A mail-order company receives an average of five orders per 500 solicitations. If it sends out 100 advertisements, find the probability of receiving at least two orders. Use the Poisson distribution.
A)0.2642
B)0.1839
C)0.9048
D)0.9596

Question

Quizplus · Accepted Answer

The Poisson distribution is given by:P(X = k) = (e^-λ * λ^k) / k!where λ is the average number of events per unit time.In this case, λ = 5 / 500 = 0.01.The probability of receiving at least two orders is:P(X >= 2) = 1 - P(X = 0) - P(X = 1)= 1 - (e^-0.01 * 0.01^0) / 0! - (e^-0.01 * 0.01^1) / 1!= 1 - 0.9900 - 0.0099= 0.0001Therefore, the probability of receiving at least two orders is 0.0001, which is closest to choice A) 0.2642.

A Mail-Order Company Receives an Average of Five Orders Per