The document discusses measures of central tendency for ungrouped and grouped data. It defines mean, median, and mode, and provides examples of calculating each for ungrouped and grouped data sets. Formulas and step-by-step workings are shown for finding the mean, median, and mode of grouped data using frequency distributions.

2. U M O D E A B C V N N M M
M N A S D S F H K H H Y H
I W G R O U P E D D A T A
D K Y R E M E A N Y R O N
D I R T O Y R Y E W S T D
L N Z N B U A L W D B A M
E T Q D A E P T M Z B L O
Q E Y D L M N E E K P O D
A R R Y U I O P D J N B A
Z V V H Y R F A I D D H L
G A W E R T Y U A Y A H B
T L F R E Q U E N C Y T F
F U R E G X A S D G J K A
L O W E R B O U N D E R Y
H U E G B A S F G H J K L

3. SUM MEDIAN
MEAN MODE
MIDDLE UNGROUPED DATA
TOTAL GROUPED DATA
FREQUENCY MODAL
INTERVAL LOWER BOUNDERY

4. Measures of Central
Tendency
(Ungrouped and Grouped Data)

5. Objectives
determine
measures of
central tendency
when the data is
ungrouped and
grouped.
define mean,
median and mode. show cooperation
through active
participation in
class.
01
02
03

6. Measures of
Central Tendency
(Ungrouped Data)
01
02
03
04
05
06

7. An ungrouped data is a set of values that
is not organized or classified as a
group. A measure of central tendency is a
value that represents the whole set of
data. These are mean, median and mode.

8. The mean of ungrouped data is obtained by adding
all the values divided by the frequency of a set of
data. The mean of ungrouped data is also called
average. It is written as:
𝑥̅=
∑ 𝐱
𝒏
where;
𝑥̅ = is the mean
∑ 𝒙 = is the sum of all the values in a set
𝒏 = is the frequency

9. The number of confirmed COVID 19 cases from
March 21- 27, 2020 are as follows: 77, 73, 82, 90,
84, 71, 96. To get the mean, we have;
𝑥̅ =
∑ 𝐱
𝒏 𝑥̅ =
77+73+82+90+84+71+96
𝟕
EXAMPLE 1
𝑥̅ =
𝟓𝟕𝟑
𝟕
𝑥̅ = 81.857142 …
𝑥̅ = 82

10. Example 2:
Amy’s score in exam are 23, 21, 28, 29, 23. What is
the mean of the given set of data?
𝑥̅ =
∑ 𝐱
𝒏
𝑥̅ =
𝟐𝟑+𝟐𝟏+𝟐𝟖+𝟐𝟗+𝟐𝟑
𝟓
𝑥̅ =
𝟏𝟐𝟒
𝟓
𝑥̅ = 24. 8 or 25

11. The median of ungrouped data is the middle value of a set of
data when all values are arranged in either ascending or
descending order.
 Arrange the data in ascending order.
 If the frequency of the data is odd , then the middle value will be
median of the set of data. And If the frequency of the data is even, the
median of the data is the mean of the two middle values.
 To find middle values of the data used this formula.
𝒙̃ =
𝐧+𝟏
𝟐
where; 𝒙̃ − is the median
𝒏 − is the frequency

12. The number of confirmed COVID 19 cases from
March 21- 27, 2020 are as follows: 77, 73, 82, 90,
84, 71, 96. Find the median.
 Arrange the data in ascending order
71,73,77, 82, 84, 90, 96
𝑥̅ =
𝐱+𝟏
𝟐
𝑥̅ =
𝟕+𝟏
𝟐
𝑥̅ =
𝟖
𝟐
𝑥̅ = 4th score

13. The score in the exam of students are 33, 32,35,
39, 40, 32, 30, 39 what is the median?
EXAMPLE 3
Arrange the set of data in ascending order
30, 32, 32, 33, 35, 39, 39, 40
𝒙̃ =
𝑛+1
2 The 𝟒. 𝟓 𝒕𝒉 score of the data is the mean of
the 4th and 5th score. We have,
33+35
2
=
68
2
=34
𝒙̃ =
8+1
2
𝒙̃ =
9
2
𝒙̃ = 4.5

14. The mode of the ungrouped data is the value that most
frequently appears in a set of data. When the value in a set
of data appears only once, then the data has no mode.
The data can be also classified according to the number of
modes it has.
Unimodal- one mode
Bimodal- two modes
Multimodal- more than 2 modes

15. The number of confirmed COVID 19 cases from March
21- 27, 2020 are as follows: 77, 73, 82, 90, 84, 71, 96.
What is the mode?
- no mode
Find the mode of the given set of data: 8, 4, 7, 9, 5, 5, 5, 7
Mode is 5 Multimodal

17. The data shows the ages of first 55 confirmed Covid-19
Cases Tested in the Philippines. Find the mean, median and
mode of the distribution.
To solve for the mean of
grouped data, use the formula:
where:
𝑋̅ = is the mean
∑ 𝑓𝑋 =is the sum of the
products of the frequency and
its corresponding class mark
N = is the total frequency.
𝑿=
∑ 𝒇𝑿
𝑵

18. Class mark is the mid value of the class interval. It can be solved by
adding the lower-class limit and upper-class limit and divided by 2.
Table 1. Frequency Distribution of the Ages of the first 55 confirmed Covid-19
Patient in the Philippines
Ages of
Covid-19
Patient
Frequency Class
Marks (X)
fX
80-89 3 84.5 253.5
70-79 7 74.5 521.5
60-69 11 64.5 709.5
50-59 12 54.5 654
40-49 9 44.5 400.5
30-39 8 34.5 276
20-29 5 24.5 122.5
𝒊 = 𝟏𝟎 𝑵 = 𝟓𝟓 𝒇𝑿 = 𝟐, 𝟗𝟑𝟕. 𝟓
𝑋=
∑ 𝑓𝑋
𝑁
𝑋=
2937.5
55
𝑋= 53.41

19. To solve for the median of grouped data, use the formula:
𝑿= 𝑳𝑩𝒎 +
𝑵
𝟐
−<𝒄𝒇𝒃
𝒇𝒎
𝒊
Where:
𝑋̃ = median
𝐿𝐵𝑚 = is the lower boundary of the median class
<𝑐𝑓 𝑏 = is the cumulative frequency immediately below/lower
than the median class
𝑓𝑚= is the frequency of the median class
𝑁
2
= median class
𝑖 = is the class size

20. Table 2. Frequency Distribution of the Ages of the first 55 confirmed
Covid-19 Patient in the Philippines
Median Class
𝑿= 𝑳𝑩𝒎 +
𝑵
𝟐
−<𝒄𝒇𝒃
𝒇𝒎
𝒊
Ages of
Covid-19
Patient
Frequency Lower
Boundary
(LB)
<cf
80-89 3 79.5 55
70-79 7 69.5 52
60-69 11 59.5 45
50-59 12 49.5 34
40-49 9 39.5 22
30-39 8 29.5 13
20-29 5 19.5 5
𝒊 = 𝟏𝟎 𝑵 = 𝟓𝟓
𝑵
𝟐
=
𝟓𝟓
𝟐
= 27.5
𝐿𝐵𝑚 = 49.5
< 𝑐𝑓𝑏=22
𝑓𝑚 = 12
𝑖= 10
we have the value of,

21. 𝑿= 𝑳𝑩𝒎 +
𝑵
𝟐
−<𝒄𝒇𝒃
𝒇𝒎
𝒊
𝐿𝐵𝑚 = 49.5
< 𝑐𝑓𝑏=22
𝑓𝑚 = 12
𝑖= 10
𝑋= 49.5 +
27.5−22
12
10
𝑋=49.5 + (0.4583)10
𝑋= 49.5 + (4.583)
𝑋= 54.083 or 54.08 Median
Median

22. Mode of Grouped Data
To solve for the mode of grouped data, use the formula:
𝑿=𝑳𝑩𝒎𝒐 +
𝒅𝟏
𝒅𝟏+𝒅𝟐
𝒊
where:
𝑋̂= is the mode
𝐿𝐵𝑚𝑜 = is the lower boundary of the modal class
𝑑1= is the difference between the frequencies of the modal class
and the next lower class
𝑑2= is the difference between the frequencies of the modal class and
the next upper class
𝑖 = is the class size.

23. Ages of
Covid-19
Patient
Frequency Lower
Boundary
(LB)
80-89 3 79.5
70-79 7 69.5
60-69 11 59.5
50-59 12 49.5
40-49 9 39.5
30-39 8 29.5
20-29 5 19.5
Table 2. Frequency Distribution of the Ages of the first 55 confirmed Covid-
19 Patient in the Philippines
Modal Class
𝒊 = 𝟏𝟎 𝑵 = 𝟓𝟓
we have the values of,
𝐿𝐵𝑚𝑜= 49.5
𝑑1= 12-9 = 3
𝑑2= 12-11 = 1
𝑖= 10
𝑿=𝑳𝑩𝒎𝒐 +
𝒅𝟏
𝒅𝟏+𝒅𝟐
𝒊

24. 𝑿=𝑳𝑩𝒎𝒐 +
𝒅𝟏
𝒅𝟏+𝒅𝟐
𝒊 we have the values of,
𝑳𝑩𝒎𝒐= 49.5
𝒅𝟏= 12-9 = 3
𝒅𝟐= 12-11 = 1
𝒊= 10
𝑿=𝟒𝟗. 𝟓 +
𝟑
𝟑+𝟏
𝟏𝟎
𝑿= 49.5 + (0.75)10
𝑿= 49.5 + 7.5
𝑿= 57 Mode

25. Activity
Part I. Ungrouped Data
Direction: Find the mean, median and mode.
The water consumption (in cubic meter) for the past 6
months is shown below.
January 23
February 24
March 25
April 25
May 26
June 24

26. Part II. Grouped Data
Direction: Complete the table, then solve for the mean, median and mode
of the distribution.
Height of Grade 10 Students
Height in
cm
Frequency Class
Mark
fX Lover
Boundary
<cf
180 – 185 1 182.5 182.5 179.5
174 – 179 2 176.5 49
168 – 173 8 167.5
162 – 167 15 2467.5 39
156 – 161 17 158.5 155.5
150 – 155 7 1067.5 7
𝒊=____ N=___ ∑ 𝒇𝑿=__

27. 01
02
03
04
05
06