/*! 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 191 Suppose that \(Y\) is a continuo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 191

Suppose that \(Y\) is a continuous random variable with distribution function given by \(F(y)\) and probability density function \(f(y) .\) We often are interested in conditional probabilities of the form for a constant \(c\) a. Show that, for \(y \geq c\) $$ P(Y \leq y | Y \geq c)=\frac{F(y)-F(c)}{1-F(c)} $$ b. Show that the function in part (a) has all the properties of a distribution function. c. If the length of life \(Y\) for a battery has a Weibull distribution with \(m=2\) and \(\alpha=3\) (with measurements in years), find the probability that the battery will last less than four years, given that it is now two years old.

Short Answer

Expert verified
a. \(P(Y \leq y | Y \geq c) = \frac{F(y)-F(c)}{1-F(c)}\). b. It satisfies all properties of a distribution function. c. \(1 - e^{-\frac{4}{3}}\).

Step by step solution

01

Understand the Conditional Probability

We need to find the conditional probability \(P(Y \leq y | Y \geq c)\). This is the probability that \(Y\) is less than or equal to \(y\), given that \(Y\) is already greater than or equal to \(c\). According to the definition of conditional probability, this is given by \(\frac{P(Y \leq y \text{ and } Y \geq c)}{P(Y \geq c)}\). Since \(Y\geq c\) is a condition already met, it translates to \(P(c \leq Y \leq y)\).
02

Use the Distribution Function

Since \(P(c \leq Y \leq y) = F(y) - F(c)\) and \(P(Y \geq c) = 1 - F(c)\), the conditional probability can be expressed as \(\frac{F(y) - F(c)}{1-F(c)}\). This is exactly what the problem asks to show.
03

Understanding Properties of a Distribution Function

A function is a distribution function if it is non-decreasing, right-continuous, and approaches 0 as input tends to negative infinity and 1 as input tends to positive infinity. For \(y \geq c\), \(\frac{F(y)-F(c)}{1-F(c)}\) is non-decreasing because \(F(y)\) is a non-decreasing function, it's continuous and approaches 1 as \(y\) becomes very large (since \(F(y)\) approaches 1).
04

Define the Weibull Distribution CDF

Given that \(Y\) follows a Weibull distribution with parameters \(m=2\) and \(\alpha=3\), the cumulative distribution function (CDF) is given by \(F(y) = 1 - e^{-(\frac{y}{\alpha})^m}\). In this case, \(F(y) = 1 - e^{-(\frac{y}{3})^2}\).
05

Calculate the Conditional Probability for Weibull

We need to find \(P(Y<4 | Y\geq 2)\). Using the derived formula \(\frac{F(y)-F(c)}{1-F(c)}\), substitute \(y = 4\) and \(c = 2\). Calculate \(F(4) = 1 - e^{-(\frac{4}{3})^2}\) and \(F(2) = 1 - e^{-(\frac{2}{3})^2}\). Substitute these into the formula: \(\frac{F(4) - F(2)}{1-F(2)}\).
06

Perform the Calculations

First, calculate \(F(4) = 1 - e^{-(\frac{4}{3})^2} = 1 - e^{-\frac{16}{9}}\) and \(F(2) = 1 - e^{-(\frac{2}{3})^2} = 1 - e^{-\frac{4}{9}}\). Substitute these into the formula to get the probability: \(\frac{1 - e^{-\frac{16}{9}} - (1 - e^{-\frac{4}{9}})}{e^{-\frac{4}{9}}} = 1 - \frac{e^{-\frac{16}{9}}}{e^{-\frac{4}{9}}}\). Simplify this to \(1 - e^{-\frac{12}{9}} = 1 - e^{-\frac{4}{3}}\).

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 91影视!

Key Concepts

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

Weibull Distribution
The Weibull Distribution is a continuous probability distribution used to model reliability and life data. It is particularly useful in assessing failure times and life spans of products. The distribution is defined by two parameters: the shape parameter \( m \) and the scale parameter \( \alpha \).

  • The shape parameter \( m \), determines the type of hazard function 鈥� whether it's increasing, decreasing, or constant over time. When \( m=2 \), often referred to as the Rayleigh distribution, it implies the hazard function is increasing.
  • The scale parameter \( \alpha \), influences the distribution's spread and scale across its domain. Essentially, \( \alpha \) adjusts the scale of time and dictates when events are likely to occur.
In analyzing life data, such as battery life, the Weibull distribution offers flexibility by adjusting these parameters to match the empirical data, which helps in predicting the time until a failure occurs. The structure of the Weibull distribution lets it handle a wide range of phenomena that require different behavior over time.
Probability Density Function
The Probability Density Function (PDF) characterizes the likelihood of a continuous random variable to take a specific value. For a random variable \( Y \) that follows a probability distribution, the PDF \( f(y) \) helps determine the probability that \( Y \) falls within a particular range.

  • The PDF must satisfy two main conditions: it is non-negative everywhere and the total area under the curve (over its range) equals 1.
  • The PDF provides a way to calculate probabilities over intervals by integrating the function over a specified range.
For the Weibull distribution, the PDF is given by:\[ f(y) = \frac{m}{\alpha} \left( \frac{y}{\alpha} \right)^{m-1} e^{-\left( \frac{y}{\alpha} \right)^m}\]This function illustrates how the likelihood of failures or events is distributed over time. It can show a peaked structure where the likelihood concentrates around a specific time or can stretch across time to show a gradual likelihood of occurrence.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) for a random variable \( Y \) describes the probability that the variable will be less than or equal to a certain value. It is an integral of the probability density function (PDF) from negative infinity to \( y \).

  • The CDF is monotonically increasing and ranges between 0 and 1, moving from the lowest to the highest values of the distribution.
  • It offers a cumulative sum of probabilities up to a point \( y \), making it useful for determining the likelihood of an event occurring up to that threshold.
For the Weibull distribution, the CDF is expressed as:\[F(y) = 1 - e^{-\left( \frac{y}{\alpha} \right)^m}\]This formula allows one to calculate the probability of obtaining a value less than or equal to \( y \). In practical terms, it helps assess the probability that a device or material will last up to a specified time frame. For instance, the CDF can give insights into the probability of a battery lasting less than four years, considering it's already two years old.

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

The time (in hours) a manager takes to interview a job applicant has an exponential distribution with \(\beta=1 / 2\). The applicants are scheduled at quarter-hour intervals, beginning at 8: 00 A.M., and the applicants arrive exactly on time. When the applicant with an 8: 15 A.M. appointment arrives at the manager's office, what is the probability that he will have to wait before seeing the manager?

Suppose that the waiting time for the first customer to enter a retail shop after 9:00 A.M. is a random variable \(Y\) with an exponential density function given by $$f(y)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta}\right) e^{-y / \theta}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the moment-generating function for \(Y\). b. Use the answer from part (a) to find \(E(Y)\) and \(V(Y)\)

If \(\theta_{1}<\theta_{2}\), derive the moment-generating function of a random variable that has a uniform distribution on the interval \(\left(\theta_{1}, \theta_{2}\right)\)

A manufacturing plant uses a specific bulk product. The amount of product used in one day can be modeled by an exponential distribution with \(\beta=4\) (measurements in tons). Find the probability that the plant will use more than 4 tons on a given day.

The length of time required by students to complete a one-hour exam is a random variable with a density function given by $$f(y)=\left\\{\begin{array}{ll}c y^{2}+y, & 0 \leq y \leq 1, \\\0, & \text { elsewhere. } \end{array}\right.$$ a. Find \(c\). b. Find \(F(y)\). c. Graph \(f(y)\) and \(F(y)\). d. Use \(F(y)\) in part (b) to find \(F(-1), F(0),\) and \(F(1)\). e. Find the probability that a randomly selected student will finish in less than half an hour. f. Given that a particular student needs at least 15 minutes to complete the exam, find the probability that she will require at least 30 minutes to finish.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation
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.