Answer: 5.59%

## Explanation This is a Poisson distribution problem where: - Average calls per hour = 2 - Time period = 8 hours - Expected calls in 8 hours (λ) = 2 × 8 = 16 - We want P(X = 20) The Poisson probability formula is: \[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\] Where: - λ = 16 (expected number of calls) - k = 20 (desired number of calls) Calculating: \[P(X = 20) = \frac{e^{-16} \cdot 16^{20}}{20!}\] Using the Poisson distribution: - e⁻¹⁶ ≈ 1.125 × 10⁻⁷ - 16²⁰ ≈ 1.208 × 10²⁴ - 20! = 2,432,902,008,176,640,000 \[P(X = 20) ≈ \frac{(1.125 \times 10^{-7}) \times (1.208 \times 10^{24})}{2.433 \times 10^{18}} ≈ 0.0559\] Converting to percentage: 0.0559 × 100% = **5.59%** This matches option A, which is the correct answer.

Author: LeetQuiz .