/*! 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 17 Find the uniform distribution of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 17

Find the uniform distribution of the continuous type on the interval \((b, c)\) that has the same mean and the same variance as those of a chi-square distribution with 8 degrees of freedom. That is, find \(b\) and \(c\).

Short Answer

Expert verified
The uniform distribution that has the same mean and the same variance as those of a chi-square distribution with 8 degrees of freedom is the interval \( (4, 12) \).

Step by step solution

01

Set up the equation for the mean

For a uniform distribution, the mean is \( \frac{b+c}{2} \). Since the mean of a chi-square distribution with 8 degrees of freedom is 8, set these equal to each other: \( \frac{b+c}{2} = 8 \). Solve this equation for one of the variables to obtain \(c = 16 - b\).
02

Set up the equation for the variance

For a uniform distribution, the variance is \( \frac{(c-b)^2}{12} \). Known that the variance of a chi-square distribution with 8 degrees of freedom is 16, we set \( \frac{(c-b)^2}{12} = 16\). Substitute \(c = 16 - b\) into the equation to obtain \( \frac{(16-2b)^2}{12} = 16 \).
03

Solve the equation for \(b\)

Solve the equation \( \frac{(16-2b)^2}{12} = 16 \) for \(b\). This will yield two possible values for \(b\), namely 4 and 12.
04

Calculate \(c\)

Substitute these two possible values of \(b\) into the equation for \(c\) to obtain the corresponding values of \(c\). When \(b = 4, c = 12\), and when \(b = 12, c = 4\).
05

Determine the valid solution

For the interval \(b, c\), \(b < c\). Therefore, only the solution \(b = 4, c = 12\) is valid.

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!

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

A certain job is completed in three steps in series. The means and standard deviations for the steps are (in minutes): $$ \begin{array}{ccc} \hline \text { Step } & \text { Mean } & \text { Standard Deviation } \\ \hline 1 & 17 & 2 \\ 2 & 13 & 1 \\ 3 & 13 & 2 \\ \hline \end{array} $$Assuming independent steps and normal distributions, compute the probability that the job will take less than 40 minutes to complete.

Determine the constant \(c\) so that \(f(x)=c x(3-x)^{4}, 0

. Let \(X\) have a conditional Burr distribution with fixed parameters \(\beta\) and \(\tau\), given parameter \(\alpha\). (a) If \(\alpha\) has the geometric pdf \(p(1-p)^{\alpha}, \alpha=0,1,2, \ldots\), show that the unconditional distribution of \(X\) is a Burr distribution. (b) If \(\alpha\) has the exponential pdf \(\beta^{-1} e^{-\alpha / \beta}, \alpha>0\), find the unconditional pdf of \(X\).

. Let $$ p\left(x_{1}, x_{2}\right)=\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)\left(\frac{1}{2}\right)^{x_{1}}\left(\frac{x_{1}}{15}\right), \begin{array}{r} x_{2}=0,1, \ldots, x_{1} \\ x_{1}=1,2,3,4,5 \end{array} $$ zero elsewhere, be the joint pmf of \(X_{1}\) and \(X_{2}\). Determine: (a) \(E\left(X_{2}\right)\). (b) \(u\left(x_{1}\right)=E\left(X_{2} \mid x_{1}\right)\). (c) \(E\left[u\left(X_{1}\right)\right]\). Compare the answers of Parts (a) and (c).

Readers may have encountered the multiple regression model in a previous course in statistics. We can briefly write it as follows. Suppose we have a vector of \(n\) observations \(\mathbf{Y}\) which has the distribution \(N_{n}\left(\mathbf{X} \boldsymbol{\beta}, \sigma^{2} \mathbf{I}\right)\), where \(\mathbf{X}\) is an \(n \times p\) matrix of known values, which has full column rank \(p\), and \(\beta\) is a \(p \times 1\) vector of unknown parameters. The least squares estimator of \(\boldsymbol{\beta}\) is $$ \widehat{\boldsymbol{\beta}}=\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \mathbf{Y} $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

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.