11
DEPARTMENT OF EDUCATION REGION III
Schools Division of Tarlac Province
Macabulos Drive, San Roque, Tarlac City
Learning Activity Sheet
in
Statistics and Probability
Quarter 3 - Week 3
Computing the Mean and Variance of a Discrete
Probability Distribution
(M11/12SP-IIIb-1, M11/12SP-IIIb-2, M11/12SP-IIIb-3, M11/12SP-IIIb-4)
Subject – Grade 11
Learning Activity Sheet
Quarter 3 – Week 3: Computing the Mean and Variance of a Discrete Probability
Distribution
First Edition, 2022
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Mathematicians usually consider the outcomes of a coin toss as a
1
random event. That is, each probability of getting a head is 2, and the
1
probability of getting a tail is 2. However, it is not possible to predict with
100% certainty which outcomes will occur. But new studies question this
theory. During World War II, a South African mathematician named John
Kerrich tossed a coin 10,000 times while he was interned in a German prison
camp. Unfortunately, the results of his experiment were never recorded. With
this, you will learn how to determine the likeliness of the happening of an
event as you go along with the lesson.
Going back to the lessons in Mathematics in your Junior High School
year, you have learned how to compute and describe for mean and variance
of ungrouped and grouped data. In this lesson, we shall discuss the mean
and variance associated with probabilities, called mean and variance of a
discrete probability distribution. And at the end of the lesson, you are able to
illustrate, calculate, interpret, and solve problems involving the mean,
variance, and standard deviation of a discrete probability distribution.
Kindly, follow the directions and answer the questions honestly in your
notebook.
Always remember, knowledge is a valuable asset that no one can take away
from you. Goodluck and enjoy learning!
Recalling that the mean of ungrouped data is known as average. The
Σ𝑥
formula for the mean of ungrouped data is mean (𝑥̅ ) = 𝑁 .
Directions: The table shows the grades of the 5 random STEM learners in
the first quarter. Find the average of each learner.
Subjects
Oral Commnunication
General Mathematics
Personal Development
Understanding Culture,
Society, and Politics
Student Student Student Student Student
A
B
C
D
E
94
88
92
96
97
97
94
92
94
93
88
95
92
82
88
83
94
87
91
81
Komunikasyon at
Pananaliksik sa Wika at
Kulturang Pilipino
Physical Education and
Health
English for Academic and
Professional Purposes
General Biology 1
Precalculus
Average
90
97
94
85
91
85
96
91
91
85
88
96
92
93
86
90
96
95
98
90
96
93
96
89
81
(1)
(2)
(3)
(4)
(5)
Suppose, you rolled a die once, what do you think is the average number of
spots that would appear? The following activity will help you answer this
question.
Mean of a Discrete Probability Distributions
Steps in Finding the Mean
Step 1:
Construct the probability distribution for the random variable X.
Step 2:
Multiply the value of the random variable X by the corresponding
probability.
Step 3:
Add the results obtained in Step 2. And the result in step 3 will
be the mean when simplified.
Let 𝑥 be the number of spots in a die and 𝑝(𝑥) be the probability of a certain
value of 𝑥. (Note, the symbol asterisk “ ∗ ” means multiplication and mu “𝜇” is the symbol
for the population mean.)
𝒙
𝒑(𝒙)
𝒙 ∗ 𝒑(𝒙)
1
1
6
1
6
1
6
1
6
1
6
1
6
1
6
2
2∗
6
3
3∗
6
4
4∗
6
5
5∗
6
6
6∗
6
21
Σ[x ∗ 𝑝(𝑥)] 𝑜𝑟 𝜇 =
𝑜𝑟 3.5
6
2
3
4
5
6
Total
Σ𝑝(𝑥) =
1∗
6
=1
6
1
=
6
1
=
6
1
=
6
1
=
6
1
=
6
1
=
6
As you can see on the table above, the p(x) in every spot is
and the sum of the p(x) is 1 which only means that it is a probability
6
distribution. The mean of the probability distribution tells us that the average
number of spots that would appear is 3.5. That is in rolling a die many times,
theoretically the mean will be 3.5.
1
Formula for the Mean of the Probability Distribution
𝜇 = [𝑥1 ∗ 𝑝(𝑥1 )] + [𝑥2 ∗ 𝑝(𝑥2 )] + ⋯ , +[𝑥𝑛 ∗ 𝑝(𝑥𝑛 )]
or
𝜇 = Σ [𝑥 ∗ 𝑝(𝑥 )]
where:
𝜇
𝑥
𝑝(𝑥)
= means
= values of random variable x; and
= the corresponding probabilities
Example 1.
Cellular Phone Sales
The probability that a cellular phone kiosk in Waltermart Capas sells 𝑥
number of new phone contracts per day which is shown in the table below.
What is the average number of cellular phone that a kiosk will be sold?
Solution:
𝒙
4
5
6
8
10
Total
𝒑(𝒙)
8
20
6
20
2
20
3
20
1
20
20
Σ𝑝(𝑥) =
=1
20
𝒙 ∗ 𝒑(𝒙)
8
32
4∗
=
20
20
6
30
5∗
=
20
20
2
12
6∗
=
20
20
3
24
8∗
=
20
20
1
10
10 ∗
=
20
20
108
Σ[x ∗ 𝑝(𝑥)] 𝑜𝑟 𝜇 =
𝑜𝑟 5.4
20
Therefore, the mean of the probability distribution is 5.4. This suggests
that the average number of cellular phone which will be sold for the day is 5.4
or approximately 5.
Always Remember
A discrete random variable 𝒙, the mean, denoted by 𝝁, is the sum of
the products formed from multiplying the possible values of 𝒙 with their
corresponding probabilities.
Variance of Discrete Probability Distributions
The Variance and Standard Deviation describe the amount of spread,
dispersion, or variability of the items in a distribution.
Steps in Finding the Variance and Standard Deviation
Step 1. Find the mean of the probability distribution.
Step 2. Subtract the mean from each value of the random variable X.
Step 3. Square the results obtained in Step 2.
Step 4. Multiply the results obtained in Step 3 by the corresponding
probability
Step 5. Get the sum of the results obtained in Step 4 to get the variance.
Step 6. Get the square root of the result in Step 5 to get the standard
deviation.
Example 1. Face Mask Consumed in a Day
The number of face mask consumed by Mr. Lee’s household for the last
three months is listed below. Estimating the number of face mask to be
bought for the next month is necessary. To get that, let’s compute for the
mean, variance, and standard deviation of probability distribution using the
following steps.
Let 𝑥 be the number of face mask consumed per month and 𝑝(𝑥) be the
corresponding probability.
13
1
5
𝒙
𝒑(𝒙)
15
2
5
16
2
5
Solutions: Step 1. Find the mean of the probability distribution.
𝒙
13
15
16
𝒑(𝒙)
1
5
2
5
2
5
𝒙 ∗ 𝒑(𝒙)
13
5
30
5
32
5
𝝁 = Σ[𝑥 ∗ 𝑝(𝑥)] =
75
= 𝟏𝟓
5
Step 2. Subtract the mean from each value of the random variable X.
𝒙
13
𝒑(𝒙)
1
5
2
5
2
5
15
16
𝒙 ∗ 𝒑(𝒙)
13
5
30
5
32
5
Step 3. Square the results obtained in Step 2.
𝒙
𝒑(𝒙)
𝒙 ∗ 𝒑(𝒙)
1
13
13
5
5
2
30
15
5
5
2
32
16
5
5
𝒙−𝝁
13 − 15 = −2
15 − 15 = 0
16 − 15 = 1
𝒙−𝝁
13 − 15 = −2
(𝒙 − 𝝁)𝟐
(−2)2 = 4
15 − 15 = 0
(0)2 = 0
16 − 15 = 1
(1)2 = 1
Step 4. Multiply the results obtained in Step 3 by the corresponding
probability.
𝒙
13
15
16
𝒑(𝒙)
1
5
2
5
2
5
𝒙 ∗ 𝒑(𝒙)
13
5
30
5
32
5
𝒙−𝝁
(𝒙 − 𝝁)𝟐
(𝒙 − 𝝁)𝟐 ∗ 𝒑(𝒙)
4
5
13 − 15 = −2
(−2)2 = 4
15 − 15 = 0
(0)2 = 0
0
16 − 15 = 1
(1)2 = 1
2
5
Step 5. Get the sum of the results obtained in Step 4 to get the variance.
𝒙
13
15
16
𝒑(𝒙)
1
5
2
5
2
5
𝒙 ∗ 𝒑(𝒙)
13
5
30
5
32
5
𝒙−𝝁
(𝒙 − 𝝁)𝟐
13 − 15 = −2
(−2)2 = 4
15 − 15 = 0
(0)2 = 0
16 − 15 = 1
(𝒙 − 𝝁)𝟐 ∗ 𝒑(𝒙)
4
5
0
2
5
6
𝝈𝟐 = Σ[(𝑥 − 𝜇)2 ∗ 𝑝(𝑥)] = = 𝟏. 𝟐
5
(1)2 = 1
Step 6. Get the square root of the result in Step 5 to get the standard
deviation.
𝜎 = √Σ[(𝑥 − 𝜇 )2 ∗ 𝑝(𝑥 )] = √1.2 ≈ 𝟏. 𝟎𝟗𝟓
So, the mean of the probability distribution is 15, the variance and standard
deviation of the probability distribution is 1.2 and 1.095 respectively.
Formula for the Variance and Standard Deviation of the Probability
Distribution
For the Variance of the Probability Distribution, the formula is
𝝈𝟐 = Σ[(𝑥 − 𝜇 )2 ∗ 𝑝(𝑥 )]
For the Standard Deviation of the Probability Distribution, the
formula is
𝝈 = √Σ[(𝑥 − 𝜇 )2 ∗ 𝑝(𝑥 )]
where:
𝜇
𝑥
𝑝(𝑥)
= means
= values of random variable x; and
= the corresponding probabilities
Example 2. Pizza Deliveries
Pizza shop owner determines the number of pizzas that are delivered each
day. Find the mean, variance, and standard deviation of the probability
distribution.
Let 𝑥 be the number of pizzas delivered in a day and 𝑝(𝑥) be the corresponding
probability.
35
0.1
𝒙
𝒑(𝒙)
Solution:
𝒙
35
36
37
38
39
𝒑(𝒙)
0.1
0.2
0.3
0.3
0.1
𝒙 ∗ 𝒑(𝒙)
3.5
7.2
11.1
11.4
3.9
𝜇 = 37.1
36
0.2
𝒙−𝝁
−2.1
−1.1
−0.1
0.9
1.9
37
0.3
38
0.3
(𝒙 − 𝝁)𝟐
4.41
1.21
0.01
0.81
3.61
39
0.1
(𝒙 − 𝝁)𝟐 ∗ 𝒑(𝒙)
0.441
0.242
0.003
0.243
0.361
2
𝜎 = 1.29
Since the variance (𝜎 2 ) is 1.29, then the standard deviation of the probability
distribution is
𝜎 = √Σ[(𝑥 − 𝜇 )2 ∗ 𝑝(𝑥 )] = √1.29 ≈ 𝟏. 𝟏𝟑𝟔.
A. Direction. Complete the table below and find the mean of the following
probability distribution.
1.
𝒙
𝒑(𝒙)
𝒙 ∗ 𝒑(𝒙)
1
0.15
2
0.25
3
0.05
4
0.35
5
0.20
𝝁=
2.
𝒙
3
6
9
12
15
18
𝒑(𝒙)
1
5
1
10
1
10
1
5
1
10
3
10
𝒙 ∗ 𝒑(𝒙)
𝝁=
B. Direction. Solve the following problems.
1. The probabilities of a machine manufacturing 0, 1, 2, 3, 4, or 5
defective parts in one day are 0.75, 0.17, 0.04, 0.025, 0.01, and
0.005, respectively. Find the mean, variance, and the standard
deviation of the probability distribution.
2. The number of laptops sold per day at a local computer store, along
with its corresponding probabilities, is shown in the table. Find the
mean, variance, and standard deviation of the distribution.
Number of Laptops Sold (𝑥)
0
1
2
3
4
Probability 𝑝(𝑥)
0.1
0.2
0.3
0.2
0.2
Direction: Based on the discussion above, answer the following questions.
1. What are the steps in finding the mean of a probability distribution?
2. What are the steps in finding the variance of a probability
distribution?
3. How to get the standard deviation of a probability distribution based
on its variance?
•
•
Formula for the Mean of the Probability Distribution
𝝁 = Σ [𝑥 ∗ 𝑝(𝑥 )]
Formula for the Variance of the Probability Distribution
𝝈𝟐 = Σ[(𝑥 − 𝜇 )2 ∗ 𝑝(𝑥 )]
•
Formula for the Standard Deviation of the Probability Distribution
𝝈 = √Σ[(𝑥 − 𝜇 )2 ∗ 𝑝(𝑥 )]
Direction: Read and analyze the problem below, write your answer on your
notebook.
Dr. Yap administered a group of 20,000 students to developed a test to
measure boredom tolerance. The possible scores were 0, 1, 2, 3, 4, 5, and 6,
with 6 indicating the highest tolerance for boredom. The results are shown
below. Complete the following table and then find the mean, variance, and
standard deviation of the probability distribution to measure boredom
tolerance of the students.
Score
Frequency
𝑓
1400
2600
3600
6000
4400
1600
400
𝑥
0
1
2
3
4
5
6
Probability
𝑝(𝑥)
Direction: Record the hours you spent in studying each day, from Monday to
Friday. Construct a table that shows your x as the number of hours, frequency
as how many times you obtain that given hour, and the probability of each.
Then, determine the mean, variance, and standard deviation of the probability
distribution. From the result, analyze it and create a conclusion.
Learning Activity Sheet in Statistics and Probability
Quarter 3 - Week 3: Computing the Mean and Variance of a Discrete
Probability Distribution
(M11/12SP-IIIb-1, M11/12SP-IIIb-2, M11/12SP-IIIb-3, M11/12SP-IIIb-4)
Key to Corrections
Let’s
1.
2.
3.
4.
5.
Recall
89.56
96.22
92.22
91.67
86.11
Let’s Practice
A. 1.
𝒙
1
2
𝒑(𝒙)
0.15
0.25
𝒙 ∗ 𝒑(𝒙)
0.15
0.50
3
4
5
0.05
0.35
0.20
𝒙
3
𝒑(𝒙)
1
5
1
10
1
10
1
5
1
10
3
10
0.15
1.40
1.00
𝝁 = 3.20
2.
6
9
12
15
18
B. 1.
𝒙
𝒑(𝒙)
0
0.75
1
0.17
2
0.04
3
0.025
4
0.01
5
0.005
𝒙 ∗ 𝒑(𝒙)
0
0.17
0.08
0.075
0.04
0.025
𝜇 = 0.39
𝒙−𝝁
−0.39
−0.61
1.61
2.61
3.61
4.61
𝒙 ∗ 𝒑(𝒙)
3
5
6
10
9
10
12
5
15
10
54
10
𝟓𝟕
𝝁=
= 𝟏𝟏. 𝟒
𝟓
(𝒙 − 𝝁)𝟐
0.152
0.372
2.592
6.812
13.032
21.252
(𝒙 − 𝝁)𝟐 ∗ 𝒑(𝒙)
0.114
0.063
0.104
0.170
0.130
0.106
2
𝜎 =0.687
𝜎 = √0.687 ≈ 0.829
2.
0
1
2
3
4
𝒙
0
1
2
3
4
𝒑(𝒙)
0.1
0.2
0.3
0.2
0.2
𝒙 ∗ 𝒑(𝒙)
0
0.2
0.6
0.6
0.8
0.1
0.2
0.3
0.2
0.2
𝒙−𝝁
−2.2
−1.2
−0.2
0.8
1.8
(𝒙 − 𝝁)𝟐
4.84
1.44
0.04
0.64
3.24
(𝒙 − 𝝁)𝟐 ∗ 𝒑(𝒙)
0.484.
0.288
0.012
0.128
0.648
𝜇 =2.2
𝜎 2 = 1.56
𝜎 = √1.56 ≈ 1.249
Let’s Enrich
Answers may vary
Let’s Perform
Answers may vary
Let’s Assess
Answers may vary
References:
Belecina, Rene R, Elisa S Baccay, and Efren B Mateo. Statistics and
Probability. Firsted. 2019., 2-8. Manila, Philippines: Rex Bookstore, 2019.
“Compendium of Notes in Statistics and Probability.” Tarlac: Schools
Division of Tarlac Province, 2020.
For inquiries or feedback, please write or call:
Department of Education – Region III
Schools Division of Tarlac Province
Office Address:
Telefax:
Email Address:
Macabulos Drive, Brgy. San Roque,
Tarlac City 2300 Tarlac
(045) 982 0345 | 982 0374
tarlac@deped.gov.ph