statisitcs 11

1. Senior High School
STATISTICS AND PROBABILITY
Core Subject
JAYSON M. MAGALONG
Subject Teacher

2. TOPIC: Mean and
Variance of a Discrete
Random Variable

3. After going through this lesson, you are
expected to:
1. apply the important concepts of mean
and variance of a discrete random variable;
and
2. calculate the mean and variance of a
discrete random variable.
 OBJECTIVES

4.  Mean is considered as a measure of the `central
location' of a random variable. It is the weighted
average of the values that random variable X can
take, with weights provided by the probability
distribution. It is also called as the Expected
Value.
 The term variance refers to a statistical
measurement of the spread between numbers in a
data set. More specifically, variance measures
how far each number in the set is from the mean
(average), and thus from every other number in
the set.
 The Standard Deviation is a measure of how
spread out numbers are.
 DEFINITION OF TERMS

5. 𝑽𝒂𝒓 𝑿 = 𝝈𝟐
𝒙 = (𝒙 − 𝝁𝒙)𝟐
𝒂𝒍𝒍 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆
𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒙
∙ 𝑷(𝒙)
𝑬 𝑿 = 𝝁𝒙 = 𝑋1 ∗ 𝑃 𝑋1 + 𝑋2 ∗ 𝑃 𝑋2 + ⋯ + 𝑋𝑛 ∗ 𝑃 𝑋𝑛
= 𝑿 ∗ 𝑷(𝑿)
𝜎𝑥 = 𝑽𝒂𝒓 𝑿 or 𝝈𝒙 = 𝝈𝒙
𝟐
Mean Formula
Variance Formula
Standard Deviation Formula

6. Mr. Umali, a Mathematics
teacher, regularly gives a
formative assessment
composed of 5 multiple-choice
items. After the assessment,
he used to check the
probability distribution of the
correct responses, and the
data is presented below:
EXAMPLE 1.
Test
Item 𝑿
Probability
𝑷(𝑿)
0 0.03
1 0.05
2 0.12
3 0.30
4 0.28
5 0.22
1. What is the average or mean of the given probability
distribution?
2. What are the values of the variance and the standard
deviation of the probability distribution?

7. Test Item,
𝒙
Probability
𝑷(𝑿)
𝒙 ∙ 𝑷(𝑿) 𝒙 − 𝝁 (𝒙 − 𝝁)𝟐
(𝒙 − 𝝁)𝟐
∙ 𝑷(𝑿)
0 0.03 0 -3.41 11.6281 0.3488
1 0.05 0.05 -2.41 5.8081 0.2904
2 0.12 0.24 -1.41 1.9881 0.2386
3 0.30 0.90 -0.41 0.1681 0.0504
4 0.28 1.12 0.59 0.3481 0.0975
5 0.22 1.10 1.59 2.5281 0.5562
𝑬 𝑿
= 𝝁𝒙
= 𝟑. 𝟒𝟏
(𝒙 − 𝝁)𝟐
∙ 𝑷 𝑿 = 𝟏. 𝟓𝟖𝟏𝟗

8. 𝐸 𝑋 = 𝜇𝑥 = 𝑥 ∙ 𝑃 𝑋
= 0 0.03 + 1 0.05 + 2 0.12
𝑬 𝑿 = 𝝁𝒙 = 𝟑. 𝟒𝟏
𝑽𝒂𝒓 𝑿 = 𝝈𝟐
𝒙 = 𝒙 − 𝝁 𝟐
∙ 𝑷 𝑿
=
0 − 3.41 2
.03 + 1 − 3.41 2
.05 + 2 − 3.41 2
0.12 + 3 − 3.41 2
0.30
+ 4 − 3.41 2 0.28 + 5 − 3.41 2 0.22
𝑽𝒂𝒓 𝑿 = 𝝈𝟐
𝒙 = 𝟏. 𝟓𝟖𝟏𝟗
The Variance is 1.5819, and the Standard
Deviation is 𝟏. 𝟓𝟖𝟏𝟗 , and it is equivalent to 𝝈𝒙 =
1.26

9. Suppose that a coin is tossed twice so that the
sample space is S = {𝐻𝐻, 𝑇𝐻, 𝐻𝑇, 𝑇𝑇}. Let X
represent the “number of heads that can come up”,
Based on the prepared discrete probability
distributions of the random variable X below, calculate
the mean, variance, and standard deviation.
Outcome or
Sample Point
HH HT TH TT
𝒙 2 1 1 0

10. 𝒙 𝑷 𝑿 𝒙
∙ 𝑷 𝑿
𝒙
− 𝝁
𝒙 − 𝝁 𝟐 𝒙 − 𝝁 𝟐
∙ 𝑷 𝑿
0 ¼ or 0.25 0 -1 1 0.25
1 ½ or 0.5 0.5 0 0 0
2 ¼ or 0.25 0.5 1 1 0.25
𝝁𝒙 = 𝒙 ∙ 𝑷 𝑿 = 𝟏 (𝒙 − 𝝁)𝟐
∙ 𝑷 𝑿 = 𝟎. 𝟓𝟎
The expected value or mean is 1.
The Variance is 0.50, and
The Standard Deviation is 𝟎. 𝟓0, and it is equivalent to 𝝈 = 0.71.

11. Independent Activity : Study and analyze
The number of patients seen in the Emergency
Room in any given hour is a random variable
represented by x. The probability distribution for x
is
X 10 11 12 13 14
P(X) 0.4 0.2 0.2 0.1 0.1
SOLVE FOR THE MEAN, VARIANCE, AND
STANDARD DEVIATION.