Problem 124 - Applied Statistics and Probability for Engineers

Question

The distance between major cracks in a highway follows an exponential distribution with a mean of five 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 five-mile stretches of the highway? (f) Given that there are no cracks in the first five miles inspected, what is the probability that there are no major cracks in the next 10 miles inspected?

Step-by-Step Solution

Answer

(a) \(0.1353\); (b) \(0.2707\); (c) \(5\) miles; (d) \(0.0616\); (e) \(0.1353\); (f) \(0.1353\).

Step 1: Identify Distribution Parameters

The problem follows an exponential distribution with a mean (\(\lambda^{-1}\)) of 5 miles. Thus, the rate parameter \(\lambda = \frac{1}{5} = 0.2\) (cracks per mile).

Step 2: Solve Part (a) Using Exponential Distribution

The probability that there are no major cracks in a 10-mile stretch is calculated using the exponential distribution formula, \( P(X > x) = e^{-\lambda x} \). For \(x = 10\), \( P(X > 10) = e^{-0.2 \times 10} = e^{-2} \approx 0.1353 \).

Step 3: Solve Part (b) Using Poisson Distribution

The probability of encountering exactly two major cracks in 10 miles is found using the Poisson distribution formula, \( P(k) = \frac{(\lambda x)^k e^{-\lambda x}}{k!} \) with \(x = 10\) and \(k = 2\). It calculates to \( P(2) = \frac{(2)^{2} e^{-2}}{2!} = \frac{4 e^{-2}}{2} \approx 0.2707 \).

Step 4: Calculate Standard Deviation (Part c)

The standard deviation for an exponential distribution is equal to its mean: \( \sigma = \lambda^{-1} = 5 \) miles.

Step 5: Solve Part (d) Using Exponential Distribution

The probability of the first major crack occurring between 12 and 15 miles is given by \( P(12 < X < 15) = P(X > 12) - P(X > 15) \). This is calculated as \( e^{-0.2 \times 12} - e^{-0.2 \times 15} \approx 0.0616 \).

Step 6: Solve Part (e) Using Independence of Sections

The probability of no major cracks in a 5-mile stretch is \( e^{-\lambda \times 5} = e^{-1} \approx 0.3679 \). Since the stretches are independent, the probability for two stretches is \( (e^{-1})^2 = e^{-2} \approx 0.1353 \).

Step 7: Solve Part (f) Using Conditional Probability

With no cracks in the first 5 miles, the probability of no major cracks in the next 10 miles, independent of the first, remains \( P(X > 10) = e^{-2} \approx 0.1353 \).

Key Concepts

Probability Calculation

Probability calculation is a fundamental concept in statistics essential for determining the likelihood of an event occurring. In many problems, it is about finding the probability of one or more events happening in a given scenario.

When dealing with the exponential distribution, which is prominent in the probability of time between events, the formula used can be expressed as \( P(X > x) = e^{-eta x} \).

• Here, \( e \) is the base of the natural logarithm.
• \( \beta \) is the rate parameter, the reciprocal of the mean.
• \( x \) is the value we are calculating the probability for.

Understanding these parameters helps calculate probabilities for sections of time or length in which an event, like finding no cracks on a highway, does not occur.

For instance, calculating the probability of no cracks in a 10-mile stretch, involves finding \( P(X > 10) = e^{-0.2 imes 10} \). This results in approximately 0.1353, or a 13.53% chance.

Poisson Distribution

The Poisson distribution is frequently used to model the number of events occurring within a fixed interval of time or space. It is particularly useful when these events are independent and occur with a known constant mean rate. The formula for the probability of observing \( k \) events is:\[ P(k) = \frac{(eta x)^k e^{-\beta x}}{k!} \]

• \( k \) represents the number of events.
• \( \beta \) is the rate parameter.
• \( x \) is the interval length, whether it is time or space.

When applied to our problem, the Poisson distribution helps calculate the probability of exactly two major cracks within a 10-mile stretch. Plugging the values \( \beta = 0.2 \), \( x = 10 \), and \( k = 2 \) into the formula, we get \( P(2) = \frac{4 e^{-2}}{2} \). This simplifies to approximately 0.2707, or a 27.07% chance. This demonstrates how the Poisson distribution effectively models counts of occurrences over a certain interval.

Standard Deviation

The concept of standard deviation represents the amount of variation or dispersion in a set of values. In an exponential distribution, this concept holds a unique characteristic. Notably, for the exponential distribution, the standard deviation is equal to the mean.

Given that the mean is 5 miles, the standard deviation is thus also 5 miles. This equality reflects the unique nature of exponential distributions and can simplify calculations in certain contexts.

The standard deviation helps in understanding how spread out the distances between cracks are around the mean distance of 5 miles. A low standard deviation would indicate that the distances are clustered closely around the mean, whereas a high standard deviation would show a wider range of distances between cracks.

Independence in Probability

Independence in probability refers to scenarios where the occurrence of one event does not affect the probability of another event. This principle is vital in simplifying probability calculations, especially in scenarios involving multiple sections or periods.

For instance, consider calculating the probability of no cracks in two separate 5-mile stretches. If these events are independent, the probability of no cracks in each section can multiply to provide the overall probability.

• The probability of no cracks in a 5-mile stretch is \( e^{-1} \).
• Since both stretches are independent, the combined probability is \( (e^{-1})^2 \), or \( e^{-2} \), approximately 0.1353.

Moreover, this independence allows us to assess situations like inspecting two successive sections without overlaps in event occurrence. It highlights how one section of inspection being crack-free does not alter the probability for the succeeding section.