Probability Distributions

Probability Distributions

A probability distribution describes how probabilities are distributed over the values of a random variable. It provides a mathematical model for predicting outcomes and calculating probabilities in business situations.

Random Variables

A random variable assigns numerical values to outcomes. Discrete: countable values (defective items: 0,1,2...). Continuous: any value in a range (weight, revenue). Expected value E(X) = ΣxP(x). Variance Var(X) = E(X²) − [E(X)]².

Binomial Distribution

Models successes in n independent trials with constant probability p. P(X=r) = C(n,r) × p^r × (1−p)^(n−r). Mean: μ = np. Variance: σ² = np(1−p). Applications: defective items, customer responses. Example: 70% satisfaction, survey 10, P(exactly 8) = C(10,8)(0.7)⁸(0.3)² = 0.2335.

Poisson Distribution

Events in fixed interval at constant rate. P(X=r) = (e^−λ × λ^r)/r!. Mean = Variance = λ. Applications: calls per hour, accidents per month, arrivals per minute. Approximates Binomial when n>20 and p<0.05.

Normal Distribution

Bell-shaped curve. Properties: symmetric, mean=median=mode, determined by μ and σ. Empirical rule: 68.27% within μ±1σ, 95.45% within μ±2σ, 99.73% within μ±3σ.

Standard Normal (Z)

μ=0, σ=1. Convert: Z = (X−μ)/σ. Z-tables give probabilities. Example: scores μ=60, σ=10, P(above 75) = P(Z>1.5) = 6.68%.

Applications

Quality control (defect rates), finance (stock returns, VaR), marketing (market potential), operations (inventory). Bolts μ=10mm, σ=0.05mm, P(outside 9.9-10.1) = P(|Z|>2) = 4.56%.

Normal Approximation to Binomial

When np≥5 and n(1−p)≥5: use normal with μ=np, σ=√np(1−p). Continuity correction: adjust by ±0.5.

Summary

Binomial (success/failure trials), Poisson (rare events), and normal (foundation of inference) model random phenomena for probability calculations, quality control, and risk assessment.

Worked Example: Binomial Distribution

A quality inspector checks 10 items from a production line where the defect rate is 5%. Find: (a) P(exactly 1 defective), (b) P(no defective), (c) P(at most 2 defective).

Solution: n = 10, p = 0.05, q = 0.95

(a) P(X=1) = C(10,1) × (0.05)¹ × (0.95)⁹ = 10 × 0.05 × 0.6302 = 0.3151

(b) P(X=0) = C(10,0) × (0.05)⁰ × (0.95)¹⁰ = 1 × 1 × 0.5987 = 0.5987

(c) P(X≤2) = P(0) + P(1) + P(2) = 0.5987 + 0.3151 + C(10,2)(0.05)²(0.95)⁸ = 0.5987 + 0.3151 + 45 × 0.0025 × 0.6634 = 0.5987 + 0.3151 + 0.0746 = 0.9884

Interpretation: There is a 59.87% chance of finding no defects, 31.51% chance of exactly one defect, and 98.84% chance of at most 2 defects in a sample of 10. This helps the factory set acceptable quality levels and decide whether to accept or reject batches.

Worked Example: Normal Distribution

The monthly income of employees in a company is normally distributed with mean Rs 40,000 and SD Rs 8,000. Find: (a) P(income above Rs 52,000), (b) P(income between Rs 32,000 and Rs 48,000).

Solution:

(a) Z = (52000 − 40000)/8000 = 12000/8000 = Z = 1.5. From Z-table: P(Z < 1.5) = 0.9332. Therefore P(X > 52000) = 1 − 0.9332 = 0.0668 or 6.68%. About 6.68% of employees earn more than Rs 52,000.

(b) Z₁ = (32000 − 40000)/8000 = −1.0. Z₂ = (48000 − 40000)/8000 = 1.0. P(−1 < Z < 1) = P(Z < 1) − P(Z < −1) = 0.8413 − 0.1587 = 0.6826 or 68.26%. About 68.26% of employees earn between Rs 32,000 and Rs 48,000. (This confirms the empirical rule: ~68% within μ ± 1σ.)

Quick Reference: Z-Table Values

Memorise these key Z-values — they appear in almost every exam on normal distribution and hypothesis testing.

Exam Tips

Tip 1: For Binomial, always verify the four conditions (fixed n, two outcomes, constant p, independent trials). Tip 2: For Poisson, state that Mean = Variance = λ — examiners specifically look for this. Tip 3: Always draw a normal curve sketch showing the area you are calculating. Tip 4: Remember to apply continuity correction (±0.5) when using normal to approximate binomial. Tip 5: If P(X > a) is asked, calculate 1 − P(X ≤ a) using the Z-table.