This document discusses various measures of central tendency and variability. It provides details on calculating the median, mean, and mode from both raw data and grouped data. The median is the middle value of a data set and is not affected by outliers. The mean is the average and is more affected by outliers. The mode is the most frequent value. Formulas and step-by-step processes are provided to compute each measure from ungrouped and grouped data using methods like class intervals and frequency distributions.

2. Measures of Central
Tendency
and
Measures of
variability
Lyra Vanessa Sales
Ma. Theresa C. Seno
Liezel Vitto

3. Measures of Central Tendency
 In simple terms, a measure of central tendency is either a
midpoint an average, or the most frequent score in
distribution of scores. The most common measures of central
tendency are:

4. The Median
The median is a point in a scale which divides
the scale into two equally.

5. Characteristic of the Median
1. The median computed from the ungrouped
scores is called a crude, rough or counting
median; from a class frequency distribution
refined median.
2. The median is a midpoint of a scale and so it
theoretically divides a scaled group into two
equal subgroups.

6. 3. The median is the most stable measures of
central tendency because it is not affected by
the meaningful of the scores. For instance, the
series 5, 6, 7, 8, 9, the median is 7. Even if 9
becomes 50 the median is still 7.
Characteristic of the Median

7. Characteristic of the Median
4. The value of the median depends upon the
magnitude of the middlemost scores or scores. If the
middlemost sore is high the median is high, if it is low
the median is low.

8. Uses of Median
 The median is computed and used when
1. The exact midpoint of a scale or scaled distribution is
needed.
2. A group with a quantified scaled characteristic, for
example ability, is to be divided into two equal
subgroups.
3. A stable measure of central tendency is needed.

9. 4. A group is too heterogeneous, that is, there are
extremely high scores of the group and there are also
extremely low scores. For instance, if the scores are 8, 10,
15, 25, and 35 the median is preferably used.
5. A class frequency distribution is badly skewed.
Uses of Median

10. Uses of Median
6. A scaled distribution is open-ended.
X F
90 & ABOVE 3
85 6
80 8
75 5
70 & BELOW 4
N = 26

11. Computation of the median from
Ungrouped Scores

12. Scale
Is a succession of numbers, steps,
classes, degrees, gradations, or
categories with a fixed interval.

13. Computation of the median from
Ungrouped Scores
 When there are few scores, say less than 30, the ordered
arrangement of scores may be used to determine the median.
The method follows:
A. Data Given: The following test scores in arithmetic:
Odd number of scores: 56 88 90 76 72 82 73 63 75 63
Even numbers of scores: 84 77 82 86 78 79 60 71 75 59
82

14. B. Procedure:
1. Arrange the scores in a descending order,
that is, from the highest to lowest score.
Examples:

15. Example 1
90
88
82
76
75
73
72
63
63
62
56
N =11
The median is 73,
the middlemost
score.
Example 2
86
84
82
82
79
78
77
75
71
70
60
59
N= 12
The middlemost
scores are 78 and 77.
Mdn = 78 + 77
2
= 155
2
= 77.5

16. 2. If the number of scores is odd, the middlemost
score is the median. In example 1, the middlemost
score is 73. Hence the median is 73.
3. If the number of scores is even, the median is the
average of the two middle most scores. In example
2, the two middlemost scores are 78 and 77. The
average is (78 + 77)/2=155/2=77.5. Hence, the median
is 77.5.

17. Computation of the Median from a class
Frequency Distribution
 When there are many scores, say more than 30, class
frequency distribution may be utilized in computing the
median. The procedures follows:
A. Data given: The following test scores in English
7 4 5 6 8 8 9 0 7 6 7 7 7 2 8 2 7 5 6 3 6 2 7 5 6 3 8 4
8 2 8 6 7 6 7 7 6 0 7 1 8 2 7 5 7 0 5 9 8 1 8 3 7 9 7 3
6 7 6 4 6 8 6 9 7 1 7 8 7 5 7 3 7 0 6 6 6 9 7 7 8 0

18. B. Procedure
1. Group the scores into a class frequency description
using eight classes.
2. Compute the cumulative frequencies upward from the
lowest class frequency to the highest class frequency.
Table 10
Table 10

19. TABLE 10
CLASS FREQUENCY DISTRIBUTION
X / CLASS TALLIES FREQUENCY
CUMULATIVE
FREQUENCY
90 – 94 I 1 41
85 – 89 II 2 40
80 – 84 IIIII-II 7 38
75 – 79 IIIII-IIIII-I 11 31
70 – 74 IIIII-III 8 20
65 - 69 IIIII 5 12
60 – 64 IIIII 5 7
55 - 59 II 2 2
TOTAL N = 41

20. FREQUENCY CUMULATIVE FREQUENCY
1 41
2 40
7 38
11 31
8 20
5 12
5 7
2 2
N = 41
The highest cumulative
frequency checks with
the number of scores,
N. They should be
equal.

21. B. Procedure
3. Use the formula
Mdn = E1 +
𝒏
𝟐
−𝑪𝒇 𝑰
𝑭
in which
Mdn = the median
E1 = the exact lower limit of the median class
N = the number of cases or scores
Cf = the cumulative frequency equal to or next lower than N/2
F = the frequency of the median class
I = the interval

22. 4. Find the values of the symbol I the formula starting with N/2 as follows:
Mdn = E1 +
𝒏
𝟐
−𝑪𝒇 𝑰
𝑭
a.
𝑁
2
=
41
2
= 20.5
b. Cf = 20, the cumulative frequency lower than 20.5
c. E1= 74.5 the exact lower limit of the median class, the class which is just
above the Cf of 20, 75-79
d. F = 11, the frequency of the median class.
e. I = 5, the interval which is the difference between two adjacent lower
limits.

23. 5. Substitute the values for the symbols
in the formula and solve.
a. Substract 20 from 20.5 = .5
Mdn = 74. 5 +
𝟐𝟎.𝟓 −𝟐𝟎 𝟓
𝟏𝟏
Mdn = 74.5 +
.𝟓 𝟓
𝟏𝟏
b. Multiply .5 by 5
Mdn = 74.5 +
2.5
11
c. Divide 2.5 by 11 to 3
decimal places
Mdn = 74.5 + .227
d. Add 74.5 and .227
Mdn = 74.727
e. Round off 74.727 to 2
decimal places
Mdn = 74.73

24. Example:
Table II
X F Cf
87 2 67
84 4 65
81 7 61
78 9 54
75 11 45
72 12 34
69 7 22
66 8 15
63 5 7
60 1 2
57 1 1
N=67
Mdn =
E1 +
𝑵
𝟐
−𝑪𝒇 𝑰
𝑭

25. The Arithmetic Mean
Arithmetic mean is the average of a
group of scores. Like the median, it is
a measure of central tendency.

26. Characteristic of the
mean
1. The arithmetic mean or simply mean, being the average of the
scores, is the center of gravity or balance point of the scores.
2. The mean is easily affected by the magnitudes of the scores.
For onstance, the mean of the scores 1, 2, 3, 4 and 5 is 3. If 5
becomes 6, the mean becomes 3.2.
3. The mean is the most reliable among the measures of central
tendency.

27. Uses of the Mean
The mean is computed and used when
1. The average of the scores is wanted.
2. The center of gravity of the scores is desired.
3. It is desired that every score has an effect upon the
measure of central tendency.
4. The most reliable measure of central tendency is
needed.

28. 5. The group from whom the scores have been
derived is more or less homogeneous. The mean is
not realistic if the group is heterogeneous. For
instance, the mean of the scores 2, 3, 5, 7 and 50 is
13.4 which is very far from all the scores.
6. Other statistical measures in which the mean is
involved such as the standard deviation, t—ratio,
critical ratio, etc., are computed.

29. Computation of Mean by
Averaging
The mean is composed by averaging when the
scores are not too many, and they are not
grouped or classified.
The method of averaging (long or absolute
method) follows:

30. A. Data given : The following are the test
scores in arithmetic:
25 38 41 68 71 52 64 30 45 35 58
B. Procedure:
1. Use the formula
M =
𝑻𝑿
𝑵
In which
M = the mean
TX = the total of the scores
N = the number of cases or scores
X
25
38
41
68
71
52
64
30
45
35
58
TX= 527

31. 2. Find the values of the symbol in the formula as follows:
a. Add the scores to find their total, TX.
TX = 527
3. Substitute the values of the symbols in the formula and solve:
M =
527
11
a. Divide 527 by 11 to 3 decimal places
M = 47. 909
b. Round off 47. 91

32. Computation of the Mean
by the Midpoint Method
The mean may be computed by the midpoint method
if the scores are many, say more than 30, and are
grouped into a class frequency distribution. The
method follows:

33. A.Data Given : The following are scores in a test in
grammar
50 91 45 60 51 54 83 56 68 71 55 73 85 66 74 65
87 45 84 70 69 59 61 67 68 77 72 80 67 71 74
57 54 64 64 75 62 69 56 77 79 66 65 63 70

34. B. Procedure
1. Use the formula
M =
𝑻𝑭𝑴𝒑
𝑵
In which
M = the mean
TFMp = the total of the products of the class midpoints
and their corresponding frequencies.
N = the number of cases or scores

35. 2. Find the value of the symbols in the formula as follows:
a. Group the scores into a class frequency distribution
X Tallies F Mp FMp
90 – 94 I 1 92 92
85 – 89 II 2 87 174
80 – 84 III 3 82 246
75 – 79 IIII 4 77 308
70 – 74 IIIII- III 8 72 576
65 – 69 IIIII -IIIII 10 67 670
60 - 64 IIIII -II 7 62 434
55 – 59 IIIII 5 57 285
50 – 54 IIII 4 52 208
45 - 44 II 2 47 94
N = 46 TFMp= 3087

36. b. Determine the midpoints of the classes. Use the formula
Mp = E1 +
𝐼
2
c. Multiply the midpoints, Mp, by their corresponding
frequencies to find
Fmp
92 x 1 =92
87 x 3 = 246
d. Add the FMp’s to find TFMp.
TFMp = 3067

37. e. Add the frequencies to find N.
N=46
3. Substitute the values for the symbol in the
formula and solve:
M =
𝑻𝑭𝑴𝒑
𝑵
M =
3087
46
M = 67.108 / 67.11

38. Another Method
By the use of lower limit
1. Use the formula
M=
𝑇𝐹𝐿
𝑁
+
𝐼
2
− .5
In which,
M = the mean
TFL = the total of the products of the class lower limits and
their corresponding frequencies.

39. I = the interval
N = the number of cases or scores
2. Find the values of the symbols in the formula as
follows:
a. Multiply the lower limits of the classes by their
corresponding frequencies.
b. Add the FL’s to find TFL

40. Table 13
Class Frequency Distribution
X F FL
90 1 90
85 2 170
80 3 240
75 4 300
70 8 560
65 10 650
60 7 420
55 5 275
50 4 200
45 2 90
N= 46 TFL = 2995
M=
𝑇𝐹𝐿
𝑁
+
𝐼
2
− .5

41. The Mode
 The mode may be defined as the score
occurring the most number of times. It is the
score with the highest frequency.

42. Determining the Mode from an
Ordered Arrangement of
scores
 The mode determined from an ordered arrangement of
scores is a rough or crude mode. The method of
determining the mode from an ordered arrangement of
scores follows:
A. Data Given: The following are the test scores in Geometry:
25 30 37 52 52 30 37 30 42 37

43. B. Procedure
1. Arrange the scores in a descending order,
that is, from the highest to the lowest
score.
2.Look for the scores that occurs the most
number of times. This is the crude mode.

44. Example
X
52
52
42
41
37
37
37
37
30
30
30
25
Mode = 37

45. Determining the Crude Mode
from a Score Frequency
Distribution
 The method of determining the rough or crude mode from a
score frequency distribution follows:
A. Data Given: The following are the test scores in Geometry:
25 30 37 52 52 30 37 30 42 37

46. B. Procedure
25 30 37 52 52 30 37 30 42 37
1. Arrange the scores in a descending order, but write the score
only once although the score occurs several times.
2. Tally the scores. For each score, write a tally or short bar
opposite the score to which it is equal. Count the tallies and
the score with the highest number of tallies is the rough
mode.

47. X Tally F
52 II 2
42 I 1
41 I 1
37 IIII 4
30 III 3
25 I 1

48. Determining the Crude Mode from
a Class Frequency Distribution
 The method of determining the crude mode from a class
frequency distribution follows:
x F
95 2
90 5
85 10
80 12
75 9
70 4
65 2

49. B. Procedure
1. Use the formula
Mo = E1 +
𝑰
𝟐
In which:
Mo = the mode
E1 = the exact lower limit of the class with the highest frequency
I = interval

50. 2. Look for the class with the highest frequency.
3. The interval of the classes is 5.
4. Substitute the values for their corresponding
symbols in the formula and solve.

51. Mo = 79.5 +
𝟓
𝟐
= 79.5 + 2.5
Mo = 82

52. Computation of the
Refined Mode
1. When the refined mode, called the Pearson theoretical mode,
is to be computed, use the formula
Mo = 3Mdn – 2M
In which
Mo = mode
Mdn = the median
M = the mean

53. 2. If the median and the mean of a frequency distribution are 84.5 and 83.75
respectively the values for their corresponding symbols in the formula and
solve:
Mo = 3Mdn – 2M
= 3 (84.5) – 2(83.75)
= 167.50 – 253.50
Mo = 86

54. Computation of the
Percentile
Percentile are points which divide a scale into 100
equal parts. A percentile indicates that a certain
percent of scores under consideration are at or
below the point and a certain percent above.
The methods of computing the percentile follows:

55. A. Data Given:
x F CF
95 2 50
90 4 48
85 7 44
80 9 37
75 10 28
70 8 18
65 6 10
60 3 4
55 1 1
N = 50

56. B. Procedure
 Use the general formula
Pp = E1 + I
𝑝𝑁
100
− 𝐶𝑓
𝐹
In which
N = the number of scores
Pp = the percentile desired, p indicates the percentile rank desired
Cf = the cumulative frequencyjust below Pn/100
E1 = the exact lower limit of the class just above Cf, the path percentile class
F= the frequency of the class containing the E1 or the pth percentile class
I = the interval, the difference between any two adjacent classes

57. 2. Find the values of the symbols in the formula. Suppose
the desired percentile
is: 35
Pp = 35
N=50
𝑝𝑁
100
=
35 (50)
100
=
1,750
100
= 17.50
Cf = 10
E1 = 69.5
F = 8
I = 5
x F CF
95 2 50
90 4 48
85 7 44
80 9 37
75 10 28
70 8 18
65 6 10
60 3 4
55 1 1
N = 50

58. Substitut
e:
Pp = E1 + I
(
𝑃𝑛
100
−𝐶𝑓)
𝐹
P35 = 69.5 +
5(
17.5
100
−10)
8
P35 = 69.5 +
5(7.5)
8
P35 = 69.5 +
37.5
8
P35 = 69.5 + 4.687
P35 = 74.187 / 74.19

59. Example: Suppose P87 is desired
P87 = E1 + I
(
𝑝𝑁
100
−𝐶𝑓)
𝐹
N = 50
Cf = 37
E1 = 84.50
F = 7
I = 5

60. P87 = E1 + I
(
𝑝𝑁
100
−𝐶𝑓)
𝐹
P87 = 84.50 +
5(43.50 −37)
7
P87 = 84.50 +
5(6.50)
7
P87 = 84.50 +
32.50
7
P87 = 84.50 + 4.642
P87 = 89.14