Problem 26 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that commercial airplane crashes in a certain country occur at the rate of \(2.5\) per year. (a) Is it reasonable to assume that such crashes are Poisson events? Explain. (b) What is the probability that four or more crashes will occur next year? (c) What is the probability that the next two crashes will occur within three months of one another?

Step-by-Step Solution

Answer

It is reasonable to consider airplane crashes as Poisson events. For part (b) the exact numerical value can be calculated following step 2 above, and for part (c), the exact numerical estimation should be calculated following step 3.

Step 1: Understanding Poisson events

Poisson events are independent events which occur at a constant average rate. In the case of commercial airplane crashes, it can be assumed they're independent events, as one crash doesn't influence the occurrence of another. Unless there's reason to believe that the rate of crashes is changing, it would be reasonable to assume that such crashes can be considered Poisson events.

Step 2: Calculating Probability of 4 or More Crashes

For a Poisson distribution, the probability of observing exactly \(k\) events in an interval is given by \(P(k;λ) = e^{-λ}λ^{k}/k!\) where \(λ\) is the average rate of occurrence. However, to find the probability of 4 or more crashes, we should calculate the complement: the probability of observing 3 or fewer crashes and subtracting it from 1. First, calculate the probabilities for 0, 1, 2 and 3 events respectively and add these up. With \(λ = 2.5\), the respective probabilities are \(P(0;2.5), P(1;2.5), P(2;2.5), P(3;2.5)\). Then calculate \(1-P(X \leq 3)\) to get P(X \geq 4) which is the probability of observing 4 or more crashes.

Step 3: Calculating the probability of next two crashes within three months

In time \(T\) months, the event rate is \(λT\). Over 3 months the event rate is \(2.5*(3/12) = 0.625\). We need to find the probability of observing 2 or more events in T months. This can be done by calculating the complement as in step 2. First, calculate probabilities for 0 and 1 events, then subtract these from 1. Thus, \(P(2+ events in 3 months) = 1- (P(0;0.625) + P(1;0.625))\).

Key Concepts

Understanding Airplane Crashes as a Poisson Event

When talking about commercial airplane crashes, it's essential to treat them as independent events occurring at a certain average rate. This average rate is known as the Poisson rate. The rate of airplane crashes in our example is \(2.5\) per year. This means, on average, there are \(2.5\) crashes annually.

In the real world, the occurrence of one airplane crash typically does not directly cause another, and external influencing factors remain minimal for crashes to be considered independent. This independence in the rate of crash occurrence makes it reasonable to model such events using the Poisson distribution.

A key characteristic of Poisson events is that they should occur at a consistent average rate over the given interval. Thus, unless there is a change in external conditions affecting the crash rate, adopting the Poisson model is fitting.

Calculating the Probability of an Event

The probability calculation for events follows a clear pattern with the Poisson distribution. In our exercise, we're interested in finding the probability of observing four or more airplane crashes in a given year.
To work this out, we use the formula for Poisson probabilities:

• \(P(k;λ) = e^{-λ}λ^{k}/k!\) where \(λ\) is the average rate, and \(k\) is the number of occurrences.

To calculate the probability of four or more crashes, begin by calculating the probability for zero through three crashes. This includes calculating \(P(0;2.5)\), \(P(1;2.5)\), \(P(2;2.5)\), and \(P(3;2.5)\).

Next, sum these probabilities to find the probability of three or fewer crashes.

Finally, subtract this from one to obtain the probability of four or more crashes occurring, which is \(1 - P(X \leq 3)\). This method helps us determine the likelihood of observing events that are four or more within the given time frame.

Grasping Independent Events

The assumption of independence is crucial when working with Poisson processes. In the context of airplane crashes, independence means that each crash is a separate event and the occurrence of one does not affect the likelihood of another.

This property greatly simplifies probability calculations because you don't need to consider interactions between past and future events. The occurrence of events over different time spans can be analyzed independently, but proportionally based on the same average rate. For example, if there are \(2.5\) crashes a year, the methodology scales down for different periods like months.

For a specific period such as three months, we calculate using the modified rate for that duration. If identifying the probability of two crashes in three months, you'd first adjust the rate to fit the three-month span ( \(2.5 * \frac{3}{12} = 0.625\) ). From there, calculate the probability of zero or one event over this shorter period, and use this to find the probability of two or more crashes, utilizing the complement of calculated probabilities.