Question

The number \(x\) of people entering the intensive care unit at a particular hospital on any one day has a Poisson probability distribution with mean equal to five persons per day. a. What is the probability that the number of people entering the intensive care unit on a particular day is two? Less than or equal to two? b. Is it likely that \(x\) will exceed \(10 ?\) Explain.

Short answer

a. Calculate the probabilities we obtained in Steps 2 and 3: $$ P(X=2)=\frac{e^{-5}5^2}{2!}=\frac{e^{-5}\cdot 25}{2} \approx 0.0842 $$ $$ P(X\leq 2)=e^{-5} + 5e^{-5}+ \frac{e^{-5}\cdot 25}{2} \approx 0.2650 $$ The probability that exactly 2 people enter the ICU on a particular day is approximately 0.0842, and the probability that 2 or fewer people enter the ICU on a particular day is approximately 0.2650. b. Calculate the probability we found in Step 4: $$ P(X>10)=1-P(X\leq 10) \approx 1-0.9863 = 0.0137 $$ The probability of more than 10 people entering the ICU on a particular day is approximately 0.0137. Since this value is relatively small, we can say that it is not likely for more than 10 people to enter the ICU on a particular day.

Step-by-step solution

Recall the Poisson probability formula

The probability mass function (PMF) of a Poisson distribution is given by: $$ P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!} $$ where \(P(X=k)\) denotes the probability of having \(k\) events, \(\lambda\) is the mean number of events, and \(k!\) is the factorial of \(k\). In this exercise, \(\lambda = 5\).

Calculate the probability of exactly 2 people entering the ICU

Using the Poisson formula, we can find the probability of exactly 2 people entering the ICU on a particular day: $$ P(X=2)=\frac{e^{-5}5^2}{2!}=\frac{e^{-5}\cdot 25}{2} $$

Calculate the probability of 2 or fewer people entering the ICU

To find the probability of 2 or fewer people entering the ICU, we'll need to find the sum of the probabilities when \(X=0\), \(X=1\), and \(X=2\). That is, we need to calculate: $$ P(X\leq 2)=P(X=0)+P(X=1)+P(X=2) $$ Using the Poisson formula, we calculate \(P(X=0)\) and \(P(X=1)\): $$ P(X=0)=\frac{e^{-5}5^0}{0!} = e^{-5} $$ $$ P(X=1)=\frac{e^{-5}5^1}{1!} =5e^{-5} $$ Now we can find the cumulative probability for \(X \leq 2\): $$ P(X\leq 2)=e^{-5} + 5e^{-5}+ \frac{e^{-5}\cdot 25}{2} $$

Determine if it's likely for \(x\) to exceed 10 people

To determine if it's likely for more than 10 people to enter the ICU in a particular day, we can calculate the cumulative probability of having 10 or fewer people enter the ICU and then find the complement. First, we calculate the cumulative probability of \(X\leq 10\): $$ P(X\leq 10)=\sum_{k=0}^{10}\frac{e^{-5}5^k}{k!} $$ To find the probability of more than 10 people entering the ICU, we'll take the complement: $$ P(X>10)=1-P(X\leq 10) $$ If \(P(X>10)\) is significantly small, we can say it is not likely for more than 10 people to enter the ICU on a particular day. By calculating the values in each step, we can now answer parts a and b of the exercise.

Key concepts

Probability Mass Function

The Probability Mass Function (PMF) is a fundamental concept in probability theory used to describe the distribution of a discrete random variable. For the Poisson Distribution, the PMF helps us determine the likelihood of a certain number of events occurring within a fixed interval of time or space.In the case of the Poisson distribution, the PMF is mathematically expressed as:\[P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!}\]where:

• \(P(X=k)\) is the probability of observing \(k\) events.
• \(\lambda\) is the average number of events in the interval.
• \(k!\) represents the factorial of \(k\).
• \(e\) is the base of the natural logarithm, approximately equal to 2.71828.

This formula allows us to calculate the probability of exactly \(k\) events happening, as seen in the example where the probability of 2 people entering the ICU was computed using \(\lambda = 5\). The PMF is crucial for solving problems concerning discrete events like the one we are examining.

Cumulative Probability

Cumulative Probability involves calculating the probability of a random variable taking on a value less than or equal to a specific threshold. For the Poisson distribution, this means summing up individual probabilities from the PMF.In our example, to find the probability of 2 or fewer people entering the ICU, we computed:\[P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)\]This approach provides insights into the likelihood of a set of outcomes rather than a single occurrence. It helps answer questions like the probability of at most a certain number of events happening.Calculating cumulative probability is particularly useful when determining the probability thresholds in real-world situations. Such calculations allow us to draw meaningful conclusions from the aggregated probability of multiple event outcomes.

Factorial Calculation

Factorial calculation is a mathematical operation crucial in understanding and computing probabilities in Poisson and other distributions. A factorial, denoted by \(k!\), is the product of an integer \(k\) and all the positive integers less than \(k\).For example:

• \(0! = 1\)
• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)

Factorials grow rapidly as numbers increase, and they are vital in probability calculations because they account for the different ways events can be arranged or occur. In the Poisson distribution PMF, factorials are used to compute the probabilities of discrete events.

Complement Rule

The Complement Rule is a straightforward yet powerful concept in probability theory. It is used to find the probability of the complement of an event, i.e., the event not happening.Mathematically, the complement of an event \(A\) is written as \(A^c\), and the rule states:\[P(A^c) = 1 - P(A)\]In the context of the Poisson example, to determine if it's likely for more than 10 people to enter the ICU, we used the complement rule:\[P(X>10)=1-P(X\leq 10)\]This method is efficient especially when it's easier to calculate \(P(X\leq 10)\) than \(P(X>10)\) directly. The complement rule simplifies probability calculations by allowing us to focus on calculating the easier side of a probability, making it a handy tool in many statistical analyses.