Business Statistics – Discrete Distributions (MDB2013 Chapter 4)

Teaching & Learning Outcomes

  • Compute probabilities from various discrete distributions.
  • Utilize discrete probability plots to determine financial applications.
  • Calculate probabilities specifically for the Binomial, Poisson, and Hypergeometric distributions.

Probability Distribution – Definition & Characteristics

  • Definition: A listing of all experiment outcomes with their associated probabilities.
  • Key characteristics:
    • Probabilities lie in the interval [0,1].
    • Outcomes are mutually exclusive.
    • Outcomes are exhaustive → \sum P(x)=1.

Illustrative Example: 3 Coin Tosses

  • Experiment: Count heads when a coin is tossed 3 times.
  • Possible values of X: 0,1,2,3 heads.
  • Probabilities:
    • P(0)=\frac{1}{8}=0.125
    • P(1)=\frac{3}{8}=0.375
    • P(2)=\frac{3}{8}=0.375
    • P(3)=\frac{1}{8}=0.125
  • Check: 0.125+0.375+0.375+0.125=1.0 (distribution validated).

Random Variables

  • Outcome values produced by a chance experiment.
  • Examples:
    • Number of employees absent on Monday (quantitative discrete).
    • Grade level of team members (qualitative categorical variable but treated as a random variable).

Types

  • Discrete: Assumes clearly separated values (e.g., number of credit cards a customer holds).
  • Continuous: Infinite possible values in an interval (e.g., flight time KLIA–Langkawi; snowfall depth).

Mean, Variance & Standard Deviation of a Discrete Distribution

  • Mean (Expected value): \mu = \sum xP(x).
  • Variance: \sigma^{2}= \sum (x-\mu)^{2}P(x).
  • Standard deviation: \sigma = \sqrt{\sigma^{2}}.

Example: Johan’s Car Sales on Saturday

  • Distribution (condensed): P(0)=0.1,\ P(1)=0.2,\ P(2)=0.3,\ P(3)=0.3,\ P(4)=0.1.
  • Calculations:
    • \mu = 2.1 cars.
    • Tabled variance computation (\sum (x-\mu)^2 P(x)) → \sigma^{2}=1.29.
    • Interpretation: On a typical Saturday Johan expects about 2.1 cars sold with moderate variability (SD \approx1.136).

Binomial Distribution

  • Definition: Widely occurring discrete distribution meeting 4 criteria:
    1. Only two outcomes (success/failure).
    2. Fixed, known number of trials n.
    3. Constant success probability \pi across trials.
    4. Trials are independent.
  • Probability mass function (PMF):
    P(X=x)=C^{n}_{x}\pi^{x}(1-\pi)^{n-x} for x=0,1,\dots,n.
  • Mean & Variance:
    • \mu = n\pi
    • \sigma^{2}=n\pi(1-\pi)

Debit-Card Coffee-Shop Example (Zus Coffee, n=5,\pi=0.28)

  • (a) Exactly one debit-card purchase:
    P(1)=C^{5}_{1}(0.28)^{1}(0.72)^{4}.
  • (b) Full distribution: Evaluate PMF for x=0\rightarrow5.
  • (c) P(X\ge 3)=P(3)+P(4)+P(5).
  • (d) P(X\le 2)=1-P(X\ge3) (or directly sum).
  • (e) \mu =5(0.28)=1.4, \sigma^{2}=5(0.28)(0.72)=1.008.
  • Verify all 4 binomial requirements (yes – meets each).

Binomial Tables Example (Dropped Calls, n=6,\pi=0.05)

  • Use Table 6-2.
    • (a) P(0)=0.7351 (value from table).
    • (b) P(1)=0.2326, etc. through P(6)=1.6\times10^{-8}.

Hypergeometric Distribution

  • Used when sampling without replacement from a finite population where n/N>0.05 (lack of independence).
  • Conditions:
    1. Two outcome categories.
    2. Fixed sample size n.
    3. Trials are dependent.
    4. Population finite and sampling w/out replacement.
  • PMF: P(X=x)=\dfrac{C^{S}{x}\,C^{N-S}{n-x}}{C^{N}_{n}}.
    • N = population size, S = total successes in population.

Classroom Committee Example

  • N=50,\ S=40,\ n=5,\ x=4.
  • Probability:
    P(4)=\dfrac{C^{40}{4}\,C^{10}{1}}{C^{50}_{5}} (compute via combinations).

Poisson Distribution

  • Describes number of occurrences in a specified interval (time, distance, area, volume).
  • Assumptions:
    1. Probability proportional to interval length.
    2. Intervals are independent.
  • PMF (parameter \lambda = mean number of events per interval):
    P(X=x)=\dfrac{\lambda^{x}e^{-\lambda}}{x!}.
  • Relation to Binomial: Limiting case when n\to\infty, \pi\to0, n\pi=\lambda.

Lost-Bags Example (Amal by Malaysia Airlines)

  • Sample: 500 flights, 20 lost bags → overall rate \lambda_{500}=20.
  • (a) Valid Poisson: counting occurrence per non-overlapping flight, independence presumed.
  • (b) Mean per flight: \lambda =20/500=0.04 bags.
  • (c) No bags lost on a flight: P(0)=e^{-0.04}=0.9608.
  • (d) At least one lost: 1-P(0)=0.0392.

Poisson Table Example (Truck Breakdowns)

  • Mean breakdowns per run: \lambda=0.30.
  • From table:
    • P(0)=0.7408
    • P(1)=0.2222 (matches \lambda e^{-\lambda} check).

Conceptual Connections & Applications

  • Financial modeling often uses binomial trees for option pricing; discrete plots help visualize risk.
  • Quality-control auditing (hypergeometric) models defect sampling without replacement.
  • Operations management (Poisson) forecasts arrival rates—e.g., call centers, traffic, insurance claims.
  • Ethical note: Accurate statistical modeling prevents misallocation of resources, fosters fair decisions.

Formulas Consolidated

  • Discrete Mean: \mu=\sum xP(x)
  • Discrete Variance: \sigma^{2}=\sum (x-\mu)^{2}P(x)
  • Binomial PMF: P(X=x)=C^{n}_{x}\pi^{x}(1-\pi)^{n-x}
  • Binomial Mean/Var: \mu=n\pi,\ \sigma^{2}=n\pi(1-\pi)
  • Hypergeometric PMF: P(X=x)=\dfrac{C^{S}{x}C^{N-S}{n-x}}{C^{N}_{n}}
  • Poisson PMF: P(X=x)=\dfrac{\lambda^{x}e^{-\lambda}}{x!}

Study Tips

  • Always check distribution requirements before applying a formula.
  • For small n without replacement, prefer hypergeometric; for large n with small \pi, Poisson may approximate Binomial.
  • Use tables or software to avoid factorial/combinatorial arithmetic mistakes.
  • Draw probability plots to see skewness and assess variance vs. mean relationships.