NOTRE DAME OF DADIANGAS UNIVERSITY
Integrated Basic Education Department
Senior High School
Lagao, General Santos City
Statistics and Probability
Quarter: Midterm
Week No.: 1
Damean’s Beat: Quality Education
NDDU’s 4Cs: ☐ Christian Leaders
 Competent Professionals
☐ Community-Oriented Citizens
☐ Culture-Sensitive Individuals
Teacher/s: Arvin Eric A. Caga
Ever Joy P. Hinampas
Rachel Lou P. Paranga
Date: 2021.01.11 – 2021.01.16
21st Century Skills:
 Critical Thinking
 Computing/ICT Literacy
☐ Communication
☐ Creativity
☐ Collaboration
☐ Cross Cultural Understanding
☐ Career and Learning Self Reliance
WEEK: 1
I. Topic: Key Concepts of Random Variables
Figure 1.1
Variable
Qualitative
Quantitative
Discrete
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Continuous
Definition
Statistics is a science of collecting, presenting, analyzing and interpreting data to arrive
at an effective decision. It as a way of transforming raw data into useful, significant
information.
Data are the quantities (numbers) or qualities (attributes) measured or observed that are
to be collected and/or analyzed.
Data set is a collection of data.
Variable is a characteristic or attribute that has no specific numerical values.
Random variable (X) is variable whose possible values are determined by chance.
Qualitative variable is when the characteristics of data is Non-numeric. One whose
categories are simply used as labels to distinguish one group from another.
Example : color, status of a person, the type of the car a person owns
Quantitative Variable is when the characteristics of data are numeric, which can be
discrete or continuous data. One whose categories can be measured and ordered
according to quantity.
Example : number of children in the family, age
Discrete variables have a finite number of possible values that can be counted. The word
counted means that they can be enumerated using the numbers 1, 2, 3, etc.
Continuous variables can assume an infinite number of values in an interval between
two specific values. These are obtained from data that can be measured rather than
counted. They can be decimal and fractional values.
Example 1.1
Determine whether the random variable X or Y is discrete or continuous variable.
1) X = number of birds in a nest
Answer: Since the number of birds can be counted, then it is a discrete variable.
2) Y = the weights in kg of randomly selected dancers after taking up aerobics
Answer: Since the weight is measured rather than counted, then it is a continuous variable.
3) Y = number of defective light bulbs among the randomly selected light bulbs
Answer: Since the number of defective light bulbs can be counted, then it is a discrete variable.
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4) X = the height of daisy plants in the backyard
Answer: Since the height is measured rather than counted, then it is a continuous variable.
5) Y = the hourly temperature last Saturday
Answer: Since the temperature is measured rather than counted, then it is a continuous variable.
Example 1.2
Identify the possible values of the random variable for each scenario.
1) Number of absent students in a class with 50 students
Possible Values: 0,1,…,50
2) Temperature of a baby with fever
Possible Values: 37.1° C - 40 ° C
3) Weight of a person
Possible Values: 40 kg – 45 kg
4) Number of members in a family
Possible Values: 3,4,5,…
5) Number of malls in a city
Possible Values: 0,1,2,…
Definition
Discrete Probability Distribution is the listing of all possible values of a discrete random
variable along with their corresponding probabilities. The probabilities are determined
theoretically or by observation.
PROPERTIES
1) The probability of each value of discrete random variable is
between 0 and 1 inclusive.
0 ≤ P (x) ≤ 1
2) The sum of all probabilities is 1.
∑ P (X) = 1
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Example 1.3
1) A coin is tossed thrice. Let the variable X be the number of heads. A) Identify the probability
distribution for each random variable and its table; and B) verify if the two properties are
satisfied.
Solution:
1st toss
2nd Toss
3rd Toss
Outcome
Number of Heads (X)
H
HHH
3
T
HHT
2
H
HTH
2
T
HTT
1
H
THH
2
T
THT
1
H
TTH
1
T
TTT
0
H
H
T
H
T
T
TOTAL NUMBER
OF OUTCOMES = 8
Probabilities for the values of X can be determined as follows:
No Head
TTT
1
8
HTT
1
8
1
8
1 Head
THT
1
8
3
8
TTH
1
8
2 Heads
HHT
HTH
THH
1
1
1
8
8
8
3 Heads
HHH
1
8
3
8
1
8
Note
The total number of outcomes serves as the denominator, thus it is equal to 8.
1
3
3
Hence, the probability of getting no head is 8, one head is 8, two heads is 8, and three
1
heads is 8.
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A) Probability Distribution:
1
𝑃 (0) = 8
3
𝑃(1) = 8
3
1
𝑃(2) = 8
𝑃 (3) = 8
Probability Distribution Table:
Number of heads (X)
Probability P(x)
0
1
8
1
3
8
2
3
8
3
1
8
Note
In constructing the probability distribution, identify the outcomes and the
probability of each outcome.
B) Properties:
1st Property:
Since all the individual values of probability are from 0 to 1,
then the first property is satisfied.
Note
1
P(0) and P(3): 0 ≤ 8 ≤ 1 is true.
3
P(1) and P(2): 0 ≤ 8 ≤ 1 is true.
2nd Property:
∑ P(X) = 1
𝑃(0) + 𝑃 (1) + 𝑃 (2) + 𝑃(3) = 1
1
3
3
1
+8+8+8 =1
8
1=1
Since the summation of all probabilities is 1, then the second property is satisfied.
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2) The spinner is divided into 10 sections. Let X be the score where the arrow will stop
(numbered as 1, 2,3 and 4)
A) Find the probability that the arrow will stop at 1, 2,3 and 4.
B) Construct the discrete probability distribution of the random variable X.
3
1
2
2
1
Solution:
A) P(1) =
P(2) =
P(3) =
P(4) =
2
1
3
4
2
3
10
4
10
2
10
1
10
Note
Since the total number of outcomes is 10, then the value of the denominator is
10. From the figure above, you can see that there are three “1” in the wheel, thus the
3
probability that the arrow will stop at “1” is
. The same logic will be applied in getting
10
the probability that the arrow will stop at “2”, “3” and “4”.
B)
Number on Spinner (X)
Probability P(x)
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1
3
10
2
4
10
3
2
10
4
1
10
Example 1.4
Consider the following tables and identify if it is a discrete probability distribution. Otherwise,
identify the property or properties that are not satisfied.
1)
X
P(X)
0
0.2
1
0.3
2
0.2
3
0.3
Solution:
1st Property:
Since all the individual values in P(X) are from 0 to 1, then the first property is satisfied.
2nd Property:
∑ P(X) = 1
𝑃(0) + 𝑃 (1) + 𝑃 (2) + 𝑃(3) = 1
0.2 + 0.3 + 0.2 + 0.3 = 1
1=1
Since the summation of all probabilities is 1, then the second property is satisfied.
Conclusion:
Therefore table 1 is a discrete probability distribution since the 2 properties are satisfied.
2)
X
P(X)
2
0.2
3
0.5
4
0.4
5
0.4
6
-0.5
Solution:
1st Property:
Since P(6) = −0.5 is not from 0 to 1, then the first property is NOT satisfied.
2nd Property:
∑ P(X) = 1
𝑃(2) + 𝑃 (3) + 𝑃 (4) + 𝑃(5) + 𝑃(6) = 1
0.2 + 0.5 + 0.4 + 0.4 − 0.5 = 1
1=1
Since the summation of all probabilities is 1, then the second property is satisfied.
Conclusion:
Therefore table 2 is NOT a discrete probability distribution since the first property is not
satisfied.
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VI. References/Materials:
Blumman, A.G., (2004). Elementary Statistics, A Step by Step Approach, Fifth Edition. Published
by McGraw – Hill Companies Inc. pp. 6-8
Mann, P.S., (2010). Introductory Statistics. Seventh Edition. John Wiley & Sons: Hoboken pp. 192
- 193
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