5.3 Mean,
Median and Mode for ungrouped data:
You must
have heard your teachers saying that the average number of students in a class
in their school is 45. What does it mean?
Normally a
school has several sections for a particular standard (Ex: Sections A, B and C
in 8^{th} standard). Most of the times,all the
sections in a school will not have same number of students.
Class A Class
B Class C
5.3 Example 1: In
the above picture ‘A’ section has 47, ‘B’ section has 42 and ‘C’ section has 46
students. If you add the number of students of all these sections you will get
135(47+42+46). If you divide this by 3 (number of sections), you get
45(=135/3). This number (=45) is what teachers mean by average number of
students in a class in their school. Similarly you can arrive at the average
number of students in a class for the entire school (considering all sections
of all standards). If there is only one section and the section has 40
students, then average and class strength are one and the same.
We have seen in statistics
that the individual figures (47,42,46) are called
scores and the number 3 (no of sections in this case ) is called Number of
scores.
5.3 Example 2: You must have heard your parents saying that
you have scored an average of 68 marks in an examination. Since you have
several subjects, it is difficult to remember the marks in each of the
subjects. Because, it is easy to remember one number, people express the marks
as average. Let the marks scored be: English: 65, Hindi: 60, Social Science:
65, General science: 70, Mathematics: 80.If you add all the marks, you get 340.
Since there are 5 subjects we divide the total marks by this number to get 68.
This is the average marks you have got. In this method you must have observed
that the average does not reflect your highest marks(got
in mathematics) or lowest marks(got in Hindi).
Note:
As it can be seen from the above example, expressing average figure
could be misleading in some cases. But it is very useful in many cases
particularly when we have large number of scores (Ex: average rain fall in a
place, average height of students in a class, etc).
5.3 Example 3: Let
your favorite cricketer score the following runs in few one day matches
27,45,40,18,80,55, 47,105,46, 40,47. What is his average?
Workings:
Total runs
scored = 27+45+40+18+80+55+ 47+105+46+40+47=550
Number of
matches played = 11
Average
Runs per match = 550/11=50
Average
can be calculated by applying the following formula:
Definitions:
‘Mean’ = Sum of all scores/Number of Scores
Let x_{1},x_{2},x_{3},x_{4} ….x_{n} be scores( thus there are
‘n’ number of scores)
Then
Average = (x_{1}+x_{2}+x_{3}+x_{4} ….+x_{n})/n
Mean
= (_{})/Number of scores
The
symbol _{}is pronounced as sigma and is used to represent the sum of
numbers.
The mean
is also called average’ or ‘Arithmetic mean’.
Let
us learn few more concepts in statistics.
Let
us arrange the runs of Example 5.3.3 in ascending order. Then the runs are
18,27,40,40,45,46,47,47,55,80,105.
Middle
figure (6^{th} out of 11) in this arrangement is 46 and is called Median.
The
middle most figure in an orderly distribution of scores is called ‘median’.
It
is interesting to note that in this Cricketer’s case, his Mean (Average) (50)
and Median (46) are very close to each other.
Let
us find the most occurrences of same number of runs scored, in case of this
cricketer. We find that he scored 40 runs and 47 runs two times each. They are
called Mode.
The
‘Mode’ is the most often repeated score in a given
set of scores.
In
the above example the number of matches was 11.
Since 11 is an
odd number, it is easy to find Median
(middle figure). What if we have even number of scores ?
5.3 Example 4:
Let us take the case of temperatures recorded (in centigrade) for 10 days at
your place.
Let
they be 25^{0} C, 30^{0}
C, 31^{0} C,34^{0} C,32^{0}
C,31^{0} C,30^{0} C,28^{0} C,30^{0} C,31^{0}
C_{.}
Working:
When
we arrange them in ascending order, we get
25^{0}C,28^{0}C,30^{0}
C,30^{0} C,30^{0} C,31^{0} C,31^{0} C,31^{0}
C,32^{0} C,34^{0} C.
Let
us tabulate them in a table:
Scores (x) 
Occurrences (frequency)(f) 
fx=x*f 
25^{0}C^{} 
1 
25 
28^{0}C^{} 
1 
28 
30^{0}C^{} 
3 
90 
31^{0}C^{} 
3 
93 
32^{0}C^{} 
1 
32 
34^{0}C^{} 
1 
34 
Total(_{}) 
10 
302 
Average
For ungrouped data (when scores and frequency are given) can also be calculated
by the formula
Average
(mean) = (_{})/ (_{} )=
302/10 = 30.2^{0}C
Median
is (30+31)/2 =30.5^{0}C (Average of 5^{th} and 6^{th}
term as there are even number of scores which is 10)
Modes
are 30^{0}C and
31^{0}C both of
which occur three times which is the most number of occurrences.
Note:
In this Example there is not much difference between Average (30.2^{0}C),Median (30.5^{0}C) and Modes(30^{0}C,31^{0}C).
When we have large number of data (scores) we may notice Average, Median, and
Mode being nearer to each other.
5.3 Problem 1:
Mean of 9 observations was found to be 35. Later on it was found that an
observation 81 was misread as 18. Find the correct mean of observations.
Solution:
Mean of 9 observations = 35
_{}Sum of all observations = 35*9 = 315
In the observations 81 was
misread as 18
The correct sum of all
observations = 31518+81 = 378
_{}The correct mean = 378/9 = 42
5.3
Summary of learning
No 
Points to remember 
1 
Mean=
(Sum of all scores)/Number of scores 
2 
The middle most figure in an orderly distribution
of scores is called median 
3 
Mode is the most often repeated score in a given
set of scores 
Additional
Points:
5.3.2 Assumed mean method for
calculation of mean.
This
method is very useful when data and their frequencies are very large. In this
method we assume one of the data to be mean and find the deviation from that
number and hence this method is called ‘assumed mean method’.
Let us
take the example
worked out earlier (5.3 Example 4) to illustrate this method.
5.3 Example 4:
Let us take the case of temperatures recorded (in centigrade) for 10 days at
your place.
Let
they be 25^{0}C, 30^{0}C, 31^{0}C, 34^{0}C, 32^{0}C,
31^{0}C, 30^{0}C, 28^{0}C, 30^{0}C, 31^{0}C_{.}
Let 30
be the assumed mean (any score can be assumed to be the mean but we normally take the score which is in the middle part of
the distribution as assumed mean)
The
Deviation D (D = Score Assumed mean) is calculated for each of the score.
Then
Average (mean) = A + (_{})/Number of scores
Scores (x) 
Frequency(f) 
Deviation D= AM 
fD = f*D 
25^{0}C^{} 
1 
5(=2530) 
5 
28^{0}C^{} 
1 
2(=2830) 
2 
30^{0}C=A^{} 
3 
0(=3030) 
0 
31^{0}C^{} 
3 
1(=3130) 
3 
32^{0}C^{} 
1 
2(=3230) 
2 
34^{0}C^{} 
1 
4(=3430) 
4 
Total(_{}) 
10 

2 
Average
(Mean) = A +(_{})/Number of scores = 30+ (2/10)
= 30.2
Is
this not the same mean value which we got earlier?
This
method is less time consuming and hence less chances of one making mistakes.