Answer to Question #230070 in Statistics and Probability for Teddy

Question

Phone calls enter the ”support desk” of an electricity supplying company on the
average two every 3 minutes. If one assumes an approximate Poisson process:
(i) What is the probability of no calls in 3 minutes?

Answer 1

$P(X=k) = \frac{λ^k \times e^{-λ}}{k!} \\ Mean \; λ=2 \\ P(X=0) = \frac{2^0 \times e^{-2}}{0!} = e^{-2} = 0.1353$