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Statistics and

Probability

Quarter 3 – Module 2 :

Mean and Variance of Discrete Random

Variable

Statistics and Probability 11

Alternative Delivery Mode

Quarter 3 – Module 1: Mean and Variance of Discrete Random Variable

First Edition, 2020

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Published by the Department of Education

Secretary: Leonor Magtolis Briones

Undersecretary: Diosdado M. San Antonio

SENIOR HS DEVELOPMENT TEAM

AUTHOR : Amethel C. Saplala

Co-Author – Language Editor : Carla Mae M. Soriano

Co-Author – Content Evaluator : Jocelyn R. Miranda

Co-Author – Illustrator : Alvin A. De Leon

Co-Author – Layout Artist : Abbie A. Tumbokon

TEAM LEADERS:

School Head : Reycor E. Sacdalan, PhD

LRMDS Coordinator : Pearly V. Villagracia

SDO-BATAAN MANAGEMENT TEAM:

Schools Division Superintendent : Romeo M. Alip, PhD, CESO V

OIC- Asst. Schools Division Superintendent : William Roderick R. Fallorin, CESE

Chief Education Supervisor, CID : Milagros M. Peñaflor, PhD

Education Program Supervisor, LRMDS : Edgar E. Garcia, MITE

Education Program Supervisor, AP/ADM : Romeo M. Layug

Education Program Supervisor, Senior HS : Danilo C. Caysido

Project Development Officer II, LRMDS : Joan T. Briz

Division Librarian II, LRMDS : Rosita P. Serrano

REGIONAL OFFICE 3 MANAGEMENT TEAM:

Regional Director : May B. Eclar, PhD, CESO III

Chief Education Supervisor, CLMD : Librada M. Rubio, PhD

Education Program Supervisor, LRMS : Ma. Editha R. Caparas, EdD

Education Program Supervisor, ADM : Nestor P. Nuesca, EdD

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-by-

step 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.

What I Need to Know

At the end of the lesson, the students are expected to:

1. Illustrate the mean and variance of a discrete random variable; (M11/12SP-IIIb-

2)

2. Calculate the mean and the variance of a discrete random variable; (M11/12SP-

IIIb-2)

3. Interpret the mean and the variance of a discrete random variable;

(M11/12SP-IIIb-3) and

4. Solve problems involving mean and variance of probability distributions.

(M11/12SP-IIIb-4)

6. The random variable X, representing the number of nuts in a chocolate bar

has the following probability distribution. What is the mean?

Y 0 1 2 3 4

P(y)

1

10

3

10

3

10

2

10

1

10

A. 1.5 C. 1.

B. 1.3 D. 1.

7. These are two values that describe how scattered or spread out the scores

are from the mean value of the random variable.

A. variance and standard deviation

B. mean and probability distribution

C. probability and statistics

D. mean and variance

For numbers 8 and 9.

Complete the table below using mean (μ) = 5. Write the answer in your answer

sheets.

Probability Distribution Table of Random Variable X

X 2 4 6 8

P(x) 0.25 0.25 0.25 0.

(x-μ)????

(x-μ)

2

????

10. Compute for the variance and standard deviation of the Probability

Distribution table of the Random Variable X.

A. Variance = 5; Standard Deviation = 2.

B. Variance = 4; Standard Deviation = 3.

C. Variance = 3; Standard Deviation = 4.

D. Variance = 2; Standard Deviation = 5.

Covid-19 is continuously spreading around the world, that is why reports

regarding average infected people per country is being updated every day. For this

kind of report, experts used Statistics and Probability to show reliable analysis in

their data. In this lesson, you will learn how to compute the average or mean of a

discrete probability distribution as well as the variance and standard deviation of a

discrete random variable.

What’s In

Let’s find out if you are ready to learn this new lesson. Do the following, write your

answer in your answer sheet.

A. Given the values of the variables X and Y, evaluate the following summations.

X 1

= 5, Y 1

= 3 X 3

= 3, Y 3

= 1

X 2

= 4, Y 2

= 2 X 4

= 2, Y 4

= 0

( 1 )

∑ 𝑋 (3)

∑ (𝑋 + 𝑌) (5)

∑ 4 𝑋𝑌

( 2 )

∑ 𝑋𝑌 (4)

∑ 𝑌

B. Construct a probability distribution of W representing the square of the number

when a die is rolled once. Copy the table in your answer sheet then write your

answer.

W

P(w)

Lesson

Mean of the Discrete

Random Variable

Examples:

1. Determine the mean or Expected Value of random variable below.

X 0 1 2 3 4

P(x) 1

Solution:

μ =

∑ ⟮𝑥𝑃(𝑥)⟯

=

1

5

1

5

1

5

1

5

1

5

=

1

5

2

5

3

5

4

5

=

10

5

or 2

Therefore, mean is 2 for the above random variable.

2. Find the mean of the random variable Y representing the number of red color

chocolates per 160-gram pack of colored chocolate packages that has the following

probability distribution.

Y 4 5 6 7

P(y) 0.10 0.37 0.33 0.

Solution

μ =

∑ ⟮𝑋𝑃(𝑥)⟯

=

∑ ⟮ 4 ( 0. 10 ) + 5 ( 0. 37 ) + 6 ( 0. 33 ) + 7 ( 0. 20 )⟯

= ∑ ⟮ 0. 40 + 1. 85 + 1. 98 + 1. 40 ⟯

= 5.

So, the mean of the probability distribution is 5.63. This implies that the average

number of red chocolates per 160-gram is 5.63.

3. The probabilities that a customer will buy 1, 2, 3, 4, or 5 items in a grocery

store are

3

10

1

10

,

1

10

,

2

10

, and

3

10

, respectively. What is the average number of items

that a customer will buy?

What is It

To solve the above problem, we will follow 3 steps below.

STEPS IN FINDING THE MEAN

Step 1: Construct the probability distribution for the random variable X

representing the number of items that the customer will buy.

Step 2: Multiply the value of the random variable X by the corresponding

probability.

Step 3: Add the results obtained in Step 2. Results obtained is the mean of the

probability distribution.

Solution:

Steps Solution

1. Construct the probability

distribution for the random variable X

representing the number of items that

the customer will buy.

Number of Items

X

Probability

P(x)

1

3

10

2

1

10

3

1

10

4

2

10

5

3

10

2. Multiply the value of the random

variable X by the corresponding

probability.

Number of

Items X

Probability

P(x)

X  P(x)

1

3

10

3

10

2

1

10

2

10

3

1

10

3

10

4

2

10

8

10

5

3

10

15

10

What’s New

Let’s try!

Let’s have examples:

1. The number of cars sold per day at a local car dealership, along with its

corresponding probabilities, is shown in the succeeding table. Compute the

variance and the standard deviation of the probability distribution by following the

given steps. Write your answer in your answer sheets.

Number of Cars Sold X Probability P(x)

0 1 0%

1 2 0%

2 3 0%

3 2 0%

4 2 0%

What is It

Variance and Standard Deviation of a Random Variable

The variance and standard deviation are two values that describe

how scattered or spread out the scores are from the mean value of

the random variable. The variance, denoted as σ

2

, is determined

using the formula:

σ

2

=

∑ (𝑥 − μ)

2

p(x)

The standard deviation σ is the square root of the variance, thus,

σ = ඥ

∑ (𝑥 − μ)²𝑝(𝑥)

σ

2

• variance σ – standard deviation

μ - mean p(x) – probability of the outcome

In solving the problem, let’s follow the steps below.

STEPS IN FINDING THE VARIANCE AND STANDARD DEVIATION

1. Find the mean of the probability distribution.
2. Subtract the mean from each value of the random variable X.
3. Square the result obtained in Step 2.
4. Multiply the results obtained in Step 3 by the corresponding probability.
5. Get the sum of the results obtained in Step 4. Results obtained is the value of

the variance of probability distribution.

Now let’s solve the problem.

STEPS SOLUTION

1. Find the mean of the probability

distribution using the formula:

μ = ∑ 𝑋𝑃(𝑥)

Number of

Car Sold

X

Probability

P(x)

X  P(x)

0 0.1 0

1 0. 2 0.

2 0.3 0.

3 0.2 0.

4 0.2 0.

TOTAL 2.

μ = ∑ 𝑿𝑷(𝒙) = 𝟐. 𝟐

2. Subtract the mean from each

value of the random variable X.

X P(x)

X

·

P(x)

X - μ

0 0.1 0 0 – 2.

= - 2.

1 0.2 0.2 1 – 2.

= - 1.

2 0.3 0.6 2 – 2.

= - 0.

3 0.2 0.6 3 – 2.

= 0.

4 0.2 0.8 4 – 2.

= 1.

3. Square the results obtained in

Step 2.

X P(x)

X

·

P(x)

X - μ (X - μ)

2

0 0.1 0 - 2.2 4.

1 0.2 0.2 - 1.2 1.

2 0.3 0.6 - 0.2 0.

3 0.2 0.6 0.8 0.

4 0.2 0.8 1.8 3.

STEPS SOLUTION

1. Find the mean of the probability

distribution using the formula

μ = ∑ 𝑋𝑃(𝑥)

Number of

Heads

X

Probability

P(x)

X ·

P(x)

0

0

1

2

3

TOTAL

12

8

μ =

𝟏𝟐

𝟖

2. Subtract the mean from each

value of the random variable X.

X P(x)

X

·

P(x)

X - μ

0

0 0 – 1.

= - 1.

1

1 – 1.

= - 0. 5

2

2 – 1.

= 0. 5

3

3 – 1.

= 1.

3. Square the results obtained in

Step 2.

X P(x) X·P(x) X - μ (X - μ)

2

0

0 - 1.5 2.

1

• 0.5 0.

2

0.5 0.

3

1.5 2.

4. Multiply the results obtained in

Step 3 by the corresponding

probability.

X P(x) X·P(x) X-μ (X - μ)

2

(X - μ)

2

· P(x)

0

0 - 1.5 2.25 2.25/8 =

1

• 0.5 0.25 0.75/8 =

2

0.5 0.25 0.75/8 =

3

1.5 2.25 2.25/8 =

2. Get the sum of the results

obtained in Step 4. The result is

the value of the variance ( σ

2

). The

formula is:

σ

2 =

∑(𝑥 − μ)

2

p(x)

X P(x) X·P(x) X-μ (X - μ)

2

(X - μ)

2

· P(x)

0

0 - 1.5 2.25 0. 28

1

• 0.5 0.25 0. 09

2

0.5 0.25 0.0 9

3

1.5 2.25 0. 28

TOTAL 0.

σ

2 =

∑ (𝒙 − μ)

2

p(x) = 0.

To solve for Standard Deviation

σ

2

=

∑( 𝑥 − μ

)

2

p(x)

= 0.

σ = √

1. 74

= 0.

The mean in tossing 3 coins with probability of Head will show up is 0.86 and

the variance is 0.74, then the standard deviation is 0.86.

4. The random variable X, representing the number of nuts in a chocolate bar has

the following probability distribution. Compute the variance and standard

deviation.

X 1 2 3 4 5

P(x)

1

10

3

10

3

10

2

10

1

10

5. The number of items sold per day in a sari-sari store, with its corresponding

probabilities, is shown in the table below. Find the variance and standard deviation

of the probability distribution.

Number of Items Sold

X

Probability

P(x)

19 0.

20 0.

21 0.

22 0.

23 0.

.

What I Have Learned

Answer the following questions in your own understanding.

1. How to compute the mean of a discrete random variable? State

the 3 steps. Write your answer in your answer sheets.

2. How to find the variance and standard deviation of a discrete

random variable? Write your answer in your answer sheets.

Make a study about how many sheets of paper you consumed weekly in

answering your Self Learning Modules. Record the quantity (total number of sheets)

per subject, then construct a probability distribution. Compute the mean, variance,

and the standard deviation of the probability distribution you made. Interpret the

result, then find out how many weeks you will consume 50 sheets of pad paper.

Assessment

Find the mean, variance, and standard deviation of the following probability

distribution then interpret the computed values. Write your answer in your answer

sheets.

1. Variable z representing the number of male teachers per Elementary school.

z 2 3 4 5 6

P(z) 40% 32% 11% 9% 8%

2. The number of mobile phones sold per day at a retail store varies as shown in

the given probability distribution below. Find the expected number of mobile

phones that will be sold in one day.

x 30 33 38 40 50

P(x) 0.2 0.2 0. 35 0. 23 0.0 2

3. Number of monthly absences of Juan Dela Cruz based on his previous records of

absences as presented in the probability distribution below.

Number of Absences (X) Percent P(x)

3 25%

4 30%

5 30%

6 15%

What I Can Do