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.