variance ( STAT).pptx

Lesson 3
Variance of a
Discrete Random
Variable

Objectives
At the end of this lesson, the learner should be able to
● correctly calculate the variance of a discrete
random variable;
● correctly interpret the variance of a discrete
random variable; and
● accurately apply the concept of variance of a
discrete random variable in real-life situations.

Essential Questions
● What is the difference between the variance of a raw data
to the variance of a probability distribution?
● How does the interpretation of the variance affects our
understanding of the probability distribution?

Warm Up!
Before we discuss about the variance of a discrete random
variable, let us review the concept of the standard deviation
through an instructional video.
(Click the link below to access the instructional video.)
Jeremy Jones. “Standard Deviation – Explained and
Visualized.” YouTube video, 3:42. Posted 05 April 2015.
Retrieved 14 July 2019 from
https://www.youtube.com/watch?v=MRqtXL2WX2M

Guide Questions
● What concepts about standard deviation have you learned
or remembered while watching the video?
● What is the standard deviation? How do we apply it in a set
of data?
● The video we watched was about a set of data commonly
known as raw data. What do you think does the standard
deviation of a random probability distribution describe?

Learn about It!
Variance (𝝈𝟐
) and Standard Deviation (𝝈) of a
Discrete Random Variable
describes the dispersion or the variability of the distribution
1
The variance can be computed using the formula
𝜎2
= Σ 𝑋2
⋅ 𝑃 𝑋 − 𝜇2
.
The standard deviation is the square root of the variance.
Thus, its formula is expressed as 𝜎 = Σ 𝑋2 ⋅ 𝑃 𝑋 − 𝜇2.

Learn about It!
Properties of the Variance
Let 𝑋 and 𝑌 be random variables, and let 𝑎 and 𝑏 be constants. The following
properties hold:
1. 𝑉𝑎𝑟 𝑎𝑋 + 𝑏 = 𝑎2𝑉𝑎𝑟(𝑋)
2. 𝑉𝑎𝑟 𝑋 + 𝑌 = 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟 𝑌 = 𝑉𝑎𝑟(𝑋 − 𝑌), provided that 𝑋 and 𝑌 are
independent of each other.
2

Try It!
Example 1: Find the variance and the standard deviation of
the discrete random variable 𝑋 with the following probability
distribution.
𝑿 𝑷(𝑿)
3 1
8
2 3
8
1 3
8
0 1
8

Try It!
Solution:
1. Compute for the mean
of the probability
distribution.
𝑿 𝑷(𝑿) 𝑿 ∙ 𝑷(𝑿)
3 1
8
3
8
2 3
8
6
8
1 3
8
3
8
0 1
8
0
TOTAL 𝟏. 𝟓
distribution.

Try It!
Solution:
Thus, the mean of the probability distribution is 𝝁 = 𝟏. 𝟓.
distribution.

Try It!
Solution:
2. Construct the column for 𝑋2
and 𝑋2
∙ 𝑃(𝑋).
Determine the values by squaring the value of the random
variable and multiplying it to its corresponding probability
value 𝑃(𝑋).
distribution.

Try It!
Solution:
distribution.
𝑿 𝑷(𝑿) 𝑿𝟐 𝑿𝟐 ∙ 𝑷(𝑿)
3 1
8
9 9
8
2 3
8
4 12
8
1 3
8
1 3
8
0 1
8
0 0
TOTAL 𝟐𝟒
𝟖
= 𝟑

Try It!
Solution:
3. Find the variance and the standard deviation using their
corresponding formula.
distribution.

Try It!
Solution:
For the variance:
𝜎2
= 𝛴 𝑋2
∙ 𝑃 𝑋 − 𝜇2
= 3 − 1.5 2
= 𝟎. 𝟕𝟓
distribution.

Try It!
Solution:
For the standard deviation:
𝜎 = 𝛴 𝑋2 ∙ 𝑃 𝑋 − 𝜇2
= 0.75
= 𝟎. 𝟖𝟕
distribution.

Try It!
Solution:
Thus, the variance of the discrete random variable 𝑋 is
𝝈𝟐
= 𝟎. 𝟕𝟓 while its standard deviation is 𝝈 = 𝟎. 𝟖𝟕.
distribution.

Try It!
Example 2: Find the variance and standard deviation of the
probability distribution shown below.
𝑿 𝑷(𝑿)
0 0.09
1 0.19
2 0.37
3 0.23
4 0.12

Try It!
𝑿 𝑷(𝑿) 𝑿 ⋅ 𝑷(𝑿)
0 0.09 0
1 0.19 0.19
2 0.37 0.74
3 0.23 0.69
4 0.12 0.48
TOTAL 𝟐. 𝟏
Solution:
1. Compute for the mean of
the probability distribution.
Thus, the mean of the
probability distribution is
𝝁 = 𝟐. 𝟏.

Try It!
Solution:
2. Construct the column for 𝑋2
and 𝑋2
⋅ 𝑃(𝑋).
Determine the values by squaring the value of the random
variable and multiplying it to its corresponding probability
value.

Try It!
Solution:
𝑿 𝑷(𝑿) 𝑿𝟐
𝑿𝟐
⋅ 𝑷(𝑿)
0 0.09 0 0
1 0.19 1 0.19
2 0.37 4 1.48
3 0.23 9 2.07
4 0.12 16 1.92
TOTAL 𝟓. 𝟔𝟔

Try It!
Solution:
3. Find the variance and the standard deviation using their
corresponding formula.

Try It!
Solution:
For the variance:
𝜎2
= Σ 𝑋2
⋅ 𝑃 𝑋 − 𝜇2
= 5.66 − 2.1 2
= 𝟏. 𝟐𝟓

Try It!
Solution:
For the standard deviation:
𝜎 = Σ 𝑋2 ⋅ 𝑃 𝑋 − 𝜇2
= 1.25
= 𝟏. 𝟏𝟐

Try It!
Solution:
Thus, the variance of the discrete random variable 𝑋 is
𝝈𝟐
= 𝟏. 𝟐𝟓, and its standard deviation is 𝝈 = 𝟏. 𝟏𝟐.

Let’s Practice!
Individual Practice:
1. Given the probability distribution of a random variable 𝑋
below, compute for the variance and standard deviation.
𝑿 1 2 3 4
𝑷(𝑿) 1
5
1
10
2
5
3
10

Let’s Practice!
Individual Practice:
2. Find the expected outcome when a die is rolled many
times. Compute for the variance and SD.

Let’s Practice!
Group Practice: To be done in groups of three
1. Jeremiah tosses an unbiased coin. He receives ₱ 50.00 if a
head appears and he pays ₱ 30.00 if a tail appears. Find
the:
a. Expected Value and,
b. Variance of his gain
c. Standard Deviation

Let’s Practice!
Group Practice: To be done in groups of two to five.
1. Trixie Anne is comparing the data retrieved from the sales
of computers last year and her organized data this year.
Last year, the expected value for computer sales was 4
computers per day with a standard deviation of 1.21
computers. Given the set of data below, which year yields
better sales?

Let’s Practice!
Number of Computers
(𝑿)
Probability
𝑷(𝑿)
0 0.12
1 0.32
2 0.08
3 0.14
4 0.21
5 0.13

Key Points
Variance (𝝈𝟐
) and Standard Deviation (𝝈) of a
Discrete Random Variable
describes the dispersion or the variability of the distribution
1
Properties of the Variance
Let 𝑋 and 𝑌 be random variables, and let 𝑎 and 𝑏 be constants. The following
properties hold:
1. 𝑉𝑎𝑟 𝑎𝑋 + 𝑏 = 𝑎2𝑉𝑎𝑟(𝑋)
2. 𝑉𝑎𝑟 𝑋 + 𝑌 = 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟 𝑌 = 𝑉𝑎𝑟(𝑋 − 𝑌), provided that 𝑋 and 𝑌 are
independent of each other.
2

Synthesis
● How do we determine the variance and the standard
deviation of a probability distribution?
● Why is it important to measure the dispersion of a
probability distribution?
● How can we construct a probability distribution for a
continuous random variable?

variance ( STAT).pptx

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variance ( STAT).pptx