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Chapter 5: Problem 1

In analyzing hits by V-1 buzz bombs in World War II, South London was partitioned into 576 regions, each with an area of \(0.25 \mathrm{km}^{2}\). A total of 535 bombs hit the combined area of 576 regions. Assume that we want to find the probability that a randomly selected region had exactly two hits. In applying Formula \(5-9,\) identify the values of \(\mu, x,\) and \(e .\) Also, briefly describe what each of those symbols represents.

Short Answer

Expert verified
The probability is approximately 0.17.

Step by step solution

01

- Understand the parameters

In this problem, we analyze the hits by V-1 buzz bombs and need to find the probability of exactly two hits in a given region. This can be modeled using the Poisson distribution.
02

- Determine \(\mu\)

The mean number of hits per region, denoted as \(\mu\), is calculated by dividing the total number of hits by the total number of regions. Given 535 bombs over 576 regions: \[\mu = \frac{535}{576} 鈮� 0.928\]
03

- Identify x

Here, \(x\) represents the exact number of hits we are interested in, which is 2: \(x = 2\)
04

- Define e

The constant \(e\), representing the base of the natural logarithm, is approximately 2.71828: \(e 鈮� 2.71828\)
05

- Apply the Poisson Formula

Using the Poisson probability formula: \[ P(X = x) = \frac{\mu^x e^{-\mu}}{x!} \]
06

- Substitute the values into the formula

Plug the values into the Poisson formula: \[ P(X = 2) = \frac{0.928^2 \cdot e^{-0.928}}{2!} 鈮� \frac{0.861 \cdot 0.395}{2} 鈮� \frac{0.34}{2} 鈮� 0.17\]
07

- Describe the outcome

The probability of having exactly two hits in a randomly selected region is approximately 0.17.

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Key Concepts

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

Mean Number of Occurrences
The mean number of occurrences, in this case, represents the average number of bombs hitting a region. To calculate it, we divide the total number of hits by the number of regions. Given data from the exercise:
  • Total hits = 535
  • Total regions = 576
We get the mean, \( \mu \), as: \( \mu = \frac{535}{576} 鈮� 0.928 \). This indicates that, on average, each region was hit by roughly 0.928 bombs. This value of the mean is crucial for calculating probabilities using the Poisson distribution.
Poisson Probability Formula
The Poisson distribution helps us determine the likelihood of a given number of events occurring within a fixed interval. The formula used to calculate the Poisson probability is: \[ P(X = x) = \frac{\mu^x e^{-\mu}}{x!} \]
In this formula:
  • \( \mu \) = Mean number of occurrences
  • \( x \) = Exact number of occurrences
  • \( e \) = Base of the natural logarithm (approximately 2.71828)
  • \( x! \) = x factorial
This formula is particularly useful in scenarios where events happen independently and the average rate of occurrence is constant.
Probability Calculation
To find the probability that a randomly selected region had exactly two hits, we substitute the values into the Poisson formula:
\[ P(X = 2) = \frac{0.928^2 \cdot e^{-0.928}}{2!} \]
Breaking this down:
  • \(0.928^2 \) calculates the mean to the power of 2, which is approximately 0.861.
  • \( e^{-0.928} \) evaluates the exponential part, which is approximately 0.395.
  • 2! (2 factorial) equals 2.
Combining these values: \( \frac{0.861 \cdot 0.395}{2} 鈮� \frac{0.34}{2} 鈮� 0.17\)

Thus, the probability of exactly two hits is approximately 0.17, or 17%.
World War II Statistics
In World War II, analyzing bomb hits provided valuable insights into attack patterns and effectiveness. The data from V-1 buzz bomb impacts in South London were crucial for understanding and predicting enemy behavior. By partitioning the city into smaller regions, statisticians could apply models like the Poisson distribution to forecast future attacks. This method of statistical analysis ensured better strategic planning and resource allocation during the war, highlighting the importance of statistics in wartime decision-making.

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Most popular questions from this chapter

Assume that the Poisson distribution applies; assume that the mean number of Atlantic hurricanes in the United States is 6.1 per year, as in Example \(I\); and proceed to find the indicated probability. Hurricanes a. Find the probability that in a year, there will be no hurricanes. b. In a 55 -year period, how many years are expected to have no hurricanes? c. How does the result from part (b) compare to the recent period of 55 years in which there were no years without any hurricanes? Does the Poisson distribution work well here?

Find the probabilities and answer the questions. A survey sponsored by the Vision Council showed that \(79 \%\) of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 20 adults are randomly selected, find the probability that at least 19 of them need correction for their eyesight. Is 19 a significantly high number of adults requiring eyesight correction?

Refer to the accompanying table, which describes the numbers of adults in groups of five who reported sleepwalking (based on data from "Prevalence and Comorbidity of Nocturnal Wandering In the U.S. Adult General Population," by Ohayon et al., Neurology, Vol. 78, No. 20). $$\begin{array}{|c|c|} \hline x & P(x) \\ \hline 0 & 0.172 \\ \hline 1 & 0.363 \\ \hline 2 & 0.306 \\ \hline 3 & 0.129 \\ \hline 4 & 0.027 \\ \hline 5 & 0.002 \\ \hline \end{array}$$ a. Find the probability of getting exactly 4 sleepwalkers among 5 adults. b. Find the probability of getting 4 or more sleepwalkers among 5 adults. c. Which probability is relevant for determining whether 4 is a significantly high number of slecpwalkers among 5 adults: the result from part (a) or part (b)? d. Is 4 a significantly high number of sleepwalkers among 5 adults? Why or why not?

The binomial distribution applies only to cases involving two types of outcomes, whereas the multinomial distribution involves more than two categories. Suppose we have three types of mutually exclusive outcomes denoted by A, B, and C. Let \(P(\mathrm{A})=p_{1}, P(\mathrm{B})=p_{2},\) and \(P(\mathrm{C})=p_{3} .\) In \(n\) independent trials, the probability of \(x_{1}\) outcomes of type \(A, x_{2}\) outcomes of type \(B\), and \(x_{3}\) outcomes of type \(C\) is given by $$\frac{n !}{\left(x_{1}\right) !\left(x_{2}\right) !\left(x_{3}\right) !} \cdot p_{1}^{x_{1}} \cdot p_{2}^{x_{2}} \cdot p_{3}^{x_{3}}$$ A roulette wheel in the Venetian casino in Las Vegas has 18 red slots, 18 black slots, and 2 green slots. If roulette is played 15 times, find the probability of getting 7 red outcomes, 6 black outcomes, and 2 green outcomes.

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied. When conducting research on color blindness in males, a researcher forms random groups with five males in each group. The random variable \(x\) is the number of males in the group who have a form of color blindness (based on data from the National Institutes of Health). $$\begin{array}{|c|c|} \hline x & P(x) \\ \hline 0 & 0.659 \\ \hline 1 & 0.287 \\ \hline 2 & 0.050 \\ \hline 3 & 0.004 \\ \hline 4 & 0.001 \\ \hline 5 & 0+ \\ \hline \end{array}$$
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