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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

Suppose that the amount of time spent by a statistical consultant with a client at their first meeting is a random variable that has a normal distribution with a mean value of 60 minutes and a standard deviation of 10 minutes. a. What is the probability that more than 45 minutes is spent at the first meeting? b. What amount of time is exceeded by only \(10 \%\) of all clients at a first meeting?

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Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part.): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts

Let \(x\) denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of \(x\) are 266 days and 16 days, respectively. Suppose that the probability distribution of \(x\) is (approximately) normal. a. What is the probability that the duration of a randomly selected pregnancy is between 250 and 300 days? b. What is the probability that the duration is at most 240 days? c. What is the probability that the duration is within 16 days of the mean duration? d. A "Dear Abby" newspaper column dated January 20, 1973 , contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the navy, occurred 310 days before the birth of her child.) What is the probability that the duration of pregnancy is at least 310 days? Does this probability make you skeptical of the claim? e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months ( 275 days). This restriction is designed to ensure that benefits are only paid if conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275 -day requirement?
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