lOMoARcPSD|18274273
Chapter 4 Analysis Interpretation of Assessment REsults
13B53C
Secondary education (Eastern Visayas State University)
StuDocu is not sponsored or endorsed by any college or university
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
CHAPTER 4
Analysis and Interpretation of Assessment Results
Introduction
Statistics plays a very important role in assessing the performance of students,
most especially in interpreting and analyzing their scores through assessment activities.
Teachers should know how to utilize these data, particularly in decision-making. Hence, a
classroom teacher should have the necessary background in statistical procedures in order
for him to give a correct description, and interpretation of student’s performance in a certain
test.
This lesson is a review of the important tools needed in describing, analyzing and
interpreting assessment results. The topics discussed in this module are presentations of
data through textual, tabular and graphical, measures of tendency, measures of dispersion,
measures of relative positions, other measure and the level of measurement.
Learning Outcomes
At the end of the module, the student
should be able to:
a. Interpret assessment results accurately and utilize them to help
learners improve their performance and achievement; and
b. Utilize assessment results to make informed-decisions to
improve instruction.
Lesson 1 - Presentation
The study of statistics begins with the collection of data or measurements. Data
collected should be organized systematically for easier and faster interpretation. Data can be
presented in three forms: textual, tabular, and graphical.
The tabular and graphical forms are used when more detailed information about the
data is to be presented. A table is used when you want to present a data in a systematic and
organized manner so that reading and interpretation will be simpler and easier.
A.1 Textual Presentation
Ungrouped data can be presented in textual form, as in paragraph form. This
involves enumerating the important characteristics, giving emphasis on significant figures
and identifying important features of the data.
Example 1. Below are the test scores of 50 students in Statistics:
25
30
43
18
35
17
40
50
9
12
33 37
46
28 19
27
41
10
18
28
21
36
13
31
20
31
35
28
16
42
40 48
13
40
3
39
50
32
32
26
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
41
lOMoARcPSD|18274273
Arranging the scores from the lowest to the highest will facilitate the enumeration of
important characteristics of the data. The test scores of the 50 students in Statistics arranged
from lowest to highest are shown below:
3
13
32
17
35
20
40
27
43
30
9
13
18
21
28
30
33
36
40
46
10
14
18
25
28
31
34
37
40
48
10
15
19
26
28
31
35
With the data now arranged according to magnitude, we can easily see the
38
41
50
important features worth mentioning in the text. One way of describing the data using the
12
16
20
26
29
32
35
textual form is as follows:
39
42
50
The highest score obtained is 50 and the lowest is 3. Ten students got a score
of 40 and above, while only 4 got ten and below. Generally, the students performed well
in the test with 33 students of 66% getting a score of 25 and above.
Arranging a mass of data manually is quite tedious, but using computers for this
purpose is so easy. In the absence of a computer, the process is made easy by putting the
data in a stem-and-leaf plot.
Stem-and-leaf plot is a table which sorts data according to a certain pattern.
It involves separating a number into two parts. In a two-digit number, the stem consists
of the first digit, and the leaf consists of the second digit. While in a three-digit number,
the stem consists of the first two digits, and leaf consists of the last digit. In a one-digit
number, the stem is zero.
Using the unarranged test scores in Statistics of 50 students as
data, stem-and-leaf plot can be used to arrange from lowest to highest. The
stems are as follows: 0, 1, 2, 3, 4, and 5, three being the lowest and 50 the
highest.
Table 1
Stem-and-Leaf Plot of Unarranged Test Scores in
Statistics of 50 Students
Stem
Leaves
0
3, 9
1
0, 0, 2, 3, 3, 4, 5, 6, 7, 8, 8, 9
2
0, 0, 1, 5, 6, 6, 7, 8, 8, 8, 9
3
0, 0, 1, 1, 2, 2, 3, 4, 5, 5, 5, 6, 7, 8, 9
4
0, 0, 0, 1, 2, 3, 6, 8
5
0, 0
looking at the stem-and-leaf plot, we can easily rank the data or
put them in order. Thus, the ten lowest scores are: 3, 9, 10, 10, 12, 13, 13, 14,
15, and 16, while the ten highest scores are: 40, 40, 40, 41, 42, 43, 46, 48, 50,
and 50.
By
A. 2. Tabular Method
Sometimes, we cold hardly grasp information from a textual presentation data.
Thus, we may present data by using tables.
By organizing the data in tables, important features about the data can be readily
42
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
understood and comparisons can be easily made. Thus, a table shows complete information
regarding the data. A table has the following parts:
1. Table number : This is for easy reference to the table.
2. Table title: It briefly explains the content of the table.
3. Column header: It describes the data in each column.
4. Row classifier: It shows the classes or categories.
5. Body: This is the main part of the table.
6. Source note: This is placed below the table when the data written are not
original.
Below is a table with all its parts indicated:
Table Number
Table 2
Distribution of Students in XYZ High School
According to Year Level
Table Title
Column
Number of Students
Header
300
250
Body
285
215
N = 1,050
Source Note
Source: XYZ High School Registrar
Year Level
First Year
Second Year
Third Year
Fourth Year
Row classifier
Another type of tabular presentation is the frequency table also known as a
frequency distribution. It is an arrangement of the data that shows the frequency of
occurrence of different values of the variables.
A frequency distribution table is a table which shows the data arranged into
different classes and the number of cases which fall into each class.
The frequency distribution table for ungrouped data is simply an arrangement of
data from lowest to highest which shows the frequency of occurrence of each value in a set.
This is best used when the range of values is not too wide.
Example:
Table 3
Ungrouped Frequency Distribution for the Ages of 50
Students Enrolled in Statistics
Age
14
15
16
17
18
18
Frequency
4
13
25
5
2
1
N = 50
1. Table number is ____.
2. Table title is Ungrouped Frequency Distribution fort he Ages of 50
Students Enrolled in Statistics
3. Column headers are:
a) Age
b) Frequency
4. Row classifiers: 14, 15, ..., 19
43
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Notice that the range of the ages is 5, that is subtracting 14, the lowest age, from
19, the highest age. However, if the range is more than 15, the best way is to group the data
into classes using the grouped frequency distribution table.
The frequency distribution for grouped data is an arrangement of data into different
classes or categories. It involves counting the data which fall into each class.
Below are the steps in constructing a frequency distribution table:
1. Find the range of scores: Range = Highest score - Lowest score
= 99 - 67 = 32
2. Decide on the number of class interval k
Maximum = 20
Minimum = 7
Ideal = 10 - 15
k n
Estimate: (rounded to the nearest whole number)
where n, is the total number of scores or cases
k  40 6
Thus, for the above scores, or 6.
3. Determine the class size i of the interval.
44
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Range 32
i
 
k
6
or i = 5. (An odd value of i is preferred).
4. Determine the lower limit LL and upper limit of the lowest class interval (the
class interval containing the lowest score).
LL = score or number closest to but less than the lowest score and
preferably a multiple of the class size i.
In the given set of scores, the lowest score is 67. The number closest
to 67 that is divisible by the class size i = 5 is 65. Thus, LL = 65.
UL = LL + (i - 1). Thus, UL = 65 + (5 - 1) = 69.
5. Determine the other class intervals by consecutively adding the class size i to
LL and UL until the interval containing the highest score is contained and
make a tally. Thus,
Illustration:
Performance ratings of government employees
76
67
99
82
86
92
85
95
86
93
87
93
79
83
98
Class Interval
95 - 99
90 - 94
85 - 89
80 - 84
75 - 79
70 - 74
65 - 69
78
91
85
87
71
87
85
81
79
81
Tally
////
//// - ///
//// - //// - //
//// - //
//// - /
//
/
88
79
96
92
86
85
92
75
80
80
92
82
88
74
94
f
4
8
12
7
6
2
1
The
limits that define the class intervals as indicated above are called apparent limits. To reflect
the continuity of scores, the true limits or class boundaries are indicated. These are obtained
by adding all upper limits and subtracting all lower limits on-half of the difference between
successive adjacent lower and upper limits.
For this particular data, the number to be added and subtracted is 0.5
Other information usually included in a frequency distribution are:
45
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
LL  U
2
X = the class mark or class midpoint of the class interval =
<cf = the less than cumulative frequency = the frequency of the interval
plus all frequencies below the interval
>cf = the greater than cumulative frequency = the frequency of the
interval plus all frequencies above the interval
46
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
rf = the
relative
f
n
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
47
lOMoARcPSD|18274273
frequency of the interval =
The same frequency distribution with additional information as cited
above is shown below:
Table 4
Performance Rating of Government Employees
Class Interval
f
X
rf
<cf
95 - 99
4
97
0.100
40
90 - 94
8
92
0.200
36
85 - 89
12
87
0.300
28
80 - 84
7
82
0.175
16
75 - 79
6
77
0.150
9
70 - 74
2
72
0.050
3
65 - 69
1
67
0.025
1
Remarks:
1. The class mark is assumed to be the average of the scores that fall within
the interval.
2. The <cf tells the number of scores or data falling below the true upper limit
of the interval. Thus, for instance, 16 scores are lower than a score of
84.5; 79.5 is the score below which we find 9 out of the 40 scores.
3. The >cf tells the number of scores or data falling above the true limit of
the interval. Thus, for instance, 37 scores are higher than a score of
74.5; 84.5 is the score above which we find 24 out of the 40 scores.
4. The relative frequency tells about the proportion or percentage of cases
within the class interval. Thus, 15% of all the scores are found between
75 - 79.
>cf
4
12
24
31
37
39
40
A.3 Graphical Presentation
A graph is a diagram which makes a systematic presentation of a class frequency
distribution together with comparison and relationship of the classes. As a graph is usually
perceptible, it is easily understood.
There are two most common methods for graphing frequency distribution:
Histogram and the frequency polygon. Histogram represents a pictorial presentation of a
frequency distribution. It may be thought of as a series of rectangles and frequencies,
respectively. In histogram, the bases is equal to the length of the interval, and the height is
equal to the frequency. It resembles a bar graph. Frequency polygon is another method of
graphing frequency distribution. It is also pictorial but it is constructed by joining with straight
lines a series of points which are the midpoints of the steps as against their corresponding
frequencies. It looks like a zigzag line.
48
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Lesson 2 - Quantitative Analysis and Interpretation
2.1 Levels of Measurement
Statistics deals mostly with measurements. We define measurement as the
assignment of symbols or numerals to objects or events according to some rules. Since
different rules are used for the assignment of symbols, then this would yield different
scales of measurement. There are four measurement scales, namely, nominal, ordinal,
interval, and ratio.
1. Nominal Scale
This is the most primitive level of measurement. The nominal level of
measurement is used when we want to distinguish one object from another for
identification purposes. In this level, we can say that one object is different from
another, but the amount of difference between them cannot be determined. We
cannot tell that one is better or worse than the other. Gender, nationality, and civil
status are of nominal scale.
2. Ordinal Scale
In the ordinal level of measurement, data are arranged in some specified
order or rank. When objects are measured in this level, we can say that one is
better or greater than the other. But we cannot tell how much more or how much
less of the characteristic one object has than the other. The ranking of contestants
in a beauty contest, of siblings in the family, or of honor students in the class are of
ordinal scale.
3. Interval Scale
If data are measured in the interval level, we can say not only one object
is greater or less than another, but we can also specify the amount of difference.
The scores in an examination are of the interval scale of measurement. To
illustrate, suppose Maria got 50 in a Math examination while Martha got 40. We can
say that Maria got higher than Martha by 10 points.
4. Ratio Scale
The ratio level of measurement is like the interval level. The only
difference is that the ratio level always starts from an absolute or true zero point. In
addition, in the ratio level, there is always the presence of units of measure. If data
are measured in this level, we can say that one object is so many times as large or
as small as the other. For example, suppose Mrs. Reyes weighs 50 kg, while her
daughter weighs 25 kg. We can say that Mrs. Reyes is twice as heavy as her
daughter. Thus, weight is an example of data measured in the ratio scale.
2.2 Measures of Central Tendency
2.2.1 Measures of Central Tendency for Ungrouped Data
2.2.1.1 The Mean
The mean (also known as the x arithmetic mean) is the most commonly used
measure of central position. It is the sum of measures divided by the number of
measures in a variable. It is symbolized as (read as x bar).
The mean is used to describe a set of data where measures cluster or
concentrate at a point. As the measure cluster around each other, a single value
appears to represent distinctively the total measures. It is, however, affected by extreme
measures, that is, very high or very low measures can easily change the value of the
mean.
To find the mean of ungrouped data, use the formula
where = the summation of x (sum of the measure)
N = number of values of x
49
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Example:
The grades in Chemistry of 10 students are 87, 85, 85, 86, 90,
79, 82, 78, 76. What is the average grade of the 10
students?
Solution:
87  84  85  85  86  90  79  82  78  76 
x
10
2.2.1.2 The Weighted Arithmetic Mean
859
.
xW

x
x
W
.58
x 10
81
Occasionally, we want to find the mean of a set of values wherein each value or
measurement has a different weight or degree of importance. We call this the weighted
mean and the formula for computing it is as follows:
where:
120

123

83

162

80

127
80(1.5)  82(1.5)  83(1)  81(2)  80(1)  85


10
10 55
50
Subject
Units
Grade
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
2.2.1.3 The Median
The median is the middle entry or term in a set of data arranged in either increasing
or decreasing order.
The median is a positional measure. Thus, the values of the individual measures in
a set of data do not affect it. It is affected by the number of measures and not by the size of
the extreme values.
To find the median of a given set of data, take note of the following:
1. Arrange the data in either increasing or decreasing order.
2. Locate the middle value. If the number of cases is odd, the middle value is the
median. If the number of cases is even, take the arithmetic mean of the two
middle measures.
Example 1: The number of books borrowed in the library from Monday to Friday last week
were 58, 60, 54, 35, and 97 respectively. Find the median.
Solution: Arrange the number of books borrowed in increasing order.
35, 54, 58, 60, 97
The median is 58.
Example 2: Cora’s quizzes for the second quarter are 8, 7,6, 10, 9, 5, 9, 6, 10,
and 7. Find the median.
Solution: Arrange the scores in increasing order.
5, 6, 6, 7, 7, 8, 9, 9, 10, 10
Since the number of measures is even, then the median is the average of the
two middle scores.
2.2.1.4 The Mode
7 8
Md 

2
The mode is another measure of position. The mode is the measure or value which
occurs most frequently in a set of data. It is the value with the greatest frequency. To find the
mode for a set of data 1. select measure that appears most often in the set;
2. if two or more measures appear the same number of items, and the
frequency they appear is greater than any of other measures, then
each of these values is a mode;
3. if every measure appears the same number of items, then the set of
data has no mode.
51
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
1
4
2
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
52
lOMoARcPSD|18274273
Example 1: The shoe size of 10 randomly selected students in a class are 6, 5,
4, 6, , 5, 6, 7, 7 and 6. What is the mode?
Answer: The mode is 6 since it is the shoe size that occurred the most
number of times.
Example 2: The sizes of 9 classes in a certain school are 50, 52, 55, 50, 51, 54,
55, 53 and 54.
Answer: The modes are 54 and 55 since the two measures occurred the
same number of times. The distribution is bimodal.
2.2.2 Measures of Central Tendency for Grouped Data
2.2.2.1 The Mean of Grouped Data Using the Class marks
When the number of items in a set of data is too big, items are grouped for
convenience. The manner of computing for the mean of grouped data is given by the
formula:
Mean 
 ( fX
 f
)
where: f is the frequency of each
class
X is the class mark of class
fX
f


53
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
The Greek symbol (sigma) is the mathematical symbol for summation. This means
that all items having this symbol are to be added. Thus, the symbol means the sum of all
frequencies, and means the sum of all the products of the frequency and the
corresponding
class mark.
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 -15
E
es:
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
f
1
5
11
12
11
5
2
1
Frequency
1
5
11
12
11
5
2
1
X
48
43
38
33
28
23
18
13
Compute the mean of the scores of the students in a Mathematics
fX
48
215
418
396
308
115
36
13
xampl
test.
The frequency distribution for the data is given below. The columns X and fX are
added.
54
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273


f 4
fX 1,5
55
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273

Mean 

1,5
Mean 
4
56
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Mean 32.
The mean score is 32.27.
2.2.2.2 The Mean of Grouped Data Using the Coded Deviation
An alternative formula for computing the mean of grouped data makes use of coded
deviation.

Mean  A.M .  
 
where:
A.M. is the assumed mean
f is the frequency of each class
d is the coded deviation from A.M.
i is the class interval
Any class mark can be considered as assumed mean. But it is convenient to
choose the class mark with the highest frequency. The class chosen to contain A.M. is
given a 0 deviation.
Subsequently, consecutive positive integers are assigned to the classes upward
and negative integers to the classes downward.
This is illustrated in the next examples using the same data in the previous
example.
Examples: Compute the mean of the scores of the students in Mathematics
test.
Class
46 - 50
41 - 45
Frequency
1
5
57
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
11
12
11
5
2
1
The frequency distributor for the data is given below. The columns X,
d and fd are added.
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
Solution:
f
1
5
11
12
11
5
2
1
X
48
43
38
33
28
23
18
13
d
3
2
1
0
-1
-2
-3
-4
fd
3
10
11
0
-11
-10
-6
-4
A
.
M
.

3
fd

f
4

i 5
,,,
58
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
  fd 
Mean  A.M .  
i
  f 
 33 
2,750     5
 30 
=
2,750  5
Mean 3,3
The mean gross sale is Php3,300.
2. 2.2.3 The Median of Grouped Data
The median is the middle value in a set of quantities. It separates an ordered set of
data into two equal parts. Half of the quantities found above the median and the other half is
found below it.
In computing for the median of grouped data, the following formula is used:
59
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
where:
lbmc is the lower boundary of the
median class
f is the frequency of each class
cf is the cumulative frequency of the
lower class next to the median
class
fmc is the frequency of the median
class
i is the class interval
The median class is the class that
median must be within the median class.
f
2
contains the
quantity. The computed
th
Examples:
1. Compute the median of the scores of the students in a Mathematics test.
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
Frequency
1
5
11
12
11
5
2
1
The frequency distribution for the data is given below. The columns for lb and
“less than” cumulative frequency are added.
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
f
1
5
11
12
11
5
2
1
lb
45.5
40.5
35.5
30.5
25.5
20.5
15.5
10.5
<cf
48
47
42
31
19
8
3
1
60
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
f
2

48
24
2
24
th
Since , the quantity is in the class 31 - 35. Hence, the median class is 31 - 35.
61
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273

i mc5
lb

30
f
4
f mc 
11
cf
Solution:
 f
Median lbmc  
2

62
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 48

19
 2
30.5  
12

 24  19

30.5  
 12
63
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 5
30.5  
 12 
2
30.5  
1

30.5  2.0
64
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Median 32.
The median score is 32.58.
2. 2.2.4 The Mode of Grouped Data
The mode of grouped data can be approximated using the following formula:
 D1 
Mode lbmo  
i
 D1  D2 
where:
lbmo is the lower boundary of the modal
class
D1 is the difference between the
frequencies of the modal class D2
is
the difference between the
frequencies of the modal class
and the next upper class
and the next lower class
i is the class interval
The modal class is the class with the highest frequency. If binomial classes
exist, any of these classes may be considered as modal class.
Examples: Compute the mode of the scores of the students in a Mathematics
test.
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
Frequency
1
5
11
12
11
5
2
1
The frequency distribution for the data given below. The column for lb
is added.
65
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Class
46 - 50
41 - 45
36 - 40
31 - 35
26 - 30
21 - 25
16 - 20
11 - 15
f
1
5
11
12
11
5
2
1
lb
45.5
40.5
35.5
30.5
25.5
20.5
15.5
10.5
Solution:
Since class 31 - 35 has the highest frequency, the modal class is 31 - 35.
lbmo = 30.5
D1 = 12 - 11 = 1
D2 = 12 - 11 = 1
i=5
 D
Mode lbmo  
D


 1
30.5  
 1 1
66
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 1
30.5  
 2

30.5  

30.5  2
67
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Mode 3
The mode score is 33.
Measures of Central Tendency
Measure
Mean
Common Name
Arithmetic
Average


Median
Middle score


Mode
Typical score

When to use
There are no
extreme scores
When the data
are interval or
ratio
The distribution is
skewed
When the data
are ranks
When a quick
estimate of the
typical score is to
be determined
Advantage
Most reliable
Stable and
less variable
from sample
to sample
 Easy
to
compute
 Not affected
by extreme
scores
 Easy
to
compute


Disadvantage
 Affected by
extreme
scores

Less stable
from sample
to sample

The most
unstable
measure
especially
when the
number of
scores is
small
Source: Dr. Gabino Petilos’ Hand-out
Relationship of the Three Measures of Central Tendency
A. For symmetric distributions
Mean = Median = Mode
68
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Source: Dr. Gabino Petilos’ Hand-out
B. For skewed distributions
a) Negatively skewed distributions
Mean Median M
Source: Dr. Gabino Petilos’ Hand-out
b) Positively skewed distributions
Mean Median M
Source: Dr. Gabino Petilos’ Hand-out
2. 3 Measures of Dispersion
The three measures of central tendencies that you have learned on the previous
lesson do not give an adequate description of the data. We need to know how the
observations spread out from average or mean. It is quite possible to have two sets of
observations with the same mean and median that differs in the variability of their
measurements about the mean. Measures of dispersion are sometimes called measures of
variability.
The measures of dispersion can be utilized in determining the size of the
distribution of scores or a portion of it. They can be used to find the deviation of scores from
the mean scores. Measures of dispersion can also be sued to establish the actual similarities
or the difference(s) of the distribution. In general, these measures are employed to further
characterize the distributions of scores.
Consider the following measurements, in liters, for two samples of apple juice in a
tetra packed by companies A and B.
Sample A
0.97
1.00
Sample B
1.06
1.01
69
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
0.94
1.03
1.11
0.88
0.91
1.14
Both samples have the same mean. It is quite obvious that company A packed
apple juice with a more uniform content than company B. We say that the variability or the
dispersion of the observations from the mean is less for sample A than for sample B.
Therefore, in buying apple juice, we would feel more confident that the tetra pack we select
will be closer to the advertised mean if we buy from company A.
Statistics other than the mean may provide additional information from the same
data. This statistics are the measure of dispersion.
Measures of dispersion or variability refer to the spread of the values about the
mean. These are important quantities used by statisticians in evaluation. Smaller dispersion
of scores arising from the comparison often indicates more consistency and more reliability.
2.3.1 The Range
The range is the simplest measure of variability. It is the difference between the
largest ad smallest measurement.
where:
R=H-L
R = Range,
H = Highest measure,
L = Lowest measure
The main advantage of the range is that it does not consider every measure in the
data.
Examples:
1. The IQs of 5 members of a family are 108, 112, 127, 118 and 113. Find the
range.
Solution:
The range of the IQs is 127 - 108 = 19.
2. The range of each of the set of scores of the three students is as follows:
Student A
Student B
Student C
H=98
H=97
H = 97
L=92
R = 98 - 92 = 6
L=90
R = 97 - 90 = 7
L = 90 R = 97 - 90 = 7
70
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Observe that two students are “tie”. This indicates that the range is not
a reliable measure of dispersion. It is a poor measure of dispersion,
particularly if the size of the sample or population is large. It considers only
the extreme values and tells us nothing about the distribution of numbers
in between.
3. Consider the following two sets of data, both with a range of 12.
In set A the mean and the median are both 8, but the numbers vary
over the entire interval from 3 to 15. In set B the mean and the median are also
8, but most of the values are closer to the center of the data. Although the
range fails to measure the dispersion between the upper and lower
observation, it does have some useful applications. In industry the range
for measurements on items coming off an assembly line might be specified
in advance. As long as all measurements fall within the specified range,
the process is said to be in control.
Disadvantages of the Range
1. It makes use of very little information: that is, it ignores but two items only.
2. It is totally dependent on the two extremes values, so it is greatly affected
by any changes in these values.
3. It should be used with caution, particularly with data that contain a single
extremely large value as this value would have a considerable effect on
the range.
4. It cannot identify the difference between two sets of data with the same
extreme values, example, the two sets of data are
2, 4, 6, 8, 10, 12, 14, 16, 18 and
2, 2, 2, 2, 2, 2, 2, 2, 18
both have the same range 16.
2.3.2 Range of a Frequency Distribution
The range of a frequency distribution is simply the difference between the upper
class boundary of the top interval an lower class boundary of the bottom interval.
Example: Scores in Midterm Exam of BEEd First Year Students
Scores
Frequency
46 - 50
1
41 - 45
10
36 - 40
10
31 - 35
16
26 - 30
9
21 - 25
4
Upper class boundary (UCB) = 50.5
Lower class boundary (LCB) = 20.5
UCB - LCB = 50.5 - 20.5 = 30
2.3.3 The Interquartile Range
Another measure of dispersion is the range of scores of specified part/s of the total
group usually the middle 50 percent of the cases lying between Q1 and Q2.
This measure is called the interquartile range. Thus, the interquartile range is the
difference between the third and the first quartiles, that is, Q 3 - Q1.
71
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Example 1: Consider the following sample raw scores of students in Statistics:
17, 17, 26, 28, 30, 30, 31, 37.
Solution:
nk (8)(1) 8
(2  3) th
Q1  
 2  Q1 
scores 
4
4
4
2

nk (8)(3) 24
6  7
30 
Q3  
 6  Q3 

4
4
4
2
2
th
I.Q.R = Q3 - Q1 = 30.5 - 21.5 = 9
Example 2: Consider the frequency distribution of test scores of 40 students in Teaching
Strategies below. Calculate the interquartile range.
Class interval
70 - 74
65 - 69
60 - 64
55 - 59
50 - 54
45 - 49
40 - 44
35 - 39
30 - 34
25 - 29
f
2
2
3
2
8
9
2
4
5
3
n = 40
Solution:
72
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
n = 40
k=1
nk

nk (40)(1) 40F


4
4
4
Q41 LQ1 
f Q1


Q1 
i mQ
5
11
L

34
Ff 

73
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 10 
34.5  
 4
2
34.5  
4

74
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
1
34.5  
4

34.5  2
75
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Q1 3
76
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
k=3
n = 40
Median class = 50 - 54
Fm-1 = 23
fQ3 = 8
LQ3 = 49.5
i=5
 nk
 F

nk  40  3 4 120
Q3 
LQ 3  
f4Q1
4
4 


77
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 (40)(3)


4
49.5  
23



78
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 30  2
49.5  
 8
I.Q. R. = Q3 - Q1
= 53.88 - 37
I.Q.R. = 16.88
7
49.5  
8

79
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
49.5 
3
Example 1: The test scores of nine students in Educational Psychology are as
follows: 15, 19, 20, 24, 28, 30, 32, 32, and 40. Compute the quartile deviation.
nk  9 1 9
Q1  
 2.25  Q1 3rd
4
4
4
nk  9 31 27
9
Q13  
 26.25
.75
 QQ1337rd
th
4
4
44
nk  9 3 27
Q3   Q Q 6.75
 Q20
3 7th
32
3
1
4
4
4
th
th
Q.D
.



nk  8 813 8 24
30
nk
(2  36)  7 
17
3

Q 
Q  
 2
6Q
score
Solution:
Q3 = 49.5 + 4.38 = 53. 88
Example 1: The test scores of nine students in Educational Psychology are as
2.3.4 The follows:
Quartile15,
Deviation
19, 20, 24, 28, 30, 32, 32, and 40. Compute the quartile deviation.
The quartile deviation (Q.D.) is another measure of dispersion that divides the
difference
of third and first quartiles into halves. It is the average distance from the median
Solution:
to the two (2) quartiles, i.e., it tells how far the quartile points (Q 1 and Q3) lie from the
median, on the average. When Q.D. Is small the set of scores is more or less
homogeneous but when Q.D. Is large, the set of scores is more or less heterogeneous.
This measure is used when there are extremely high and low scores especially when there
are big gaps between scores. It is also essentially used when the main concern is the
concentration of the middle 50% of the scores around the median. Mathematically,
Q.D. 
Q3  Q1
2
Example 2: The test scores of eight students in Statistics are as follows: 17, 17, 26,
13
44
44
4 42
1
3
2
by and
Jeconi37.
Joice
Tanggan-Paler
28, Downloaded
30, 30, 31,
Compute
the(jjstanggan@usm.edu.ph)
quartile deviation.
2
2
80
22
lOMoARcPSD|18274273
In interpreting the quartile deviation of any distribution of scores, the size of the
value is always the indicator. That is, if the value of the quartile deviation is high, the test
scores in the distribution can be interpreted as “heterogeneous” or the scores are more
scattered away from the mean; if the value of the quartile deviation is low, the scores are said
to be “homogeneous” or less scattered around the mean.
The interpretation of the quartile deviation result can be best achieved when
comparison between two quartile deviation values of any two distributions is considered.
Thus, between the distributions of scores in Statistics having a Q.D. of 4.5 and Educational
Psychology, Q.D. = 6, the scores in Statistics can be interpreted as less scattered around the
mean (homogeneous), while scores of students in Educational Psychology are more
distributed away from the mean.
81
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Example 3: The frequency distribution of 40 students in Assessment of Learning
are shown below. Calculate the quartile deviation.
Class Interval
70 - 74
65 - 69
60 - 64
55 - 59
50 - 54
45 - 49
40 - 44
35 - 39
30 - 34
25 - 29
f
2
2
3
2
8
9
2
4
5
3
n = 40
Solution:
n = 40
k=1
nk

nk (40)(1) 40F


4
4
4
Q41 LQ1 
f Q1


Q1 
i mQ
5
11
L

34
Ff 

82
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
  401


4
34.5  
4



83
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
 10 
34.5  
 4
2
34.5  
4

84
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
34.5 
1
Q1 34.5  2.5
2.3.5 The Average Deviation
The average deviation (A.D.) is a measure of absolute dispersion that is affected by
every individual score. It is the mean of the absolute deviations of the individual scores from
the mean of all the scores.
A large average deviation would mean that a set of scores is widely dispersed
about the mean, while a small average deviation would imply that the set of scores is closer
85
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
to the mean.
where: A.D. = average deviation
Ʃ = symbol for “summation”
n = total number of scores
X = individual scores
Steps in determining the average deviation:
1. Compute the mean from the given scores.
X
X X
2. Subtract the mean from the individual scores to get the deviation. That is, .
3. Get the sum of the deviation regardless of signs and divide it by (n-1), where n is the
total number of scores. The quotient is the average deviation.
86
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Example 1: The raw scores of eight students in Statistics are given as follows: 17,
17, 26, 28, 30, 30, 31, and 37. Compute the average deviation.
X
X  X
Solution:
17
-10
17
-10
26
-1
28
1
30
3
30
3
31
4
37
10
ƩX=216
X X 


A.D. 

42
X
42
42
57
.
3

X

7
8

A
.
D
.
9
8
A.D. 
7.1
n 1
For the scores organized in the form of frequency distribution, the average
deviation
is computed as follows:
where:
A.D = average deviation
Ʃ = symbol for “summation”
fi = frequency of the ith class interval
n = total number of scores
Xi = midpoint of the ith class interval
X

240

2
9
The computed average deviation (A.D.) of scores in Statistics is 6
while test scores in Psychology is 7.17. This can be interpreted as the scores
in Statistics are less dispersed about the mean while the scores in
Psychology are more dispersed around the mean.
X
In other words, the scores in Statistics having an A.D. of 6 are closely
distributed near the mean (homogeneous) while the scores in Psychology
having an A.D. of
7.17 are dispersed away from the mean (heterogeneous).
n
X
A.D. 
X 
X
n
n 1
87

216
27
8
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
The steps in determining the average deviation from frequency distribution are as
follows:
1. Find the mean of the frequency distribution.
2. Get the midpoint of each class interval.
3. Subtract the mean from the midpoint of each class interval to get the deviations
and
then, take their absolute values.
4. Multiply the frequency of each class  f i X i  X
interval to the corresponding
absolute deviation to get .
5. Get the sum of Step 4 and then divide it by (n - 1), where n is the total number of
frequencies. The quotient is the average deviation.
Example 1. Below is a frequency distribution of test scores of 40 students in Assessment
of Learning. Calculate the average deviation.
Class Interval
70 - 74
65 - 69
60 - 64
55 - 59
50 - 54
45 - 49
40 - 44
35 - 39
30 - 34
25 - 29
fi
2
2
3
2
8
9
2
4
5
3
n = 40
Solution:
Table ____
Calculation of Average Deviation from Frequency Distribution
Of the Sample Test Scores of 40 Students in
Assessment of Student Learning
88
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)

 i
Class Interval
fi
Xi
lOMoARcPSD|18274273
fiXi
fX
X

X
i
i X
70 - 74
65 - 69
60 - 64
55 - 59
50 - 54
45 - 49
40 - 44

35 - 39
30 - 34
25 - 29
2
2
3
2
8
9
72
67
62
57
52
47
144
134
186
114
416
423
2
4
5
3
n = 40
42
37
32
27
84
148
160
81
ƩfiXi= 1,890
24.75
19.75
14.75
9.75
4.75
-0.25
-5.25
49.5
39.5
44.25
19.5
38
-2.25
-10.5

-10.25
-15.25
-20.25
-41
-76.25
-60.75
f i X i  X 38
89
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Solving for the mean value:

X
fX
f

A.D. 
Xi  X
n
i
1
1,890


40
381.5 38


40 1
3
2.3.6 Variance and Standard Deviation
Standard deviation is the most important measure of variation or dispersion. It is the
average distance of all the scores that deviates from the mean value. It shows variation
about the mean. It is also known as the square root of the variance.
Variance is one of the most important measures of variability or dispersion. It shows
variation about the mean.
Population Variance
Sample Variance
Steps in Solving Variance of Ungrouped Data
1. Solve the mean value
2. Subtract the mean value from each score.
3. Square the difference between the mean and each score.
4. Find the sum pf step 4.
5. Solve for the population variance or sample variance using the formula of ungrouped data.
Population Standard Deviation
Deviation
Sample Standard
90
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Steps in Solving Standard Deviation of Ungrouped Data
1. Solve for the mean value.
2. Subtract the mean value from each score.
3. Square the difference between the mean and each score.
4. Find the sum of step 4.
5. Solve for the population standards deviation or sample standard deviation using
formula of ungrouped data.
the
Note: If the variance is already solved, take the square root of the variance to get the value
of the standard deviation.
Example: Below are the scores of 10 students in Mathematics quiz consists of
20 items. Compute the population and sample variance and population
and sample standard deviation.
Solution:
91
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
xx  XX
lOMoARcPSD|18274273
x
6
8
9
10
13
15
16
16
17
20
6 - 13 = -7
8 - 13 = -5
9 - 13 = -4
10 - 13 = -3
13 - 13 = 0
15 - 13 = 2
16 - 13 = 3
16 - 13 = 3
17 - 13 = 4
(-7)2 = 49
(-5)2 = 25
(-4)2 = 16
(-3)2 = 9
(0)2 = 0
(2)2 = 4
(3)2 = 9
(3)2 = 9
(4)2 = 16
20 13 = 7
(7)2 = 49



x

13

x
130

x

X

X

n
10
2
92
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Population Variance
Sample Variance
93
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273

X



x

X

186
N1
186

10
s 

2
2
10
9
94
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
s 20

18.6
186

X

186

x



s


N

10
4
.
3
s
4
.
5
n
1
18
10
9
X
22
2.3.7 Relative Measures of Variation
Coefficient of Variation shows a variation relative to the mean. It is used to compare
two or more groups of distribution of scores. Usually expressed in percent, the smaller the
value of the coefficient of variation, the more homogeneous the scores are. On the other
hand, the higher the value of the coefficient of variation, the more dispersed the scores are in
that particular distribution.
The formula in computing the coefficient of variation is,
Example: Find the coefficient of variation of the given data below.
95
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Section A
10
10
10
14
14
16
17
18
18
19
19
20
Section B
10
13
15
15
15
15
16
16
16
16
17
20
Section C
10
10
11
11
12
12
12
15
17
20
20
20
x
17
18

Solution:
ss
243..04
35
75

CV

x
100


x
100
CVCBA  100
x
14
17
15
33
15 42x
n = 12
n = 12
n = 12
CVC 28.51%
CV

24
.
32
15
33
A
B
X 14
15.1
43
2.4
96
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Measures of Relative Positions
An individual score has meaning only in relation to the rest of the scores. Thus, to
interpret a score, we have to use the entire distribution as basis for interpreting individual
scores.
We learned that the median is that point in the distribution below which lie 50% of
the scores. In exactly the same manner, we calculate the values on the scale below which lie
a certain percentage of the scores. These values are called quantiles.
The value that divide the distribution into 100 equal parts are called
percentiles. Thus, Px = xth percentile is the value on the scale which lie x% of the scores.
Examples:
P90 = the 90th percentile value is the value in the distribution below which
lie 90% of all the the scores
In a class consisting of 50 pupils, a pupil whose final grade corresponds
to P90 is said to belong to the upper 10% of the entire pupils in the class.
This also means that his grade is better than 90% (50) = 45 pupils in the class.
P10 = the 10th percentile value is the value below which lie 10% of all the
scores in the distribution.
P50 = the 50th percentile value below which lie 50% of all the other scores in
the distribution. Thus, P50 is the same as the median.
Other Quantiles
Deciles - values on the scale that divide the distribution into ten equal parts.
D1 - the first decile = the value on the scale below which lie 10% of the
scores in the distribution
D5 - fifth decile = P50 = Median
D9 - 9th decile = P90
Quartiles - values on the scale that divide the distribution into four equal
parts.
Q1 - first quartile = P25 = the value on the scale below which lie 25% of
the scores in the distribution. Thus, 75% of all the scores are
higher than Q1.
Q2 = D5 = P50 = Median
Q3 = P75
97
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Computation of Quantiles for Ungrouped Data
Steps:
1. Arrange the data in ascending order.
2. To determine the location of Px, we first get x% of (n+1) rounded to the next
whole number if the result is not a whole number.
Example 1. Find Q1 and Q3 for the following data:
95 81 59
68 100
71 88 100 94 87
Solution:
59 65 67 68 71
85 87 88 91 92
92 75 67 85 79
65 93 72 83 91
72 75 79 81
83
93 94 95 100 100
To find the Q1 which is the same as P25, we get
25%(21) = 0.25 x 21 = 5.25
Hence, Q1 is the 6th score or Q1 = 72.
Similarly, to find Q3, we get 75%(20+1) = 0.75 x 21 = 15.75. Thus Q3 is
the 16th score or Q3 = 93.
Instead of rounding the value to the next whole number, we can get a
more accurate value of quantiles through interpolation. For instance, since
25% (21) = 0.25 x 21 = 5.25, this means that Q1 is the score that is 1/4 of the
way from the 5th and the 6th score. To get Q1, we add to the 5th score 1/4 of
the difference between the 6th and the 5th score. Thus,
Q1 = 71 + 0.25(72 - 71) + 0.25 = 71.25
Example 2. Find P11 and P93 for the data given above.
Solution:
Since x = 11, we get 11%(20+1) = 0.11x 21 = 2.31. The value 2.31 suggests
that P11 is a value that is 0.31 of the way from the 2 nd score to the third score. To get
P11, we got 0.31 of the difference between the 2 nd and 3rd score and add the result to
the 2nd score. Thus we have,
0.31 (67 - 65) = 0.31 (2) = 0.62
And since the 2nd score is 65, we have P11 = 65 + 0.62 = 65.62
Similarly, 93%(20+1) = 0.93 x 21 = 19.53. P93 is the score that is 0.53 of the
way from the 19th score to the 20th score. Thus,
P93 = 100 + 0.53 (100-100) = 100 + 0 = 100
Computation of Quantiles for Grouped Data
98
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
The computation of quantiles for grouped data is similar to the computation of the
median (P50) for grouped data.
The formula for finding the median is given by
Example 1. Find a) Q1 and b) Q3 for the data given below:
Class Interval
f
<cf
95 - 99
1
30
90 - 94
4
29
85 - 89
8
25
80
84
10
17
Since Px is defined as the value below which lie x% of the total number of
75 - 79
4
7
cases, we can revised the above formula by merely changing to x% (n). Thus
70 - 74
3
3
30
where:
LL = the true limit of the class interval containing Px
Solution:
Fb = the <cf below the interval containing Px
a) Since
Q1 = Pof
weinterval
first getcontaining
25%(n) to
25, the
f = frequency
Pxdetermine the interval containing
Q
n = 1.number of cases
= 0.25 x 30 = 7.5. We next look at the <cf column. Since 7.5 is
i =25%(30)
class size
between 7 and 17, we identify the interval 80 - 84 as containing P25.
Thus,
LL = 79.5
; Fb = 7;
f = 10;
i=5
Therefore,
 x7
%(
n
)

F
.
5

7

P

LL

x
79.5  5f.5
0

10
22
.
5

17


x
%(
n
)

F

84
.
5

79


.5 
Px 84
LL
x   10
f
88
2

27


79
.
5



84
.
5

P 79
.
7
1
 8
99
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
= 79 5 + 0 25
lOMoARcPSD|18274273
Assessment
Directions: Answer the following. Show all pertinent solutions.
1. Find the mean, median, and mode/modes of each of the following sets of data:
a. 103, 234, 156, 365, 234, 268, 333, 103, 256, 365
b. 18, 24, 25, 16, 35, 21, 24, 33, 34, 25, 45,33,28, 17, 18, 16, 21, 45
2.Find the range, average deviation, variance, and standard deviation of the following
sets of data:
a. 70, 65, 69, 73, 90, 87, 81, 89.
b. 24, 27, 32, 29, 31, 35, 27, 32, 23, 25, 30, 24.
3.The salaries of all the 130 employees of a company are tabulated in a frequency
distribution as shown below
Salaries (in thousand pesos)
Number of
Workers
33 - 36
4
29 - 32
10
100
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
25 - 28
12
21 - 24
24
17 - 20
38
13 - 16
18
9 - 12
15
5-8
6
1-4
3
Find the range, average deviation, variance, and standard deviation.
5. The results of the midterm examination in Differential Calculus of BS Math Junior
students were taken and are presented in a frequency distribution. Find the
mean, median, and mode.
Class Interval
f
94- 99
2
88- 93
7
82 - 87
19
76- 81
8
70- 75
10
64- 69
28
58- 63
37
52- 57
19
46- 51
8
40- 45
1
6. The test scores of 18 students in Analytic Geometry and Calculus I are as follows:
27, 48, 33, 39, 52, 25, 50, 47, 42, 32, 21, 28, 42, 45, 55, 20, 37, and 38.
Determine the following:
a. Q1
b. P68
c. D6
7. The table below gives the age distribution of 100 individuals living in the vicinity of
Escolta.
Age
Frequency
55 - 59
2
50 - 54
5
45 - 49
10
40 - 44
12
35 - 39
15
30 - 34
16
101
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
Solve for the following:
a. P85
25 - 29
13
20 - 24
10
15 - 19
4
10 - 14
4
b. Q3
d. D3
Note: WRITE YOUR ANSWERS on A SHORT BOND PAPER. Compile your
exercises/assessments and placed it in a short FOLDER (not sliding folder)
labelled with your NAME, STUDENT NUMBER, COURSE and SECTION.
References:
Buendicho, Flordeliza C. (2013). Assessment of Student Learning I. Manila, Philippines: REX
Book Store
Christian Brothers University (2016). Wrting perfect learning outcomes. Available online:
https://www.cbu.edu/assests/2091/writing_perfect_learning_outcomes.pdf
Department Order No. 73, series of 2012 - Guidelines on the Assessment and Rating of
Learning Outcomes under the K to 12 Basic Education Curriculum. Available
online:
http://www.deped. gov.ph/wp-content/uploads/2018/07/DO_s2012_73.pdf
Garcia, Carlito D. (2013). Measuring and Evaluating Learning Outcomes: A Textbook in
Assessment of Learning 1 & 2 2nd edition. Mandaluyong City, Philippines: Books
Atbp. Publishing Corp
Krathwohl, D. R. (2002). A revision of Bloom’s taxonomy: An Overview. Theory into practice,
41(4), 212-218. Retrieved from https://cmapspublic2.ihmc.us/rid=1Q2PTM7HL26LTFBX-9YN8/Krathwohl%202002.pdf
Navarro, Rosita L., et.al. (2019). Assessment of Learning 4th edition. Manila, Philippines:
LORIMAR Publishing, Inc..
https://www.statisticshowto.com/probability-and-statistics/coefficient-of-determination-r-squared/
https://www.statisticshowto.com/inter-rater-reliability/
https://chfasoa.uni.edu/reliabilityandvalidity.htm
Prepared by:
(SGD) MARIA MILAGROS D. DAIZ
Asso Prof I
Approved by:
(SGD) ALVIN B. LACABA, Ph.D.
College Dean
MERRY CHRISTMAS
102
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)
lOMoARcPSD|18274273
and
A PROSPEROUS NEW YEAR...
KEEP SAFE...
103
Downloaded by Jeconi Joice Tanggan-Paler (jjstanggan@usm.edu.ph)