The document outlines the process of presenting data in an organized manner, specifically through statistical tables, charts, and graphs. It details the construction of frequency distributions, including determining class intervals, frequencies, and cumulative distributions, as well as methods for computing means and medians for both ungrouped and grouped data. Examples are provided to illustrate these concepts, alongside methods for weighted means and unit deviation calculations.

1. Data presentation Data collected should
be presented in an organized manner to
make a meaningful interpretation. After
organizing the data, it must be presented by
constructing statistical tables, charts or
graphs. When data are collected in original
form, the data are called raw data.
Frequency Distribution

2. Frequency distribution table are the most
common presentation of gathered data. It
consists of columns indicating the variables
measured the corresponding frequency and
the equivalent percent. For the large amount
of data, presentation can be summarized in
classes and groups.

3.  Class limits refers to the lowest and the highest value that can be
entered in each class.
 The lowest value that can go in each class known as the lower
limits and the highest value that can go in each class is called the
upper class limits frequency. The number of values that fall in a
given interval.
 In frequency distribution, it is assumed that the values are evenly
distributed within the interval. There are some instances, however,
where an interval has to be summarized and be represented by
single value. This value called that the midpoint or class mark
serves as the represented of the given interval. Generally, the
midpoint is obtained by adding the class limits and then dividing
the sum by two. .

4.  Construction of a Frequency Distribution
1. Get the lowest and the highest value in the
distribution. We shall let H and L be the highest and
lower value in the distribution.
2. Get the value of the range. The range denoted by R,
refers to the difference between the highest and the lowest
value in the distribution. Thus
R = H – L

5.  3. Determine the number of classes. In the determination of the
number of classes, it should be noted that there is no standard
method to follow. Generally, the number of classes must not be less
than 5 and should not be more than 15. In the some instances,
however, the number of classes can be approximately by using the
relation.
K = 1+3.3 log n
Where: k =is the number of classes
n=is the sample size
 4. Determine the size of the class interval. The value of c can be
obtained by dividing the range by the desired number of classes.
Hence,
C = R/K

6. 5. Construct the classes. In constructing the classes, we
first determine the lowest lower limit of the distribution.
The value of this lower limit can be chosen arbitrarily as
long as the lowest value shall fall on the first interval and
the highest value to the last interval.
6. Determine the frequency of each class. The
determination of the number of the frequencies is done by
counting the number of items that fall in each interval.

7.  Example:
 A Mathematics Class with 60 students was given an
examination and the results are the following:
48 73 57 57 69 88 11 80 82
47
46 70 49 45 75 81 33 65 38
59
94 59 62 36 58 69 45 55 58
65
30 49 73 29 41 53 37 35 61
48
22 51 56 55 60 37 56 59 57

8. Solution:
1. Get the lowest and the highest value
H = 94 and L = 11
2. Get the range
R = H-L
= 94-11
= 83
3. Determine the number of class interval
K = 1+3.3 log n
= 1+3.3 log60
= 6.867899126 or 7

9. 4. Determine the size of the class interval
 C = R/K
 = 83 / 7
= 11.86 or 12
5. Construct class
Classes
11 – 22
23 – 34
35 – 46
47 – 58
59 – 70
71 – 82
83 – 94

10. 6. Determine the frequencies
Classes f
11 – 22 3
23 – 34 5
35 – 46 11
47 – 58 19
59 – 70 14
71 – 82 6
83 – 94 2
n = 60

11. 
Relative Frequency Distribution

12. Classes f %f
11 – 22 3 5
23 – 34 5 8.33
35 – 46 11 18.33
47 – 58 19 31.67
59 – 70 14 23.33
71 – 82 6 10
83 – 94 2 3.33
n = 60 99.99%

13. The Cumulative Frequency Distribution can also be
derived from frequency distribution. This distribution can
be obtained by simply adding the class frequencies.
There are two types of cumulative frequency distribution.
These are as follows;
1. Less than cumulative frequency refers to the
distribution whose frequencies are less than or below
the upper class boundary they correspond to. We shall
let <cumf be the less than cumulative frequency.

14.  Greater than cumulative frequency
refers to the distribution whose
frequencies are greater than or above the
lower class boundary they correspond to.
We shall let >cumf be the greater than
cumulative frequency

15. Classes f <cumf >cumf
11 – 22 3 3 60
23 – 34 5 8 57
35 – 46 11 19 52
47 – 58 19 38 41
59 – 70 14 52 22
71 – 82 6 58 8
83 – 94 2 60 2
Example

16.  Mean - one of the simplest and most efficient
measures of the central tendency is the mean.
It is the value obtained by adding the values in
the distribution and dividing the sum by the
total number of values . To compute the mean
for ungrouped data, we shall let X be the value
of the mean, then by definition, we have:
n
In symbols, we have x
̄ =
Measure of Central Tendency

17. Example: Consider the following values.
21 10 36 42 39 52 30 25
26
Compute the value of the mean.
Solution:
To compute for the mean, we shall use the
formula:
x
̄ =
x
̄ = (21 + 10 + 36 + 42 + 39 + 52 + 30 + 25 + 26) ÷ 9
= 281 ÷ 9
= 31.22

18. Weighted mean
 There are some instances
where, in the computation of
the mean of a set of data,
each value in the distribution
is associated with a certain
weight or degree of
importance.

19. Subject Units Grade WX
1 3 2.0 6.0
2 3 3.0 9.0
3 5 1.25 6.25
4 1 3.0 3.0
5 2 2.5 5.0
6 3 2.5 7.5
Example:
A student in a certain college is enrolled in 6
subjects where not all of the subjects carry a tree
unit load. Assuming further, that the said student
was able to obtain the following grades as shown
below.

20. If each subject carries a three unit load, then
the mean grade can be obtained by simply
adding the grade in the third column and then
divide the sum by six.
The method discussed above can be
represented by the formula:
x
̄ =
Where: x = represents the item value
w = represents the weighted associated
to x.

21. To compute the value of the weighted mean, we
have
 x
̄ =
= 3(2.0) + 3(3.0) + 5(1.25) + 1(3.0) +2(2.5)
+3(2.5)
3 + 3 + 5 + 1 + 2 + 3
= 36.75 ÷ 17
= 2.16

22. Mean for grouped data
 To compute the mean of a data represented in a
frequency distribution, we shall consider two methods:
Midpoint Method and Unit deviation Method
In using the midpoint, the midpoint of each class interval
is taken as the representative of each class. The
midpoints are multiplied by their corresponding
frequencies. The products are added and the sum is
divided by the total number of frequencies. The value
obtained is considered the mean of the grouped data.
x
̄ =
Where: f = represents the frequency of each class.
x = the midpoint of each class
n = the total number of frequencies or sample size

23. To compute the midpoint methods use the
following step:
1. Get the midpoint of each class
2. Multiply each midpoint by its corresponding
frequency.
3. Get the sum of the products in step 2
4. Divide the sum obtained in step 3 by the total
number of frequencies. The result shall be
rounded off to two decimal places.

24. Example:
A Mathematics Class with 60 students was given
an examination and the results are the following:
48 73 57 57 69 88 11 80 82
47 46 70 49 45 75 81 33 65
38 59 94 59 62 36 58 69 45
55 58 65 30 49 73 29 41 53
37 35 61 48 22 51 56 55 60
37 56 59 57 36 12 36 50 63
68 30 56 70 53 28

25. Classes f x
11-22 3 16.5
23-34 5 28.5
35-46 11 40.5
47-58 19 52.5
59-70 14 64.5
71-82 6 76.5
83-94 2 88.5
Solution:
To be able to compute the value of the mean, we
shall follow the steps discussed above.
•Get the midpoint of each class

26. Multiply each midpoint by its corresponding
frequency.
Classes f x fx
11-22 3 16.5 49.5
23-34 5 28.5 142.5
35-46 11 40.5 445.5
47-58 19 52.5 997.5
59-70 14 64.5 903.0
71-82 6 76.5 459.0

27. Classes f x fx
11-22 3 16.5 49.5
23-34 5 28.5 142.5
35-46 11 40.5 445.5
47-58 19 52.5 997.5
59-70 14 64.5 903.0
71-82 6 76.5 459.0
83-94 2 88.5 177.0
n = 60 ∑fx = 3,174
•Get the sum of the products in step 2.

28. Divide the result in step 3 by the
sample size.
x
̄ =
=
= 52.9

29.  The alternative method of
computing the value of the mean
for grouped data is the Unit
Deviation Method. Instead of
using midpoints, this method
uses unit deviations.

30. Example:
If the distribution has 9 classes and the fifth class
interval is the assumed class mean, then the
entries in the unit deviation column shall be -4, -3,
-2, -1, 0, 1, 2, 3, 4. However, if the assumed class
means is the 4th
class interval, then the entries in
the unit deviation column will be -3, -2, -1, 0, 1, 2, 3
respectively. The unit deviations are usually
represented by d.

31.  The third step is implemented
by multiplying the frequencies
by their corresponding unit
deviations. The products are
added and the sum is divided
by the sample size. The result
is multiply by the size of the
size of the class interval.

32. Finally, the value of the mean is determined
by adding the product to the assumed
mean.
The formula as follows:
X =xa + ( ) c
Where: xa = represents the assumed mean
f = the frequency of each class
d = the unit deviation
c = the size of the class interval
n = the sample size.

33.  Step in solving unit deviation
1. Choose an assumed mean by getting the
midpoint of any interval
2. Construct unit deviation
3. Multiply the frequencies by their corresponding
unit deviation. Add the products.
4. Divide the sum in step 3 by the sample size.
5. Multiply the result in step 4 by the size of the
class interval.
6. Add the value obtained in step 5 to the assumed
mean. the obtained result which is the mean should
be rounded off to two decimal places.

34. Example:
A Mathematics Class with 60 students was given
an examination and the results are the following:
48 73 57 57 69 88 11 80 82
47 46 70 49 45 75 81 33 65
38 59 94 59 62 36 58 69 45
55 58 65 30 49 73 29 41 53
37 35 61 48 22 51 56 55 60
37 56 59 57 36 12 36 50 63
68 30 56 70 53 28

35. Classes f
11-22 3
23-34 5
35-46 11
47-58 19
59-70 14
71-82 6
Solution:
To be able to compute the value of the mean, we shall
follow the steps discussed above.
•Choose an assumed mean by getting the midpoint of
any interval

36. Construct Unit deviation
Classes f d
11-22 3 -3
23-34 5 -2
35-46 11 -1
47-58 19 0
59-70 14 1
71-82 6 2

37. Classes f d fd
11-22 3 -3 -9
23-34 5 -2 -10
35-46 11 -1 -11
47-58 19 0 0
59-70 14 1 14
71-82 6 2 12
83-94 2 3 6
•Multiply the frequencies by their corresponding
unit deviation.

38. Step 4, 5, and 6, we will
now apply.
x
̄ = xa + ( ) c
= 52.5 + ( ) 12
= 52.9

39.  Median
 In the process of computing the mean, we
observed that all the values are taken into
consideration. Thus, if a distribution
contains extreme values, then the value of
the mean is usually pulled either to the right
or the left depending on the position of
these extreme values. This measure, called
the median, is a positional measure defined
as the middlemost value in the distribution.
Hence the value divides a given set of data
into two equal parts.

40. Median for ungrouped data
In the determination of the median of
ungrouped data, it is always a must that
the values be arranged in terms of
magnitude either from lowest to highest
or vice versa.
Let x be the median
x
̄ = (n + 1) / 2 if n is odd
x
̄ = (n/2 +1) if n is even

41. Example:
Find the median of the following values.
21, 10, 36, 42, 39, 52, 30, 25, 26
Solution: before identifying the value of the
median, it is necessary that the values be
arranged in terms of magnitude. Thus we have
10, 21, 25, 26, 30, 36, 39, 42, 52
Since n = 9 and is odd, then we use the formula:
x
̄ = (n+1)/2 =(9 + 1)/2 =10/2 = 5
Therefore the median is in fifth value is equal
to 30.

42. Another example:
The following values are the number of students of
the first 8 classes in a certain college taken for
inspection:
21, 25, 26, 30, 36, 39, 42, 55
Determine the median.
Solution:
x
̄ = (x/2) +1 = (8/2) + 1 = 4+1 = 5
x
̄ = (x4 + x5) / 2 = (30 +36) / 2= 66/2 = 33

43. Median for Grouped data
Just like the mean, the
computation of the value of the
median is done through
interpolation. The procedure
requires to construction of the less
than cumulative frequency column
(<cumf).

44. The step in finding the value of the median
1.Get 1/2 the total number of values.
2.Determine the value of cumulative frequency
before (cumfb)
3.Determine the median class.
4.Determine the lower boundary and the frequency
of the median class and the size of the class
interval.
5.Substitute the values obtained in step 1-4 to the
formula. Round off the final result to two
decimal places.

45. Formula for grouped data
x
̄ = xlb
Where: xlb = refers to the lower boundary
of the median class
cumfb = the cumulative frequency
before the median class
fm = the frequency of the median
class.

46. Example:
A Mathematics Class with 60 students was given
an examination and the results are the following:
48 73 57 57 69 88 11 80 82
47 46 70 49 45 75 81 33 65
38 59 94 59 62 36 58 69 45
55 58 65 30 49 73 29 41 53
37 35 61 48 22 51 56 55 60
37 56 59 57 36 12 36 50 63
68 30 56 70 53 28

47. Classes f <cumfb
11-22 3 3
23-34 5 8
35-46 11 19 cumfb
47-58 19 fm 38 Median class
59-70 14 52
71-82 6 58
83-94 2 60
Solution: we shall first construct the less than frequency column.
Using the steps indicated, we have

48. Steps:
1. n/2 = 60/2 = 30
2. cumfb = 19
3. median class = 47-58
4. xlb = 46.5; fm = 19; c = 12
5. x = xlb
x
̄ = 46.5
x
̄ = 46.5 + (0.578947368)12
x
̄ = 46.5 + 6.947368421
x
̄ = 53.45

49.  Mode
The mode is referred to as the most frequent value in the
distribution.
Mode for ungrouped data, the value of the mode can be
obtained through inspection, thus, no computation is
needed. In some instances, the mode might exist or it
might not exist. If it exist, it can be more than one value.
Let us consider the following set of measurements as
example:
A = 31, 21, 16, 15, 21, 27, 19, 18
B = 17, 25, 34, 25, 27, 19, 19, 24

50. In set A, notice that the value 21 appeared twice. Since
this value has the most number of occurrence, there we
may say that x = 21. In set B, there are two most
frequent values in the distribution. Hence, we can say
that the distribution contains two values representing
the mode. These values are 19 and 25.
 Mode for grouped data
In the computation of the value of the mode for grouped
data, it is necessary to identify the class interval that
contains the mode. This interval, called the modal class,
contains the highest frequency in the distribution.

51. We are consider the following step in getting the
mode of grouped data
1. Determine the modal class.
2. Get the value of d1
3. Get the value of d2
4. Get the lower boundary of the modal class
5. Apply the formula by substituting the values obtained in
the preceding steps.
Formula:
x
̄ = xlb

52. Example:
A Mathematics Class with 60 students was given
an examination and the results are the following:
48 73 57 57 69 88 11 80 82
47 46 70 49 45 75 81 33 65
38 59 94 59 62 36 58 69 45
55 58 65 30 49 73 29 41 53
37 35 61 48 22 51 56 55 60
37 56 59 57 36 12 36 50 63
68 30 56 70 53 28

53. Classes f
11-22 3
23-34 5
35-46 11
47-58 19 Modal class
59-70 14
71-82 6
83-94 2
Solution: Construct the frequency distribution

54. To get the value of d1 and d2 we have
d1 = 19 – 11 =8
d2 = 19 – 14 = 5
Substituting these values to the formula:
x
̄ = xlb
x
̄ = 46.5
x
̄ = 46.5 + (0.615384615)12
x
̄ = 46.5 +7.384615385
x
̄ = 53.88