Step-by-Step Solution

Topic: Poisson Distribution Approximation

Given Parameters

• • Sample Size (n) = 100
• • Probability of Defect (p) = 6% = 0.06

Since n is large and p is small, we use the Poisson Approximation to the Binomial Distribution.

Parameter (λ) = np = 100 × 0.06 = 6

Formula: P(X = x) = (e^-λ × λ^x) / x!

Given: e^-6 = 0.0024

(I) PROBABILITY OF NO DEFECTIVE BULBS

We need P(X = 0):

P(0) = (e^-6 × 6⁰) / 0!

Since 6⁰ = 1 and 0! = 1:

P(0) = e^-6 = 0.0024

(II) PROBABILITY OF EXACTLY TWO DEFECTIVE

We need P(X = 2):

P(2) = (e^-6 × 6²) / 2!

P(2) = (0.0024 × 36) / 2

P(2) = 0.0024 × 18

Calculation: 24 × 18 = 432. Adjust decimals.

(III)(A) NOT MORE THAN ONE DEFECTIVE

“Not more than one” means X = 0 or X = 1.

P(X ≤ 1) = P(0) + P(1)

1. P(0) = 0.0024 (Calculated above)

2. P(1) = (e^-6 × 6¹) / 1! = 0.0024 × 6 = 0.0144

Total: 0.0024 + 0.0144

(III)(B) MEAN AND VARIANCE

For a Binomial Distribution B(n, p):

1. Mean (μ) = n × p

μ = 100 × 0.06 = 6

2. Variance (σ²) = n × p × q

Where q = 1 – p = 1 – 0.06 = 0.94

σ² = 100 × 0.06 × 0.94

σ² = 6 × 0.94

Note: While Poisson approximation uses λ for both mean and variance (6), the exact Binomial Variance (5.64) is technically more accurate for a finite sample.

⚠️ Silly Mistake Audit: Variance Confusion

Don’t Confuse Models!
When using the Poisson approximation for probabilities, it’s easy to assume the Variance is also just λ (which would be 6). However, because we have a finite sample size (n=100) and probability (p), the distribution is actually Binomial. Therefore, the Variance must be calculated as npq (5.64), not just np.