting the curve at P.
Draw PM  x-axis, intersecting the x-axis at M.Then, OM = 55 units and hence,
median = 55.

Finding Median from Double Ogives
When the both Ogives are given or drawn on a single graph then the point of intersection of both the
Ogives shows the Median dirtectly. In such case we need to find the corresponding observation value of
the class from the graph with respect to the point of intersection of the Less than and More than Ogives.

Example-5:The table given below shows the frequency distribution of the scores
obtained by 200candidates in a BBA entrance examination. Draw cumulative
frequency curves by using
(i)'less than' series
(ii)'More than' series
(iii)Hence, find the median
Solution:
(i)Less than' series :Prepare the 'less than' series as under (1 mark)
(ii) 'More than' series:Prepare the 'more than' series as under: (1 mark)
Score Number of candidates
200 - 250 30
250 - 300 15
300 - 350 45
350 - 400 20
400 - 450 25
450 - 500 40
500 - 550 10
550 - 600 15
Score Number of Candidates
Less than 200 0
Less than 250 30
Less than 300 45
Less than 350 90
Less than 400 110
Less than 450 135
Less than 500 175
Less than 550 185
Less than 600 200
Score Number of candidates
More than 200 200
More than 250 170
More than 300 155
More than 350 110
More than 400 90
More than 450 65
More than 500 25
More than 550 15
More than 600 0

THE DOUBLE OGIVE FROMTHE ABOVETWOTABLES:
(1.5 marks)
The two curves intersect at the point P(375, 100)
Hence, median = 375 (1/2 mark)

Example-5
Find the median from the following ogive curve. Also Identify the "Less than" and "More than" curves.
Question figure: Solution Figure:
Answer: 17.5 is the median of the given graph of less than ogive and more than ogive.

SUMMARY
In this chapter we have studied the following points :
The mean of grouped data can be found by
(i) Direct method : =
(ii)Assumed Mean Method : ( ) = a +
(iii) Step Deviation method: = a +
•With the assumption that the frequency of a class is centered at its mid point, called its class
mark.
•The mode of the group data can be found by using the formula :
Mode = l + , where symbols have their usual meaning .
•The cumulative frequency of a class is the frequency obtained by adding the frequencies of all
classes preceding the given class.
•The median of a grouped data is found by the formula:
Median = + , where symbols have their usual meaning.
•Representing a cumulative frequency distribution graphically as a cumulative curve , or an
ogive of less than type and of more than type.
•The median grouped data can be obtained graphically as the x coordinate of the point of
intersection of the two ogives for a data.

CONCEPT MAPPING OFTHETOPIC :
STATISTICS

WORKSHEETS LINK FORTEST OF DIFFERENT
LEVEL
BASIC LEVEL
STANDARD LEVEL
ADVANCED LEVEL

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