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

Concepts and reason

The concepts used to solve this problem is application of Poisson distribution.

The occurrence of the event within a fixed interval of time with a known mean rate and independently of the occurrence of the last event is called the Poisson distribution.

Poisson distribution has following conditions,

First is $n \to \infty$ that is the number of trials is indefinite.

Second is $p \to 0$ represents the probability of success is small for each trial.

Third is $np = \lambda$ which is finite.

Where, $\lambda$ is a positive real number.

Fundamentals

For a random variable X, the probability density function of Poisson distribution can be denoted as:

$f\left( x \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}$

Where, $\lambda {\rm{ is the mean or expected value of }}X$

(a)

Consider X to be a random variable denoting the arrival of people in the room.

As per the provided information the number of people arriving for treatment at an emergency room follows Poisson distribution with a rate parameter of four per hour. So, the random variable X will follow Poisson distribution with the parameter $\lambda$ as 4.

That is,

$X \sim P\left( 4 \right)$

The probability of exactly three people arriving during a particular hour can be calculated by substituting the value of X as 3 in the probability density function of Poisson distribution.

$\begin{array}{c}\\P\left( {X = 3} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\\\\ = \frac{{{e^{ - 4}}{{\left( 4 \right)}^3}}}{{3!}}\\\\ = 0.195\\\end{array}$

(b)

The probability of at least 3 people arriving in the room is calculated as,

$\begin{array}{c}\\P\left( {x \ge 3} \right) = 1 - P\left( {x \le 2} \right)\\\\ = 1 - \left[ {P\left( {x = 0} \right) + P\left( {x = 1} \right) + P\left( {x = 2} \right)} \right]\\\\ = 1 - \left( {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}} + \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}} + \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \right)\\\\ = 1 - \left( {\frac{{{e^{ - 4}}{{\left( 4 \right)}^0}}}{{0!}} + \frac{{{e^{ - 4}}{{\left( 4 \right)}^1}}}{{1!}} + \frac{{{e^{ - 4}}{{\left( 4 \right)}^2}}}{{2!}}} \right)\\\end{array}$

$\begin{array}{l}\\ = 1 - \left( {0.018 + 0.073 + 0.147} \right)\\\\ = 0.762\\\end{array}$

(c)

By the property of Poisson distribution, it can be denoted that mean or expected number of a random variable X following Poisson distribution is equal to the parameter $\lambda$ .

Therefore,

${\rm{Mean or }}E\left( X \right) = \lambda$

As per the provided information the value of parameter $\lambda$ is 4, that is, the excepted number of people arriving in the room in an hour is 4. Therefore, the expected number of people arriving in 30 minutes is the half of the mean or $\lambda$ . So,

$\begin{array}{c}\\\left( \begin{array}{l}\\{\rm{Expected number of }}\\\\{\rm{people arriving in the}}\\\\{\rm{ room in 30 mins}}\\\end{array} \right) = \frac{{\left( \begin{array}{c}\\{\rm{Expected number of people }}\\\\{\rm{arriving in the room in 1 hr }}\\\end{array} \right)}}{2}\\\\ = \frac{\lambda }{2}\\\\ = \frac{4}{2}\\\\ = 2\\\end{array}$

Ans: Part a

The probability of exactly three people arriving during a particular hour is 0.195.

Part b

The probability of at least three people arriving during a particular hour is 0.762.

Part c

The expected number of people arriving in 30 minutes is 2.