Question

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of six per hour. (a) What is the probability that exactly two arrivals occur during a particular hour? (Round your answer to three decimal places.) (b) What is the probability that at least two 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? people

Answer 1

Answer:

a) 0.045 = 4.5% probability that exactly two arrivals occur during a particular hour

b) 0.983 = 98.3% probability that at least two people arrive during a particular hour

c) 4.5 arrivals 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 time interval.

Poisson process with a rate parameter of six per hour.

This means that $\mu = 6$. So

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

This is P(X = 2).

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

$P(X = 2) = \frac{e^{-6}*(6)^{2}}{(2)!} = 0.045$

0.045 = 4.5% probability that exactly two arrivals occur during a particular hour

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

Either less than two people arrive during a particular hour, or at least two do. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 2)$

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

In which

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

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

$P(X = 0) = \frac{e^{-6}*(6)^{0}}{(0)!} = 0.002$

$P(X = 1) = \frac{e^{-6}*(6)^{1}}{(1)!} = 0.015$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.002 + 0.015 = 0.017$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.017 = 0.983$

0.983 = 98.3% probability that at least two people arrive during a particular hour

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

An hour has 60 mins.

45/60 = 3/4.

So

$\mu = \frac{3*6}{4} = 4.5$

4.5 arrivals during a 45-min period.