/*! 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 90 Suppose \(X\) has an exponential... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose \(X\) has an exponential distribution with \(\lambda=2\). Determine the following: (a) \(P(X \leq 0)\) (b) \(P(X \geq 2)\) (c) \(P(X \leq 1)\) (d) \(P(1

Short Answer

Expert verified
(a) 0. (b) \(e^{-4}\). (c) \(1-e^{-2}\). (d) \(e^{-2}-e^{-4}\). (e) \(x \approx 0.0256\).

Step by step solution

01

Understand the Exponential Distribution

An exponential distribution is often used to model the time until an event occurs. The probability density function (PDF) for an exponential distribution with rate parameter \( \lambda \) is defined as \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). The cumulative distribution function (CDF) is given by \( F(x) = 1 - e^{-\lambda x} \). Given \( \lambda = 2 \), we can solve each part of the question.
02

Calculate (a) P(X ≤ 0)

For an exponential distribution, \( P(X \leq 0) = F(0) \). Using the CDF, we find that \( F(0) = 1 - e^{-2 \cdot 0} = 1 - 1 = 0 \). Therefore, \( P(X \leq 0) = 0 \).
03

Calculate (b) P(X ≥ 2)

To find \( P(X \geq 2) \), use the property \( P(X \geq a) = 1 - F(a) \). Thus: \[P(X \geq 2) = 1 - (1 - e^{-2 \cdot 2}) = 1 - (1 - e^{-4}) = e^{-4}.\]
04

Calculate (c) P(X ≤ 1)

For \( P(X \leq 1) \), use the cumulative distribution function:\[P(X \leq 1) = F(1) = 1 - e^{-2 \cdot 1} = 1 - e^{-2}.\]
05

Calculate (d) P(1 < X < 2)

Use the properties of the CDF for both endpoints:\[P(1 < X < 2) = F(2) - F(1) = (1 - e^{-4}) - (1 - e^{-2}) = e^{-2} - e^{-4}.\]
06

Solve (e) for x such that P(X < x) = 0.05

Set up the equation using the CDF for \( x \):\[1 - e^{-2x} = 0.05.\]Solve for \( e^{-2x} = 0.95 \), and take the natural logarithm:\[-2x = \ln(0.95),\]which gives:\[x = -\frac{\ln(0.95)}{2}.\]Evaluate to find \( x \approx 0.0256 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability
In the context of exponential distribution, probability helps us understand the likelihood of a variable falling within a particular range. When dealing with continuous variables, like those in an exponential distribution, we use probability density functions (PDFs) to describe how likely a value is to occur. For instance, the PDF for an exponential distribution with rate parameter \( \lambda \) is \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \).When calculating probabilities such as \( P(X \leq 1) \) or \( P(X \geq 2) \), we often transition from the PDF to the cumulative distribution function (CDF), which gives the total probability up to a certain value. This allows us to easily determine how likely it is for the random variable \( X \) to be less than or greater than a specified value.The general concept of probability in exponential distributions involves understanding that as time progresses, the probability of an event happening can either increase or decrease based on \( \lambda \). The exponential factor \( e^{-\lambda x} \) showcases the decay of probability with increasing \( x \).
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) for an exponential distribution is a powerful tool for analyzing probabilities. The CDF, denoted as \( F(x) \), represents the probability that the random variable \( X \) will take a value less than or equal to \( x \).For an exponential distribution with rate parameter \( \lambda \), the CDF is defined as:\[ F(x) = 1 - e^{-\lambda x} \]This formula provides us with an easy way to calculate probabilities. For example, in the original exercise, calculating \( P(X \leq 1) \) involves using the CDF: \[ P(X \leq 1) = 1 - e^{-2 \cdot 1} = 1 - e^{-2} \]. By plugging values into the CDF, we can find the probability up to any point \( x \). Additionally, when seeking values where \( P(X < x) = 0.05 \), such as in part (e) of the exercise, we rearrange the CDF formula to solve for \( x \). In general, the CDF is instrumental in converting exponential models into practical probability calculations.
Exponential Distribution Properties
Exponential distribution is characterized by its memoryless property. This means that the probability of an event occurring in the future is independent of how much time has already elapsed. This feature is unique to the exponential distribution and is particularly useful in fields like reliability analysis and queuing theory.Another key property is the relationship between the mean and the rate parameter \( \lambda \). It is given by the formula \( \text{mean} = \frac{1}{\lambda} \), indicating that a higher rate \( \lambda \) results in a smaller mean time to the next event. Furthermore, the exponential distribution is positively skewed, meaning that it stretches out longer on the right side. This is illustrated in its PDF and CDF, where the exponential decay \( e^{-\lambda x} \) leads to diminishing probabilities as \( x \) increases.These properties make the exponential distribution a preferred model for events occurring at a constant average rate, making it applicable in scenarios such as calculating time between arrivals in a queue, or lifetimes of electronic components. Understanding these properties helps in identifying when an exponential distribution is appropriate for modeling real-world phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Asbestos fibers in a dust sample are identified by an electron microscope after sample preparation. Suppose that the number of fibers is a Poisson random variable and the mean number of fibers per square centimeter of surface dust is 100\. A sample of 800 square centimeters of dust is analyzed. Assume a particular grid cell under the microscope represents \(1 / 160,000\) of the sample. (a) What is the probability that at least one fiber is visible in the grid cell? (b) What is the mean of the number of grid cells that need to be viewed to observe 10 that contain fibers? (c) What is the standard deviation of the number of grid cells that need to be viewed to observe 10 that contain fibers?

In an accelerator center, an experiment needs a 1.41 -cmthick aluminum cylinder (http://puhepl.princeton.edu/mumu/ target/Solenoid_Coil.pdf). Suppose that the thickness of a cylinder has a normal distribution with a mean of \(1,41 \mathrm{~cm}\) and a standard deviation of \(0.01 \mathrm{~cm}\). (a) What is the probability that a thickness is greater than \(1.42 \mathrm{~cm} ?\) (b) What thickness is exceeded by \(95 \%\) of the samples? (c) If the specifications require that the thickness is between \(1.39 \mathrm{~cm}\) and \(1.43 \mathrm{~cm}\), what proportion of the samples meet specifications?

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch. (a) What is the probability that the diameter of a dot exceeds 0.0026 inch? (b) What is the probability that a diameter is between 0.0014 and 0.0026 inch? (c) What standard deviation of diameters is needed so that the probability in part (b) is \(0.995 ?\)

The speed of a file transfer from a server on campus to a personal computer at a student's home on a weekday evening is normally distributed with a mean of 60 kilobits per second and a standard deviation of 4 kilobits per second. (a) What is the probability that the file will transfer at a speed of 70 kilobits per second or more? (b) What is the probability that the file will transfer at a speed of less than 58 kilobits per second? (c) If the file is 1 megabyte, what is the average time it will take to transfer the file? (Assume eight bits per byte \()\)

The length of stay at an emergency department is the sum of the waiting and service times. Let \(X\) denote the proportion of time spent waiting and assume a beta distribution with \(\alpha=10\) and \(\beta=1\). Determine the following: (a) \(P(X>0.9)\) (b) \(P(X<0.5)\) (c) mean and variance

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.