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Based on an analysis of sample data, the article "Pedestrians' Crossing Behaviors and Safety at Unmarked Roadways in China" (Accident Analysis and Prevention, 2011: 1927-1936) proposed the pdf \(f(x)=.15 e^{-.15(x-1)}\) when \(x \geq 1\) as a model for the distribution of \(X=\) time (sec) spent at the median line. a. What is the probability that waiting time is at most \(5 \mathrm{sec}\) ? More than \(5 \mathrm{sec}\) ? b. What is the probability that waiting time is between 2 and \(5 \sec\) ?

Short Answer

Expert verified
a. At most 5 sec: 0.4512; More than 5 sec: 0.5488. b. Between 2 and 5 sec: 0.2905.

Step by step solution

01

Understanding the Problem

We need to find probabilities related to the time a pedestrian spends at the median line, given the probability density function (pdf) \(f(x) = 0.15 e^{-0.15(x-1)}\) for \(x \geq 1\). This is an exponential distribution with a rate \(\lambda = 0.15\), and the waiting time \(X\) is shifted by 1 second.
02

Finding Probability Up to 5 Seconds

To find the probability that \(X\) is at most 5 seconds, calculate the cumulative distribution function (CDF) at \(x = 5\):\[ P(X \leq 5) = \int_{1}^{5} 0.15 e^{-0.15(x-1)} \, dx \]This integral computes the area under the curve from \(x = 1\) to \(x = 5\). The integral of an exponential function \(e^{ax}\) is \(\frac{1}{a} e^{ax}\). Thus,\[ P(X \leq 5) = \left[ -e^{-0.15(x-1)} \right]_{1}^{5} = 1 - e^{-0.6} \]Evaluating the expression gives \[ P(X \leq 5) \approx 0.4512 \].
03

Finding Probability More Than 5 Seconds

The probability that \(X\) is more than 5 seconds is simply the complement of the probability that it is at most 5 seconds:\[ P(X > 5) = 1 - P(X \leq 5) = 1 - 0.4512 = 0.5488 \].
04

Finding Probability Between 2 and 5 Seconds

To find the probability that \(X\) is between 2 and 5 seconds, we compute:\[ P(2 \leq X \leq 5) = \int_{2}^{5} 0.15 e^{-0.15(x-1)} \, dx \]This results in:\[ P(2 \leq X \leq 5) = \left[ -e^{-0.15(x-1)} \right]_{2}^{5} = e^{-0.15(2-1)} - e^{-0.15(5-1)} \]Simplifying, we get:\[ P(2 \leq X \leq 5) = e^{-0.15} - e^{-0.6} \approx 0.7417 - 0.4512 = 0.2905 \].

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Key Concepts

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

Probability Density Function
In statistics, the Probability Density Function (pdf) is a fundamental concept used to describe the likelihood of a random variable taking on a particular value. For continuous random variables, like the time a pedestrian waits at the median line, the pdf provides a way to visualize the distribution of data across various values. When we talk about a pdf, we're essentially looking at a curve that shows which values a variable is likely to take within a certain range.

A characteristic feature of the pdf is that the total area under the curve is equal to one. This implies that the combined probabilities of all the possible outcomes is 100%. In the exercise above, we're working with the pdf given by the function \( f(x) = 0.15 e^{-0.15(x-1)} \) when \( x \geq 1 \). This represents an exponential distribution, a common model for the time between events in a process where events occur continuously and independently at a constant average rate.

The rate or "lambda" (\( \lambda \)) in our function is 0.15. This indicates how quickly the probability decreases as time progresses. The exponential distribution is often used in survival analysis, reliability theory, and in this case, to model waiting times. Understanding the shape and implications of the pdf can help make predictions about the waiting time at pedestrian medians.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is a crucial tool for understanding the probability that a random variable will fall within a certain range. Unlike the pdf, which only shows the likelihood of a random variable being equal to a specific value, the CDF provides the probability that a random variable is less than or equal to a given number.

In mathematical terms, the CDF is the integral of the pdf from the lower bound of the domain to the point of interest. For instance, to find the probability that our waiting time \( X \) is at most 5 seconds, we need the cumulative probability up to \( x = 5 \). This is denoted as \( P(X \leq 5) \), and is calculated using the integral of the pdf from \( x = 1 \) (the start of our domain) to \( x = 5 \).

Through integration, the exercise tells us that \( P(X \leq 5) = 1 - e^{-0.6} \). This result, approximately 0.4512, represents the area under the pdf curve from \( x = 1 \) to \( x = 5 \), showing the likelihood of waiting time being 5 seconds or less.
  • The CDF becomes particularly useful for finding probabilities over intervals, such as the probability for times more than 5 seconds or between two specific times, by subtracting relevant cumulative probabilities from each other.
Integration of Exponential Functions
One of the vital mathematical skills when working with exponential distributions is the ability to integrate exponential functions. This helps to calculate probabilities over specific intervals, as shown in our exercise.

The process of integration allows us to find the area under the probability density curve between two points and can be quite straightforward with exponential functions because of their mathematical properties. The general integral formula for an exponential function \( e^{ax} \) is \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} \) plus a constant. However, when solving definite integrals like those encountered in probability (where specific limits are given), this constant is not included in the solution between the specified limits.

In the original exercise, integration was used to solve for several probabilities, including the probability of the waiting time being between 2 and 5 seconds. By finding \( \int_{2}^{5} 0.15 e^{-0.15(x-1)} \, dx \), the area under the curve was calculated for the interval \([2, 5]\).

Understanding how to perform integration, especially how to set up the limits and apply the correct constants, is necessary for successfully navigating problems involving exponential distributions. This integration skill is crucial for interpreting and analyzing real-world data modeled by these functions.

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Most popular questions from this chapter

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