Explanation

This is a Poisson distribution problem where:

• Average calls per hour (λ) = 2
• Time period = 8 hours
• Total average calls in 8 hours = λ_total = 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 (average calls in 8 hours)
• k = 20 (number of calls we want)

$P(X = 20) = \frac{e^{-16} \cdot 16^{20}}{20!}$

Calculating this:

• e^{-16} ≈ 1.125 × 10^{-7}
• 16^{20} ≈ 1.208 × 10^{24}
• 20! ≈ 2.433 × 10^{18}

$P(X = 20) ≈ \frac{(1.125 × 10^{-7}) × (1.208 × 10^{24})}{2.433 × 10^{18}} ≈ 0.0559$

Therefore, the probability is approximately 5.59%, which matches option A.