There are two main characteristics of a Poisson experiment.
- The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a hospital quality manager might be interested in the number of medication labeling errors found in the pharmacy. It might be that, on average, there are five errors per 100 patient charts. The interval is 100 patient charts.
- The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a "success."
The random variable X = the number of occurrences in the interval of interest.
The average number of patients admitted to an emergency room triage station in a half-hour period is 12. Of interest is the number of patients admitted in five minutes. The time interval of interest is five minutes. What is the probability that the number of patients, selected randomly, admitted in five minutes is three?
Solution
Let X= the number of patients admitted in five minutes. If the average number of patients admitted in 30 minutes (half-hour) is 12, then the average number of patients admitted in five minutes is (305)(12)=2 patients.
The probability question asks you to find \(P(x=3)\).
A hospital administration department expects to find six incorrectly coded medical records per day, on average, during their routine audits. What is the probability of the department finding fewer than five incorrectly coded records on any given day?
Of interest is the number of records the department identifies in one day, so the time interval of interest is one day. Let X= the number of incorrectly coded medical records the department identifies in one day. If the department expects to find six incorrectly coded records per day, then the average is six records per day.
- Answer
-
P (x < 5)
Notation for the Poisson: P = Poisson Probability Distribution Function
\(X \sim P(\mu)\)
Read this as " \(X\) is a random variable with a Poisson distribution." The parameter is \(\mu(\) or \(\lambda) ; \mu(\) or \(\lambda)=\) the mean for the interval of our interest. The mean is the number of occurrences that occur on average during the interval period.
A clinic receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that the clinic receives more than one call in the next 15 minutes?
Solution
Let \(X=\) the number of calls the clinic receives in 15 minutes. (The interval of interest is 15 minutes or \(\frac{1}{4}\) hour.)
\[
x=0,1,2,3, \ldots
\]
If the clinic receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then the clinic receives
\(\left(\frac{1}{8}\right)(6)=0.75\) calls in 15 minutes, on average. So, \(\mu=0.75\) for this problem.
The Poisson distribution is discrete, and thus \(>1\) includes all whole numbers through infinity. The solution is to subtract the probability less than 1 thus \(1-[P(x=0)+P(x=1)]\)
\[\begin{array}{c}
P(x)=\frac{\mu^x e^{-\mu}}{x!} \\
P(x>1)=1-P(x \leq 1)=1-[P(x=0)+P(x=1)] \\
=1-\left[\frac{0.75^0 e^{-0.75}}{0!}+\frac{0.75^1 e^{-0.75}}{1!}\right] \\
=1-\left[\frac{(1)(0.4724)}{1}+\frac{(0.75)(0.4724)}{1}\right] \\
=1-[0.4724+0.3543]=0.1733
\end{array}\]
The y-axis in Figure \(\PageIndex{1}\) contains the probability of x where X = the number of calls in 15 minutes
According to regional health data, a metropolitan Poison Control Center receives, on average, 147 calls per day. Let X = the number of calls the center receives per day. The discrete random variable X takes on the values x=0,1,2…. The random variable X has a Poisson distribution: X∼P(147). The mean is 147 calls.
- What is the probability that an email user receives exactly 160 calls per day?
- What is the probability that an email user receives at most 160 calls per day?
- What is the standard deviation?
- Answer
-
a. \(P(x=160)=\sim P(147,160) \approx 0.0180\)
b. \(P(x \leq 160)=\sim P(147,160) \approx 0.8666\)
c. Standard Deviation \(=\sigma=\sqrt{\mu}=\sqrt{147} \approx 12.1244\)
Solution
- Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is
41.5 24 41.5 24 - X ~ P(1.7292), so P(x = 2) = poissonpdf(1.7292, 2) ≈ 0.2653
- P(x > 2) = 1 – P(x ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 – 0.7495 = 0.2505
In a specialized clinical study, the probability of a patient experiencing a specific rare adverse reaction to a new medication was reported as about 1.02% (p=0.0102). Use this information for a sample of 200 patients to find the probability that exactly ten of them will experience this reaction. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
- Answer
-
Let \(X=\) the number of patients with the reaction.
Using the binomial distribution:
\[P(x=10)=\dfrac{200!}{10!(200-10)!} \times .0102^{10} \times .9898^{190}=0.000039\]
Using the Poisson distribution:
\[\begin{array}{l}
\text { Calculate } \mu=n p=200(0.0102) \approx 2.04 \\
P(x=10)=\dfrac{\mu^x e^{-\mu}}{\mathrm{x}!}=\dfrac{2.04^{10} e^{-2.04}}{10!}=0.000045
\end{array}\]We expect the approximation to be good because \(n\) is large (greater than 20) and \(p\) is small (less than \(0.05)\). The results are close-both probabilities reported are almost 0.
Solution
Let X = the number of days with low seismic activity.
Using the binomial distribution:
- P(x = 10) = binompdf(200, .0102, 10) ≈ 0.000039
Using the Poisson distribution:
- Calculate μ = np = 200(0.0102) ≈ 2.04
- P(x = 10) = poissonpdf(2.04, 10) ≈ 0.000045
We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0.