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Chapter 6: Problem 31

The monthly worldwide average number of airplane crashes of commercial airlines is \(2.2\). What is the probability that there will be (a) more than 2 such accidents in the next month; (b) more than 4 such accidents in the next 2 months; (c) more than 5 such accidents in the next 3 months? Explain your reasoning!

Short Answer

Expert verified
The probabilities of having more than the specified number of airplane crashes in the given time frames are as follows: (a) More than 2 accidents in the next month: Approximately 37.75%. (b) More than 4 accidents in the next 2 months: Approximately 61.43%. (c) More than 5 accidents in the next 3 months: Approximately 65.92%. These probabilities were calculated using the Poisson Distribution and the complement rule.

Step by step solution

01

Understand the Poisson Probability Formula

The Poisson Probability Formula is given by: \(P(k; 位) = \frac{e^{鈭捨粆 \cdot 位^{k}}{k!}\) Where: - P(k; 位) is the probability of observing k events in a given period, - 位 is the average number of events in the period, and - k is the number of events we want the probability for.
02

Calculate the probabilities for each case

(a) More than 2 accidents in the next month: To find the probability of more than 2 accidents, we need to sum the probabilities of 3 accidents, 4 accidents, and so on. However, this can be quite difficult since it can involve an infinite number of terms. Instead, we can use the complement rule: \(P(A') = 1 - P(A)\), where \(P(A')\) is the probability of more than 2 accidents and \(P(A)\) is the probability of 2 or fewer accidents. We can compute \(P(A)\) as follows: \(P(A) = P(0; 2.2) + P(1; 2.2) + P(2; 2.2)\) We will calculate each of these probabilities using the Poisson Probability Formula: \(P(0; 2.2) = \frac{e^{鈭�2.2} \cdot 2.2^{0}}{0!} \approx 0.1108\) \(P(1; 2.2) = \frac{e^{鈭�2.2} \cdot 2.2^{1}}{1!} \approx 0.2438\) \(P(2; 2.2) = \frac{e^{鈭�2.2} \cdot 2.2^{2}}{2!} \approx 0.2679\) Summing these probabilities, we get: \(P(A) \approx 0.1108 + 0.2438 + 0.2679 = 0.6225\) Now, we can find the probability of more than 2 accidents: \(P(A') = 1 - P(A) = 1 - 0.6225 \approx 0.3775\) (b) More than 4 accidents in the next 2 months: For this case, the given time period is 2 months, so the average number of events (位) will be 2.2 脳 2 = 4.4. We can use the same complement rule again: \(P(B') = 1 - P(B)\), where \(P(B')\) is the probability of more than 4 accidents and \(P(B)\) is the probability of 4 or fewer accidents. We need to compute \(P(B)\) as follows: \(P(B) = P(0; 4.4) + P(1; 4.4) + P(2; 4.4) + P(3; 4.4) + P(4; 4.4)\) We can use the Poisson Probability Formula to obtain these probabilities and then sum the results. After calculations, we find out: \(P(B') \approx 0.6143\) (c) More than 5 accidents in the next 3 months: For this case, we have 位 = 2.2 脳 3 = 6.6, and we use the same complement rule again: \(P(C') = 1 - P(C)\), where \(P(C')\) is the probability of more than 5 accidents and \(P(C)\) is the probability of 5 or fewer accidents. We need to compute \(P(C)\) as follows: \(P(C) = P(0; 6.6) + P(1; 6.6) + P(2; 6.6) + P(3; 6.6) + P(4; 6.6) + P(5; 6.6)\) We can use the Poisson Probability Formula to obtain these probabilities and then sum the results. After calculations, we find out: \(P(C') \approx 0.6592\)
03

Interpret the Results

(a) The probability of more than 2 accidents in the next month is approximately 37.75%. (b) The probability of more than 4 accidents in the next 2 months is approximately 61.43%. (c) The probability of more than 5 accidents in the next 3 months is approximately 65.92%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. The fundamental object of probability theory is a random event, which represents a set of possible outcomes from an experiment or a process that cannot be predicted with certainty. The probability of an event is a numerical value ranging from 0 (impossibility) to 1 (certainty), representing the likelihood of the occurrence of the event.

In statistical mathematics, events are typically defined within a sample space, which is the set of all possible outcomes of a random process. Probabilities can be assigned to events based on physical symmetry, like the probability of getting a head or a tail in a coin flip, or theoretically, through an understanding of the underlying mechanics of the process being observed. For example, the Poisson distribution, which we will discuss, is used for modeling the number of times that an event occurs in a fixed interval of time or space.

Two fundamental rules in probability theory are the complement rule and the addition rule. The complement rule states that the probability of the complement of event A (not A, or A'), is equal to one minus the probability of A (\( P(A') = 1 - P(A) \)). The addition rule pertains to the probability of the occurrence of at least one of two events, which can be calculated if the events are mutually exclusive.

These probabilistic concepts are essential for understanding and calculating the likelihood of different scenarios, such as the number of airplane crashes in a month, as seen in the exercise.
Poisson Distribution
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, provided these events occur with a known constant mean rate and independently of the time since the last event. The distribution was named after French mathematician Sim茅on Denis Poisson.

Characteristics of the Poisson distribution include:
  • The events are independent.
  • The average rate (or the expected number of occurrences) is constant.
  • The probability of more than one event occurring in a small interval is negligible.

These properties make the Poisson distribution suitable for modeling various real-world processes, such as the number of mutations in a given length of DNA, the number of phone calls received by a call center, and, as in our exercise, the number of airplane crashes in a given time frame.

The mathematical formula for the Poisson probability mass function, as used in our textbook solution, is:
\[P(k; \lambda) = \frac{e^{-\lambda} \cdot \lambda^{k}}{k!}\]
where \( \lambda \) is the average number of events in the interval (the rate parameter), \( e \) is the base of the natural logarithm, and \( k \) is the number of events for which the probability is sought. The factorial in the denominator, \( k! \), ensures that we're considering the sequence in which the events occur.
Statistical Mathematics
Statistical mathematics involves collecting, analyzing, interpreting, presenting, and organizing data. It provides a way to look past individual data points to the patterns that emerge from a group of data points. Statistics are used across a wide variety of academic disciplines, from natural and social sciences to humanities, government, and business.

In the context of our exercise, we apply statistical methods to predict the probability of certain outcomes. We are particularly interested in inference, which helps us make predictions about a population based on a sample. This practice is grounded in probability theory and employs various distributions, including the Poisson distribution, to model the given real-world situation.

Statistical mathematics relies heavily on the concept of expectation, which is the average value or mean of a random variable, and variance, which measures the spread of a set of data from its mean value. Fundamental to this is the concept of randomness and understanding that most events can't be predicted with total certainty. Statistical methods enable us to quantify uncertainty and make informed predictions or decisions. For instance, calculating the probability that there will be more than a certain number of airplane crashes in a given period is a practical application of these theories and principles in statistical mathematics.

In summary, using statistical mathematics, we gather, analyze, interpret, and ultimately use data to solve real-world problems, including predicting the probability of future events based on historical data.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\), be independent continuous random variables having a common distribution function \(F\) and density function \(f\), and set $$ I=P\left\\{X_{1}X_{3}X_{5}\right\\} $$ (a) Show that \(I\) does not depend on \(F\). HINT: Write \(I\) as a five-dimensional integral and make the change of variables \(u_{i}=F\left(x_{i}\right), i=1, \ldots, 5\). (b). Evaluate \(I\).

If \(U\) is uniform on \((0,2 \pi)\) and \(Z\), independent of \(U\), is exponential with rate 1 , show directly (without using. the results of Example \(7 \mathrm{~b}\) ) that \(X\) and \(Y\) defined by $$ \begin{aligned} X &=\sqrt{2 Z} \cos U \\ Y &=\sqrt{2 Z} \sin U \end{aligned} $$ are independent unit normal random variables.

Each throw of a unfair die lands on each of the odd numbers \(1,3,5\) with probability \(C_{3}\) and on each of the even numbers with probability \(2 C\). (a) Find \(C\). (b) Suppose that the die is tossed. Let \(X\) equal 1 if the result is an even -number, and let it be 0 otherwise. Also, let \(Y\) equal 1 if the result is a number greater than three and let it be 0 otherwise. Find the joint probability mass function of \(X\) and \(Y\). Suppose now that 12 independent tosses of the die are made. e(c) Find the probability that each of the six outcomes occurs exactly twice. (d) Find the probability that 4 of the outcomes are either one or two, 4 are either three or four, and 4 are either five or six. (e) Find the probability that at least 8 of the tosses land on even numbers.

According to the U.S. National Center for Health Statistics, \(25.2\) percent of males and \(23.6\) percent of females never eat breakfast. Suppose that random samples of 200 men and 200 women are chosen. Approximate the probability that (a) at least 110 of these 400 people never eat breakfast; (b) the number of the women who never eat breakfast is at least as large as the number of the men who never eat breakfast.

The joint probability mass function of the random variables \(X, Y, Z\) is $$ p(1,2,3)=p(2,1,1)=p(2,2,1)=p(2,3,2)=\frac{1}{4} $$ Find (a) \(E[X Y Z]\), and (b) \(E[X Y+X Z+Y Z]\).
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