Question

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of four per hour. (a) What is the probability that exactly three arrivals occur during a particular hour? (Round your answer to three decimal places.) (b) What is the probability that at least three people arrive during a particular hour? (Round your answer to three decimal places.) (c) How many people do you expect to arrive during a 30-min period?people

Answer 1

Answer:

a) 0.195 = 19.5% probability that exactly three arrivals occur during a particular hour.

b) 0.762 = 76.2% probability that at least three people arrive during a particular hour

c) 2 people.

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 interval.

Poisson process with a rate parameter of four per hour.

This means that $\mu = 4$.

(a) What is the probability that exactly three arrivals occur during a particular hour?

This is P(X = 3). So

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

$P(X = 3) = \frac{e^{-4}*4^{3}}{(3)!} = 0.195$

0.195 = 19.5% probability that exactly three arrivals occur during a particular hour.

(b) What is the probability that at least three people arrive during a particular hour?

Either less than three people arrive during a particular hour, or at least three does. The sum of the probabilities of these outcomes is 1. So

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3) = 1 - P(X < 3)$, in which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$. So

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

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.018$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.073$

$P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.147$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.018 + 0.073 + 0.147 = 0.238$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.238 = 0.762$

0.762 = 76.2% probability that at least three people arrive during a particular hour.

c) How many people do you expect to arrive during a 30-min period?

One hour, 4 people

Half an hour = 30 min = 4/2 = 2 people.