Discrete Random Variable (Probability Distribution)

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This presentation the statistics teachers to discuss discrete random variable since it includes examples and solutions. Content: -definition of random variable -creating a frequency distribution table - creating a histogram -solving for the mean, variance and standard deviation. References: http://www.elcamino.edu/faculty/klaureano/documents/math%20150/chapternotes/chapter6.sullivan.pdf https://www.mathsisfun.com/data/random-variables-mean-variance.html https://www.youtube.com/watch?v=OvTEhNL96v0 https://www150.statcan.gc.ca/n1/edu/power-pouvoir/ch12/5214891-eng.htm

Discrete Random Variable:
Probability Distribution

Definitions
Random Variable – a variable whose
variables are determined by
chance.
Discrete Probability Distribution –
Consist of random variables and
probabilities of the values.

Requirements of a Probability
Distribution
1. All probabilities must between 0 and 1.
2. The sum of probabilities must add up to 1.

Example: Basic Concept of Probability
Review:
What is the probability of having two tails
in flipping a coin twice?
What is the probability of getting the
numbers 1 and 5 in rolling a die.

Example: Basic Concept of Probability
Review:
In a group of 40 people, 15 have high blood
pressure and 25 have high level of cholesterol,
what is the probability that a person that will be
randomly selected have
a) has high blood pressure (event A)?
b) has high level of cholesterol(event B)?

Example: Probability Distribution of
Random Variable
1. Construct a probability distribution
for drawing a card from a deck of
40cards consisting of 10 cards
numbered 1, 10 cards numbered
2, 15 cards numbered 3, and 5
cards numbered 4.

1. Construct a probability distribution for
drawing a card from a deck of 40cards consisting
of 10 cards numbered 1, 10 cards numbered 2,
15 cards numbered 3, and 5 cards numbered 4.
Cards 1 2 3 4
Number of Cards
Probability

Draw a probability histogram.
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
P(1) P(2) P(3) P(4)
Series 1
Series 1

Example: Probability Distribution of
Random Variable
2. The following data represents the
enrollees of Ford Academy of the
Arts in grades 7 to 10 in the year
2019.
Grade Level 7 8 9 10
Enrollees 54 62 57 72

Example: Probability Distribution of
Random Variable
a) Construct a probability distribution
for a random variable.
b) Draw a probability histogram.
Grade Level 7 8 9 10
Enrollees 54 62 57 72

Mean, Variance and
Standard Deviation of
Discrete Random
Variable

Definition
Mean (μ) – average value
- Expectation of random variable or
E(X)
E(X) or μ
Variance (σ2) – measures of spread
Standard Deviation (σ) – square root of the
variance
- measure of spread

Formulas
Formula:
Mean: μ = ∑ (x)[p(x)]
Variance: σ2 = ∑ (x – μ) 2 p(x)
Standard Deviation: σ = σ2

Using the frequency distribution table of
example 1, solve for the mean.

Let the number of cards be x and probability of
the number of cards be p(x), then
Cards 1 2 3 4
Number of Cards (x) 10 10 15 5
Probability p(x) 0.25 0.25 0.375 0.125

Mean: μ = ∑ (x)[p(x)]
μ = ∑ (x)[p(x)]
Substitute the given.
μ = (10 x 0.25) + (10 x 0.25) + (15 x 0.375) + (5 x 0.125)
 Multiply
μ = 2.5 + 2.5 + 5.625 + 0.625
 Add
μ = 11.25

Using the frequency distribution table of
example 1, solve for the variance.

Variance: σ2 = ∑ (x – μ) 2 p(x)
σ2 = ∑ (x – μ) 2 p(x)
Substitute the given
= [(10 – 11.25) 2 (0.25)] + [(10 – 11.25) 2 (0.25)] +
[(15 – 11.25) 2 (0.375)] + [(5 – 11.25) 2 (0.125)]
Perform the operation
= 0.39 + 0.39 + 5.27 + 4.88
 Add
= 10.93

Using the frequency distribution table of
example 1, solve for the standard
deviation.

Standard Deviation: σ = σ2
σ = σ2
Substitute the given.
σ = 10.93
Get the root of the variance.
σ = 3.31

References
http://www.elcamino.edu/faculty/klaureano/do
cuments/math%20150/chapternotes/chapter6.s
ullivan.pdf
https://www.mathsisfun.com/data/random-
variables-mean-variance.html
https://www.youtube.com/watch?v=OvTEhNL96
v0
https://www150.statcan.gc.ca/n1/edu/power-
pouvoir/ch12/5214891-eng.htm

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Discrete Random Variable (Probability Distribution)

1. Discrete Random Variable: Probability Distribution

2. Definitions Random Variable – a variable whose variables are determined by chance. Discrete Probability Distribution – Consist of random variables and probabilities of the values.

3. Requirements of a Probability Distribution 1. All probabilities must between 0 and 1. 2. The sum of probabilities must add up to 1.

4. Example: Basic Concept of Probability Review: What is the probability of having two tails in flipping a coin twice? What is the probability of getting the numbers 1 and 5 in rolling a die.

5. Example: Basic Concept of Probability Review: In a group of 40 people, 15 have high blood pressure and 25 have high level of cholesterol, what is the probability that a person that will be randomly selected have a) has high blood pressure (event A)? b) has high level of cholesterol(event B)?

6. Example: Probability Distribution of Random Variable 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4.

7. 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4. Cards 1 2 3 4 Number of Cards Probability

8. 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4. Cards 1 2 3 4 Number of Cards 10 10 15 5 Probability

9. 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4. Cards 1 2 3 4 Number of Cards 10 10 15 5 Probability 0.25

10. 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4. Cards 1 2 3 4 Number of Cards 10 10 15 5 Probability 0.25 0.25

11. 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4. Cards 1 2 3 4 Number of Cards 10 10 15 5 Probability 0.25 0.25 0.375

12. 1. Construct a probability distribution for drawing a card from a deck of 40cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4. Cards 1 2 3 4 Number of Cards 10 10 15 5 Probability 0.25 0.25 0.375 0.125

13. Draw a probability histogram. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 P(1) P(2) P(3) P(4) Series 1 Series 1

14. Example: Probability Distribution of Random Variable 2. The following data represents the enrollees of Ford Academy of the Arts in grades 7 to 10 in the year 2019. Grade Level 7 8 9 10 Enrollees 54 62 57 72

15. Example: Probability Distribution of Random Variable a) Construct a probability distribution for a random variable. b) Draw a probability histogram. Grade Level 7 8 9 10 Enrollees 54 62 57 72

16. Mean, Variance and Standard Deviation of Discrete Random Variable

17. Definition Mean (μ) – average value - Expectation of random variable or E(X) E(X) or μ Variance (σ2) – measures of spread Standard Deviation (σ) – square root of the variance - measure of spread

18. Formulas Formula: Mean: μ = ∑ (x)[p(x)] Variance: σ2 = ∑ (x – μ) 2 p(x) Standard Deviation: σ = σ2

19. Using the frequency distribution table of example 1, solve for the mean.

20. Let the number of cards be x and probability of the number of cards be p(x), then Cards 1 2 3 4 Number of Cards (x) 10 10 15 5 Probability p(x) 0.25 0.25 0.375 0.125

21. Mean: μ = ∑ (x)[p(x)] μ = ∑ (x)[p(x)] Substitute the given. μ = (10 x 0.25) + (10 x 0.25) + (15 x 0.375) + (5 x 0.125)  Multiply μ = 2.5 + 2.5 + 5.625 + 0.625  Add μ = 11.25

22. Using the frequency distribution table of example 1, solve for the variance.

23. Variance: σ2 = ∑ (x – μ) 2 p(x) σ2 = ∑ (x – μ) 2 p(x) Substitute the given = [(10 – 11.25) 2 (0.25)] + [(10 – 11.25) 2 (0.25)] + [(15 – 11.25) 2 (0.375)] + [(5 – 11.25) 2 (0.125)] Perform the operation = 0.39 + 0.39 + 5.27 + 4.88  Add = 10.93

24. Using the frequency distribution table of example 1, solve for the standard deviation.

25. Standard Deviation: σ = σ2 σ = σ2 Substitute the given. σ = 10.93 Get the root of the variance. σ = 3.31

26. References http://www.elcamino.edu/faculty/klaureano/do cuments/math%20150/chapternotes/chapter6.s ullivan.pdf https://www.mathsisfun.com/data/random- variables-mean-variance.html https://www.youtube.com/watch?v=OvTEhNL96 v0 https://www150.statcan.gc.ca/n1/edu/power- pouvoir/ch12/5214891-eng.htm

Discrete Random Variable (Probability Distribution)

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