Question

The number of bankruptcies filed in the district court has a Poisson distribution with an average of 5 per week. a. What is the probability that there will be no bankruptcy filings during a given week? b. What is the probability that there will be at least one bankruptcy filing during a given week? c. Within what limits would you expect to see the number of bankruptcy filings per week at least \(75 \%\) of the time?

Short answer

Question: In a district court, the number of bankruptcies filed follows a Poisson distribution with an average of 5 per week. Calculate the following probabilities and limits: a. Probability of no bankruptcies during a week. b. Probability of having at least one bankruptcy filing during a week. c. Limits that include 75% of weekly bankruptcy filings. Answer: a. The probability of no bankruptcies during a week is approximately 0.0067. b. The probability of having at least one bankruptcy filing during a week is approximately 0.9933. c. Weekly bankruptcy filings will likely fall between 2 and 7 bankruptcies 75% of the time.

Step-by-step solution

Recall the Poisson Distribution PMF

Recall the probability mass function (PMF) of a Poisson distribution: \(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\), where \(\lambda\) is the average rate of the event (in this case, 5 bankruptcies per week) and \(k\) is the number of events.

Step 2a: Calculate the probability of no bankruptcies

We are asked to find the probability that there will be no bankruptcy filings during a given week. Plug in \(\lambda = 5\) and \(k=0\) into the PMF: \(P(X=0) = \frac{e^{-5}5^0}{0!} = e^{-5}\).

Step 2b: Calculate the probability of at least one bankruptcy filing

We are asked to find the probability that there will be at least one bankruptcy filing during a given week. To calculate this, we will use the complementary probability: \(P(X \geq 1) = 1 - P(X=0) = 1 - e^{-5}\).

Find the limits for 75% of bankruptcy filings

We want to find the upper and lower limits \(a\) and \(b\) such that \(P(a \leq X \leq b) = 0.75\). To do this, we will use the cumulative distribution function (CDF) of the Poisson distribution, which is the sum of the probabilities from \(k=0\) to some value \(k\). Since calculating the CDF manually can be tedious, we will approximate the limits using available Poisson CDF tables or software tools. By consulting a Poisson table or using a software tool, we find that when \(a=2\) and \(b=7\), we get \(P(2 \leq X \leq 7) \approx 0.7577\), which is close enough to 75%. Thus, within the limits 2 and 7, we would expect to see the number of bankruptcy filings per week at least 75% of the time. The results for this exercise are as follows: a. The probability that there will be no bankruptcy filings during a given week is approximately \(e^{-5} \approx 0.0067\). b. The probability that there will be at least one bankruptcy filing during a given week is approximately \(1 - e^{-5} \approx 0.9933\). c. We would expect to see the number of bankruptcy filings per week between 2 and 7 at least 75% of the time.

Key concepts

Probability Mass Function (PMF)

The probability mass function (PMF) lies at the heart of understanding discrete probability distributions, especially the Poisson distribution. It tells us the probability of our random variable, which in this context is the number of bankruptcy filings, taking on an exact integer value.

Mathematically, for a Poisson distribution with an average rate of \( \lambda \) events per time period, the PMF is given by \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \), where \( k \) is the number of events (bankruptcy filings in our case), \( e \) is the base of the natural logarithm, and \( k! \) is the factorial of \( k \).

For example, if there's an average of 5 bankruptcies per week (\lambda = 5\), the PMF will give us the probability for exactly 0, 1, 2, ... bankruptcies in that week. The event of 'no bankruptcies' (\(k=0\)) had a low probability, which was calculated by plugging the values into the PMF, highlighting the unlikely situation where an event with a reasonable expected rate does not occur at all within a given period.

Cumulative Distribution Function (CDF)

Understanding the cumulative distribution function (CDF) allows students to comprehend the likelihood of a random variable falling within a certain range. Unlike the PMF which gives the chance of a single exact outcome, the CDF provides the probability that the variable will take on a value less than or equal to a certain level.

The CDF is particularly valuable when dealing with ranges or 'at least' type questions. For the Poisson distribution, the CDF at a point \( k \) can be calculated by summing up the PMF probabilities from 0 to \( k \. In the bankruptcy filings example, finding the range within which 75% of observations fall involves identifying \( a \) and \( b \) so that \( P(a \leq X \leq b) = 0.75 \).

Though this can be complex to calculate directly, using Poisson distribution tables or software simplifies the process. With the correct thresholds, the CDF can be effectively used to capture the 'middle' 75% of observed values, enabling practical predictions about the number of weekly bankruptcy filings.

Complementary Probability

The concept of complementary probability is an essential tool in probability theory and is particularly useful for questions that involve 'at least' scenarios. The complementary probability is based on the fact that the total probability for all outcomes is 1.

To find the probability of the complement of an event—meaning, the event does not happen—you subtract the probability of the event from 1. In the case of bankruptcy filings, we calculated \( P(X \geq 1) \) by subtracting \( P(X=0) \) from 1. This method often simplifies calculations and intuitively makes sense—knowing the chance of something not happening gives immediate insight into the chance that it will happen.

For students grappling with 'fewer than' or 'more than' questions, this approach can make it easier to find answers without the need to calculate every single outcome. The use of the complementary probability illustrates the relationship between the probability of occurrence and non-occurrence, promoting a holistic understanding of probabilistic events.