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spin [16.1K]
2 years ago
7

The arrival of customers at a service desk follows a Poisson distribution. If they arrive at a rate of two every five minutes, w

hat is the probability that no customers arrive in a five-minute period? ​
Mathematics
1 answer:
vampirchik [111]2 years ago
5 0

Answer:

13.53% probability that no customers arrive in a five-minute period

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.

They arrive at a rate of two every five minutes

This means that \mu = 2

What is the probability that no customers arrive in a five-minute period?

This is P(X = 0).

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

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

13.53% probability that no customers arrive in a five-minute period

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