/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 The waiting time, in hours, betw... [FREE SOLUTION] | 91Ó°ÊÓ

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The waiting time, in hours, between successive speeders spotted by a radar unit is a continuous random variable with cumulative distribution function $$ F(x)=\left\\{\begin{array}{ll} 0_{+} & x<0, \\ 1-e^{-k x}, & x \geq 0. \end{array}\right. $$ Find the probability of waiting less than 12 minutes between successive speeders (a) using the cumulative distribution function of \(X\); (b) using the probability density function of \(X\).

Short Answer

Expert verified
The probability of waiting less than 12 minutes between successive speeders, both using the cumulative distribution function and the probability density function of \(X\), is \(1 - e^{-0.2k}\) in each case.

Step by step solution

01

Convert time to relevant units

Firstly, convert the time from minutes to hours as the distribution given in the problem is in hours. Since there are 60 minutes in an hour, you can convert 12 minutes to hours by dividing by 60 to get \(x = \frac{12}{60} = 0.2\) hours.
02

Substitute the time into cumulative distribution function

Next, you substitute \(x = 0.2\) into the cumulative distribution function (CDF). The function is given by \(F(x)=1-e^{-k x}\). The problem does not specify the value for \(k\), so the answer will be left in terms of \(k\). Thus, the probability of waiting less than 12 minutes using the cumulative distribution function of \(X\) is \(F(0.2) = 1 - e^{-k * 0.2}\).
03

Derive the probability density function from cumulative distribution function

To solve the problem using the probability density function (PDF), you have to derive the PDF from the CDF. The derivative of \(F(x) = 1 - e^{-kx}\) with respect to \(x\) is \(f(x) = k * e^{-kx}\).
04

Calculate Integral

Now the probability of waiting less than 12 minutes using the PDF is obtained by integrating the PDF between 0 and 0.2 (these limits are use because the probability that the waiting time will be less than 12 minutes is asked). Thus, \(P[X<0.2] = \int_0^{0.2} k*e^{-kx} dx\). After integrating and substitizing the limits, this simplifies to \(1 - e^{-k*0.2}\). You will see that this is the same as the result obtained using the CDF.

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function, often abbreviated as CDF, is a crucial concept in probability theory, especially in dealing with continuous random variables. In essence, the CDF, denoted as \( F(x) \), gives the probability that a random variable \( X \) will take a value less than or equal to \( x \). This makes it a non-decreasing function that ranges from 0 to 1.

For continuous random variables, the CDF is constructed by integrating the probability density function (PDF) from negative infinity to \( x \). However, when a specific functional form is provided, like in this problem where \( F(x) = 1 - e^{-kx} \) for \( x \geq 0 \), it represents the probability that waiting times will be less than a particular value of \( x \).

The CDF is particularly useful because it automatically accounts for the total probability up to point \( x \). For instance, in the original problem, using the CDF directly calculates the probability of waiting less than 12 minutes by inserting \( 0.2 \) hours into the function, resulting in \( F(0.2) = 1 - e^{-k \cdot 0.2} \), without requiring any further integration.
Probability Density Function
A probability density function, or PDF, essentially describes the likelihood of the random variable taking on a particular value. For continuous random variables, unlike discrete ones, the PDF does not give probability directly, but it indicates how dense the probability is at any given point.

In mathematical terms, the PDF is the derivative of the CDF, \( f(x) = \frac{d}{dx}F(x) \). For the example given, we compute the PDF by differentiating \( F(x) = 1 - e^{-kx} \), resulting in \( f(x) = ke^{-kx} \). This expression conveys the probability density relative to each point in the distribution line above zero.

In practical applications, the PDF is used with an integral to establish actual probabilities across intervals. In our problem, once we obtain the PDF, to find the probability of a waiting time less than 12 minutes, we integrate it from zero to 0.2 hours. Interestingly, it results in the same calculation with CDF, demonstrating a critical link between these two functions.
Integration in Probability
In probability, integration plays a pivotal role, particularly when dealing with probability density functions (PDFs). It is used to derive meaningful probabilities from a PDF by integrating it across a specified interval. This method is especially vital for continuous random variables, where probability at a singular point is zero, but non-zero over an interval.

When we integrate the PDF over an interval \([a, b]\), we obtain the probability that the random variable falls within this range, mathematically represented as \( P(a \leq X \leq b) = \int_a^b f(x) \, dx \). In the context of our problem, integration is used to determine the probability of the radar unit spotting a speeder within 0 to 0.2 hours, computed as \( \int_0^{0.2} k e^{-kx} \, dx \).

This calculated result aligns perfectly with what we get using the CDF, confirming the integral's role in ensuring the validity of probabilistic models. It unravels the cumulative effects of infinitesimal probabilities spread across a continuum and transforms them into substantial probability that decision makers and analysts can use.

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Most popular questions from this chapter

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An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let \(X\) be the number of months between successive payments. The cumulative distribution function of \(X\) is $$ F(x)=\left\\{\begin{array}{ll} 0, & \text { if } x<1 \\ 0.4, & \text { if } 1 \leq x<3, \\ 0.6, & \text { if } 3 \leq x<5, \\ 0.8, & \text { if } 5 \leq x<7, \\ 1.0, & \text { if } x \geq 7. \end{array}\right. $$ (a) What is the probability mass function of \(X ?\) (b) Compute \(\mathrm{P}(4

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