MEASURE OF
CENTRAL
TENDENCY
 After a test is administered and scored ,
the problem of interpreting the scores
crops up. In order to interpret the scores
, the teacher should first know how
tabulate
scores
and
prepare
a
frequency distribution.
 The simplest way of tabulating scores is to arrange
them from highest to lowest. In order to present the
scores in a more organized manner , a teacher must
prepare a frequency distribution.
 A frequency distribution reveals how often a particular
score occurs. The group frequency distribution uses
intervals or ranges of scores and the number of times
an interval of scores occurred.
FREQUENCY DISTRIBUTION
SCORE INTERVALS
Frequency
Steps in constructing a grouped
of frequency distribution
1. Determine the range of scores. This
is done by subtracting the lowest
score from the highest score
R= Highest Score – Lowest Score
Example: R= 40 -10
The Range score is 30
2. Determine the appropriate or
ideal number of intervals
 Most of the expert believe that the ideal
number of intervals is 10.
3. Divide the range by the number
of intervals selected. The result
will be the interval with (i)
i= R/Number of intervals
Example
i=30/10
=3
4. Determine the Lower Limit of
the interval by dividing the
Lowest Score with interval width,
Then multiply the quotient by the
same interval width,
Example: 10 = Lowest Score
3= Interval width
10/3=3.33 x3 =10
 Mr.Mercado administered
a 50 items multiple
choice exam for their 2nd Quarter examination in
Biological Science .The class has the population
of 48 . The test score result are the following ;
32
36
47
49
31
31
41
23
35
38
31
33
29
23
45
38
36
33
25
31
32
34
43
44
45
23
33
43
42
23
45
35
23
25
45
23
36
34
35
36
41
42
35
37
42
43
45
37
23-25
26-28
29-31
32-34
35-37
38-40
41-43
44-46
47-49
50R = 49 -23 = 26
i= 26 /10 = 2.6 =3
LL = (23/ 3 ) X3
7.6 X3
23
0
 What will be the frequency distribution of the
scores? If Mr . Mercado wants to use 10 intervals
, What will be the interval width And the lower
limit of interval?
RANKING
 Another way to organize test scores is ranking.
Ranking is arranging a group of scores from
highest to lowest. The highest score is designate
Rank 1st, the second highest, Rank 2nd the third
highest , Rank 3rd and so on .
 The steps in ranking the scores are given below:
 1. Arrange the score from highest to lowest . A
particular score may be written as many times as
it occurs .
 2. Designate the rank of each score successively
with number 1 for the highest , 2 for the second
highest and so on, to the last score
 3. Assign each rank in ordinal way such 1st
3rd ,4th, 5th, and so on
, 2nd
SAMPLE SCORE
50
47
42
39
32
49
47
42
39
31
48
46
40
36
30
47
45
39
35
25
 4. Average the ranks of score appearing more
than once. Thus scores which are the same share
similar rank .
 Example: Score 47 occurs thrice in the list. The
ranks to be considered are 4th ,5th, abs 6th,. To get
the average rank , we add 4 ,5, and 6 = 15 . The
result 15 / 3 = 5 . The score 47 occurring thrice
shares rank 5th .
Score
Number
Rank
50
1
1st
49
2
2nd
48
3
3rd
47
4
5th
47
5
5th
47
6
5th
46
7
7th
45
8
8th
42
42
9
10
10th
10th
RANK THE FOLLOWING SCORES
82
32
55
79
75
36
40
80
22
64
64
45
30
55
40
58
58
52
20
60
40
31
31
78
38
71
78
42
44
54
Measures of
Central
Tendency
A
measure of central
tendency indicates the
idea of the average score
in a distribution. The
three kinds of measures of
central tendency are the
mean, median and mode.
The Mean
The mean is simply the average of a group of
scores.
average = mean
The formula for computing the mean of
ungrouped scores is
X = Summation X
N
SAMPLE MEAN
Find the mean scores of the following
52
60
60
54
52
X= 52+60+60+54+50
5
= 55.2
THE MEAN OF THE GROUPED SCORE
The formula for calculating the mean of group scores is
X= A.M +( Summation fd) i
N
X= mean
A.M = Assumed mean
Summation of fd= summation of frequency
multiplied by deviation
 f= frequency
 d= deviation
 N= number of scores in a distribution
 i= class interval
LET US FIND THE MEAN OF GROUP SCORES
Interval
f
d
fd
39-41
1
6
6
36-38
2
5
10
33-35
4
4
16
30-32
4
3
12
27-29
3
2
6
26-24
5
1
5
21-23 A.M
8
0
0
18-20
6
-1
-6
15-17
3
-2
-6
12-14
3
-3
-9
9-11
1
-4
-4
X = A.M +(Sum fd/N) i
X = 21-23+ (30/40) 3
X = 22+ (0.75) 3
X = 22+ 2.25
X = 24.25
MEDIAN
 The median is the middle score in a distribution .
It divides the distribution in half : 50 percent of
the scores is found above the median, and the
other 50 percent lies below the median.
 Example ; Find the median of the following set of
ungrouped scores: 20, 35,15, 40, 50
 Arrange the ungrouped score in ascending order (
from the highest to the lowest score)
 The median is the score that is in the middle of
the distribution.
 Example;
50 ,40,30,35,20,15
THE MEDIAN OF GROUPED SCORES
 Another formula is used to compute the median
of grouped scores
 Median = LL+

(N/2- cf) i
fm
 LL= Lowest Limit
 N= Number of cases
 Cf= cumulative frequency
 fm= measure frequency
 i= interval
 Step in computing the median of grouped scores:
 1. Get the cumulative frequencies by adding the
numerical values in the f column . To get 4 in cf
column , add 1 and 3 under the column f and so
on until the last sum upwards equals the total
number of scores .
Intervals
f
cf
39-41
1
40
36-38
2
39
33-35
4
37
30-32
4
33
27-29
3
29
24-26
5
26
21-23
8
21
18-20
6
13
15-17
3
7
12-14
3
4
9-11
1
1
 2. Determine the number of cases that represent
50 percent of the total number of cases. In our
example , 50 % of 40 is 20
 40/ 2= 20
 3.Find the interval in which the cumulative
frequency (cf) is less than the required number of
cases. In our example , the cumulative frequency
is 13 , the upper most cf which is less than 20.
Intervals
f
cf
39-41
1
40
36-38
2
39
33-35
4
37
30-32
4
33
27-29
3
29
24-26
5
26
21-23
8
21
18-20
6
13
15-17
3
7
12-14
3
4
9-11
1
1
Intervals
f
cf
39-41
1
40
36-38
2
39
33-35
4
37
30-32
4
33
27-29
3
29
24-26
5
26
21-23
8
21
18-20
6
13
15-17
3
7
12-14
3
4
9-11
1
1
 4 .Compute for the formula using the formula
= LL+ (N/2- cf) i
fm
Median == LL+ (N/2- cf)
fm
= 20.5 + ( 40/2-13)
8
= 20.5+ (7/8) 3
= 20.5 + (.88) 3
= 20.5 + 2.64
=23.14
3
THE MODE OF UNGROUP SCORES
 The mode is the score that occurs most frequently

example : Find the mode of these ungrouped
scores , 34,30,45,30,33,30,50
 The most frequent score in our example is 30 .
The score 30 occurs three times in the
distribution.
THE MODE OF GROUPED SCORES
 The mode of grouped scores can be calculated
after the mean and the median have been
computed.
 Mode= ( 3xMedian)- (2xmean)
 In our previous example
using the same
frequency distribution, the mean is 24.25 while
median is 23.14
 Mode= (3X23.14)- (2X24.25)
= 69.42- 48.50
= 20.92