Discrete Random Variables and Probability Distributions

Learning Objectives

• Identification of Probability Distributions: Define and list all critical properties and characteristics that govern a valid probability distribution.

• Distinction of Random Variable Types: Differentiate between discrete random variables and continuous random variables based on their mathematical nature and measurement methods.

• Statistical Computation: Compute fundamental descriptive measures of a discrete probability distribution, including the mean (expected value), variance, and standard deviation.

Fundamentals of Random Variables

• Definition of a Random Variable: A variable measured or observed as the outcome of a chance experiment. The variable assumes different numerical values solely by chance.

• Examples of Random Variables:

  • The number of employees absent from the day shift on Monday; potential values include $0$, $1$, $2$, or $3$. Here, the number absent represents the random variable.

• Classification of Random Variables:

  • Discrete Random Variable: A random variable that can assume only certain clearly separated values. Discrete variables are typically the result of counting.

  • Example 1: Tossing a coin three times and counting the total number of heads obtained.

  • Example 2: A bank counting the exact number of credit cards owned by its customers.

  • Case Study (Bank of the Carolinas): The Bank of the Carolinas counts credit card ownership among a group of customers. The number of cards owned is the discrete random variable $X$. The empirical distribution is recorded as follows:

  * $0$ credit cards: relative frequency = $0.03$
  * $1$ credit card: relative frequency = $0.10$
  * $2$ credit cards: relative frequency = $0.18$
  * $3$ credit cards: relative frequency = $0.21$
  * $4$ or more credit cards: relative frequency = $0.48$
  * Total relative frequency sum = $1.00$

• Continuous Random Variable: A random variable that can assume an infinite number of values within a given range. Continuous variables are typically the result of measuring continuous scales.

  • Example 1: The flight duration between Atlanta and LA, taking values such as $4.67\text{ hours}$, $5.13\text{ hours}$, or any intermediate real number.

  • Example 2: The annual snowfall measured in Minneapolis, MN, expressed in inches.

Characteristics and Representation of Probability Distributions

• Definition of Probability Distribution: A complete tabular or mathematical listing of all possible outcomes of an experiment alongside the probability associated with each individual outcome.

• Three Essential Characteristics of a Probability Distribution:

  1. Bounded Probabilities: The probability $P(X)$ of any specific outcome $X$ lies strictly between $0$ and $1$, inclusive:      0 \normalfont{\text{ ≤ }} P(X) \normalfont{\text{ ≤ }} 1

  2. Mutually Exclusive Outcomes: No two outcomes can occur simultaneously during a single trial of the experiment.

  3. Exhaustive Outcomes: The list includes every possible outcome of the experiment, meaning the sum of all individual probabilities is exactly equal to $1$:      $∑ P(X) = 1.0$

• Example 1 (Course Enrollment Distribution):

  • Objective: Model the number of courses taken by university students per semester.

  • Random Variable: $X = \text{number of courses taken}$.

  • Possible outcomes: $2\text{ subjects}$, $3\text{ subjects}$, $4\text{ subjects}$, and $5\text{ subjects}$.

• Distribution Table:

  • $X = 2 → P(X) = 0.20$

  • $X = 3 → P(X) = 0.40$

  • $X = 4 → P(X) = 0.24$

  • $X = 5 → P(X) = 0.16$

  • Sum of probabilities: $0.20 + 0.40 + 0.24 + 0.16 = 1.00$

  • Example 2 (Three Coin Tosses):

• Objective: Model the number of heads appearing face up when tossing a fair coin $3$ times.

• Sample Space ($8$ equally likely outcome events):

* Outcome $1$: $\text{T, T, T} → X = 0$
* Outcome $2$: $\text{T, T, H} → X = 1$
* Outcome $3$: $\text{T, H, T} → X = 1$
* Outcome $4$: $\text{T, H, H} → X = 2$
* Outcome $5$: $\text{H, T, T} → X = 1$
* Outcome $6$: $\text{H, T, H} → X = 2$
* Outcome $7$: $\text{H, H, T} → X = 2$
* Outcome $8$: $\text{H, H, H} → X = 3$

• Probability Distribution Table and Graphical Representation:

* $X = 0 → P(X) = {1}/{8} = 0.125$
* $X = 1 → P(X) = {3}/{8} = 0.375$
* $X = 2 → P(X) = {3}/{8} = 0.375$
* $X = 3 → P(X) = {1}/{8} = 0.125$
* Total sum: ${8}/{8} = 1.000$

• Class Exercise (Vehicle Ownership Distribution):

  • Given raw household count data for car ownership:

* $0\text{ cars} → 12\text{ households}$
* $1\text{ car} → 40\text{ households}$
* $2\text{ cars} → 30\text{ households}$
* $3\text{ cars} → 15\text{ households}$
* $4\text{ cars} → 3\text{ households}$
* Total sample size = $12 + 40 + 30 + 15 + 3 = 100\text{ households}$

• Formulated Probability Distribution for $X = \text{number of cars owned}$:

  • $X = 0 → P(X) = {12}/{100} = 0.12$

  • $X = 1 → P(X) = {40}/{100} = 0.40$

  • $X = 2 → P(X) = {30}/{100} = 0.30$

  • $X = 3 → P(X) = {15}/{100} = 0.15$

  • $X = 4 → P(X) = {3}/{100} = 0.03$

Measures of Central Tendency and Dispersion

• Mean (Expected Value) of a Discrete Probability Distribution:

  • Represents the typical value used to summarize the central location of the distribution.

  • Equivalently called the expected value, denoted as $E(X)$ or ̢

  • Formula [6-1]:     ̢ = E(X) = ∑ [x ⋅ P(x)]

• Variance and Standard Deviation of a Discrete Probability Distribution:

  • Variance (̢^2 or $\text{Var}(X)$): Measures the amount of dispersion, spread, or variation in the distribution.

  • Mathematical definitions:     \text{Var}(X) = E[(X - ̢)^2] = ∑ [(x - ̢)^2 ⋅ P(x)]     \text{Var}(X) = E(X^2) - [E(X)]^2 = ∑ [x^2 ⋅ P(x)] - ̢^2

  • Standard Deviation (̢_X): The positive square root of the variance:     ̢_X = √{\text{Var}(X)}

• Systematic Steps for Computing Variance:

  1. Compute the distribution mean ̢.

  2. Subtract ̢ from each value of $x$ to determine the deviations (x - ̢).

  3. Square each deviation term to obtain (x - ̢)^2

  4. Multiply each squared deviation by its corresponding probability $P(x)$ to yield (x - ̢)^2 P(x).

  5. Sum these product values to calculate the total variance ̢^2

• Comprehensive Worked Example (John's Saturday Car Sales):

  • Scenario: John tracks his Saturday automobile sales using a discrete probability model.

• Step 1: Calculate Expected Car Sales (̢):     ̢ = ∑ [x ⋅ P(x)] = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.3) + 4(0.1)     ̢ = 0.0 + 0.2 + 0.6 + 0.9 + 0.4 = 2.1\text{ cars}     John expects to sell $2.1\text{ cars}$ on a typical Saturday.

• Step 2: Calculate Variance (̢^2):

* $x = 0 → (0 - 2.1) = -2.1 → (-2.1)^2 = 4.41 → 4.41(0.10) = 0.441$
* $x = 1 → (1 - 2.1) = -1.1 → (-1.1)^2 = 1.21 → 1.21(0.20) = 0.242$
* $x = 2 → (2 - 2.1) = -0.1 → (-0.1)^2 = 0.01 → 0.01(0.30) = 0.003$
* $x = 3 → (3 - 2.1) = 0.9 → (0.9)^2 = 0.81 → 0.81(0.30) = 0.243$
* $x = 4 → (4 - 2.1) = 1.9 → (1.9)^2 = 3.61 → 3.61(0.10) = 0.361$
* Variance ̢^2 = 0.441 + 0.242 + 0.003 + 0.243 + 0.361 = 1.290
* Standard Deviation ̢ = √{1.290} ≈ 1.1358\text{ cars}

• Class Exercise (Mean, Variance, and Standard Deviation):

  • Given distribution:

  • $X = 0 → P(X) = 0.15$

  • $X = 1 → P(X) = 0.20$

  • $X = 2 → P(X) = 0.15$

  • $X = 3 → P(X) = 0.25$

  • $X = 4 → P(X) = 0.25$

  • Mean Computation:     ̢ = 0(0.15) + 1(0.20) + 2(0.15) + 3(0.25) + 4(0.25)     ̢ = 0 + 0.20 + 0.30 + 0.75 + 1.00 = 2.25

  • Variance Computation (̢^2 = ∑ x^2 P(x) - ̢^2):     $E(X^2) = 0^2(0.15) + 1^2(0.20) + 2^2(0.15) + 3^2(0.25) + 4^2(0.25)$     $E(X^2) = 0 + 0.20 + 0.60 + 2.25 + 4.00 = 7.05$     ̢^2 = 7.05 - (2.25)^2 = 7.05 - 5.0625 = 1.9875

  • Standard Deviation Computation:     ̢ = √{1.9875} ≈ 1.4098

Calculation of Event Probabilities

• Example (Weekly Pendrive Production):

  • Past data for weekly pendrive production by a machine:

  • $X = 0 → P(X) = 0.15$

  • $X = 1 → P(X) = 0.20$

  • $X = 2 → P(X) = 0.35$

  • $X = 3 → P(X) = 0.30$

  • Solutions for Specific Events:

  • a) Exactly 2 pendrives:       $P(X = 2) = 0.35$

  • b) 0 to 2 pendrives:       P(0 \normalfont{\text{ ≤ }} X \normalfont{\text{ ≤ }} 2) = P(0) + P(1) + P(2) = 0.15 + 0.20 + 0.35 = 0.70

  • c) More than 1 pendrive:       $P(X > 1) = P(2) + P(3) = 0.35 + 0.30 = 0.65$

  • d) At most 1 pendrive:       P(X \normalfont{\text{ ≤ }} 1) = P(0) + P(1) = 0.15 + 0.20 = 0.35

• Class Exercise (Customer Complaints per Day):

  • Distribution of daily complaints $X$:

  • $X = 0 → P(X) = 0.05$

  • $X = 1 → P(X) = 0.15$

  • $X = 2 → P(X) = 0.40$

  • $X = 3 → P(X) = 0.20$

  • $X = 4 → P(X) = 0.10$

  • $X = 5 → P(X) = 0.10$

  • Solutions for Specific Events:

  • a) At least two complaints in a day:       P(X \normalfont{\text{ ≥ }} 2) = P(2) + P(3) + P(4) + P(5) = 0.40 + 0.20 + 0.10 + 0.10 = 0.80

  • b) At most two complaints in a day:       P(X \normalfont{\text{ ≤ }} 2) = P(0) + P(1) + P(2) = 0.05 + 0.15 + 0.40 = 0.60

  • c) More than three complaints in a day:       $P(X > 3) = P(4) + P(5) = 0.10 + 0.10 = 0.20$

Mathematical Laws of Expected Value and Variance

• Laws of Expected Value:

  1. Constant Rule: The expectation of a constant $a$ is the constant itself:      $E(a) = a$      Examples: $E(10) = 10$, $E(200) = 200$

  2. Linear Factor Rule: Multiplying a random variable by a constant $a$ scales its expectation by $a$:      $E(aX) = a ⋅ E(X)$      $E(aX^2) = a ⋅ E(X^2)$      Example: $E(2X) = 2 ⋅ E(X)$

  • Worked Example (Expectation Transformation):

  • Given: $E(X) = 100$ and $E(X^2) = 250$

  • Problem: Compute $E(3X - 2X^2 + 1000)$

  • Evaluation:       $E(3X - 2X^2 + 1000) = 3 ⋅ E(X) - 2 ⋅ E(X^2) + 1000$       $E(3X - 2X^2 + 1000) = 3(100) - 2(250) + 1000 = 300 - 500 + 1000 = 800$

• Laws of Variance:

  1. Constant Rule: The variance of any constant $a$ is zero:      $\text{Var}(a) = 0$      Examples: $\text{Var}(100) = 0$, $\text{Var}(3000) = 0$

  2. Scale Factor Rule: Multiplying a random variable by a constant $a$ scales its variance by $a^2$:      $\text{Var}(aX) = a^2 ⋅ \text{Var}(X)$      Examples: $\text{Var}(3X) = 3^2 ⋅ \text{Var}(X) = 9 ⋅ \text{Var}(X)$, $\text{Var}(4X) = 16 ⋅ \text{Var}(X)$

  • Worked Example (Variance Transformation):

  • Given: $\text{Var}(X) = 120$

  • Problem: Compute $\text{Var}(3X - 1000)$

  • Evaluation:       $\text{Var}(3X - 1000) = \text{Var}(3X) + \text{Var}(1000)$       $\text{Var}(3X - 1000) = 3^2 ⋅ \text{Var}(X) + 0 = 9(120) = 1080$

• Class Exercise (Linear Revenue Function):

  • Scenario: A company's daily sales (measured in hundreds of units) is represented by a random variable $X$ with expected value $E(X) = 10$. Revenue (in RM thousands) is modeled by $Y = 5X + 20$

  • Problem: Determine expected daily revenue $E(Y)$.

  • Evaluation:     $E(Y) = E(5X + 20) = 5 ⋅ E(X) + 20$     $E(Y) = 5(10) + 20 = 70$

  • Expected daily revenue is $\text{RM } 70,000$

Academic References

• Textbook: Lind, D., Marchal, W., & Wathen, S. (2022). Basic Statistics for Business & Economics (10th ed.). McGraw-Hill Education.

• Chapter Reference: Chapter 6 (Discrete Probability Distributions).