You are the only bank teller on duty and you want to take a break for 10 minutes but you don't want to miss any customers. suppose the arrival of customers can be models by a poisson distribution with mean of 2 customers per hour. what's the probability that 2 or more people arrive in the next 10 minutes?

Answer:4.47% probability that 2 or more people arrive in the next 10 minutesStep-by-step explanation:In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]In whichx is the number of sucessese = 2.71828 is the Euler number[tex]\mu[/tex] is the mean in the given time interval.Mean of 2 customers per hour.What's the probability that 2 or more people arrive in the next 10 minutes?10 minutes, so [tex]\mu = \frac{2*10}{60} = 0.3333[/tex]Either less than two people arrive, or more than two do. The sum of the probabilities of these events is decimal 1. So[tex]P(X < 2) + P(X \geq 2) = 1[/tex]We want [tex]P(X \geq 2)[/tex]. So[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]In which[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex][tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex][tex]P(X = 0) = \frac{e^{-0.3333}*(0.3333)^{0}}{(0)!} = 0.7165[/tex][tex]P(X = 1) = \frac{e^{-0.3333}*(0.3333)^{1}}{(1)!} = 0.2388[/tex][tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.7165 + 0.2388 = 0.9553[/tex][tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9553 = 0.0447[/tex]4.47% probability that 2 or more people arrive in the next 10 minutes