Intro to Business Statistics

📉Intro to Business Statistics Unit 4 – Discrete Random Variables

Discrete 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)=xP(X=x)E(X) = \sum x \cdot 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: Var(X)=(xμ)2P(X=x)Var(X) = \sum (x - \mu)^2 \cdot P(X = x)
    • Standard deviation is the square root of variance: σ=Var(X)\sigma = \sqrt{Var(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)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} 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)=eλλkk!P(X = k) = \frac{e^{-\lambda}\lambda^k}{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)=(1p)k1pP(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)=(Kk)(NKnk)(Nn)P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}, 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) = (32)(1/6)2(5/6)10.0463\binom{3}{2} (1/6)^2 (5/6)^1 \approx 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

Key Formulas and Concepts to Remember

  • PMF: P(X = x), defines the probability of each outcome for a discrete random variable
  • Expected value: E(X)=xP(X=x)E(X) = \sum x \cdot P(X = x), the average value of a discrete random variable
  • Variance: Var(X)=(xμ)2P(X=x)Var(X) = \sum (x - \mu)^2 \cdot P(X = x), measures the spread of a discrete random variable
  • Standard deviation: σ=Var(X)\sigma = \sqrt{Var(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)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} 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)=eλλkk!P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}
  • Geometric distribution: models the number of trials until the first success occurs
    • PMF: P(X=k)=(1p)k1pP(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)=(Kk)(NKnk)(Nn)P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

Practice Problems and Solutions

  1. 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
  2. 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) = e8833!0.0498\frac{e^{-8}8^3}{3!} \approx 0.0498
  3. 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
  4. 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) = (10.3)30.30.1029(1-0.3)^{3}0.3 \approx 0.1029
  5. 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) = (204)(306)(5010)0.2517\frac{\binom{20}{4}\binom{30}{6}}{\binom{50}{10}} \approx 0.2517


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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.