During late-night hours, the 10^th Street convenience store averages four customers per hour.No customers have arrived in the last half-hour.Assuming that all Poisson conditions are met, determine the probability that it will be between 15 and 30 more minutes before the next customer arrives. A).2326 B).3461 C).2784 D).3523 E).2558

Question

Quizplus · Accepted Answer

The probability of no arrivals in 30 minutes is e^(-4*0.5) = 0.368. The probability of 1 arrival in 30 minutes is 4*0.5*e^(-4*0.5) = 0.368. The probability of 2 arrivals in 30 minutes is (4*0.5)^2*e^(-4*0.5)/2! = 0.184. The probability of 3 arrivals in 30 minutes is (4*0.5)^3*e^(-4*0.5)/3! = 0.061. The probability of 4 or more arrivals in 30 minutes is 1 - 0.368 - 0.368 - 0.184 - 0.061 = 0.019. The probability of it being between 15 and 30 more minutes before the next customer arrives is 0.368 + 0.368 + 0.184 = 0.92.

During Late-Night Hours, the 10th Street Convenience Store Averages Four

During Late-Night Hours, the 10^th Street Convenience Store Averages Four