Statistical Distribution & Inferences – Lecture 01 Comprehensive Notes

Binomial Distribution

  • Discrete probability model built from a sequence of independent Bernoulli trials.

    • Four defining conditions:

    • Experiment consists of nn identical trials.

    • Each trial has exactly two possible outcomes: success or failure.

    • Constant probability of success per trial: pp; therefore probability of failure q=1−pq = 1-p.

    • Trials are mutually independent.

  • Probability Mass Function (PMF)

    • P(X=x)=(nx)pxqn−xP(X = x)= \binom{n}{x} p^{x} q^{n-x}

    • Parameter meanings:

    • nn – number of trials (sample size)

    • xx – number of observed successes

    • pp – probability of success on a single trial

    • q=1−pq = 1-p – probability of failure on a single trial

  • Moments

    • Mean (Expectation): μ=E[X]=np\mu = E[X] = np

    • Variance: σ2=np(1−p)\sigma^{2} = np(1-p)

    • Standard Deviation: σ=np(1−p)\sigma = \sqrt{np(1-p)}

  • Worked Example ① – Biased Coin

    • Scenario: Coin has p=0.3p = 0.3 for HEADS. Tossed n=6n = 6 times.

    • Required: P(X=4)P(X = 4) where XX is the number of heads.

    • Calculation: P(X=4)=(64)0.34(1−0.3)6−4=0.0595P(X=4) = \binom{6}{4} 0.3^{4}(1-0.3)^{6-4} = 0.0595

    • Interpretation: Roughly a 5.95 % chance of seeing exactly four heads.

  • Practice Exercise – Instant Lottery

    • Given: 20 % of tickets win; purchase 8 tickets (X∼Bin(8,0.2)X \sim Bin(8,0.2)).

    • Probability of exactly 2 winning tickets:
      P(X=2)=(82)0.22(1−0.2)6=0.2936P(X=2)= \binom{8}{2} 0.2^{2}(1-0.2)^{6}=0.2936

    • Significance: ~29.36 % likelihood for that outcome.

Cumulative Frequency & Cumulative Binomial Probabilities

  • Cumulative Frequency (CF) in descriptive stats:

    • Add each frequency to the sum of all prior frequencies.

    • Final CF equals total sample size.

    • Example class table provided (0–5 through 31–36) culminating in CF 21.

  • Binomial Cumulative Probability: P(X≤x)=∑k=0x(nk)pkqn−kP(X\le x)=\sum_{k=0}^{x} \binom{n}{k}p^{k}q^{n-k}

    • Pre-computed tables & software are standard; direct manual PMF evaluation is rare in practice.

  • Table-Based Worked Example ② – Bin(7,0.2)Bin(7,0.2)

    • Extracted cumulative probabilities:

    • P(X≤0)=0.2097P(X\le0)=0.2097

    • P(X≤1)=0.5767P(X\le1)=0.5767

    • P(X≤2)=0.8520P(X\le2)=0.8520

    • P(X≤3)=0.9667P(X\le3)=0.9667

    • P(X≤4)=0.9953P(X\le4)=0.9953

    • P(X≤5)=0.9996,  P(X≤6)=1.0000,  P(X≤7)=1.0000P(X\le5)=0.9996,\;P(X\le6)=1.0000,\;P(X\le7)=1.0000

    • Illustrative Queries:

    • (a) P(X=3)=P(X≤3)−P(X≤2)=0.9667−0.8520=0.115P(X=3)=P(X\le3)-P(X\le2)=0.9667-0.8520=0.115

    • (b) P(1<X<4)=P(X\le4)-P(X\le1)=0.9953-0.5767=0.419

    • (c) P(X>1)=1-P(X\le1)=0.423

  • Exercise – Bin(5,0.3)Bin(5,0.3) (answers rounded to 3 d.p.)

    • Table extract: P(X≤r)P(X\le r) for r=0…5r=0\ldots5 supplied.

    • (a) P(X≤4)=0.998P(X\le4)=0.998

    • (b) P(X=2)=P(X≤2)−P(X≤1)=0.8369−0.5282=0.309P(X=2)=P(X\le2)-P(X\le1)=0.8369-0.5282=0.309

    • (c) P(X<3)=P(X\le2)=0.837

    • (d) P(X>1)=1-P(X\le1)=0.472

    • (e) P(X≥3)=1−P(X≤2)=0.163P(X\ge3)=1-P(X\le2)=0.163

Poisson Distribution

  • Discrete model for rare events over a continuous domain (time, space, etc.).

    • Focuses on the count of occurrences in a fixed interval, not explicit successes vs. failures.

  • Probability Mass Function (PMF)

    • P(X=x)=e−λλxx!P(X = x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}

    • Single parameter: λ=average rate=np\lambda = \text{average rate}=np when derived as a binomial limit.

  • Limiting Relationship (Binomial → Poisson)

    • Conditions:

    1. nn is extremely large.

    2. pp is extremely small.

    3. Product np=λnp = \lambda remains finite.

  • Properties

    • Mean E[X]=λE[X]=\lambda

    • Variance Var(X)=λVar(X)=\lambda (mean and variance equal)

  • Worked Example ③ – House Fires

    • Data: 1 fire per 1000 houses annually (p=0.001p=0.001). District has n=2000n=2000 houses.

    • λ=np=2000×0.001=2\lambda = np = 2000\times0.001=2

    • Probability exactly 5 fires:
      P(X=5)=e−2255!=0.036P(X=5)=\dfrac{e^{-2}2^{5}}{5!}=0.036

    • Provided constant e2=0.13534e^{2}=0.13534 used for numerical substitution.

  • Exercise – Defective Bulbs

    • Manufacturer defect rate 2 % (p=0.02p=0.02); sample n=200n=200 ⇒ λ=np=4\lambda = np = 4.

    • (i) Fewer than 2 defects:
      P(X<2)=P(0)+P(1)=e^{-4}\left(1+4\right)=0.0915

    • (ii) More than 3 defects:
      P(X>3)=1-P(X\le3)=0.567

    • Table value given: e−4=0.0183e^{-4}=0.0183.

Practical & Conceptual Connections

  • Model Choice

    • Binomial appropriate when both success and failure are explicitly counted across fixed nn.

    • Poisson preferred for high-volume, low-probability events where nn is implicit or infinite.

  • Equal Means & Variances in Poisson serve as a diagnostic; empirical data with similar mean-variance equality often modeled Poisson.

  • Cumulative Techniques reduce computation and improve numerical stability—important for exam contexts lacking calculators with direct pmf functions.

Quick Formula Reference (Exam Sheet)

  • Binomial PMF: P(X=x)=(nx)px(1−p)n−xP(X=x)= \binom{n}{x} p^{x}(1-p)^{n-x}

  • Binomial Mean/Var: μ=np ,  σ2=np(1−p)\mu = np\,,\;\sigma^{2}=np(1-p)

  • Poisson PMF: P(X=x)=e−λλxx!P(X=x)= \dfrac{e^{-\lambda}\lambda^{x}}{x!}

  • Poisson Mean = Variance = λ\lambda

  • Cumulative Probability: P(X≤x)=∑k=0xP(X=k)P(X\le x)= \sum_{k=0}^{x} P(X=k)