Statistics and Probability
Quarter 3 – Module 5:
Finding the Mean and the
Variance of the Sampling
Distribution of the Sample Means
Statistics and Probability
Alternative Delivery Mode
Quarter 3 – Module 5: Finding the Mean and the Variance of the Sampling Distribution
of the Sample Means
First Edition, 2020
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Published by the Department of Education
Secretary: Leonor Magtolis Briones
Undersecretary: Diosdado M. San Antonio
SENIOR HS MODULE DEVELOPMENT TEAM
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Statistics and Probability
Quarter 3 – Module 5:
Finding the Mean and the
Variance of the Sampling
Distribution of the Sample Means
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners,
can continue your studies and learn while at home. Activities, questions, directions,
exercises, and discussions are carefully stated for you to understand each lesson.
Each SLM is composed of different parts. Each part shall guide you step-bystep as you discover and understand the lesson prepared for you.
Pre-tests are provided to measure your prior knowledge on lessons in each
SLM. This will tell you if you need to proceed on completing this module or if you
need to ask your facilitator or your teacher’s assistance for better understanding of
the lesson. At the end of each module, you need to answer the post-test to self-check
your learning. Answer keys are provided for each activity and test. We trust that you
will be honest in using these.
In addition to the material in the main text, Notes to the Teacher are also
provided to our facilitators and parents for strategies and reminders on how they can
best help you on your home-based learning.
Please use this module with care. Do not put unnecessary marks on any part
of this SLM. Use a separate sheet of paper in answering the exercises and tests. And
read the instructions carefully before performing each task.
If you have any questions in using this SLM or any difficulty in answering the
tasks in this module, do not hesitate to consult your teacher or facilitator.
Thank you.
General Note to the Teacher
This module contains helpful procedures
or strategies that will help you in guiding
the learners towards the attainment of the
objectives intended for this lesson.
What I Need to Know
After successfully completing this self-learning module, you are
expected to master essential knowledge and skills about finding the mean
and the variance of the sampling distribution of the sample means.
Specifically, you are more likely able to:
1. find the mean and variance of the sampling distribution of the
sample mean (M11/12SP-IIId-5); and
2. define the sampling distribution of the sample mean for normal
population when the variance is: (a) known; (b) unknown
(M11/12SP-IIIe-1).
These most essential learning competencies will be condensed into a
simplified user lesson that will be discussed along the path of your
academic journey with this self-learning module.
Enjoy your steps towards the attainment of our objectives. Are you
ready? If you are, let’s go!
What I Know
After you get acquainted with the objectives, it's important to assess
your knowledge and experience about the lessons you are just about to
discover.
First, let us try the next Alternate Response Diagnostic Test Type to
check something you already know about this lesson.
Consider that your score throughout this part of the module does not
actually affect your performance. So, it's all right to get bad scores. Let's
get a warm-up!
Direction: On your answer sheets, write “WE HEAL” if the statement is
correct and write “WE RECOVER” if the statement is incorrect.
1.
2.
Theoretically, a histogram demonstrating the mean of all samples
analyzed from a given population would be known as the sampling
distribution.
The review of sampling distributions of the sample means will be
the baseline for inferential statistics and the test hypothesis.
1
3.
The total value of all possible samples means that all possible
random samples of a given population size will be equal to the
population mean only if the data were normally distributed.
4. For any population, the standard deviation of the sample means is
approximately equal to the standard deviation of the population.
5. When the average of all possible values of the sample statistic is
equal to the parameter, the statistic is known as a biased estimator
of the parameter.
6. The mean of the sampling distribution of the sample means is
equal to the mean of the population mean.
7. The standard deviation of the sampling distribution of the mean is
also called as the standard error of the mean.
8. The error of sampling calculated from a small sample will always
be greater than one calculated from a large sample.
9. Sampling error is the difference between the sample mean and the
population mean.
10. To reduce the potential for extreme sampling error, the size of the
sample should be reduced.
You can now verify your answers whether they are right or not.
How's the result going? Is that good or bad? Don't worry if you have bad
scores, because that would be a reliable indicator that you're going to need
this self-learning module. That implies there are some sources of
competitive advantage that you need to develop and start exploring.
So, let's just go! Let us move ahead to the next part of this module so
that you can achieve the skills and competencies that you need to improve.
Lesson
1
Finding the Mean and the
Variance of the Sampling
Distribution of the Sample
Means
Recognize that the sampling distribution of the sample means is
definitely the probability distribution of the sample means, which also
implies that the sample is the random variable in this probability
distribution. But since the sampling distribution of the means is the
probability distribution of the random variable X, we could perhaps
calculate its mean and variance.
2
In this lesson, we shall solve for the mean in the variance of the
sampling distribution of the sample means and investigate some of its
important properties along with the definition for normal population when
the variance is known and unknown.
What’s In
In this portion, you will connect your learned concepts and skills
from previous lessons, which have much to do with the introduction to
sampling distribution of the sample means, to this current lesson, which
is finding mean and variance of the sampling distribution of the sample
means.
There are several concepts from the previous lesson that are directly
related to this lesson. Now, the activity given below will help you link those
concepts and ideas as you explore the lesson of this module.
Direction: With the given jumbled letters, complete the statements
below by arranging the letters below. Write your answers on your answer
sheets.
1. A – S – N – L – M – I – P – G
The _____________ distribution of sample means is frequency
distribution of the sample means taken from a population.
2. O – T – M – A – R – S – I – G – H
The _____________ of the sampling distribution of the sample
means is a bar graph constructed by plotting the sample means
along the horizontal axis and the problem along the vertical axis.
3. A – M – E – N
The _____________ of the sampling distribution of the sample
means is equal to the mean of the population.
4. R – O – R – E – R
Standard _____________ of the mean is the standard deviation
of the sampling distribution of the sample mean.
5. D – E – X – E – P – E – T – C
Mean is also called _____________ value.
Check your answers at this point if they are correct. If this is not the
case, you may read the given sentences and analyze why it is completed
as it was.
3
Recall the terms discussed in the previous modules and try to link them
to the new terms you encountered in this section of the module. For a
maximum of 5 minutes, reflect on their similarities and differences so that
you can get through this module smoothly.
Since you're done reflecting and familiarizing yourself with the terms
related to the previous lesson, you can proceed to the next part of this
module! Way to go there!
What’s New
Now that you have connected the previous lesson to your approach to
learning concepts and skills through this module, let us consider the
following activity that will give you a recall of the concepts of the previous
lesson.
Situation: It is Monday! Harvey is very enthusiastic and challenged for
his Modular Distance Learning (MDL) experience at Week 5 in Statistics
and Probability. His encounter with past quarters and modules justifies
his excitement in learning more about the core subject. He chose the MDL's
Digital Module Scheme because it is more practical to his condition.
On the other hand, his mobile phone notified him that Module 5 of
Statistics and Probability had been posted to their Facebook Learning
Space, ready to be accessed and downloaded. Surprisingly, there's a
problem with Harvey. He forgot his 5-digit smartphone passcode!
Let us help Harvey decode his passcode by reviewing the past lesson on
the introduction of sampling distribution of sample means.
Direction: On your answer sheets, copy the Code Table and review the
concepts of the previous lesson by completing the paragraphs and tables
provided. Link them to the corresponding number in the Code Table.
Code Table:
CODE
ANSWER
602.720
3.727
24.550
13.889
5.860
Concept 1 Reviewer:
Recall that the Variance (𝛔𝟐 ) and the Standard deviation (𝛔) of
ungrouped data are computed by using the formulas presented on the
next page, respectively.
4
𝝈𝟐 =
𝚺𝑿𝟐
𝑵
𝚺𝐗 𝟐
𝚺𝑿𝟐
𝝈=√
− (𝑵)
where:
𝑵
𝚺𝐗 𝟐
− (𝑵)
where:
σ2 = variance
σ = standard deviation
X = score or value
X = score or value
N = number of scores or values
N = number of scores or values
Concept 1 Example:
Given the set of data: X = { 2, 5, 6, 9, 11, 13 }, complete the
corresponding table and compute for the variance and standard
deviation.
X
X2
2
4
5
25
6
36
9
81
11
121
13
169
46
436
Variance:
𝜎2 =
𝜎2 =
Σ𝑋 2
𝑁
436
6
Standard Deviation:
ΣX 2
Σ𝑋 2
𝜎=√
−(𝑁)
46 2
𝑁
436
𝜎=√
−(6)
𝜎 2 = (𝟏)
6
ΣX 2
−(𝑁)
46 2
−(6)
𝜎 = (𝟐)
Concept 2 Reviewer:
Recall that the Mean (𝛍) or the Expected Value E(X) of a discreet
probability distribution is computed using the formula presented below.
𝝁 = 𝑬(𝑿) = 𝚺[𝑿 ∙ 𝑷(𝑿)]
5
where:
μ = mean
E(X) = expected value
X = value of the random variable
P(X) = probability value of the random variable
Concept 2 Example:
Find the mean or the expected value of the given probability
distribution below.
X
P(X)
X • P(X)
0
0.100
0.000
2
0.150
0.300
4
0.200
0.800
5
0.140
0.700
6
0.150
0.900
8
0.090
0.720
11
0.030
0.330
15
0.050
0.750
14
0.050
0.700
16
0.030
0.480
18
0.010
0.180
46
1.000
5.860
Mean or Expected Value:
𝜇 = 𝐸(𝑋) = Σ[𝑋 ∙ 𝑃 (𝑋)]
𝜇 = 𝐸(𝑋) = (𝟑)
Concept 3 Reviewer:
Recall that the Variance (𝛔𝟐 ) and the Standard deviation (𝛔) of
probability distribution are computed by using the formulas presented
below, respectively.
𝝈𝟐 = 𝚺 [𝑿𝟐 ∙ 𝑷(𝑿)] − 𝝁𝟐
𝝈 = √𝚺 [𝑿𝟐 ∙ 𝑷(𝑿)] − 𝝁𝟐
6
where:
where:
σ2 = variance
σ = standard deviation
X = score or value
X = score or value
P(X) = probability value of the
random variable
μ = mean or expected value
P(X) = probability value of the
random variable
μ = mean or expected value
Concept 3 Example:
Compute the variance and standard deviation of the given
probability distribution below.
X
X2
P(X)
X2 • P(X)
0
0
0.100
0.000
2
4
0.150
1.200
4
16
0.200
12.800
5
25
0.140
17.500
6
36
0.150
32.400
8
64
0.090
46.080
11
121
0.030
39.930
15
225
0.050
168.750
14
196
0.050
137.200
16
256
0.030
122.880
18
324
0.010
58.320
46
1 267
1.000
637.060
Variance:
Standard Deviation:
𝜎 2 = Σ [𝑋 2 ∙ 𝑃(𝑋)] − 𝜇 2
𝜎 = √Σ [𝑋 2 ∙ 𝑃(𝑋)] − 𝜇 2
𝜎 2 = 637.060 − (5.860)2
𝜎 = √637.060 − (5.860)2
𝜎 2 = (𝟒)
𝜎 = (𝟓)
Congratulations! You accessed Harvey’s smartphone. Now, Module 5 of
his subject, Statistics and Probability, can be made available and
downloaded. Let us continue to help Harvey explore his learning
experience through this module!
7
The next part of this module will be a brief and simple discussion of the
lesson. This will be followed by a series of formative assessment activities.
Hey, just boost it up!
What is It
It’s time to take a general tour around this module. This portion of your
journey acts a simple and brief discussion of the lesson that aims to help
you discover and understand new concepts and skills. This will really help
you understand the real-life applications of mean and variance of sampling
distribution of the sample means.
Basically, the flow of the discussion will be starting with the
presentation of the example problem about the mean of the sampling
distribution of the sample means and to be followed by the corresponding
problem about its variance and standard deviation.
By helping Harvey in this module, you are able to have a learning buddy
in order for you to reach your goals in finishing this module with learned
skills and competencies on your intellects!
First, Harvey wants to investigate the mean of the sampling distribution
of the sample means and compare it with the mean of the population. After
the comparison of the means, Harvey must compare the variance and the
standard deviation of the population to the sample means.
Let us do this!
Illustrative Example:
The following table gives the sum of tutorial rate of six teachers in
Central Luzon per month. Suppose that random samples of size 4 are
taken from this population of six teachers, do the following tasks.
Teacher
Tutorial Rate (in thousand pesos) X
A
8
B
12
C
16
D
20
E
24
F
28
8
1.
2.
3.
4.
Solve for the mean of the population μ.
Solve for the mean of the sampling distribution of the sample means μx̄.
Compare μ and μx̅ .
Solve for the variance (σ2 ) and the standard deviation (σ) of the
population.
5. Solve the variance (σ2 x̅ ) and the standard deviation (σx̅ ) of the sampling
distribution of the sample means μx̅ .
6. Compare σ and σx̅.
Solutions:
1. The population mean μ is solved as follows.
ΣX
8 + 12 + 16 + 20 + 24 + 28 108
𝜇=
=
=
= 18
𝑁
6
6
Therefore, the population mean of the tutorial rates of the select
teachers in Central Luzon is 18 thousand pesos per month.
2. To solve for the mean of the sampling distribution of the sample means,
the following steps are to be considered.
a. Identify the possible samples of size 4 and compute their
individual means.
Possible Sample
Sample Mean
̅
𝐗
8, 12, 16, 20
14
8, 12, 16, 24
15
8, 12, 16, 28
16
8, 12, 20, 24
16
8, 12, 20, 28
17
8, 12, 24, 28
18
8, 16, 20, 24
17
8, 16, 20, 28
18
8, 16, 24, 28
19
8, 20, 24, 28
20
12, 16, 20, 24
18
12, 16, 20, 28
19
12, 16, 24, 28
20
12, 20, 24, 28
21
16, 20, 24, 28
22
9
b. Construct the sampling distribution table for the sample means
and multiply the sample means to their probabilities.
Sample Mean
̅
𝐗
Frequency
F
Probability
̅)
P (𝐗
̅ • P (𝐗
̅)
𝐗
14
1
1/15
14/15
15
1
1/15
15/15
16
2
2/15
32/15
17
2
2/15
34/15
18
3
3/15
54/15
19
2
2/15
38/15
20
2
2/15
40/15
21
1
1/15
21/15
22
1
1/15
22/15
Total
15
15/15 or 1
270/15 = 18
c. Solve for the mean of the sampling distribution of the sample
̅ • 𝑃(𝑋
̅ )].
means by using the following formula 𝜇𝑥̅ = Σ[𝑋
̅ • 𝑃 (𝑋
̅ )] =
𝜇 𝑥 = Σ[ 𝑋
270
= 18
15
Therefore, the mean of the sampling distribution of the
sample means is 18.
3. The population mean μ is 18 and the mean of the sample means μx̅ is
18. Therefore, μ = μx̅ .
4. To solve for the variance (σ2 ) and the standard deviation (σ) of the
population, the given table below will be utilized.
X
X2
8
64
12
144
16
256
20
400
24
576
28
784
108
2 224
10
Variance:
𝜎2 =
𝜎2 =
Σ𝑋 2
𝑁
Standard Deviation:
ΣX 2
Σ𝑋 2
𝜎=√
−(𝑁)
2 224
6
108 2
−(
6
𝑁
2 224
𝜎=√
)
𝜎 2 = 4.67
6
ΣX 2
−(𝑁)
108 2
−(
6
)
𝜎 = 6.83
Therefore, the variance of the population is 4.67 and the standard
deviation is 6.83.
5. To solve for the variance (σ2 x̅ ) and the standard deviation (σx̅ ) of the
sampling distribution of the sample means, the following steps are
to be considered.
a. The following formula will be utilized to solve for the variance
(σ2 x̅ ) and the standard deviation (σx̅ ) of the sampling
distribution of the sample means.
Variance:
Standard Deviation:
𝜎 2 𝑥 = Σ [𝑋̅ 2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
𝜎𝑥 = √Σ [𝑋̅2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
b. Construct the sampling distribution table for the sample
means, square the sample means and multiply the result to
the probabilities.
Sample Mean
̅
𝐗
̅2
𝐗
Probability
̅)
P (𝐗
̅ 2 • P (𝐗
̅)
𝐗
14
196
1/15
196/15
15
225
1/15
225/15
16
256
2/15
512/15
17
289
2/15
578/15
18
324
3/15
972/15
19
361
2/15
722/15
20
400
2/15
800/15
21
441
1/15
441/15
22
484
1/15
484/15
Total
2 976
15/15 or 1
4 930/15 or 328.67
11
c. Solve for the variance and standard deviation of the sampling
distribution of the sample means by using the given formula.
Variance:
Standard Deviation:
𝜎 2 𝑥 = Σ [𝑋̅ 2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
𝜎𝑥 = √Σ [𝑋̅2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
𝜎 2 𝑥 = 328.67 − (18)2
𝜎𝑥 = √328.67 − (18)2
𝜎 2 𝑥 = 4.67
𝜎𝑥 = 2.16
Therefore, the variance and the standard deviation of the
sampling distribution of the sample means are 4.67 and
2.16, respectively.
6. The variance of the population and sample mean are equal with the
value of 4.67. The standard deviation of the population is 6.83 and
standard deviation of the sample means is 2.16. Therefore, σ ≠ σx̅ .
The existing relationship of the variance and standard deviation of the
population to the mean of the sample means has something to do with
Central Limit Theorem which will be discussed in the next module.
What’s More
This time help each other to discover level up skills in answering the
following activity about the mean and the variance of the sampling
distribution of the sample means.
Situation: After considering the first example on the previous part of
this module, Harvey has some questions and difficulties in solving the
mean and the variance of the sampling distribution of the sample means.
Help Harvey in acquiring desired skills by doing the given activity below.
Direction: With the given problem on the next page, complete the tables
and the solutions to have a complete and full discussion of computing the
mean and the variance of the sampling distribution of sample means. On
your answer sheets, copy and answer the problem with complete tables
and solutions.
12
Problem
A group of ABM students in Hermosa National High School
planned to have an online business as part of their Applied
Subject, Entrepreneurship. Their shares are ₱ 2 000.00, ₱ 3
000.00, ₱ 4 000.00, ₱ 5 000.00, and ₱ 6 000.00. A sample size of
2 is to be taken from this population as part of their market
analysis. Complete the following tasks to help the ABM students
in their market analysis. (A) Solve for the mean of the population
μ. (B) Solve for the mean of the sampling distribution of the
sample means μx̅ . C) Compare μ and μx̅ . (D) Solve for the
variance (σ2 ) and the standard deviation (σ) of the population.
(E) Solve the variance (σ2 x̅ )) and the standard deviation (σx̅ ) of
the sampling distribution of the sample means μx̅ . (F) Compare
σ and σx̅ .
Let us take a look at the solution below for Task A in the problem and
take note of the process of computing for the final answer.
Solution for Task A
Take note that you are going to fill in the area for the Formula to proceed
to the next tasks.
Given
X Values = ₱ 2 000.00, ₱ 3 000.00,
₱ 4 000.00, ₱ 5 000.00, ₱ 6 000.00
N=5
Unknown
Mean of the Population μ
Formula
(1)
𝜇=
Solution
Conclusion
.
2 000 + 3 000 + 4 000 + 5 000 + 6 000
5
20 000
𝜇=
5
𝜇 = 4 000
Therefore, the population mean of the shares
of the group of ABM Students to their
planned online business is ₱ 4 000.00.
At this point, let us answer Task B in the problem and familiarize
yourself to the table construction and the sample mean computations.
In here, complete the table on the next page by solving for the sample
mean of the possible samples.
13
Table 1 for Task B
Possible Samples
̅
Sample Mean 𝐗
2 000.00, 3 000.00
2 500.00
2 000.00, 4 000.00
(2)
2 000.00, 5 000.00
3 500.00
2 000.00, 6 000.00
(3)
3 000.00, 4 000.00
3 500.00
3 000.00, 5 000.00
(4)
3 000.00, 6 000.00
4 500.00
4 000.00, 5 000.00
(5)
4 000.00, 6 000.00
5 000.00
5 000.00, 6 000.00
(6)
.
.
.
.
.
Table 2 for Task B
As part of Task B, complete the sampling distribution table for the
sample means and multiply the sample means to their probabilities and
supply the missing values.
Sample Mean
̅
𝐗
Frequency
f
Probability
̅)
P (𝐗
̅ • P (𝐗
̅)
𝐗
2 500.00
1
1/10
2 500.00/10
3 000.00
1
1/10
3 000.00/10
3 500.00
2
2/10
7 000.00/10
4 000.00
2
2/10
8 000.00/10
4 500.00
2
2/10
9 000.00/10
5 000.00
1
1/10
5 000.00/10
5 500.00
1
1/10
5 500.00/10
Total
10
10/10 or 1
40 000.00/10
or 4 000.00
You are now ready to complete Task B by mean of the sampling
distribution of the sample means μx̅ .
Take note that you are going to fill in the area for the Conclusion to
proceed to the next tasks.
14
Solution for Task B
Given
̅ • P(̅
Σ[X
X)] = 40 000.00/10 or 4 000.00
Unknown
Mean of the Sampling Distribution of the
Sample Means μx̅ .
Formula
𝜇𝑥 = Σ[𝑋̅ • 𝑃(𝑋̅)]
Solution
𝜇𝑥 = ₱ 4 000.00
Conclusion
(7)
.
Based on the given answers with their corresponding solutions, Task C
is ready to be accomplished.
Answer for Task C
The population mean μ is ₱ 4 000.00 and the mean of the sample means μx̅
is also ₱ 4 000.00. Therefore, μ = μx̅ .
Table 1 for Task D
Next in line is the solution below for Task D that deals with the
computation of the variance and the standard deviation of the population.
Complete first the given table and proceed with the main solution by
squaring the values of the random variable X.
X
X2
2 000.00
4 000 000.00
3 000.00
(8)
4 000.00
16 000 000.00
5 000.00
(9)
6 000.00
(10)
20 000.00
90 000 000.00
.
.
.
Task D will be finished after the computation of the variance and the
standard deviation using the completed data from the table above.
Complete the solution on the next page by filling in the incomplete
portions.
15
Given
ƩX = 20 000.00; N = 5; ƩX2 = 90 000 000.00
Unknown
Variance (σ2 ) of the Population
Formula
Σ𝑋 2
ΣX 2
𝜎 =
−( )
𝑁
𝑁
Solution
90 000 000.00
20 000 2
)
−(
5
5
𝜎 2 = 18 000 000.00 − (4 000.00)2
𝜎 2 = 18 000 000.00 − 16 000 000.00
𝜎2 =
(11)
.
2
Solutions for Task D
𝜎2 =
Conclusion
Therefore, the variance of the population is
₱
(11)
.
Given
ƩX = 20 000.00; N = 5; ƩX2 = 90 000 000.00
Unknown
Standard Deviation ( σ ) of the Population
Formula
𝜎=√
𝜎=√
Solution
Σ𝑋 2
ΣX 2
−( )
𝑁
𝑁
90 000 000.00
20 000 2
)
−(
5
5
𝜎 = √18 000 000.00 − (4 000.00)2
𝜎 = √18 000 000.00 − 16 000 000.00
𝜎=
Conclusion
(12)
.
Therefore, the standard deviation of the
population is ₱
(12)
.
We are on our way to the finish line! Let us answer Task D in the given
problem and always be keen in observing the processes in constructing the
tables and the computations in solving for the variance and standard
deviations of the sample means.
16
To proceed to the complete solution, complete first the table below so
that the given values will be completed.
Table 1 for Task E
Square the sample means and multiply the results to its corresponding
probabilities.
Sample
Mean
̅
𝐗
̅2
𝐗
Probability
̅)
P (𝐗
̅ 2 • P (𝐗
̅)
𝐗
2 500.00
6 250 000.00
1/10
6 250 000.00/10
3 000.00
9 000 000.00
1/10
9 000 000.00/10
3 500.00
12 250 000.00
2/10
24 500 000.00/10
4 000.00
16 000 000.00
2/10
32 000 000.00/10
4 500.00
20 250 000.00
2/10
40 500 000.00/10
5 000.00
25 000 000.00
1/10
25 000 000.00/10
5 500.00
30 250 000.00
1/10
30 250 000.00/10
Total
119 000 000.00 10/10 or 1
167 500 000.00/10
or 16 750 000.00
Using the data from the above table, compute for the variance and
standard deviation of the sample means.
Complete the table on the next page so that the solutions will be useful
for a better understanding of the last part of the lesson.
By the way, note that after the comparison of the computed variances
and standard deviations of the population and the sample means, you will
be able to explore the continuity of the application of your learned concepts
from this module to the next module which has something to do with
Central Limit Theorem.
Do not stop. Way to go, leaners! Make your academic journey to this
module be continuous and productive.
Go and explore!
17
Given
Ʃ[X̅̅2 • P (X̅̅)] = 167 500 000.00/10 or 16 750
000.00; μx̅ = 4 000.00
Unknown
Variance (σ2 x̅ ) of the Sampling Distribution of
the Sample Means μx̅
Formula
𝜎 2 𝑥 = Σ [𝑋̅ 2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
Solutions for Task E
Solution
𝜎 2 𝑥 = 16 750 000.00 − (4 000.00)
2
𝜎 2 𝑥 = 16 750 000.00 − 16 000 000.00
𝜎2𝑥 =
(13)
.
Conclusion
Therefore, the variance of the sampling
distribution of the sample mean is ₱
(13) .
Given
Ʃ[X̅̅2 • P (X̅̅)] = 167 500 000.00/10 or 16 750
000.00; μx̅ = 4 000.00
Unknown
Standard Deviation (σx̅ ) of the Sampling
Distribution of the Sample Means μx̅
Formula
𝜎𝑥 = √Σ [𝑋̅2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
𝜎𝑥 = √16 750 000.00 − (4 000.00)
Solution
2
𝜎𝑥 = √16 750 000.00 − 16 000 000.00
𝜎𝑥 = √750 000.00
𝜎𝑥 =
Conclusion
(14)
.
Therefore, the standard deviation of the sampling
distribution of the sample mean is ₱
(14)
..
Given the answers on the variances and standard deviation, with their
corresponding solutions, Task F will be finalized. Complete the answer for
the last number of this portion of the module.
Answer for Task F
The variance of the population is 2 000 000.00 and variance of the sample
mean is 13. 750 000.00. On the other hand, the standard deviation of the
population is 1 414.21 and standard deviation of the sample means is
866.03. Therefore,
(15)
..
18
What I Have Learned
To generalize your learned skills and concepts, take note of the
realizations and reflections about your experience in solving the mean and
variance of the sampling distribution of sample means.
Direction: On your answer sheets, complete the table below by
supplying the missing entries and answer the reflective questions using 3
to 5 sentences referring to your answers on the table. Also, copy and be
familiar with the scoring rubric provided.
Criteria
Poor
(1 point)
Fair
Good
(2 points) (3 points)
Excellent
(4 points)
Total
Depth of
reflection
Required
components
Quality of
information
Structure
and
organization
Grammar
Total Score
No.
1
Mean of a Discrete Probability Distribution
Symbol
No.
2
Formula
Required Given
Variance of Ungrouped Data
Symbol
Formula
19
Required Given
No.
3
Standard Deviation of Ungrouped Data
Symbol
No.
4
No.
5
No.
7
No.
8
Required Given
Variance of Probability Distribution
Symbol
Formula
Required Given
Standard Deviation of Probability Distribution
Symbol
No.
6
Formula
Formula
Required Given
Mean of the Population
Symbol
Formula
Required Given
Mean of the Sampling Distribution of the Sample Means
Symbol
Formula
Required Given
Variance of the Sampling Distribution of the Sample Means
Symbol
Formula
20
Required Given
No.
9
Standard Deviation of the Sampling Distribution of the
Sample Means
Symbol
Formula
Required Given
Reflective Questions:
1. How do you solve for the mean of the sampling distribution of sample
means?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
2. How do you compute for the variance and standard deviation of the
sampling distribution of sample means?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
3. What are the significance and relationship among the mean,
variance, and standard deviation of the sampling distribution of
sample means?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
21
What I Can Do
To transfer your new knowledge and skill into real life situations and to
appreciate the problems given in this module, perform the given activity
below.
Direction: On your answer sheets, copy and complete the table below
by indicating the desired data and values, simple interpretations, and
implications related to the means and variance of the sampling distribution
of the sample means.
No.
Problem
𝛍
𝛍𝐱̅
Illustrative
Example
1
Real-Life
Interpretation
and
Implications
What’s More
Problem
2
Real-Life
Interpretation
and
Implications
Assessment
Problem
3
Real-Life
Interpretation
and
Implications
Additional
Activities
Problem
4
Real-Life
Interpretation
and
Implications
22
𝛔𝟐
𝛔
𝛔𝟐 𝐱̅
𝛔𝐱̅
Assessment
Let us see if you have mastered the skills and concepts of this lesson.
Perform the given activity below.
Direction: On your answer sheets, match Column X to Column Y and
Match Column Y to Column Z by writing the same and exact number or
letter that corresponds to your answers for the following columns: Column
X displays the Unknown Values, Column Y enumerates the Symbols of the
Unknown Values, and Column Z reveals the exact answers. Consider the
problem below for your reference.
Problem: Random samples of size N = 2 are drawn from a finite
population consisting of the number 5, 6, 7, 8, and 9. Compute for the
Mean, Variance and Standard Deviation of the Population, and the Mean,
Variance and Standard Deviation of the Sample Means.
Column X
__ A. Population
Mean
Column Y
__ 1. σ2 x̅
Column Z
a. 7
__ B. Mean of the
Sample Means
__ 2. σx̅
b. 8.24
__ C. Variance of the
Population
__ 3. μx̅
c. 7.11
__ D. Variance of the
Sample Means
__ 4. σ
d. 2.87
__ E. Standard
Deviation of the
Population
__ 5. μ
__ F. Standard
Deviation of the
Sample Means
__ 6. σ2
23
e. 7
f. 2.67
Additional Activities
Okay! This is our last wrap up. You will be able to help Harvey all
throughout this Module 5 in Statistics and Probability.
The following are just supplemental problems to enrich your knowledge
and skills of the lesson learned. Good luck and God bless!
Direction: Solve the given real – life problem below. Show the complete
solution for the following tasks.
Problem: As a Physical Education teacher of Hermosa National High
School Senior High School, Ma’am Mary Ann wants to compute for the
Body Mass Index of few students as part of her research. As a start, the
table below shows the heights of five students in meters. Supposed
samples of size 2 are taken from this population of five students, complete
the given table and answer the following tasks. Construct the desired
tables needed for your solutions and answers.
Student
Height (X)
X2
Hazel Kim
1.47
2.16
Mee Ann Mae
1.58
2.50
Grace Ann
1.67
2.79
Jethro
1.69
2.86
Chelsea Mae
1.51
2.28
TOTAL
7.92
12.59
1.
2.
3.
4.
5.
Solve for the mean of the population μ.
Solve for the mean of the sampling distribution of the sample means μx̅ .
Compare μ and μx̅ .
Solve for the variance (σ2 ) and the standard deviation (σ) of the population.
Solve the variance (σ2 x̅ ) and the standard deviation (σx̅ ) of the sampling
distribution of the sample means μx̅ .
6. Compare σ and σx̅.
24
What’s In
SAMPLING
HISTOGRAM
MEANE
ERROR
EXPECTED
1.
2.
3.
4.
5.
What’s More
Σ
𝜇=𝑁
3 000.00
4 000.00
4 000.00
4 500.00
5 500.00
Therefore, the mean of the sampling
distribution of the sample means is
₱ 4 000.00.
9 000 000.00
25 000 000.00
36 000 000.00
2 000 000.00
1 414.21
750 000.00
866.03
and σx̅
8.
9.
10.
11.
12.
13.
14.
15.
1.
2.
3.
4.
5.
6.
7.
25
1.
2.
3.
4.
5.
13.889
3.727
5.860
602.720
24.550
What’s New
* The 5-digit passcode of Harvey’s
smartphone is 42513. *
σ
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
What I Know
WE HEAL
WE HEAL
WE RECOVER
WE RECOVER
WE RECOVER
WE HEAL
WE HEAL
WE RECOVER
WE HEAL
WE RECOVER
Answer Key
Additional Activities
1.μ = 1.58
2.μx̅ = 1.59
3.μ ≠ μx̅
4.σ2 = 1.19; σ = 1.09
5.σ2 x̅ = 0.01; σx̅= 0.09
6.σ ≠ σx̅
26
Assessment
Column X
A. 5
B. 3
C. 6
D. 4
E. 1
F. 2
Column Y
1.c
2.f
3.a or e
4.d
5.a or e
6.b
What I Can do
1.Illustrative
Example
• μ = 18
• μx̅ = 18
• σ2 = 4.67
• σ = 6.83
• σ2 x̅ = 4.67
• σx̅ =2.16
2.What’s More
Problem
• μ = 4 000.00
• μx̅ = 4 000.00
• σ2 = 2 000 000.00
• σ = 1 414.21
• σ2 x̅ = 750 000.00
• σx̅ = 866.03
3.Assessment
Problem
• μ=7
• μx̅ = 7
• σ2 =8.24
• σ = 2.87
• σ2 x̅ = 7.11
• σx̅ = 2.67
4.Additional
Activities Problem
• μ = 1.58
• μx̅ = 1.59
• σ2 = 1.19
• σ = 1.09
• σ2 x̅ = 0.01
• σx̅ = 0.09
* Students’ answers on the Real-Life Interpretation and Implications may vary depending on
their learnings and experiences with this module. *
What I Have Learned
1.Mean of a Discrete
Probability Distribution
•
𝜇
•
𝜇 = 𝐸(𝑋) = Σ[𝑋 ∙ 𝑃(𝑋)]
•
X = value of the
random variable
• P(X) = probability value of
the random variable
2.Variance of Ungrouped
Data
𝜎2
•
•
𝜎2 =
Σ𝑋 2
𝑁
ΣX 2
𝑁
−( )
• X = score or value
• N = number of scores or
value
3.Standard Deviation of
Ungrouped Data
𝜎
•
Σ𝑋 2
𝑁
ΣX 2
𝑁
X = score or value
N = number of scores
•
𝜎=√
•
•
−( )
4.Variance of Probability
Distribution
• 𝜎2
• 𝜎 2 = Σ [𝑋 2 ∙ 𝑃(𝑋)] − 𝜇 2
• X = value of the random
variable
• P(X) = probability value of
the random variable
• μ = mean or expected
value
5.Standard Deviation of
Probability Distribution
• 𝜎
• 𝜎 = √Σ [𝑋 2 ∙ 𝑃(𝑋)] − 𝜇 2
• X = score or value
• P(X) = probability value of
the random variable
• μ = mean or expected
value
6.Mean of the Population
• 𝜇
ΣX
• 𝜇=
𝑁
• X = score or value
N = number of scores
•
* Students’ answers on the Reflective Questions may vary
depending on their learnings and experiences with this module. *
7.Mean of the Sampling
Distribution of the Sample
Means
• 𝜇𝑥
̅ • 𝑃(𝑋̅)]
• 𝜇𝑥 = Σ[𝑋
̅ = sample mean
• X
̅) = probability of the
• P(X
sample mean
7.Variance of the Sampling
Distribution of the Sample
Means
• 𝜎 2𝑥
̅ 2 ∙ 𝑃(𝑋̅)] − 𝜇2
• 𝜎 2 𝑥 = Σ [𝑋
𝑥
̅ = sample mean
• X
̅) = probability of the
• P(X
sample mean
• μ = population mean
8.Standard Deviation of the
Sampling Distribution of
the Sample Means
𝜎𝑥
•
̅ = sample mean
X
̅) = probability of the
P(X
sample mean
μ = population mean
•
𝜎𝑥 = √Σ [𝑋̅ 2 ∙ 𝑃(𝑋̅)] − 𝜇2 𝑥
•
•
•
References
Books
Albacea, Z. V., & Ayaay, M. J. (2016). Statistics and Probability. Quezon City,
Diliman, Philippines: Commission on Higher Education - K to 12
Transition Program Management Unit.
Department of Education. (2020). K to 12 Most Essential Learning
Competencies. Pasig City, Philippines.
Jose M. Ocampo, J., & Marquez, W. G. (2016). Senior High Conceptual Math
& Beyond: Statistics and Probability. Quezon City, Novaliches,
Philippines: Brilliant Creations Publishing Inc.
Electronic Source
Mcmilllan, T. (2017). iRubric: Reflective Essay Rubric. Reflective Essay
Rubric . R Campus. Retrieved from
https://www.rcampus.com/rubricshowc.cfm?sp=yes&code=L34935
27
For inquiries or feedback, please write or call:
Department of Education – Region III,
Schools Division of Bataan - Curriculum Implementation Division
Learning Resources Management and Development Section (LRMDS)
Provincial Capitol Compound, Balanga City, Bataan
Telefax: (047) 237-2102
Email Address: bataan@deped.gov.ph