📊Honors Statistics Unit 4 – Discrete Random Variables

Key Concepts

• Discrete random variables take on a countable number of distinct values
• Probability mass function (PMF) describes the probability distribution of a discrete random variable
• Cumulative distribution function (CDF) gives the probability that a discrete random variable is less than or equal to a specific value
• Expected value represents the average value of a discrete random variable over many trials
  • Calculated by summing the product of each possible value and its probability
• Variance measures the spread or dispersion of a discrete random variable around its expected value
• Common discrete distributions include Bernoulli, binomial, Poisson, and geometric distributions
• Discrete random variables have applications in various fields (finance, engineering, biology)

Types of Discrete Random Variables

• Bernoulli random variable takes on only two possible values (0 or 1)
  • Represents a single trial of an experiment with binary outcomes (success or failure)
• Binomial random variable counts the number of successes in a fixed number of independent Bernoulli trials
  • Characterized by the number of trials ($n$) and the probability of success ($p$)
• Poisson random variable models the number of events occurring in a fixed interval of time or space
  • Characterized by the average rate of occurrence ($\lambda$)
• Geometric random variable represents the number of trials needed to achieve the first success in a series of independent Bernoulli trials
• Hypergeometric random variable describes the number of successes in a fixed number of draws from a population without replacement
• Negative binomial random variable counts the number of failures before a specified number of successes occur in independent Bernoulli trials

Probability Mass Functions

• PMF is a function that assigns probabilities to each possible value of a discrete random variable
• Denoted as $P(X = x)$, where $X$ is the random variable and $x$ is a specific value
• Properties of a valid PMF:
  • Non-negative: $P(X = x) \geq 0$ for all $x$
  • Sum of probabilities: $\sum_{x} P(X = x) = 1$
• PMF can be represented as a table, graph, or formula
• Helps calculate probabilities of specific events and derive other properties of the random variable
• Examples of PMFs for common discrete distributions:
  • Binomial: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
  • Poisson: $P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$

Cumulative Distribution Functions

• CDF gives the probability that a discrete random variable $X$ is less than or equal to a specific value $x$
• Denoted as $F(x) = P(X \leq x)$
• Properties of a CDF:
  • Non-decreasing: $F(x_1) \leq F(x_2)$ for $x_1 < x_2$
  • Limits: $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = 1$
• CDF can be derived from the PMF by summing probabilities up to a given value
• Helps calculate probabilities of events involving inequalities
• Examples of CDFs for common discrete distributions:
  • Binomial: $F(x) = \sum_{k=0}^{\lfloor x \rfloor} \binom{n}{k} p^k (1-p)^{n-k}$
  • Poisson: $F(x) = \sum_{k=0}^{\lfloor x \rfloor} \frac{e^{-\lambda}\lambda^k}{k!}$

Expected Value and Variance

• Expected value (mean) of a discrete random variable $X$ is the weighted average of all possible values
  • Denoted as $E(X)$ or $\mu$
  • Calculated by $E(X) = \sum_{x} x \cdot P(X = x)$
• Variance measures the average squared deviation from the expected value
  • Denoted as $Var(X)$ or $\sigma^2$
  • Calculated by $Var(X) = E[(X - \mu)^2] = E(X^2) - [E(X)]^2$
• Standard deviation is the square root of the variance, denoted as $\sigma$
• Properties of expected value and variance:
  • Linearity of expectation: $E(aX + b) = aE(X) + b$
  • Variance of a constant: $Var(c) = 0$
  • Variance of a linear transformation: $Var(aX + b) = a^2Var(X)$
• Expected value and variance provide insights into the central tendency and dispersion of a random variable

Common Discrete Distributions

• Bernoulli distribution:
  • PMF: $P(X = 1) = p$, $P(X = 0) = 1 - p$
  • Expected value: $E(X) = p$
  • Variance: $Var(X) = p(1-p)$
• Binomial distribution:
  • PMF: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
  • Expected value: $E(X) = np$
  • Variance: $Var(X) = np(1-p)$
• Poisson distribution:
  • PMF: $P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$
  • Expected value: $E(X) = \lambda$
  • Variance: $Var(X) = \lambda$
• Geometric distribution:
  • PMF: $P(X = k) = (1-p)^{k-1}p$
  • Expected value: $E(X) = \frac{1}{p}$
  • Variance: $Var(X) = \frac{1-p}{p^2}$

Applications and Examples

• Bernoulli distribution: Modeling the success or failure of a single trial (coin flip, defective product)
• Binomial distribution: Number of defective items in a batch, number of successful free throws in a basketball game
• Poisson distribution: Number of customers arriving at a store per hour, number of earthquakes per year in a region
• Geometric distribution: Number of job interviews until receiving an offer, number of dice rolls until getting a six
• Hypergeometric distribution: Number of defective items in a sample drawn from a lot without replacement
• Negative binomial distribution: Number of failures before achieving a specified number of successes (number of unsuccessful sales calls before making a sale)

Practice Problems and Solutions

1. A fair coin is tossed 5 times. Find the probability of getting exactly 3 heads.

   • Solution: Binomial distribution with $n=5$, $p=0.5$, and $k=3$
   • $P(X = 3) = \binom{5}{3} (0.5)^3 (1-0.5)^{5-3} = 0.3125$

2. The average number of customers arriving at a store per hour is 6. Find the probability that exactly 4 customers arrive in a given hour.

   • Solution: Poisson distribution with $\lambda=6$ and $k=4$
   • $P(X = 4) = \frac{e^{-6}6^4}{4!} \approx 0.1339$

3. The probability of a defective product is 0.02. Find the expected number of defective products in a batch of 100.

   • Solution: Binomial distribution with $n=100$ and $p=0.02$
   • $E(X) = np = 100 \cdot 0.02 = 2$

4. A die is rolled repeatedly until a 6 appears. Find the expected number of rolls needed.

   • Solution: Geometric distribution with $p=\frac{1}{6}$
   • $E(X) = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6$