Chapter 4 - Discrete Random Variables and Binomial Distributions

Administrative Overview and Scope

• Scope of Material: Fundamentals of discrete random variables, discrete probability distribution functions (PDFs), calculation and interpretation of expected value ($\mu$), computation of variance ($\text{Var}$) and standard deviation ($\sigma$), and properties, formulas, and applications of binomial distributions and Bernoulli trials.

Foundations of Discrete Random Variables

• Definition of a Random Variable:

  • A random variable describes the numerical outcomes of a statistical experiment in words.

  • It serves as a designated placeholder or variable name for all possible experimental outcomes, allowing the formal analysis of properties, behaviors, and associated probabilities.

  • The specific numerical value resulting from a random variable can vary with each repetition of an experiment.

• Standard Notation Conventions:

  • Uppercase letters, such as $X$ or $Y$, represent the random variable itself (the conceptual description of the outcome).

  • Lowercase letters, such as $x$ or $y$, represent specific numerical values that the random variable can assume in a given realization or outcome.

• Nature of Discrete Random Variables:

  • The possible values of a discrete random variable are separate, countable numbers with distinct gaps between them (for example, counts of occurrences).

  • Intermediate fractional or continuous values (such as $2.5$ in a count of heads) do not exist within a discrete sample space.

• Three Fair Coins Toss Example:

  • Experiment: Tossing three fair coins simultaneously.

  • Definition of Random Variable: Let $X = \text{the number of heads obtained when tossing three fair coins}$.

  • Full Sample Space: $\text{Sample Space} = \{TTT, THH, HTH, HHT, HTT, THT, TTH, HHH\}$, yielding $8$ equally likely sample outcomes.

  • Possible Values of $X$: The variable $x$ can take on any integer value within the set $\{0, 1, 2, 3\}$.

Discrete Probability Distribution Functions

• Definition of a Probability Distribution Function (PDF):

  • A discrete probability distribution function assigns a probability to each possible value of a discrete random variable.

  • Common functional notations include $P(X = x)$, $P(x)$, or cumulative notations such as $P(X \le x)$.

  • An effective way to represent a discrete PDF is via a probability distribution table matching every value $x$ with its corresponding probability $P(x)$.

• Fundamental Characteristics of a Valid Discrete PDF:

  • Each individual probability must lie between zero and one, inclusive: $0 \le P(x) \le 1$ for every possible outcome $x$.

  • The sum of all probabilities across the entire sample space must equal exactly one: $\sum P(x) = 1$.

• Two Dice Rolling Sum Example:

  • Experiment: Rolling two fair six-sided dice and calculating the sum of the upward-facing numbers.

  • Definition of Random Variable: Let $X = \text{the sum of the two dice rolls}$.

  • Sample Space: $\text{Sample Space} = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

  • Total Possible Outcomes: Rolling two independent six-sided dice produces $6 \times 6 = 36$ total outcomes in the sample space.

  • Distribution Breakdown by Sum:

  • Sum $x = 2$: $1$ combination $(1, 1)$; probability $P(2) = \frac{1}{36}$.

  • Sum $x = 3$: $2$ combinations $(1, 2), (2, 1)$; probability $P(3) = \frac{2}{36}$.

  • Sum $x = 4$: $3$ combinations $(1, 3), (2, 2), (3, 1)$; probability $P(4) = \frac{3}{36}$.

  • Sum $x = 5$: $4$ combinations $(1, 4), (2, 3), (3, 2), (4, 1)$; probability $P(5) = \frac{4}{36}$.

  • Sum $x = 6$: $5$ combinations $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$; probability $P(6) = \frac{5}{36}$.

  • Sum $x = 7$: $6$ combinations $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$; probability $P(7) = \frac{6}{36}$.

  • Sum $x = 8$: $5$ combinations $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$; probability $P(8) = \frac{5}{36}$.

  • Sum $x = 9$: $4$ combinations $(3, 6), (4, 5), (5, 4), (6, 3)$; probability $P(9) = \frac{4}{36}$.

  • Sum $x = 10$: $3$ combinations $(4, 6), (5, 5), (6, 4)$; probability $P(10) = \frac{3}{36}$.

  • Sum $x = 11$: $2$ combinations $(5, 6), (6, 5)$; probability $P(11) = \frac{2}{36}$.

  • Sum $x = 12$: $1$ combination $(6, 6)$; probability $P(12) = \frac{1}{36}$.

  • Combinatorial Symmetry: The counts of combinations increase linearly from $1$ (for sum $2$) up to $6$ (for sum $7$), and subsequently mirror back down symmetrically from $5$ to $1$ (for sums $8$ through $12$).

• Nancy Class Attendance Example:

  • Problem Context: Nancy has academic classes scheduled $3$ days per week. In any randomly selected week, she attends all $3$ days $80\%$ of the time, attends $2$ days $15\%$ of the time, attends $1$ day $4\%$ of the time, and attends $0$ days $1\%$ of the time.

  • Random Variable Definition: Let $X = \text{the number of days Nancy attends class in a randomly selected week}$.

  • Possible Values: $x \in \{0, 1, 2, 3\}$.

  • Probability Distribution Function Values:

  • $P(X = 0) = P(0) = 0.01$

  • $P(X = 1) = P(1) = 0.04$

  • $P(X = 2) = P(2) = 0.15$

  • $P(X = 3) = P(3) = 0.80$

  • Caveat on Missed Classes: The probability that Nancy misses at least one class during the week is equal to the complementary sum of attending $0$, $1$, or $2$ days: $0.01 + 0.04 + 0.15 = 0.20$ ($20\%$).

  • PDF Verification: $0.01 + 0.04 + 0.15 + 0.80 = 1.00$, satisfying the mandatory sum criterion.

Measures of Central Tendency: Expected Value

• Theoretical Concept of Expected Value:

  • The expected value, denoted by the Greek letter $\mu$, represents the long-term average or mean outcome of a statistical experiment if repeated independently over many trials.

  • It describes the theoretical center of mass of the probability distribution rather than a value that must occur in any single individual trial.

• General Mathematical Formula for Expected Value:

  • $\mu = \sum (x \cdot P(X = x))$

  • Each possible outcome $x$ is weighted by its respective probability of occurrence $P(x)$, and the products are summed across all possible outcomes.

• Rolling a Single Fair Die Example:

  • Sample Space: $\text{Outcomes} \in \{1, 2, 3, 4, 5, 6\}$, each with probability $P(x) = \frac{1}{6}$.

  • Calculation of Expected Value:

  • $\mu = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right)$

  • $\mu = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5$

  • Practical Interpretation: If a fair six-sided die is rolled an immense number of times, the empirical arithmetic average of all recorded outcomes will converge very close to $3.5$.

• Water Bottle Orders Example:

  • Problem Setup: A customer orders $1$, $2$, or $3$ bottles of water. The probability of ordering $1$ bottle is $0.6$, ordering $2$ bottles is $0.3$, and ordering $3$ bottles is $0.1$.

  • Random Variable: Let $X = \text{the number of water bottles ordered by a customer}$.

  • Calculation of Expected Value:

  • $\mu = \sum (x \cdot P(x)) = 1(0.6) + 2(0.3) + 3(0.1)$

  • $\mu = 0.6 + 0.6 + 0.3 = 1.5$

  • Real-World Interpretation: While no individual customer can purchase $1.5$ bottles of water, a merchant expects to sell an average of $1.5$ bottles per customer served. Multiplying the total number of customers on a given day by $1.5$ gives a close estimate of total inventory sold.

• Biased Coin Betting Game Example:

  • Game Rules: A biased coin is tossed once. The probability of landing heads is $P(\text{heads}) = \frac{2}{3}$ and the probability of landing tails is $P(\text{tails}) = \frac{1}{3}$.

  • Financial Payoff Structure:

  • Tossing a head results in paying $\$6$ (a net loss, represented by $x = -6$).

  • Tossing a tail results in winning $\$10$ (a net gain, represented by $x = 10$).

  • Random Variable Definition: Let $X = \text{net financial payout in dollars from a single coin toss}$.

  • Tabular Expected Value Computation:

  • Outcome Win (Tails): $x = 10$, $P(x) = \frac{1}{3}$, Product $x \cdot P(x) = 10 \left(\frac{1}{3}\right) = \frac{10}{3}$.

  • Outcome Lose (Heads): $x = -6$, $P(x) = \frac{2}{3}$, Product $x \cdot P(x) = -6 \left(\frac{2}{3}\right) = \frac{-12}{3}$.

  • Expected Value Calculation:

  • $\mu = \frac{10}{3} + \left(\frac{-12}{3}\right) = -\frac{2}{3} \approx -0.67$

  • Profitability Determination: The player does not come out ahead. Because the expected value is negative ($-\frac{2}{3}$ dollars per play), the player anticipates losing approximately $\$0.67$ per play in the long run. If played $N$ times, total expected loss is $N \times \left(-\frac{2}{3}\right)$ dollars.

Measures of Dispersion: Variance and Standard Deviation

• Conceptual Foundation of Dispersion in PDFs:

  • Standard deviation measures the spread or variability of outcomes around the theoretical mean $\mu$.

  • Whereas descriptive statistics calculates dispersion across an empirical dataset collected from observations, probability distributions compute dispersion across theoretical values weighted by their likelihood.

• Mathematical Formulations:

  • Variance ($\text{Var}$):

  • $\text{Var} = \sum \left[(x - \mu)^2 \cdot P(x)\right]$

  • Standard Deviation ($\sigma$):

  • $\sigma = \sqrt{\text{Var}} = \sqrt{\sum \left[(x - \mu)^2 \cdot P(x)\right]}$

• Computational Implementation:

  • Manual evaluation of variance and standard deviation requires computing squared deviations for every possible outcome, weighting each by $P(x)$, summing them, and taking the square root.

  • Because manual calculations are highly space-consuming and prone to arithmetic error, standard practice employs computational tools, dedicated spreadsheets, or graphing calculators.

Binomial Experiments and Probability Distributions

• Three Defining Characteristics of a Binomial Experiment:

  • Fixed Number of Trials: There must be a predetermined, fixed number of trials (repetitions of an experiment), designated by the variable $n$.

  • Binary Mutually Exclusive Outcomes: Each trial can result in only one of two possible outcomes, conventionally classified as "success" and "failure".

  • Let $p$ denote the probability of a success on any individual trial.

  • Let $q$ denote the probability of a failure on any individual trial.

  • Because the two outcomes partition the probability space: $p + q = 1$ (or $q = 1 - p$).

  • Independent and Identical Trials: All $n$ trials must be mutually independent and conducted under strictly identical experimental conditions.

  • Example of Valid Independence: Drawing playing cards from a standard deck with replacement and thorough reshuffling between draws (such that drawing a heart represents success and any other suit represents failure).

  • Example of Non-Binomial Setup: Sampling without replacement violates independence because the outcome of prior draws alters the subsequent probabilities.

• Bernoulli Trial Definition:

  • A Bernoulli trial is defined as any binomial experiment that consists of exactly one trial ($n = 1$).

• Characteristics of the Binomial Probability Distribution:

  • Outcomes of any experiment satisfying the three binomial criteria follow a binomial probability distribution.

  • Binomial Random Variable: Let $X = \text{the number of successes obtained in the } n \text{ independent trials}$.

  • Standard Distribution Notation:

  • $X \sim B(n, p)$

  • This indicates that the random variable $X$ follows a binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success on each trial).

• Binomial Summary Statistics Formulas:

  • Mean (Expected Value):

  • $\mu = n p$

  • Variance:

  • $\text{Var} = n p q = n p (1 - p)$

  • Standard Deviation:

  • $\sigma = \sqrt{n p q} = \sqrt{n p (1 - p)}$

• State School Lunches Fruit Availability Example:

  • Problem Context: A state health board investigates daily fruit availability in school lunches. Exactly $48\%$ of schools within the state offer fruit daily, implying that $52\%$ do not.

  • Component Identification:

  • Success Outcome: A randomly selected school offers fruit in its lunch every day.

  • Failure Outcome: A randomly selected school does not offer fruit in its lunch every day.

  • Probability of Success: $p = 0.48$

  • Probability of Failure: $q = 1 - 0.48 = 0.52$

  • Random Variable: Let $X = \text{the number of schools that offer fruit daily out of } n \text{ sampled schools}$.

  • Binomial Parameterization: $X \sim B(n, 0.48)$, where $n$ denotes the total number of schools surveyed.

• Statistics Homework Submission Example:

  • Problem Context: Approximately $70\%$ of statistics students complete their homework on time to be collected and graded. Each student works completely independently. In a statistics class composed of $50$ students randomly selected, find the probability that at least $40$ students turn in their homework on time.

  • Experimental Verification:

  • Fixed trials: $n = 50$ students.

  • Binary outcomes: Student completes homework on time (success) versus student does not complete homework on time (failure).

  • Success probability: $p = 0.70$.

  • Failure probability: $q = 1 - 0.70 = 0.30$.

  • Independence: Homework completion by one student has no bearing on another student.

  • Random Variable and Notation:

  • Let $X = \text{the number of students turning in homework on time}$.

  • Distribution: $X \sim B(50, 0.70)$.

  • Probability Objective:

  • Calculate $P(X \ge 40)$.

  • Computational Note: Calculating cumulative binomial sums manually (evaluating the binomial formula individually from $x = 40$ up to $x = 50$) is computationally intensive; such values are evaluated using technology, including statistical graphing calculators or spreadsheet software functions (e.g., in Microsoft Excel).