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Chapter 6: Problem 27

6.27 The article "Probabilistic Risk Assessment of Infrastructure Networks Subjected to Hurricanes" (12th International Conference on Applications of Statistics and Probability in Civil Engineering, 2015) suggests a uniform distribution as a model for the actual landfall position of the eye of a hurricane. Consider the random variable \(x=\) distance of actual landfall from predicted landfall. Suppose that a uniform distribution on the interval that ranges from \(0 \mathrm{~km}\) to \(400 \mathrm{~km}\) is a reasonable model for \(x\). a. Draw the density curve for \(x\). b. What is the height of the density curve? c. What is the probability that \(x\) is at most \(100 ?\) d. What is the probability that \(x\) is between 200 and \(300 ?\) Between 50 and \(150 ?\) Why are these two probabilities equal?

Short Answer

Expert verified
In this problem, the random variable \(x\) follows a uniform distribution on the interval \((0, 400\mathrm{~km})\). The height of the density curve is \(\frac{1}{400}\). The probability that \(x\) is at most \(100\mathrm{~km}\) is \(0.25\), and the probabilities that \(x\) is between \(200\) and \(300\mathrm{~km}\) or between \(50\) and \(150\mathrm{~km}\) are both \(0.25\). These probabilities are equal because the intervals have the same length, and the uniform distribution weighs each interval of the same length with the same probability.

Step by step solution

01

Understand the uniform distribution

Let's consider a continuous uniform distribution with a range of \((a, b)\). The probability density function (pdf) of this distribution is given by: \[f(x) = \frac{1}{b-a}, \text{ for } a \le x \le b\] In our case, \(a=0\mathrm{~km}\) and \(b=400\mathrm{~km}\). Using this information, we can answer each question.
02

Draw the density curve

To draw the density curve of the uniform distribution on the interval \((0, 400\mathrm{~km})\), we need to first find the value of \(f(x)\), which is constant on this interval. We have already established that \(f(x) = \frac{1}{b-a}\) for a uniform distribution. In our case \(a=0\), \(b=400\), therefore: \[f(x) = \frac{1}{400-0} = \frac{1}{400}\] Now, we can plot the density curve which is a horizontal line at height \(\frac{1}{400}\) between \(x=0\) and \(x=400\). The area under the curve is equal to \(1\).
03

Determine the height of the density curve

As discussed above, the height of the density curve is given by the function \(f(x)\). For the given problem, we've found that: \[f(x) = \frac{1}{400}\] So, the height of the density curve is \(\frac{1}{400}\).
04

Calculate the probability (\(x \le 100\))

To find the probability that \(x\) is at most \(100\mathrm{~km}\), we need to calculate the area under the density curve from \(x=0\) to \(x=100\). Since this is a uniform distribution, the area can be calculated as: \[P(x \le 100\mathrm{~km}) = f(x) \times (100 - 0) = \frac{1}{400} \times 100\] \[P(x \le 100\mathrm{~km}) = 0.25\] So, there's a \(25\%\) chance that \(x\) is at most \(100\mathrm{~km}\).
05

Calculate probabilities for given intervals

We have been asked to find the probabilities that \(x\) is between \(200\) and \(300\mathrm{~km}\), and between \(50\) and \(150\mathrm{~km}\). Using the same method as in step 4, we can calculate these probabilities: (a) For the interval \(200 \le x \le 300\mathrm{~km}\): \[P(200\mathrm{~km} \le x \le 300\mathrm{~km}) = f(x) \times (300 - 200) = \frac{1}{400} \times 100\] \[P(200\mathrm{~km} \le x \le 300\mathrm{~km}) = 0.25\] (b) For the interval \(50 \le x \le 150\mathrm{~km}\): \[P(50\mathrm{~km} \le x \le 150\mathrm{~km}) = f(x) \times (150 - 50) = \frac{1}{400} \times 100\] \[P(50\mathrm{~km} \le x \le 150\mathrm{~km}) = 0.25\] These two probabilities are equal because the intervals have the same length (100 km), and the uniform distribution weighs each interval of the same length with the same probability.

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Key Concepts

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

Probability Density Function
In the context of uniform distribution, the probability density function (pdf) provides a mathematical description of the relative likelihood for a continuous random variable to take on a given value.

Imagine the pdf as a formula that gives the height of a line along the range of possible values a variable can assume. In the case of a uniform distribution, this line is flat, indicating that all values are equally likely within the specified range. The formula for the pdf of a uniform distribution is given by \( f(x) = \frac{1}{b-a} \) for \( a \le x \le b \), where \(a\) is the minimum value, and \(b\) is the maximum value of \(x\).
  • For our exercise, \(a=0\) km and \(b=400\) km, implying a flat line at \( \frac{1}{400} \) over this interval.
It's crucial to grasp that the area under the entire density curve always equals 1, reflecting the total probability of all possible outcomes. This area represents the certainty that the variable will assume some value within the bounds of \(a\) and \(b\).
Continuous Random Variable
A continuous random variable is a variable that can take on an infinite number of possible values within a given range. Unlike discrete random variables that have countable outcomes, continuous random variables deal with measurements and can include every fraction or decimal within an interval.

Within our hurricane landfall example, \(x\)鈥攖he distance of actual landfall from predicted landfall鈥攊s a continuous random variable because it can be any value between 0 km and 400 km.

Characteristics of a Continuous Random Variable

Here, the continuity reflects the real-world phenomena where precise distances are not limited to whole numbers but can take any fractional value within the possible range.
Statistical Probability
The concept of statistical probability involves quantifying the likelihood of an event occurring based on data, analysis, and a given statistical model.

Probability values are expressed between 0 and 1, with 0 indicating that an event cannot occur, and 1 indicating certainty that an event will occur.

Calculating Probability in Uniform Distribution

In uniform distribution, calculating the probability of an event, such as \(x\) being between two values, is straightforward鈥攕imply multiply the height of the pdf by the width of the interval of interest. For instance, the probability of \(x\) being less than 100 km (\( P(x \le 100\,\text{km}) \) ) is 0.25 due to the flat nature of the distribution curve and the equal likelihood of every 100 km segment within the 0 to 400 km range.

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Most popular questions from this chapter

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection? b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail the inspection? c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation? d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (randomly select one of the five answers) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Calculate the variance and standard deviation of \(x\). d. Based on your answers to Parts \((\mathrm{b})\) and \((\mathrm{c}),\) is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

Twenty-five percent of the customers of a grocery store use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. Calculate \(p(2),\) that is, \(P(x=2)\). b. Calculate \(P(x \leq 1)\). c. Calculate \(P(x \geq 2)\). (Hint: Make use of your answer to Part (b).) d. Calculate \(P(x \neq 2)\).

A coin is flipped 25 times. Let \(x\) be the number of flips that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. Suppose that a coin is not fair and that \(P(\mathrm{H})=0.9\) What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(\mathrm{H})=0.1 ?\) c. What is the probability of judging a coin fair if \(P(\mathrm{H})=0.6 ?\) if \(P(\mathrm{H})=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and \((b)\) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

An automobile manufacturer is interested in the fuel efficiency of a proposed new car design. Six nonprofessional drivers were selected, and each one drove a prototype of the new car from Phoenix to Los Angeles. The resulting fuel efficiencies \((x,\) in miles per gallon \()\) are: $$ \begin{array}{llllll} 27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6 \end{array} $$ The normal scores for a sample of size 6 are $$ \begin{array}{llllll} -1.282 & -0.643 & -0.202 & 0.202 & 0.643 & 1.282 \end{array} $$ a. Construct a normal probability plot for the fuel efficiency data. Does the plot look linear? b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the fuel efficiency distribution is approximately normal.
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