Problem 86 - Calculus for Biology and Medicine

Question

Suppose the number of phone calls arriving at a switchboard per hour is Poisson distributed with mean 3 calls per hour. (a) Find the probability that at least one phone call arrives between noon and \(1 \mathrm{P.M}\). (b) Assuming that phone calls in different hours are independent of each other, find the probability that no phone calls arrive between noon and 2 P.M.

Step-by-Step Solution

Answer

(a) 0.9502. (b) 0.00248.

Step 1: Understanding the Poisson Distribution

The problem states that the number of phone calls per hour follows a Poisson distribution with a mean (rate parameter, \( \lambda \)) of 3 calls per hour. The Poisson distribution is used to model the probability of a given number of events happening in a fixed interval of time.

Step 2: Calculating Probability for at Least One Call (Part a)

For part (a), we want to find the probability of at least one call arriving between noon and 1 PM. The complement of this event is that no calls come in during this period, which we denote as \( X = 0 \). The probability mass function for a Poisson distribution is given by \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \). Therefore, \( P(X = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3} \). The probability of at least one call is then \( 1 - P(X = 0) = 1 - e^{-3} \).

Step 3: Solving Part (a)

Now we calculate \( e^{-3} \) using a calculator to find \( P(X \geq 1) \). \( e^{-3} \approx 0.0498 \), thus \( 1 - e^{-3} \approx 0.9502 \). This is the probability that at least one phone call will arrive between noon and 1 PM.

Step 4: Calculating Probability for No Calls in 2 Hours (Part b)

For part (b), we need to find the probability of no calls arriving between noon and 2 PM. This period covers two hours, so the mean number of calls is \( \lambda = 3 \times 2 = 6 \). Using the Poisson formula, \( P(X=0) = \frac{e^{-6} 6^0}{0!} = e^{-6} \).

Step 5: Solving Part (b)

Calculate \( e^{-6} \) using a calculator. \( e^{-6} \approx 0.00248 \). Thus, the probability that no phone calls arrive between noon and 2 PM is approximately 0.00248.

Key Concepts

Understanding Probability Calculation with Poisson Distribution

When dealing with the Poisson distribution, it's all about determining the likelihood of a certain number of events occurring within a fixed time frame. This is the essence of probability calculation. The Poisson distribution specifically handles situations where events happen independently at a constant mean rate.

To calculate these probabilities, we rely on the probability mass function of a Poisson distribution:

• The formula is given as \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where:
• \( \lambda \) (lambda) is the mean rate of occurrence within the given time period.
• \( k \) is the number of events we want the probability for.

This formula allows us to calculate probabilities for different scenarios, such as having no events (like no phone calls) or at least one event happening. To find the probability of at least one event, we often use the complement rule because it's generally easier to calculate the probability of zero events and subtract it from one.

Remember, calculating these probabilities involves using the constant \( e \), known as the base of the natural logarithms, approximately equal to 2.71828.

The Concept of Independent Events in Poisson Distribution

Independent events are a pivotal component when working with Poisson distributions, especially regarding how we calculate probabilities over different periods or experimental units. For the Poisson process, events in distinct intervals do not influence each other's occurrence.

This characteristic is captured mathematically by stating that if two intervals of time (or space) are non-overlapping, then the events occurring in one interval are independent of the events in the other interval. This property is crucial when assessing periods longer than the basic given interval, as seen when transitioning from one hour to two hours in our examples.

For example, when calculating the probability of no phone calls within a two-hour window based on our initial one-hour mean, we multiply the average rate by the number of hours. Since the events are independent, the same Poisson distribution formula holds. A Poisson distribution allows us to scale from one interval to another seamlessly, maintaining the independence of each segment.

Significance of the Mean Rate Parameter \( \lambda \)

In Poisson distribution, the mean rate parameter, denoted by \( \lambda \), is a central element. It defines the expected number of events occurring in a fixed period. For our problem, \( \lambda \) is 3, representing an average of 3 phone calls per hour.

This parameter not only sets the expectation but also describes the variability within the process. In a Poisson distribution, both the mean and the variance are equal to \( \lambda \). This dual role helps in understanding the spread and central tendency of the distribution.

• A higher \( \lambda \) means more frequent events, leading to a shift in the probability distribution towards higher values of events.
• A lower \( \lambda \) implies infrequent occurrences, concentrating probabilities near zero, indicating fewer events.

Therefore, the mean rate parameter is integral in defining the behavior and shape of the Poisson distribution, dictating both where most outcomes lie and how scattered they might be within the probability landscape.