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.