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?

The probability that 2 or more people arrive in the next 10 minutes is;

P(X ≥ 2) =  0.0447

This is a poisson distribution problem with the formula;

P(X = x) = [e^(-μ) × μˣ]/x!

Where;

x is number of successes

μ is the mean in the time interval given

Since we have a mean of 2 customers per hour and want to find the probability that 2 or more people arrive in the next 10 minutes. Thus;

μ = (2 × 10)/60

μ = 0.333

Thus,  probability that 2 or more people arrive in the next 10 minutes is;

P(X ≥ 2) = 1 - (P(X = 0) + P(X = 1))

P(X = 0) =  [e^(-0.333) × 0.333⁰]/0!

P(X = 0) = 0.7165

P(X = 2) =  [e^(-0.333) × 0.333²]/2!

P(X = 2) = 0.2388

Thus;

P(X ≥ 2) = 1 - (0.7165 + 0.2388)

P(X ≥ 2) =  0.0447

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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:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

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

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

We want . So

In which

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

0.6

39mph

step-by-step explanation: