All Study Guides Intro to Business Statistics Unit 4
📉 Intro to Business Statistics Unit 4 – Discrete Random VariablesDiscrete random variables are a fundamental concept in statistics, describing outcomes that can be counted. They form the basis for many probability distributions used in business and data analysis. Understanding discrete random variables is crucial for making informed decisions based on data.
This unit covers the essentials of discrete random variables, including probability mass functions, expected values, and variance. It also introduces common distributions like Bernoulli, binomial, and Poisson, which are widely used to model real-world scenarios in business and other fields.
What are Discrete Random Variables?
Discrete random variables take on a countable number of distinct values
Typically represented by uppercase letters (X, Y, Z)
Outcomes are often integers or whole numbers (0, 1, 2, 3...)
Probability of each outcome is between 0 and 1, inclusive
Sum of all probabilities must equal 1
Examples include:
Number of defective items in a batch (0, 1, 2...)
Number of customers arriving at a store per hour (0, 1, 2...)
Contrast with continuous random variables, which can take on any value within a range
Discrete random variables are the foundation for many probability distributions
Probability Mass Functions (PMF)
PMF defines the probability of each possible outcome for a discrete random variable
Denoted as P(X = x), where X is the random variable and x is a specific outcome
Must satisfy two conditions:
P(X = x) ≥ 0 for all x
Sum of P(X = x) over all possible x values equals 1
Can be represented as a table, graph, or formula
Example: PMF for the number of heads in two coin flips (0, 1, or 2)
P(X = 0) = 0.25, P(X = 1) = 0.5, P(X = 2) = 0.25
Helps calculate probabilities of specific outcomes or ranges of outcomes
Foundation for calculating expected value and variance
Expected Value and Variance
Expected value (mean) is the average value of a discrete random variable over many trials
Denoted as E(X) or μ
Calculated as the sum of each outcome multiplied by its probability: E ( X ) = ∑ x ⋅ P ( X = x ) E(X) = \sum x \cdot P(X = x) E ( X ) = ∑ x ⋅ P ( X = x )
Variance measures the spread or dispersion of a discrete random variable around its expected value
Denoted as Var(X) or σ^2
Calculated as the sum of squared deviations from the mean, weighted by their probabilities: V a r ( X ) = ∑ ( x − μ ) 2 ⋅ P ( X = x ) Var(X) = \sum (x - \mu)^2 \cdot P(X = x) Va r ( X ) = ∑ ( x − μ ) 2 ⋅ P ( X = x )
Standard deviation is the square root of variance: σ = V a r ( X ) \sigma = \sqrt{Var(X)} σ = Va r ( X )
Both expected value and variance provide insights into the behavior of a discrete random variable
Used to compare different distributions and make informed decisions
Common Discrete Distributions
Bernoulli distribution: models a single trial with two possible outcomes (success or failure)
P(X = 1) = p, P(X = 0) = 1 - p, where p is the probability of success
Binomial distribution: models the number of successes in a fixed number of independent Bernoulli trials
Denoted as X ~ Bin(n, p), where n is the number of trials and p is the probability of success
PMF: P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} P ( X = k ) = ( k n ) p k ( 1 − p ) n − k , where k is the number of successes
Poisson distribution: models the number of events occurring in a fixed interval of time or space
Denoted as X ~ Poisson(λ), where λ is the average rate of events
PMF: P ( X = k ) = e − λ λ k k ! P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} P ( X = k ) = k ! e − λ λ k , where k is the number of events
Geometric distribution: models the number of trials until the first success occurs
Denoted as X ~ Geo(p), where p is the probability of success
PMF: P ( X = k ) = ( 1 − p ) k − 1 p P(X = k) = (1-p)^{k-1}p P ( X = k ) = ( 1 − p ) k − 1 p , where k is the number of trials
Hypergeometric distribution: models the number of successes in a fixed number of draws without replacement from a finite population
Denoted as X ~ HGeom(N, K, n), where N is the population size, K is the number of successes in the population, and n is the number of draws
PMF: P ( X = k ) = ( K k ) ( N − K n − k ) ( N n ) P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} P ( X = k ) = ( n N ) ( k K ) ( n − k N − K ) , where k is the number of successes in the sample
Solving Problems with Discrete Random Variables
Identify the appropriate discrete random variable and distribution for the given problem
Determine the parameters of the distribution (e.g., p for Bernoulli, n and p for Binomial)
Write the PMF or use built-in functions in software (Excel, R, Python) to calculate probabilities
Calculate the expected value and variance using the formulas or built-in functions
Interpret the results in the context of the problem
Example: A fair die is rolled 3 times. Let X be the number of times a 6 appears. Find P(X = 2) and E(X).
X ~ Bin(3, 1/6) since there are 3 trials and the probability of success (rolling a 6) is 1/6
P(X = 2) = ( 3 2 ) ( 1 / 6 ) 2 ( 5 / 6 ) 1 ≈ 0.0463 \binom{3}{2} (1/6)^2 (5/6)^1 \approx 0.0463 ( 2 3 ) ( 1/6 ) 2 ( 5/6 ) 1 ≈ 0.0463
E(X) = np = 3 × (1/6) = 0.5
Adapt problem-solving strategies to different scenarios and distributions
Real-World Applications in Business
Quality control: model the number of defective items in a batch using the Binomial or Poisson distribution
Customer service: model the number of customer complaints per day using the Poisson distribution
Inventory management: model the demand for a product using the Poisson or Geometric distribution
Marketing: model the number of people who respond to a promotional offer using the Binomial distribution
Finance: model the number of defaults in a loan portfolio using the Binomial distribution
Project management: model the number of tasks completed on time using the Binomial distribution
Supply chain: model the number of supplier deliveries that arrive late using the Poisson distribution
Applying discrete random variables in real-world scenarios helps make informed business decisions and manage risks
PMF: P(X = x), defines the probability of each outcome for a discrete random variable
Expected value: E ( X ) = ∑ x ⋅ P ( X = x ) E(X) = \sum x \cdot P(X = x) E ( X ) = ∑ x ⋅ P ( X = x ) , the average value of a discrete random variable
Variance: V a r ( X ) = ∑ ( x − μ ) 2 ⋅ P ( X = x ) Var(X) = \sum (x - \mu)^2 \cdot P(X = x) Va r ( X ) = ∑ ( x − μ ) 2 ⋅ P ( X = x ) , measures the spread of a discrete random variable
Standard deviation: σ = V a r ( X ) \sigma = \sqrt{Var(X)} σ = Va r ( X ) , square root of variance
Bernoulli distribution: models a single trial with two possible outcomes
Binomial distribution: models the number of successes in a fixed number of independent Bernoulli trials
PMF: P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} P ( X = k ) = ( k n ) p k ( 1 − p ) n − k
Poisson distribution: models the number of events occurring in a fixed interval of time or space
PMF: P ( X = k ) = e − λ λ k k ! P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} P ( X = k ) = k ! e − λ λ k
Geometric distribution: models the number of trials until the first success occurs
PMF: P ( X = k ) = ( 1 − p ) k − 1 p P(X = k) = (1-p)^{k-1}p P ( X = k ) = ( 1 − p ) k − 1 p
Hypergeometric distribution: models the number of successes in a fixed number of draws without replacement from a finite population
PMF: P ( X = k ) = ( K k ) ( N − K n − k ) ( N n ) P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} P ( X = k ) = ( n N ) ( k K ) ( n − k N − K )
Practice Problems and Solutions
A fair coin is tossed 5 times. Let X be the number of heads. Find P(X ≥ 3) and E(X).
X ~ Bin(5, 0.5)
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) ≈ 0.5
E(X) = np = 5 × 0.5 = 2.5
A call center receives an average of 4 calls per minute. Find the probability of receiving exactly 3 calls in a 2-minute period.
X ~ Poisson(4 × 2 = 8)
P(X = 3) = e − 8 8 3 3 ! ≈ 0.0498 \frac{e^{-8}8^3}{3!} \approx 0.0498 3 ! e − 8 8 3 ≈ 0.0498
A machine produces defective items with a probability of 0.02. Find the probability of observing at most 2 defective items in a batch of 100.
X ~ Bin(100, 0.02)
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) ≈ 0.9999
The probability of a customer making a purchase is 0.3. Find the probability that the 4th customer is the first to make a purchase.
X ~ Geo(0.3)
P(X = 4) = ( 1 − 0.3 ) 3 0.3 ≈ 0.1029 (1-0.3)^{3}0.3 \approx 0.1029 ( 1 − 0.3 ) 3 0.3 ≈ 0.1029
A box contains 20 red marbles and 30 blue marbles. If 10 marbles are drawn at random without replacement, find the probability of drawing exactly 4 red marbles.
X ~ HGeom(50, 20, 10)
P(X = 4) = ( 20 4 ) ( 30 6 ) ( 50 10 ) ≈ 0.2517 \frac{\binom{20}{4}\binom{30}{6}}{\binom{50}{10}} \approx 0.2517 ( 10 50 ) ( 4 20 ) ( 6 30 ) ≈ 0.2517