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4.7 .8 WP The distance between major cracks in a highway follows an exponential distribution with a mean of 5 miles. a. What is the probability that there are no major cracks in a 10 -mile stretch of the highway? b. What is the probability that there are two major cracks in a 10 -mile stretch of the highway? c. What is the standard deviation of the distance between major cracks? d. What is the probability that the first major crack occurs between 12 and 15 miles of the start of inspection? e. What is the probability that there are no major cracks in two separate 5 -mile stretches of the highway? Given that there are no cracks in the first 5 miles inspected, what is the probability that there are no major cracks in the next 10 miles inspected?

Short Answer

Expert verified
a) 0.1353; b) 0.2707; c) 5 miles; d) 0.0323; e) 0.1353, remaining 10 miles with no cracks: 0.1353.

Step by step solution

01

Understand the Exponential Distribution

The problem states that the distance between cracks follows an exponential distribution with a mean of 5 miles. The mean of an exponential distribution is given by \( \frac{1}{\lambda} \). Therefore, \( \lambda = \frac{1}{5} \) or 0.2 cracks per mile.
02

Calculate Probability of No Major Cracks in 10 Miles

We use the exponential distribution formula \( P(X > x) = e^{-\lambda x} \). For the probability of no major cracks in a 10-mile stretch, we set \( x = 10 \): \( P(X > 10) = e^{-\frac{1}{5} \cdot 10} = e^{-2} \approx 0.1353 \).
03

Apply Poisson Distribution for Two Cracks in 10 Miles

Since cracks are rare events in a specific interval, we use the Poisson distribution. The parameter is \( \lambda = 10 \cdot \frac{1}{5} = 2 \). The probability formula is \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \). Thus, for \( k = 2 \): \( P(X = 2) = \frac{e^{-2} \cdot 2^2}{2!} = e^{-2} \cdot 2 \approx 0.2707 \).
04

Determine Standard Deviation of Distance Between Cracks

For an exponential distribution, the standard deviation is equal to the mean. Thus, the standard deviation is 5 miles.
05

Probability of First Crack Between 12 and 15 Miles

Using the exponential distribution formula: \( P(12 < X < 15) = e^{-\frac{12}{5}} - e^{-\frac{15}{5}} \). Calculating each: \( e^{-\frac{12}{5}} \approx 0.0821 \) and \( e^{-\frac{15}{5}} \approx 0.0498 \), resulting in \( 0.0821 - 0.0498 = 0.0323 \).
06

No Cracks in Two Separate 5-Mile Stretches

Assuming independence, for each 5-mile stretch, \( P(X > 5) = e^{-1} \approx 0.3679 \). So, \( (0.3679)^2 \approx 0.1353 \) for the probability of no cracks in both stretches.
07

No Cracks in Next 10 Miles Given No Cracks in First 5 Miles

Use the memoryless property of the exponential distribution. The probability of no cracks in the 10 miles is independent of the previous 5 miles. \( P(X > 10) = e^{-2} \approx 0.1353 \).

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Key Concepts

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

Poisson Distribution
The Poisson distribution is a helpful model for probability calculations, especially when dealing with rare events within a fixed interval. For instance, the probability of finding cracks in a highway over a set distance can be modeled by this distribution.

It is characterized by the parameter \( \lambda \), which is the average number of occurrences within the interval. In our example, the average number of cracks in a 10-mile stretch is 2, calculated by multiplying the rate of cracks per mile (\( \lambda = 0.2 \) cracks per mile) by the interval in miles (10 miles).

The formula for the Poisson probability mass function is:
\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Where:
  • \( e \) is the base of the natural logarithm (approximately 2.718)
  • \( \lambda \) is the average number of occurrences in the interval
  • \( k \) is the actual number of occurrences you're calculating for
  • \( k! \) is the factorial of \( k \)
Probability Calculations
Probability calculation is the process of determining the likelihood of one or more events occurring. In the context of our problem, probability can be calculated using various models such as the exponential and Poisson distributions.

For instance, when calculating the probability that no major cracks will occur over a stretch of highway, the exponential distribution formula can be used:
\[ P(X > x) = e^{-\lambda x} \]
This formula helps in understanding the probability of an event exceeding a certain value, such as "no cracks in 10 miles," by determining the joint likelihood using the decay rate represented by \( \lambda \).

Probability calculations allow us to make informed predictions about random events, essential for decision-making and risk assessment.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It gives us insight into how much measurements deviate from the mean of the distribution. In the context of exponential distributions, like the one describing the distance between cracks, the standard deviation is conveniently equal to the mean.

So, for our highway example, with a mean distance of 5 miles between cracks, the standard deviation is also 5 miles.

Knowing the standard deviation helps in assessing the spread or variability of the dataset. It provides a quantitative idea of the expected deviation from the average, allowing for better-informed engineering assessments when planning road maintenance, for instance.
Memoryless Property
One of the unique features of the exponential distribution is its memoryless property. This implies that future probabilities are independent of past events. If you haven’t encountered a crack over a 5-mile stretch, the probability of encountering or not encountering one over the next 10 miles remains unchanged.

Mathematically, this is expressed as:
\[ P(X > s + t | X > s) = P(X > t) \]
This property is especially useful in real-world scenarios where the likelihood of an event isn't influenced by previous durations, which simplifies the calculation of probabilities over subsequent intervals immensely.

In our example, knowing the memoryless property allows us to compute the probability of no further cracks over an additional 10 miles regardless of the absence of cracks in previous stretches, maintaining focus on future projections alone.

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