Question

The number of people arriving for treatment at an emergency room can be modeled by a Poisson Distribution with a rate parameter of seven 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 45-min period?

Answer 1

Answer:

a) 0.052 = 5.2% probability that exactly three arrivals occur during a particular hour

b) 0.971 = 97.1% probability that at least three people arrive during a particular hour

c) 5.25 people are expected to arrive during a 45-min 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 interval.

Poisson Distribution with a rate parameter of seven per hour.

This means that $\mu = 7$

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

This is P(X = 3).

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

$P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.052$

0.052 = 5.2% 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? (Round your answer to three decimal places.)

Either less than three people arrive, or at least three does. The sum of the probabilities of these events is 1. So

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

We want $P(X \geq 3)$, which is

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

In which

$P(X \geq 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.001$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.006$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.022$

So

$P(X \geq 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.001 + 0.006 + 0.022 + 0.029$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.029 = 0.971$

0.971 = 97.1% probability that at least three people arrive during a particular hour

(c) How many people do you expect to arrive during a 45-min period?

During an hour(60 minutes), 7 people are expected to arrive. So, using proportions:

$\frac{45*7}{60} = \frac{3*7}{4} = \frac{21}{4} = 5.25$

5.25 people are expected to arrive during a 45-min period