/*! 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 112 Suppose that \(X\) has an expone... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(X\) 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) \(-\frac{\ln(0.95)}{2}\).

Step by step solution

01

Understanding Exponential Distribution

The exponential distribution is defined by the probability density function (PDF) \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \) and \( \lambda > 0 \). In our case, \( \lambda = 2 \). The cumulative distribution function (CDF) is given by \( F(x) = 1 - e^{-\lambda x} \) for \( x \geq 0 \). This will help determine the probabilities.
02

Calculate P(X ≤ 0)

Since \( X \) is non-negative in an exponential distribution, \( P(X \leq 0) = 0 \). This is because the exponential PDF is defined for \( x > 0 \).
03

Calculate P(X ≥ 2)

To find \( P(X \geq 2) \), use the complement of the CDF. We have \( P(X \geq 2) = 1 - F(2) = e^{-2 \times 2} = e^{-4} \).
04

Calculate P(X ≤ 1)

Use the CDF to find \( P(X \leq 1) = F(1) = 1 - e^{-2 \times 1} = 1 - e^{-2} \).
05

Calculate P(1 < X < 2)

This requires finding the difference of the CDFs: \( P(1 < X < 2) = F(2) - F(1) = (1 - e^{-4}) - (1 - e^{-2}) = e^{-2} - e^{-4} \).
06

Solve for x in P(X < x) = 0.05

Set the CDF \( F(x) = 1 - e^{-2x} = 0.05 \), which leads to \( e^{-2x} = 0.95 \). Taking the natural logarithm, we have \( -2x = \ln(0.95) \), so \( x = -\frac{\ln(0.95)}{2} \).

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 Density Function
In the context of the exponential distribution, the probability density function (PDF) is crucial to understand. The PDF describes the likelihood of a random variable falling within a specific range of values. For an exponential distribution with rate parameter \( \lambda \), the PDF is given by:\[ f(x) = \lambda e^{-\lambda x} \text{ for } x \geq 0 \] This function is only defined for non-negative values of \( x \); meaning that, in practical terms, the distribution is primarily used to model the time until an event occurs, like the time until a radioactive particle decays or the time between arrivals of buses. Here, \( \lambda \) is the rate per unit time, and it shapes how steep or flat the PDF will be. A higher \( \lambda \) will result in a steeper decline in probability as \( x \) increases. This exponential model is apt for describing scenarios where events occur continuously and independently at a constant average rate. Remember, the area under this PDF over the domain from \(0\) to \( \infty \) always sums to 1, ensuring it's a valid probability distribution.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept when dealing with exponential distributions. It provides the probability that a random variable \(X\) is less than or equal to a certain value \(x\). For an exponential distribution, the CDF is given by:\[ F(x) = 1 - e^{-\lambda x} \text{ for } x \geq 0 \] This formula reveals how probabilities accumulate as we consider larger intervals starting from zero. Essentially, \( F(x) \) tells us the area under the probability density function (PDF) from 0 to \( x \).
  • When \( x = 0 \), we find \( F(0) = 0 \), illustrating that the probability of \( X \) being no more than 0 is zero.
  • For large values of \( x \), \( F(x) \) approaches 1, indicating near certainty that \( X \) will fall below such an extreme value.
Utilizing the CDF can simplify calculations significantly, particularly through the use of complementary probabilities, as observed in calculations like \( P(X \geq 2) \). This approach involves recognizing that \( P(X \geq x) = 1 - F(x) \). Having a firm grasp on both PDF and CDF is vital for correctly interpreting and solving problems involving exponential distributions.
Non-negative Random Variable
A defining feature of an exponential distribution is its restriction to non-negative random variables. This characteristic makes it ideal for modeling situations where a variable cannot take on negative values, such as time or distance. In the case of the exponential distribution, \(X\) is always non-negative, meaning:- There is zero probability of \(X\) being negative.- Probability calculations such as \( P(X \leq 0) \) directly result in zero, as non-negative random variables exclude zero and all values below it.This mirrors many real-world phenomena where negative measurements are not logically possible.
  • Think of lifetime of machine components which cannot be negative.
  • Similarly, the delay between successive traffic lights turning green can never be a negative time span.
Thus, it's crucial to consider this attribute when applying the exponential distribution to any practical scenario.
Complementary Probability
Complementary probability is a concept often used interchangeably alongside the cumulative distribution function (CDF), and especially so in contexts like the exponential distribution. When calculating probabilities such as \( P(X \geq 2) \) for an exponential distribution, the complementary probability is an efficient technique:To compute \( P(X \geq 2) \), we observe that it's the complement of \( P(X \leq 2) \).- Thus, \( P(X \geq 2) = 1 - P(X \leq 2) = 1 - F(2) = e^{-4} \).This manipulation relies on the principle of complementarity, where all possible outcomes must sum to 1.
  • The idea is straightforward: either the event occurs or it doesn't.
  • Finding the probability of an event not happening (e.g., being greater than \( x \)) can be as simple as subtracting the probability of the event happening (i.e., being less than or equal to \( x \)) from 1.
Complementary probabilities offer computational ease, reducing potential errors in more complex calculations, and are a staple in exponential distribution problem-solving.

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

An article in Electronic Journal of Applied Statistical Analysis ["Survival Analysis of Dialysis Patients Under Parametric and Non-Parametric Approaches" \((2012,\) Vol. \(5(2),\) pp. \(271-288\) ) ] modeled the survival time of dialysis patients with chronic kidney disease with a Weibull distribution. The mean and standard deviation of survival time were 16.01 and 11.66 months, respectively. Determine the following: (a) Shape and scale parameters of this Weibull distribution (b) Probability that survival time is more than 48 months (c) Survival time exceeded with \(90 \%\) probability

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 that 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?

Suppose that \(X\) has a lognormal distribution with parameters \(\theta=2\) and \(\omega^{2}=4 .\) Determine the following in parts (a) and (b): (a) \(P(X<500)\) (b) Conditional probability that \(X<1500\) given that \(X>1000\) (c) What does the difference between the probabilities in parts (a) and (b) imply about lifetimes of lognormal random variables?

Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the following: (a) \(P(X<10,000)\) (b) \(P(X>5000)\)

A square inch of carpeting contains 50 carpet fibers. The probability of a damaged fiber is \(0.0001 .\) Assume that the damaged fibers occur independently. (a) Approximate the probability of one or more damaged fibers in one square yard of carpeting. (b) Approximate the probability of four or more damaged fibers in one square yard of carpeting.

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.