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
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
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
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
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
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
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

Discrete Random Variable (Probability Distribution)

  • 1.
  • 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 aProbability Distribution 1. All probabilities must between 0 and 1. 2. The sum of probabilities must add up to 1.
  • 4.
    Example: Basic Conceptof 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 Conceptof 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 Distributionof 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 aprobability 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 aprobability 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 aprobability 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 aprobability 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 aprobability 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 aprobability 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 probabilityhistogram. 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 Distributionof 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 Distributionof 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 StandardDeviation 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 frequencydistribution table of example 1, solve for the mean.
  • 20.
    Let the numberof 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 frequencydistribution 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 frequencydistribution 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.