Question

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 1

Ten billion trillion

Answer 2

Answer:

4.47% probability that 2 or more people arrive in the next 10 minutes

Step-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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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 $\mu = \frac{2*10}{60} = 0.3333$

Either less than two people arrive, or more than two do. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$. So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.3333}*(0.3333)^{0}}{(0)!} = 0.7165$

$P(X = 1) = \frac{e^{-0.3333}*(0.3333)^{1}}{(1)!} = 0.2388$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.7165 + 0.2388 = 0.9553$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9553 = 0.0447$

4.47% probability that 2 or more people arrive in the next 10 minutes