This PDF file is a lesson plan about the mean of grouped.

1. DEPED
Mean of Grouped Data
Lesson Plan For Demonstration-Senior High
Elton John Balignot Embodo
Applicant
This Lesson Plan is intended for the demonstration in applying a position Teacher I

2. I. Objectives: At the end of the lesson, students are expected to:
a. enumerate the steps in calculating the mean;
b. solve for the mean of grouped data;
c. describe the importance of the mean in real life
situations.
II. Subject Matter: The Mean of Grouped Data
III. Reference: Introduction to Statistics page 39 – 49 by: Francisco A. Febre, Jr.
Skills: analysing and solving
Values: center of life
IV. Materials: power point (multimedia presentation), projector, and laptop
V. Procedure: Developmental Method
Teacher’s Activity Students’ Activity
A. Preparation
a. Review
Before we formally start our lesson this
afternoon, let’s have first a review. To refresh
your knowledge about calculating the mean of
ungrouped data, do this short activity in your
seats.
Directions: Calculate the mean of the
following final grades obtained by Mr.
Embodo in his Education subjects in College.
97 92 97 93
91 92 93 93
94 97 88 94
Very good!
Do you still have any clarification in
computing the mean of ungrouped data class?
b. Motivation
We’ve done already discussing about the
calculation of the mean of ungrouped data.
Since, there is an ungrouped data; do you know
that there is also a grouped data?
Do you know on how to calculate the mean of
grouped data?
B. Presentation
So, be with me this afternoon class as I discuss
on how to calculate the Mean of Grouped Data.
Mean = Ʃx = 97 + 91 + 94 + . . . + 94
n 12
Mean = 92. 4267
None, Sir
Yes/No, Sir
No, Sir
The Mean of Grouped Data

3. a. Statement of the Aim
 enumerate the steps in calculating the
mean;
 solve for the mean of grouped data;
 describe the importance of the mean in
real life situations
C. Development Proper
Mean – The mean of grouped data frequency
distribution may be obtained in almost the
same way as the mean of ungrouped data is
computed. All the scores included in a class
interval are represented by the midpoint or
class mark of that class interval. This class
mark is multiplied by its corresponding
frequency; the product is summed, and divided
by n.
̅ = ∑
n
∑ – the sum of the product of the
frequency and the class mark or class midpoint.
n – The sum of the frequencies or the size of
the sample.
Since in the given grouped data, on the class
interval and the frequency are given.
So, additionally we have to get its class mark,
and multiply it by its corresponding frequency
and the sum of the products.
So, I have here a grouped data.
Class Interval Frequency
20 – 24 2
25 – 39 6
30 – 34 9
35– 49 10
40 – 44 12
45 – 49 7
50 – 54 4
In a grouped data class, the class interval and
the frequency are already given. So, to get its
mean, we have to obtain the things in the
formula.
Since we only have the class interval and the
frequency, we will be the one to get the class
marks, the product of the frequency and the
class mark and the sum of the product of the
frequency and the class mark.

4. To get the class mark of each class interval, we
will just get the midpoint of the lower limit and
upper limit; we will just add the lower limit
and upper limit and divide the sum by 2.
Since, we have already had necessary
information to calculate the mean; we can now
use the midpoint formula.
̅ = ∑
n
̅ = 1905
50
̅ = 38.10 is the mean of the given grouped
data.
From the frequency distribution, based on the
formula, what are the things that we need to
have to calculate the mean?
What else aside from the class marks?
What do we have to do with the product of the
frequencies and the class marks?
What will we do with the product of the
frequencies and the class marks to finally get
the mean of the given grouped data?
Very good!
As we all know that the mean is one of the
measures of central tendency, it simply means
that it is a single value which can represent the
whole quality of the given data.
The smaller the mean, the smaller are the data.
But the mean has also some limitations, it is
not the best representation of the data when the
values are extreme, it happens when there are
some values which are far distant from the
other.
Class
Interval
Frequency
f
Class
mark
(Xm)
fXm
20 – 24 2 22 44
25 – 39 6 27 162
30 – 34 9 32 288
35– 39 10 37 370
40 – 44 12 42 504
45 – 49 7 47 329
50 – 54 4 52 208
i = 5 n = 50 ƩfXm = 1, 905
We have to obtain the class mark of each class
interval.
We also need to the product of the frequency
and the class mark.
We have to get the sum of the product of the
frequencies and the class marks.
We will the divide the product of the
frequencies and the class marks by the size of
the given data (n).

5. How are we going to find the mean of grouped
data?
Group Task
So now, we are going to have a grouped
activity. I will group in two five groups and
each will do the same task. The group which
can first before the given allotted time will be
declared as the winner
Directions: Calculate the mean of the grouped
data in your group.
Class Interval Frequency
0 – 19 6
20 – 39 12
40 – 59 18
60 – 79 20
80 – 99 36
100 – 119 28
120 – 139 21
140 – 159 9
Values Integration
A while ago, we have discussed about the
Mean of grouped data, as one of the measures
of central tendencies.
We can associate the word central to center. A
center value of the data.
In connection to our lives class, who is the
center of your lives?
Who is would like to answer?
To find the mean of grouped data first: we
have to make a column for class marks, next is
we will multiply each class mark to its
corresponding frequency, next is we have to
get the sum of the product of the frequencies
and class marks and last we will divide the
product by the size of the sample.
Students do as told
Expected output from the groups
Class
Interval
Frequency
f
Class
mark
(Xm)
fXm
0 – 19 6 9.5 27
20 – 39 12 29.5 254
40 – 59 18 49.5 531
60 – 79 20 69.5 1390
80 – 99 36 89.5 3222
100 – 119 28 109.5 3066
120 – 139 21 129.5 2719.5
140 – 159 9 149.5 1345.5
n = 150 Ʃfxm =
12555
Ʃfxm = 12555 = 83.7
n 150
The center of my life is my family because
they are the ones who are always at my side in
times of happiness and sadness.
The center of my life is God because He is the
source of everything I need in this world.

6. Activity 1
Directions: Read, analyze and solve the
problem below.
The following is a distribution for the number
of employees in 40 companies belonging to a
certain industry. Calculate the mean of the
grouped data below.
No. of Employees No. of Companies
14 – 20 1
21 – 27 3
28 – 34 6
35 – 41 11
42 – 48 8
49 – 55 7
56 - 62 4
Evaluation
Directions: Read, analyze and solve the
problem below.
The distribution of the hourly rates of 70
professors from different schools who attended
a convention is indicated below. Calculate the
mean.
Hourly rate No. of Professors
20.00 – 27.45 12
27.50 – 34.95 18
35.00 – 42. 45 11
42.50 – 49.95 9
50.00 – 57. 45 8
57.50 – 64.95 6
65.00 – 72. 45 3
72. 50 – 79. 95 3
IV. Assignment
Directions: Read in advance about the Median
of Grouped Data.
̅
No. of
Employees
No. of
Companies
Class
marks
fXm
14 – 20 1 17 17
21 – 27 3 24 72
28 – 34 6 31 186
35 – 41 11 38 418
42 – 48 8 45 360
49 – 55 7 52 364
56 - 62 4 59 236
n = 40 ƩfXm =
1653
̅ = 41.325
Hourly
rate
No. of
Professors
Class
marks
fXm
20.00 –
27.45
12 23.725 284.7
27.50 –
34.95
18 31.225 562.05
35.00 –
42. 45
11 38.725 425.975
42.50 –
49.95
9 46.225 416.025
50.00 –
57. 45
8 53.725 429.8
57.50 –
64.95
6 61.225 367.35
65.00 –
72. 45
3 68.725 206.175
72. 50 –
79. 95
3 76.225 228.675
n = 70 2920.75